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FOR THE YEAR 1860;
WITH SOLUTIONS.
BT
The Rev. H. W. WATSON, M.A.
LATE FBIXOW OF TKINITY OOLLBOB.
AJif)
E. J. EOUTH, M.A.
WMUJOW AKD AflSmANT .TUTOR OF ST PETEB'S COLLBOB. CAJfBKIDOK
AjrO BXAlOirEfi IN THE aMVEBSITY OF LUSDON.
Cambrilige:
MACMILLAN AND CO.
AND a, HENRIETTA STREET, COVENT GARDEN,
Hontion.
I860.
CITambntigt:
fKINTED BY C J. CLAY, M.A.
AT THE UiaVBBSITY TBESS.
PREFACE.
The value of a Collection of Solutions depends in great
measure on the fact that every Problem is solved by the
framer of the question, thus showing the student the manner
in which he was expected to proceed in the Senate-House.
The Moderators desire therefore to thank the Examiners for
the many valuable Solutions of the Problems set by them,
by which the book has been made more complete than it
would othei'wise have been.
The Senior Moderator also acknowledges his obligation to
Mr Droop, Fellow of Trinity College, for much valuable
assistance, and particularly for the suggestion and the solu-
tion of the three following Problems, viz. No. \'i. of Tuesday
Morning, Jan. 17, and Nos. 3 and 5 of Wednesday Moniing,
Jan. 18.
SOLUTIONS OF SENATE-HOUSE
PROBLEMS AND RIDERS
FOR THE YEAR EIGHTEEN HUNDRED AND SIXTY.
Tuesday, Jan. 3. 9 to 12.
JUNIOB MODEBATOB. Arabic numbers.
Seniob ExAiiiNEB. Roman numbers.
1. If a straight line DME be drawn through the middle i^
point M of the base of a triangle ABC, so as to cut off equal
parts AD, AE from the sides AB, A C, produced if necessary,
respectively, then shall BD be equal to CE.
Through C draw CJ?' parallel to AB, and cutting BE in F.
Then the two triangles DMB, FMG are clearly equal, and
therefore CF= BD. Again, CF being parallel to AB, the
angle CFE = the angle ADE, and because AD = AE, the
angle ADE = angle AED ; whence it easily follows that
CF= CE.
2. Shew how to construct a rectangle which shall be equal *^
to a given square ; (I) when the sum and (2) when the differ-
ence of two adjacent sides is given.
The first case is too obvious to require any solution. In the
second case, refer to the figure in Euclid, Book ii. Prop. 14.
A little consideration will shewthat GE is twice the differ-
ence between the two sides BE, ED. Whence the following
construction. Take QE=\ia\i the given difference, describe
2 SENATE-HOUSE PROBLEMS [Jan. 3,
a circle BIIF with radius equal to the side of the given
square, and cutting GE produced in B and F. Tlien BE,
EF are the sides of the rectangle required.
3. If two chords AB, AC ha drawn from any point A of
a circle, and be produced to D and E, so that the rectangle
A C, AE is equal to the rectangle AB, AD, tlien if 0 be the
centre of the circle, AO is perpendicular to DE.
Since AB . AD = AC . AE, a circle may be described about
BCED. Therefore the angle BDE = BCA. Hence if A and
B be fixed while C moves round the circle, the angle ADE
will be constant and tlie locus of E will be a straight line.
Take AC to pass through 0 and cut the circle in C and DE
in P. Then as before the angle APD = ABC = a right
angle.
iv. Describe an isosceles triangle having each of the
angles at the base double of the third angle.
If tI be the vertex, and BD the base of the constructed
triangle, D being one of the points of intersection of the two
circles employed in the construction, and E the other, and AE
be drawn meeting BD produced in F, prove that FAB is
another isosceles triano-le of the same kind.
*o*
For ADE is an isosceles triangle, and the angle AED at
the base is the supplement of the angle A CD in the opposite
segment of the circle. Hence AED = BCD and therefore bv
Euclid = ABD, and also the angles ADE, ADB are equal,
therefore the third angle DAE= the third angle BAD. Hence
the whole angle BAE is double the angle BAD, and therefore
equal to ABD. Hence the triangle FAB is isosceles, and
each of the angles at the base is equal to the angles at the
base of ABD. Therefore, &c.
V. Prove that the straight lines bisecting one angle of
a triangle internally and the other two externally pass
through the same point.
Let the exterior angles A and C of the triangle ABC be
bisected by AD, CO, meeting each other in 0; tben BO will
bisect the angle ABC. Because AD bisects the exterior
9 — 12.] AND RIDERS. 3
angle A, BA : BD :: AC \ CD. And because CO bisects
tlie angle ACD, therefore AC -. CD :: AO : OD, therefore
BA : BD :: AO : ODy and therefore BO bisects the angle
ABD. See fig. 1.
vi. If three straight lines, which do not all lie in one
plane, be cut in the same ratio bj three planes, two of which
are parallel, shew that the third will be parallel to the other
two, if its intersections with the three straight lines are not
all in one straight line.
This may be easily proved by a " reductio ad absurdum."
vii. Define a parabola: and prove from the definition
that it cannot be cut by a straight line in more than two
points.
For if possible let a straight line cut the parabola in three
points P, Q, R, and let it cut the directrix in T. Draw Pp,
Qq, Br perpendiculars to the directrix, and let ;S^ be the focus.
Then since SP=Pp, SQ= Qq, it follows that SP: SQ :: PT
: QT, and therefore ST bisects the exterior angle to PSQ.
Similarly ST also bisects the exterior angle to PSE. Which
is absurd.
viii. P, Q are points in two confocal ellipses, at which the
line joining the common foci subtends equal angles ; prove
that the tangents at P, Q are inclined at an angle which is
equal to the angle subtended by PQ at either focus.
Let the normals at P and Q meet in G, join QP and pro-
duce it to any point R. Then the angle between the tangents
is equal to the angle PG Q which is
= RPG -RQG = {RPS- RQS) + {SPG - SQG).
Now SPG = SQG, being the halves of equal angles, and the
difference RPS -RQS = PSQ. Similarly the angle PGQ
may be proved = PUQ.
ix. If a circle, passing through Y and Z, touch the major
axis in Q, and that diameter of the circle, which passes through
Q, meet the tangent in P, then PQ = BC.
b2
4 SENATE-HOUSE PROBLEMS AND EIDERS. [Jan. 3.
Let the tangent YZ cut the major axis in T. Then by
similar triangles
PQ _SY .PQ_nZ P(^ SY.HZ
QT" YT QT~ ZT'' •*• QT""" TY.TZ'
But TY. TZ= Tqt by Euclid, iii. 36, and 8Y , HZ=BG*;
.'. PQ = BC.
11. In an hyperbola, supposing the two asymptotes and
one point of the curve to be given in position, shew how to
construct the curve ; and find the position of the foci.
Let OX, 0 F be the two asymptotes, and P the given point.
Draw PN parallel to OY cutting OX in N. Measure OB
= OJE along the asymptotes, such that OB' = 4 . ON. NP.
Bisect the angle BOE by OA cutting BE in A. Then OA,
AB are equal to the axes; and the remainder of the con-
struction is obvious.
12. Given a right cone and a point within it, there are
but two sections which have this point for focus; and the
f)lanes of these sections make equal angles with the straight
ine joining the given point and the vertex of the cone.
Let V be the vertex, VCO the axis of the given cone,
and P the given point. Then, if two spheres be inscribed in
the cone and passing through P, the tangent planes to these
spheres will evidently be the only two sections whose foci are
at P. Let G and 0 be the centres of the two spheres, then
VCi VO :: CL : OM :: CP: PO; therefore VP bisects the
angle exterior to CPO in the triangle CPO. But the raSii
CP, PO are perpendicular to the sections AB, BE, therefore
VP bisects the angle between these sections. See fig. 2.
( 5 )
Tuesday, Jan. 3. 1^ to 4.
Sbkiob Modebatob. Arabic numbers.
JONIOB ExAMiNEB. Eoman numbers.
4. (3) Solve the equations,
Multiplying the second and third of these equations
together, and subtracting the square of the first, we get
X {3xyz — o^—y^ — a') = 5V — a* ;
therefore by symmetry
Hence, substituting in the first equation,
therefore x—±
y = ±
2 = +
(o' + J' + c'-Sa'JV)*'
(a' + ^^' + c'-Sa'iV)**
6 SENATE-HOUSE PROBLEMS [Jan. 3,
m xi 1 • • p sin (tt COS ^) " ^
viii. Trace the cnansres in sign oi 7 — ■. — ^i , as 0 vanes
° ° cos (tt sin 0}
from 0 to TT.
The numerator is positive when 6 lies between 0 and - ,
TT
and negative when ^ lies between - and tt.
TT
The denominator is positive when 6 lies between 0 and - ,
negative when 6 lies between — and — , and positive when B
lies between -— and tt.
6
If
Hence the expression is positive from 0 to - ,
negative
11
TT
6
to
TT
2'
positive
»
TT
2
to
6 '
negative
>j
h-ir
6
to
TT.
Prove that,
sin 3 (^ - 15) = 4 cos [A - 45) cos {A + 15) sin (^ - 15),
and find sin A and sin B from the equations,
a sin' -4 + 6 sin" B=G,
a sin 2-4—6 sin 2B= 0,
(1) sin 3-4 = 3 sin -4- 4 sinM, ^
cos 3-4 = 4 cos' -4 — 3 cos vl ;
.'. sin 3-4 — cos 3-4
= (sin A + cos -4) [3 — 4 (sin' A + cos' -4 — sin -4 cos -4)},
= 2 sin 45 cos {A - 45) {2 sin 2^-1],
= 4 sin 45 cos (-4 — 45) {sin 2-4 — sin 30},
= 8 sin 45 cos [A — 45) sin (-4 — 15) cos (-4 + 15).
But sin ZA - cos 3.4 = sin 3-4 - sin (90 - ZA)
= 2 cos 45 sin (3-4 - 45) ;
1^—4.] AND RIDERS. 7
.-. sin 3(^ - 15) = 4 cos {A - 45) sin {A - 15) cos (^ + 15).
(2) a sin' A + b sin'J? = c,
a^ sin' A cos' J. = J' sin' B cos' -B,
a' sin' A-b^ sin' 5= a' sin* ^ - i' sin* B
= c{a sin' A — b sin' 5) ;
.*. a (a — c) sin' A = b{b — c) sin' -B ;
T /7 \ 7 / \ 6c (J — c) ac(a — c)
^ '^ ^ ' sm'^ sin^j^ '
• « J 5c (5 — c) _ e{b — c)
.-. sin^ -ab{a + b-2c)~a{a + b-2c)'
sin' 5
_ ac (a — c) c (a — c)
~ ab{a + b-2c) ^b{a+b-2c)
whence, &c.
xii. A railway passenger seated in one corner of the car-
riage looks out of the windows at the further end and observes
that a star near the horizon is traversing these windows in the
direction of the train's motion and that it is obscured by the
partition between the corner window on his own side of the
carriage and the middle window while the train is moving
through the seventh part of a mile. Shew that the train is
on a curve the concavity of which is directed towards the
star, and which, if it be circular, has a radius of nearly three
miles; the length of the carriage being seven feet and the
breadth of the partition four inches.
Owing to the great distance of the star, the motion of the
carriage parallel to itself has no effect upon the point in which
the line joining the star and the passenger's eye meets the
window. Hence since this line meets the window in points
which move in the direction of the carriage's motion, the
direction of the carriage must be continually varying, and
the carriage must be on a curve concave to the star. The
rest of the question is too obvious for explanation.
xiii. If a, &, and B be given, shew under what circum*
8 SENATE-HOUSE PROBLEMS AND RIDEES. [Jan. 3.
Stances there will be two triangles satisfying the conditions
of the problem.
Prove that the circles circumscribing both triangles are
equal in magnitude, and that the distance between their
centres is »J{b^cosec^B—a').
The radius of the circumscribing circle = - -: — f^, and this
is the same for both triangles. See fig. 3.
The centres of the circles 0' and 0 are situated in the line
ODO', which bisects BG at right angles, and since the per-
pendiculars from 0' and 0 upon BA'A meet this line at
distances -—— and —^ from B, it follows that
00' = \{BA- BA') cosec B.
Now BA and BA are the values of p determined from the
equation
(? — 2ac cos B-^a' = ¥;
.'. BA-BA'=r2^{h*-a'8iR'B);
.'. I {BA - BA') cosec B= >^{V cosec' B-a").
( 9 )
Wednesday, Jan. 4. 9 to 12.
Seniob Moderatob. Arabic niimbers.
Jdniob Exakineb. Boman numbers.
1. Enunciate the proposition of the parallelogram of
forces; and, assuming its truth for the magnitude, prove it
also for the direction, of the resultant.
Let AB and A C represent two forces acting on the point A.
Complete the parallelogram AD, then by hypothesis AD re-
presents the resultant in magnitude ; it is required to prove
that AD represents the resultant in direction also. Fig. 4.
Draw AD' in the direction of the resultant and equal to it
and therefore also equal to AD. Complete the parallelogram
AD' EC and draw the diagonal AE.
AD', AB and AC represent three forces in equilihrium,
each of them is therefore equal and opposite to the resultant
of the other two. But by hypothesis AE is equal to the
resultant of AD' and A C.
Therefore AE is equal to AB and therefore to CD,
Also EC is equal to AD' and therefore to AD.
Hence the quadrilateral AECD has its opposite sides equal,
it is therefore a parallelogram; therefore EC is parallel to
AD.
But EC is also parallel to AD' ; therefore AD' and AD
are in the same straight line.
2. Two equal particles, each attracting with a force varying
directly as the distance, are situated at the opposite extremrties
10 SENATE-HOUSE PROBLEMS [Jan. 4,
of a diameter of a liorizontal circle, on whose circumference a
small smooth ring is capable of sliding ; prove that the ring
will be kept at rest in any position under the attraction of
the particles.
Let A and B be the particles at opposite extremities of the
diameter AB, and let P be the attracted ring. Join AP, BP.
These lines represent the forces on P, and the resultant force is
therefore in the direction of that diagonal of the parallelogram
on AP, PB which passes through P. Hence the resultant
passes through 0 the centre of the circle since the diagonals
of parallelogi-ams bisect each other, and therefore the reaction
of the smooth curve is capable of counteracting the resultant
force wherever P may be situated.
3. Two equal heavy particles are situated at the extremi-
ties of the latus rectum of a parabolic arc without weight,
which is placed with its vertex in contact with that of an
equal parabola, whose axis is vertical and concavity down-
wards ; prove that the parabolic arc may be turned through
any angle without disturbing its equilibrium, provided no
sliding be possible between the curves.
In the figure, fig. 5, it follows from the equality of the
parabolas that the arcs AP and A'P, and the angles ASP
and A'S'P are equal, and that the tangent at P bisects the
angle SPS'. But the tangent at P bisects the angle between
SP and the line through P parallel to the axis AS. Hence
PS' is parallel to AS and therefore vertical.
Hence the perpendiculars from P upon the verticals through
L and L' the extremities of the latus rectum are always equal,
and therefore equal weights at these points always balance
about P.
4. Find the position of equilibrium when a common
balance is loaded with given unequal weights.
If the tongue of the balance be very slightly out of adjust-
ment, prove that the true weight of a body is the arithmetic
mean of its apparent weights, when weighed in the opposite
scales.
^ „ /, (P-Q)a
Generally tan S = ppL^^^-^ .
9—12.] AND RIDERS. 11
In this case the beam appears horizontal when it is really
inclined at a very small angle a to the horizon. If then P
be the true weight, W^and W' the apparent weights,
P- W= {{P+ W)h + Wh] tana,
W'-P= {(P+ W')h + WJc] tan a,
and we may consider W and W as equal in the coefficients of
the very small quantity tan a ;
.*. P^ W= W — P approximately,
2
5. In the figure of Euclid, Book i. Prop. 47, if the perime-
ters of the squares be regarded as physical lines uniform
throughout, prove that the figure will balsuice about the mid-
dle point of the hypothenuse with that line horisontal the lines
of construction having no weight.
Let D be the middle point of AB, and DNE, DMF per-
pendiculars from D bisecting the sides AG and CB in N
and M. See fig. 6.
Then DE=EN+NI)
=^AC+^CB
= DF.
Therefore the perpendiculars from D upon the verticals
through E and F are proportional to the cosines of the angles
EBA and FBB respectively, and are therefore proportional
to BG and A C respectively, i. e. to the weights of the peri-
meters of the squares acting at F and E respectively.
Hence these weights balance about B when ^P is hori-
zontal, and it is clear that the weight of the perimeter on AB
passes through B; therefore the whole balances about this
point.
6. A uniform heavy rod, having one extremity attached
to a fixed point, about which it is free to move in all direc-
tions, passes over the circumference of a rough ring whose
12 SENATE-HOUSE PROBLEMS [Jan. 4,
centre is at the fixed point and whose plane is inclined at a
given angle to the horizon ; find the limiting position of equi-
librium.
Let 0 be the centre of the ring, fig. 7. OH the line of
greatest slope through 0 in the plane of the ring and inclined
to the horizon at the angle a, OB the rod, acting on the circle
at a
Let 2a he the length of the rod,
b radius of the ring,
0 angle HOB,
fi the coefficient of friction, W the weight of the rod, and B
the normal action at C.
XI' (ft* _ yjr J fv
We may replace TF acting at G by — ^^ at 0 and W~
at C, and this latter by TF-cosa, perpendicular to the plane
of ring, and W- sin a along this plane ;
.*. B = W- cos a and aB = W- sin a sin 0,
T T
resolving perpendicular to rod ; therefore by division
sin ^ = /i cot a.
vii. A point, moving with a uniform acceleration, describes
20 feet in the half-second which elapses after the first second
of its motion ; compare its acceleration with that of a falliug
heavy particle; and give its numerical measure, taking a
minute as the unit of time, and a mile as that of space.
Let a be the numerical measure of the acceleration of the
point, taking a foot as the unit of space, and half-a-second as
that of time.
Then it will describe - feet in the first half-second.
At the end of this time it will have a velocity, which, if
it continued uniform during the next half-second, would carry
it over a feet.
9—12.] AND EIDERS. 13
At the end of the first second, its velocity will be twice as
great as at the end of the first half-second ; that is, a velocity
which if it continued uniform during the next half-second,
would carry the point over 2a feet. In consequence of the
acceleration, it will move over - feet more, or — feet Hence
1 = 20,
or a = 8.
Hence, taking a second as the unit of time, and a foot as
the unit of space, the numerical measure of the acceleration
will be 32 ; that is,
acceleration of this point : acceleration of falling particle
:: 32 : 32*2
:: 160 : 161.
If a minute be the unit of time, and a mile that of space,
the acceleration will be measured by
32 X 60' 32 X 15
5280 22
_240
~ 11 •
ix. A heavy particle slides down a smooth inclined plane
of given height ; prove that the time of its descent varies as
the secant of the inclination of tlie plane to the vertical.
Let h be the height of the plane, a its inclination to the
vertical, then its length will be h sec a, and the acceleration
down the plane g cos a ;
.'. if T be the time of descent
gcosa. T* ,
= h sec a,
or T* = 2gh.Bec*a;
.-. T={2gh)^8eca,
or Tex: sec a,
14 SENATE-HOUSE PROBLEMS [Jan. 4,
X. A heavy particle is projected from a given point with a
given velocity, so as to pass through another given point;
prove tliat, in general, there will be two parabolic paths which
the particle may describe ; and give a geometrical construction
to determine their foci. Also find the locus of the second
point in order that there may be only one parabolic path.
Since the velocity of projection is given, the directrix of the
parabolic path is given.
Let MN be the directrix, P the point of projection, Q the
point through which the particle is to pass. Fig. 8.
Draw PM, QN, perpendicular to MN. With P, Q as
centres and PM, QNas radii, describe two circles ; these will
in general intersect in two points S, S', which will be the foci
of the two parabolic paths.
If, however, the two circles toueh one another, there will be
but one parabolic path. In order that this may be the case,
we must have
QN+PM=PQ.
Hence if QN be produced to K, so that NK may be equal
to PMf and KK' be drawn parallel to MN, we shall have
distance of Q from KK'^PQ,
or, the locus of <? is a parabola, of which P is the focus, and
KK ' the directrix.
xi. A series of perfectly elastic balls are arranged in tlie
same straight line, one of them impinges directly on the next,
and so on ; prove that, if their masses form a geometrical pr^
gression of which the common ratio is 2, their velocities after
impact will form a geometrical progression of which the com-
mon ratio is f .
Let M, 2il/, be the masses of two adjacent balls, v the velo-
city of M before impact, u^, ii^, the respective velocities of
2
M, 23/, after impact, then we have to prove that %=nV.
o
Now the momentum is the same before and after impact ;
.'. Mu^-VlMu^ = Mv (1),
9_12.] AND RIDERS.
15
and since tlie balls are perfectly elastic, the vis viva is un-
altered: , ,
.-. Mu* + 2Mu^ = Mv^ (2).
Squaring (1) and multiplying (2) by if, we get
Hence eliminating u^ by (1),
2v
•••««= 3"-
( 16 )
Wednesday, Jan. 4. 1^ to 4.
JuHlOB MODEBATOB. Arabic numbers.
Senior Examines. Boman numbers.
1. A UNIFORM tube is bent into the form of a parabola, and
placed with its vertex downwards and axis vertical : supposing
any quantities of two fluids of densities p, p to be poured into
it, and r, r to be the distances of the two free surfaces respec-
tively from the focus, then the distance of the common surface
TO "-• T O
from the focus will be — — , - .
P-P
This follows at once from the two principles :
(1) If two fluids be placed in a bent tube, the altitudes of
the free surfaces above the common surface are inversely as
their densities.
(2) The distance of any point of a parabola from the fd^s
is equal to its distance fi'om the directrix.
2. A parallelogram is immersed in a fluid with one side in
the surface ; shew how to draw a line from one extremity of
this side dividing the parallelogram into two parts on which
the pressures are equal.
Let ABCD be the parallelogram, AE the line drawn from
one angle to the base. Then the pressure on the triangle
is = - the pressure on the whole parallelogram. Fig. 9.
1^—4.] SENATE-HOUSE PROBLEMS AND RIDERS. l7
Now the pressure cc area . depth of centre of gravity ; the
CE
areas of the two figures are as — - : CD ; the depths of the
centres of gravity are aa - : - ;
O it
. 2 CE_l 1
•• 3- 2 -2*2^^'
.-. CE=^-CD.
4
3. A heavy hollow right cone, closed by a base without
weight, is immersed in a fluid, find the force that will sustain
it with its axis horizontal.
Let AB = A be the axis of the cone, A the vertex. The forces
acting on the cone are, its weight TF acting downwards at G the
2
centre of gravity of the surface, AG= -h; and W the weight
of the water displaced acting upwards at H the centre of
3
gravity of the volume, An=-h.
The resultant of these isW—W acting at a point C, where
(W-W")AG=W.lh-W','^h.
^ o 4
4. A given weight of heavy elastic fluid of uniform
temperature is confined in a smooth vertical cylinder by a
piston of given weight; shew how to find the volume of
the fluid.
It is proved in Goodwin's Course^ lipp be the pressures of
the atmosphere at two points whose vertical distance is x, that
X = -^ r . log ^ ,
,o,(,-f) P
where t is the thickness of the very small layers into which
the atmosphere was supposed divided.
c
la* SENATE-HOUSE PROBLEMS [Jan. 4,
If W= weight of the piston, A its area, and W = weight
of air contained in the cylinder, then pA = W, p'A = W+ W',
and Ax becomes the volume of the nuid. Hence we have
T , W
X = . log
In this equation t is any very small quantity. If t = 0 it
is proved in the Course cited above, that the expression for x
becomes
5. If A be the area of the section of each pump of the fire
engine, I the length of the down stroke, n the number of
strokes per minute, B the area of the hose, then it is obvious
that the average velocity from the hose, when both pumps
, 2Aln
work, = — j5— .
JD
6. Supposing some light material, whose density is p, to
be weighed by means of weights of density p , the density
of the atmosphere when the barometer stands at 30 inches
being unity ; shew that, if the mercury in the barometer
fall one inch, the material will appear to be altered by
P'-P
(p-l)(30p'-29)
of its former weight. Will it appear to weigh more or less ?
Let V = volume of the material, Fthe volume of the weightj'
then V measures the apparent weight of the material, and
we have
V{p'-l) = v{p-l).
AVlien the barometer has fallen one inch, the density of the air
29
lias become — , and in this state let V be the volume of the
weight required to balance the same material. Then
1^—4.] AND RIDERS. 19
V (30/)'-29) (p-l)*
ix. A bright point is at the bottom of still water, and an
eye is vertically above it, at the same distance from the sur-
face ; if a small isosceles prism, of which the refractive angle
i is nearly two right angles, be interposed so as to have its
base in contact with the water, prove that the angular distance
between the images of the point in the two faces is
fi, fi being the refractive indices for water and for the prism
respectively.
Let a ray diverge from the bright point Q, and after pass-
ing through the prism enter the centre of the eye E. Then
this ray makes equal angles {<f)) with the vertical before and
after refraction, because the eye and Q are at equal distances
from the surface of the water. Imagine a very thin layer of
air to be placed between the prism and the surface, the de-
viation of a ray on entering this layer will be — (/i'— 1) <f).
The deviation on passing through the prism will be
Hence the total deviation will be
0^-i)^'-(/-i)<^.
But the total deviation of a ray is the sura of the acute angles
it makes with the vertical before and after refraction, which is
2<f>. Hence equating these, we get
X. Prove that, as the focus of an incident convergent pencil
moves from a concave lens, the distance between the conju-
gate foci always increases, except when the focus of incident
rays passes between the distances /and 2/ from the lens.
C2
20 SENATE-HOUSE PROBLEMS [Jan. A,
In a convergent pencil converging to a distance u,
1_1_1
v~f u'
As u increases from 0 to/,
V 0 ... 00 ,
(— V)— W 0...CX3.
As u increases from / to 2/j
V is positive and decreases from co to 2/,
and v + u co ...4/.
As u increases from 2f to qo ,
V decreases 2/'...0,
v + u increases 4/*... qo .
xi. If the focal length of a convex lens be 3 inches, and
the shortest distance of distinct vision be 6 inches, prove that,
when the eye is always placed so as to see distinctly under the
greatest possible angle, the lens magnifies when within 6 inches
of the object, and diminishes at greater distances.
If PQ be the object on the axis, CQ of the lens. Fig. 10.
PQ
—^ = greatest angle with the naked eye.
I. When the image is on the same side as PQ, \eXpq be
the image ;
J_^J 1 ^
•*• Gq~ CQ 3'
(1) Let C^<6; ... J^> 1 + 1 or 1;
.'. ratio of apparent angles = ^k = 7777 > 1.
(2) Let Cq > 6, or CQ > 2,
then the greatest angle, with lens, = ^ ■ ,
1^—4.] AND EIDERS. 21
ratio of apparent angles — pk- yr = /yT^ ^ ^'
II. When the image is on the opposite side, SLSjy'q'j
ratio of apparent angles = ^-r. = -jSi >
3
.*. the ratio = -j^rri — - > 1 if (7^ < 6,
< 1 if 0<3 > 6.
xii. If the object-glass be divided, so as to form two semi-
circular lenses, and these be displaced along the line of di-
vision, what must be the displacement of the centres in order
that a double star may appear as three stars ?
This is the combination devised to cause a duplication of
an image, and called the Heliometer.
Each half forms an image of each star, 8, S', and 8, 8",
and if one of each pair coincide at 8, the double star ap-
pears as three stars. See Figs. 11 and 12.
If a be the number of seconds in the angular distance of the
stars, F the focal length of each semi-lens, the distance of
the centres (7, C will subtend at the middle of the three
images of the stars an angle 08C ' = a" ;
,'. the distance of the centres = Fa sin 1".
( 22 )
Thursday, Jan, 5. 9 to 12.
Seniob Modebatob. Koman numbers.
JUKIOB MODBBATOB. Arabic numbers.
i. Three concentric circles are drawn in the same plane.
Draw a straight line, such that one of its segments between
the inner and outer circumference may be bisected at one of
the points in which the line meets the middle circumference.
Let 0 be the common centre. Take any point P on the
circumference of the middle circle; join OF and produce to
Q making FQ = OF. With centre Q and radius equal to
that of the smallest circle describe a circle, and let one of
the points in which it meets the outermost circle be B.
Again, with centre Q and radius equal to that of the largest
circle describe a circle, and let one of the points in which it
meets the innermost circle be S. Then if E and ;S^ be taken
properly HP and FS shall be in one straight line which
line will also satisfy the required condition. Fig. 13.
Join OB, QS, OS, QR,
Then ORQSia a parallelogram, because its opposite sides
are equal, and from this, together with the property that
diagonals of a parallelogram bisect each other, the truth of
the proposition is obvious.
ii. If a quadrilateral circumscribes an ellipse, prove that
cither pair of opposite sides subtends supplementary angles
at either focus.
Let ABCD be the quadrilateral ; P, Q, B, T the points of
contact of the respective sides, and S one of one foci. Fig. 14.
9—12.] SENATE-HOUSE PROBLEMS AND RIDERS. 23
Join S with the angles of the quadrilateral and the points
of contact.
By a property of the ellipse ASP= TSA, applying this to
the eight angles at S taken in pairs, we get
A8B + DSG=ASD + B8G;
and since all the angles at 8 are together equal to four right
angles the truth of the proposition is evident.
iii. If a polygon of a given number of sides circumscribes
an ellipse, prove that, when its area is a minimum, any
side is parallel to the line joining the points of contact of the
two adjacent sides.
The polygon of minimum area and given number of sides
circumscribing a circle is the regular polygon, and any side is
therefore parallel to the line joining the points of contact of
the two adjacent sides.
Hence by projecting this circle into an ellipse the truth of
the proposition is obvious.
Or we may prove it thus : Fig. 15.
Let EA, AB, BFhe three consecutive sides of the polygon.
Then if the area is a minimum a small displacement given
to AB while KA and DF remain fixed cannot alter the area
of the polygon. Let AB be displaced to A'B". The point
of intersection of these lines will ultimately coincide with tlie
point of contact P, and we have
AAPA' = ABPB ultimately ;
.-. AP. A'P sin APA' = BP. BP sin BPB,
AP.A'P=BP.B'P;
.'. AP=BP,
since A'P and B'P are ultimately equal to AP and BP respec-
tively.
Hence the diameter which bisects chords parallel to EF
meets the ellipse in P, and therefore the tangent at P is
parallel to EF.
24 SENATE-HOUSE PROBLEMS [Jan. 5,
4. If the tangent at any point P of an hyperbola cut an
asymptote in T, and if SP cut the same asymptote in Q,
then SQ ^QT. See Fig. 16.
If from any point T, two tangents are drawn to a conic,
they subtend equal angles at either focus. One of these tan-
gents in this problem becomes an asymptote, the other is TP.
Therefore if SR be drawn parallel to QT, the angles TSQ,
TSR are equal, and therefore the angles QST, QTS are equal,
or SQ^QT.
5. Prove that the sum of the products of the first n natural
numbers taken two and two together is
{n-l)n{n + l){3n+2)
24
Since
{a + h + c + ...y = a' + h''-\-c''+... + 2{ab + hc+...),
let a = 1, i = 2, c = 3, &c. ... , then we have the sum of the
products
= H("-"-^)'-"-""e'"'1'
which reduces to — — ' ^ — — .
24
6. The centres of the escribed circles of a triangle must
lie without the circumscribing circle, and cannot be equidistant
from it unless the triangle be equilateral.
It may be proved, as in Todhunter's Trigonometry (Art^.^
253), if Q be the centre of the circumscribing circle, P of
any one of the escribed circles, and R, r, their radii, that
PQ^ = I^+2Rr^; whence it follows that PQ must be greater
than R, and the tliree distances cannot be equal, unless the
radii of the escribed circles are equal. The formulte for these
radii are respectively -^ — , -^ — -. , -^ — , where a, b, c are the
)j — a o — O a — C
sides and A the area of the triangle. Whence it follows that
a, i, c are all equal.
This may also be proved independently of the proposition
quoted from Todhunter.
9—12.] AND EIDERS. 26
vii. If perpendiculars be drawn from the angles of an equi-
lateral triangle upon any tangent to the inscribed circle,
prove that the sura of the reciprocals of those perpendiculars
which f.dl upon the same side of the tangent is equal to the
reciprocal of that perpendicular which falls upon the opposite
side.
Let ABC be the triangle, 0 the centre of the inscribed
circle, and P be the point of contact of the tangent in ques-
tion so that OP makes the angle 6 with AO, {0 being the
centre of the circle). Fig. 17.
Then the inclinations of OB and OC to OP produced are
|-^, and| + ^.
Hence remembering that in this case the radius of the cir-
cumscribing circle = 2 the radius of the inscribed = 2r, suppose
perpendicular from A = '2r cos 6 — r = r {2 cos ^ — 1),
and
for
B =2rco3(^-0\+r=r i2QoaC^-e)+l
0 = 2rcos('|+^)+r = r|2cos^| + ^]+ll
+ '
o f"^ , a\ , t o f"^ n\ , 1 2 cos ^ — 1 '
2cosf-+^j+l 2cos(-— ^j+I
+
(2co8^+l)-V38in^ (cos^+1) + V3 sin^
2 (cos 0 + 1)
~l + co3'^-3 8in''^+ 2 co8^
2(C08^ + 1) cos^+1
4cos''^ + 2cos^-2 2cos' ^+008^-1
cos ^ + 1 1
~(2co8^-l) (cos^+l)~2co8^-l ■
viii. Four equal particles are mutually repulsive, the law
of force being that of the inverse distance. If they be joined
26 SENATE-HOUSE PROBLEMS [Jan. 5,
together by four strings of given length so as to form a
quadrilateral, prove that, when there is equilibrium, the four
particles lie in a circle. Fig. 18.
When there is equilibrium, the action of C on -4 : action
of 5 on -4
:: ein DAB : aia DAC,
also action of D on JB : action of -4 on 5
:: Bin ABC : ain BBC;
.'. action of 0 on ^ : action of i> on B
:: sin BAB . sin BBC : sin BAC. sin ABC;
.-. BB : AC :: sin BAB sin BBC : sin BA C sin ABC.
(It is to be observed that the action between A and B is
the difference between the repulsive force and the tension, it
therefore follows no law) ;
AB sin BAB , AB sin ABC
•** sin^Z?^ • sin ACB
:: sin BAB . sin BBC : sin BA C sin ABC,
sin A CB : sin ABB :: sin BBC : sin BA C ;
BO.ainBBC AG sin BAC . ^^^ . „,^
.'. Y)7i * nn '• ^^^ BBC : sin BA C ;
.'. BO.OB = AO.OC, whence, &c.
9. A heavy rod is placed in any manner resting on two
points A and B oi & rough horizontal curve, and a string
attached to the middle point C of the chord is pulled in any
direction so that the rod is on the point of motion. Prove
that the locus of the intersection of the string with the
directions of the frictions at the points of support is an arc
of a circle and a part of a straight line.
Find also how the force must be applied that its inter-
sections with the frictions may trace out the remainder of the
circle. See Fig. 19.
9—12.] AND EIDERS. 27
First, let the rod be on point of slipping at both A and
B, and let F, F' be the frictions at the two points. Then
F, F' are both known, and depend only on the weight and the
position of the centre of gravity of the rod. Since there la
equilibrium the two frictions and the tension must meet in
one point, let this be P. Then since AC= CB, it is evident
that CP is half the diagonal of the parallelogram whose sides
are AP, PB ; hence by the triangle of forces, AP, PB, and
2 . PC will respectively represent the forces in those directions.
Hence AP : PB :: F : F' and are therefore in a constant
ratio. Therefore the locus of P is a circle.
The string CP cuts the circle in two points, but the forces
can meet in only one of these. It is evident that the rod must
be on the point of turning about some one point as a centre,
which point 0 is the intersection of the perpendiculars drawn
to PA, PB at A and B. Now the frictions, in order to balance
the tension must act towards P and therefore the directions of
motion of ^ and B must hejrom P. This clearly cannot be
the case unless the point 0 is on the same side of the line AB
as P. Therefore the angle PAB is greater than a right angle.
Thus the point P cannot lie on the dotted part of the circle.
Secondly. Let the rod be on the point of slipping at one
point of support only. Then since the centre of gravity is
nearer B than A, the rod will slip at A, and turn round B as
a fixed point. Thus the friction acts along QA, and the locus
of P is the fixed straight line QA.
But P cannot lie on the dotted part of the straight line, for
if possible let it be at R. Then if AR represents F, RB must
be less than F' because there is no slipping at B. But because
R lies within the circle, the ratio ^775 is < ^rrr, i.e. <r=; and
' RB PB F
therefore RB > F' and therefore the rod has moved at B which
is contrary to supposition. Thus the string being produced
will always cut the arc of the circle and the part of tlie
straight line in one point and one point only, and the frictions
always tend towards that point when the rod is on the point
of motion.
In order that the locus of P may be the dotted part of the
circle, it is necessary that the frictions should tena one from
28 SENATE-HOUSE PROBLEMS [/an. 5,
Pand the other to P, and the tension must therefore act in
the angle between PA and FB produced. By the triangle of
forces APB we see that the tension must act parallel to AB
and be proportional to it.
X. A rigid wire without appreciable mass is formed into
an arc of an equiangular spiral and carries a small heavy
particle fixed in its pole. If the convexity of the wire be
placed in contact with a perfectly rough horizontal plane, prove
that the point of contact with the plane will move with uniform
acceleration, and find this acceleration.
Let P be the point of contact at any instant and S the
corresponding position of the pole.
Since the curve rolls on the line the instantaneous direction
of the motion of S is perpendicular to SP, i. e. SP is the
normal to the path described by S at S.
But SP is always parallel to itself by the property of the
spiral, and therefore the path of /S' is a straight line inclined
at the angle „ — a to the horizon, where a is the constant
angle between the tangent and radius vector.
Hence the heavy particle is constrained to move along this
line, and the acceleration of gravity resolved in the direction
of the particle's motion is ff cos a.
If S' and P' be consecutive positions of S and P respectively,
it is clear that PP' : >S'^' :: 1 : sin a ;
.*. acceleration of P is —. = g cot a. ^
sm a ^
11. If two parabolas be placed with their axes vertical,
vertices downwards, and foci coincident, prove that there are
three chords down which the time of descent of a particle
under the action of gravity from one curve to the other is
a minimum, and that one of these is the principal diameter
and the other two make an angle of 60° with it on either
side. See Figs. 20 and 21.
The chord PQ do^vn which a particle will slide in tlie
least time from a given point P to a given curve CD,
makes equal angles with the vertical and the normal to the
9—12.] AND BIDERS. 29
given curve at the point Q where the chord cuts the curve.
For the chord PQ will clearly be found by describing a circle
to touch the curve in Q and the centre 0 of which shall be
vertical]y under P. Then it is evident that P Q makes equal
angles with the normal QO and with the vertical PO.
Similarly it may be proved, that the chord of shortest
descent PQ from any curve AB io & fixed point Q, makes
equal angles with the normal at P and with the vertical.
Again, if PQ be the chord of quickest descent from any
curve AB to any other curve CD, by considering P fixed
and Q variable, it is evident that PQ makes equal angles
with the normal at Q and with the vertical. Also by con-
sidering P variable and Q fixed, it is evident that PQ makes
equal angles with the normal at P and with the vertical.
Hence the normals at P and Q must be parallel.
Now the parabolas in the problem are similar, and have
their foci coincident, therefore the normals to the two para-
bolas at the extremities of any radius vector through the
focus are parallel, and no others are parallel. Hence the
chord of quickest descent passes through the focus.
First. To find the chord of quickest descent from the
outer to the inner. We must have the angle SQO = the
angle GSQ, and therefore GQ=GS. But SG = 8Q;
therefore the triangle GSQ is equilateral, and the angle
GSQ = eO\ Fig. 22.
Secondly. To find the chord of quickest descent from the
inner to the outer. We must have the angle GQS= the ex-
terior angle QSA, which is impossible imless SQ coincides
with the axis.
12. If a particle slide along a chord of a circle under
the action of a centre of force varying as the distance, the
time will be the same for all chords provided they terminate
at either extremity of the diameter through the centre of
force. See Figs. 23 and 24.
If a particle describe an ellipse about a centre of force in
the centre C, the time of describing any arc AP from the
vertex A is known to be measured by the angle ACQ, where
30 SENATE-HOUSE PROBLEMS [Jan. 5,
QPN is a common ordinate of the ellipse and the auxiliary
circle. This proposition is still true when the ellipse de-
generates into its major axis and the particle describes the
straight line AC. Thus the time of describing AN is
measured by cob'^yta •
Let AB be any chord, S the centre of force, then, drawing
SC perpendicular to AB, the resolved part of the attraction of
S on any point P is proportional to CF, and therefore the
time of describing AB is measured by cos"^ y^ . But by
. I 1 . CB SB 1 • 1
similar triangles the ratio 77-4= "on which is constant.
Therefore the time down all chords through B is the same.
13. A hollow cone floats with its vertex downwards in a
cylindrical vessel containing water. Determine the equal
quantities of water that may be poured into the cone and
into the cylinder that the position of the cone in space may
be unaltered.
Let AB, CD be the old and new planes of floatation,
cutting the cone in EF, HC. The condition that the posi-
tion of the cone may be unaltered is the volume HF= ^ vol.
CB. Fig. 25.
Let h = OL the part originally immersed, x = LMa = radius
of cylinder, 2a = angle of cone. Then
o
.'. x=-lh± V(ia' cof a - f A").
The lower sign makes x negative and is inadmissible ; this
determines the required quantity of water.
xiv. A hemispherical bowl is filled to the brim with fluid,
and a rod specifically heavier than the fluid, rests with one
end in contact with the concave surface of the bowl, and
passes over the rim of the bowl, find an equation for de-
termining the position of equilibrium.
9—12.] AND RIDERS. 31
In this case let 2a be the length of the rod A C, 2h the
radius of the bowl whose centre is 0, 6 the inclination of
the rod to the horizon, p the relative specific gravity of the
rod and the fluid, A and B the points where the rod rests
against the concave surface and the rim of the bowl respec-
tively.
The forces acting on the rod are,
(1) A force proportional to AB{p— 1) vertically down-
wards through G the middle point of AB.
(2) A force proportional to BC.p vertically downwards
through H the middle point of BC.
(3) A force B along A 0.
(4) A force B' perpendicular to AB at B.
These two last forces obviously intersect in I) the other
extremity of the diameter through A. Let the vertical through
I) meet the rod in E. Then for equilibrium taking moments
about £,
{p-l)AB:GE=p.BC.EH.
AB = 2bcosd, BC =2 {a -b cos 6),
BD" _hs{n^e
AB~cosd '
Substituting, we get
cos'6'-|p|cos'^ + i(/>^! + l)cos<9 + g = 0,
a cubic equation, with its last term positive, whence the
positions of equilibrium may be found, and from which it
appears that the equilibrium can only be possible when all
the roots of the equations are real.
XV. A ray of light passes through a medium of which
the refractive index at any point is inversely proportional
to the distance of that point from a certain plane. Prove
that the path of the ray is a circular arc of which the centre
is in the above-mentioned plane.
The medium is obviously arranged in planes of equal re-
fracting power and parallel to the plane mentioned in the
question.
32 SENATE-HOUSE PROBLEMS [Jan. 5,
And it is clear that the path of a ray is in one plane
perpendicular to the above-mentioned plane.
Let the plane in which the ray's path lies be the plane
of the paper, and let a small portion of the path be RPQ,
AB being the intersection of the plane of reference by the
plane of the paper and HP, and FQ elementary portions of
the path before and after passing through the plane at P,
parallel to AB, and which may tlierefore be considered as
small straight lines. PJfand ^iV perpendiculars on AB^ QN
being produced to L, draw PO perpendicular to PR to meet
the line AB in 0, and join Q0\ See Fig. 26.
1
sin P03f_QiV PM ^
'*• smPQL _1_ QN'
PM
" smPQL Bin POM
But if the perpendicular to PQ met AB in 0' then
am PQL ^^ '
.'. QO'=POy
which is impossible unless 0 and O coincide.
Therefore the normal to the ray at every point of its path
meets the line AB in the same point. Whence, &c.
IG. A small bead is projected with any velocity along a
circular wire under the action of a force varying inversely as
the fifth power of the distance from a centre of force situated
in the circumference. Prove that the pressure on the wire is
constant.
This is a particular case of a more general proposition.
Let a wire be of such a form that a particle, if projected with
velocity F', would freely describe it without causing any
pressure on the wire ; then if the particle be projected with
9—12.] AND EIDERS. 33
velocity V, the pressure at any point where the radius of
curvature is p will be m . The pressure will there-
P
fore vary as the curvature.
For divide the arc into small elements s^ Sg • • • *«» ^^^ ^^^
t\ v^ ...v^he the velocities acquired at the end of those arcs ;
let F^F^... Fn be the resolved parts of the impressed forces
along the respective tangents. Then when the arcs are very
small, we have
v,'-r=2F,s,-]
v^ — v* = 2Fs I
&c. = &c. J
Let r/ v^...vj be the corresponding velocities of the particle
when freely describing the wire, then by similar reasoning
v:'-V'' = 2F,s,+ ... + 2F„s^;
... v:-v^''=T-V'\
Now the pressure on the wire = Statical Pressure + the
centrifugal force = P+ m -^ . But when the particle describes
the curve freely, the pressure = 0, therefore P= — m-^.
.*. Pressure = m ~ ^
P
= m .
P
In the case of a circle described under the action of a
central force varying as the inverse fifth power, we know by
Newton, that the particle if properly projected would not exert
any pressure on the wire. Therefore, when otherwise pro-
jected, the pressure varies inversely as the radius of curvature,
that is, it is constant.
17. A bright spot of white light is viewed through a
right cone of glass, the vertex of which is pointed directly
D
34 SENATE-HOUSE PROBLEMS [Jan. 5,
towards the spot. Describe the appearances seen ; and prove
that, if a section of the locus of the images corresponding
to different values of the refractive index be made by a
plane through the axis of the cone, it will be a rectangular
hyperbola.
Let AB be the axis of the cone, fig. 27, A the vertex, 2a the
angle of the cone, h the height, Q the bright spot, AQ = u.
By the ordinary optical formulae it can be easily proved
that the image of Q, formed by light of refractive inaex fi,
will be a ring whose radius and position is given by the
formulse
qn = (jj,— 1) u sin a cos a ]
„ _u+h+ {fi—l)u sin' a> •
^~ ^i J
Hence
{qn + w sin a cos a) {Bn — u sin' a) = (w cos' a + h)u sin a cos a.
Therefore, by Goodwin's Conies, Prop, ix., q lies on a
rectangular hyperbola whose asymptotes are parallel and per-
pendicular to BA. And since the position and magnitude of
this hyperbola is independent of //., all the coloured rings will
lie on the surface formed by the revolution of this hyperbola.
xviii. An elastic string passes through a smooth straight
tube whose length is the natural length of the string. It is
then pulled out equally at both ends until its length is
increased by \/2 times its original length. Two equal per-
fectly elastic balls are attached to the extremities and pro^
jected with equal velocities at right angles to the string, and
so as to impinge upon each other. Prove that the time of
impact is independent of the velocity of projection, and that
after impact each ball will move in a straight line, assuming
that the tension of the string is proj)Oi1;ional to the extension
throughout the motion.
Let AB be the tube, C and D the positions to which the
ends of the string are extended.
Each particle describes an ellipse round the coiTcsponding
extremity of the tube as centre, the absolute force depending
on the material of which the string is composed.
9—12.] AND KIDERS. 35
The line CD will coincide with the major or minor axes
of the ellipses according to the magnitude of the initially
impressed velocity, and the particles will impinge at a point P
in the line PE bisecting AB at right angles. See fig. 28.
(1) If CD be the direction of the major axes the arc
of the auxiliary circle described by either particle is
, -xa EP
_,a h >^{2.AE'-AE') ^ _. Stt
or IT — tan r • - • — ^ Trf ='jr— tan 1 = — .
0 a AE 4
(2) If CD be the direction of the minor axes, then the
corresponding arc of the auxiliary circle is
or
or
or
therefore in both cases the time of impact is independent of the
velocity of projection. Let the tangents to the two curves
before impact at P be PT and PT' meeting AB in T
and T';
then AT.AE=2AE^;
.'. AT=2AE,
= AB;
therefore T and T' coincide with B and A respectively, and
therefore since the velocities parallel to AB are reversed at
impact, those perpendicular to AB remaining imaltered, it is
d2
2+**° i
AE
EP'
TT , _, a
AE
~,^{b'-AE')
l + tan-l;
(V 5 = V2.
Stt
4 '
36 SENATE-HOUSE PROBLEMS AND RIDERS. [Jan. 5.
clear tliat the direction of each particle's motion after impact
passes through A and B respectively.
xix. A particle is projected along a chord of an ellipse
from any point in the curve, and wlien it again meets the
ellipse has a certain impulse towards the centre of the ellipse
impressed upon it, causing it again to describe a chord, and
so on for any number of times. Prove that, if after a given
number of such impulses the particle pass through another
given point on the circumference of the curve, the polygonal
area so described about the centre is a maximum, when the
successive chords are described in equal times.
Since the particle leaves one given point on the curve, and
Yjasses through another given point after touching a given
number of points on the curve, (see fig. 29); then in order
that the polygonal area described about the centre should be
a maximum every such triangle as FQB must also be a
maximum, P and M being fixed and Q variable. Hence, if
we take a point Q' near to Q, the triangle RQ'P must be
equal to RQP, and therefore the tangent to the ellipse at
Q must be parallel to PR. Hence by the property of the
ellipse if QT' be the direction of central impulse at Q, QT
bisects RP.
Produce PQ to S, making QS equal to PQ. Then RS is
parallel to QT.
Now QS represents the original velocity at Q in direction
and magnitude, RS represents the direction of the impressed
velocity, and QR of the resultant velocity after the central,
impulse; therefore QR is proportional to the magnitude or
the resultant velocity, and therefore time through QR equals
time through PQ, and so on.
( 37 )
Thursdaf, Jan. 5. 1 to 4.
Senior Exahineb. Roman numbers.
Junior Examiner. Arabic numbers.
2. Enunciate and prove Newton's tenth Lemma.
If the curve employed in the proof of this Lemma be an
arc of a parabola, the axis of which is perpendicular to the
straight line on which time is measured, prove that the ac-
celerating effect of the force will vary as the distance from
the axis of the parabola.
Let time be measured along the line AN from the point
N, and let V be the vertex of the parabola. Then, at the
instant corresponding to F, the time is represented by AN^
and the velocity by FN. See fig. 30.
Now, if Z be the latus rectum of the parabola,
FM' = L.VM,
AK' = L.VK;
.: FM'-AK' = L.FN,
or AN{AK+AN) = L.FN;
.'. when F approaches indefinitely near to A
,. . FN 2AK
^'"^''AN-^-ir^
or varies as the distance of A from the axis of the parabola.
38 SENATE-HOUSE PROBLEMS [Jan. 5,
Again, if P be not indefinitely near to A, and P' a point
contiguous to P, it may be shewn that the force at P
^^''^'' NN'
= tan FXN, if FX be the tangent at P,
_2.VK
~ MF
2. MP
~ L '
or varies as the distance of P from the axis.
3. One circle rolls uniformly within another of twice its
radius; prove that the resultant acceleration of a particle
situated on the circumference of the rolling circle tends to the
centre of the fixed circle, and varies as the distance from that
centre.
Let 0 be the centre of the fixed, C of the moving circle, P
the point, the acceleration of which is required. See fig. 31.
Now P describes a circle uniformly round C while C de-
scribes a circle of equal radius, and in the same time, uniform-
ly round 0.
Hence the acceleration of P is made up of a constant accele-
ration in the direction (7P, and of an equal constant acceleration
in the direction CO. Therefore its whole acceleration will be
represented in magnitude and direction by 0F\ or tends to 0^
and varies as the distance from 0.
Note. If the point P be fixed relatively to, though not on
the circumference of the moving circle, it may be proved in a
similar manner that its acceleration will still tend to 0, and be
proportional to OF. In this case, it may be geometrically
proved that the path of P will be an ellipse of which 0 is the
centre ; hence we learn that if the acceleration of a moving
point tend to a fixed point, and vary as the distance from it,
its path will be an ellipse of which the fixed point is the
centre; the converse of Newton, Sect. ii. Prop. 9.
1 — 4.] AND RIDERS. 39
iv. Prove that, when a body moves along a smooth tube
under the action of any force tending to a point and varying
as the distance from the point, the difference of the squares
of the velocities at the beginning and end of an arc varies
as the difference of the squares of the distances of the ex-
tremities of the arc from the fixed point.
The acceleration in PQ = fi . SP. -pr-. , ultimately, fig. 32 ;
An
.-. (vel.)' at ^ - (vel.)' at P= 2fi8P. —^ . PQ
= 2fj,SP . Pm, ultimately,
= ^l{SP-\-SQ){SP-SQ)
= fi {SP^- SQ'), ultimately;
.'. by Lemma IV.
(vel.)* at SA - (vel.)» at 5 = /i {SA' - SB').
V. A body is revolving in an ellipse under the action
of such a force, and when it arrives at the extremity of the
major axis the force ceases to act until the body has moved
through a distance equal to the semi-minor axis, it then
acts for a quarter of the periodic time in the ellipse ; prove
that, if it again ceases to act for the same time as before,
the body will have arrived at the other extremity of the
major axis.
The velocity a.t A = <^fiCB = velocity at D. Fig. 33.
The body on arriving at I) proceeds to describe an ellipse of
which CD and CSsire semi-conjugate diameters, and in a quar-
ter of the periodic time it arrives at B and moves with velo-
city tJfiCB in direction parallel to DC and therefore towards
o, and arrives at a in time -; — 7^7^= -, — ttp^, the time from
*/fi CJJ */fiBu
Ato D.
vi. When a body revolves in an ellipse under the action
of a force tending to the focus, find the velocity at any point
of its orbit, and the periodic time.
40 SENATE-HOUSE PROBLEMS AND RIDERS. [Jan. 5.
If on arriving at the extremity of the minor axis, the
force has its law clianged, so that it varies as the distance,
the magnitude at that point remaining the same, the periodic
time will be unaltered, and the sum of the new axes is to
their difference as the sum of the old axes to the distance
between the foci.
-^ = fi'SB, SB = a;
27r _ 27rrt*
{\e\.YsitB=-=fjL'a^, where a = semi-diameter parallel to the
major axis of the old orbit ; if a, /S be the semi-axes of the
new orbit. Since SB and the semi-conjugate diameter each
equals a
a/3 = ab,
{a±^y = 2a{a±b);
.'. a 4-/8 : a-/3 :: ^/{a + b) : s/{a-b)
:: a + b : V(a*-&*)
:: 2{AC + BC) : SH.
( 41 )
Monday, Jan. 16. 9 to 12.
Seniok Modkbatob. Arabic numbers.
Seniob Examixeb. Roman numbers.
1. A UNIFORM heavy ellipsoid has a given point in contact
with a smooth horizontal plane. Find the plane of the
couple necessary to keep it at rest in this position ; and
investigate its equation referred to the principal axes of the
ellipsoid.
The ellipsoid is acted on by its weight vertically down-
wards, through the centre, and the normal action of the
horizontal plane vertically upwards, through the point of
contact.
The plane of the required couple must therefore be the
vertical plane which passes through the centre and the point
of contact or be parallel to this plane. If ar, y, z be the co-ordi-
nates of the point of contact referred to the principal axes
of the ellipsoid, the equations of the normal at tnat point are
^-a? ^ v-y ^ .y-g
X y z '
a* h* ?
Hence the plane sought must contain this line, and pass
through the centre, or be parallel to the plane thus deter-
mined.
42 SENATE-HOUSE PROBLEMS [Jan. 16,
Its equation is therefore easily found to be
where d is any arbitrary constant.
2. An oblong taible has the legs at the four comers alike
in all respects, and slightly compressible. Supposing the
floor and top of the table to be perfectly rigid, find the
pressures on the legs, when the table is loaded in any given
manner, supposing the compression to be proportional to the
pressure ; and prove that, when the resultant weight lies in
one of four straight lines on the surface of the table, the
table is supported by three legs only.
Let ABGD be the top of the table, the sides AB and AD
being 2a and 26 respectively. See fig. 31.
Let the natural length of each leg be c, and let P be the
position of the resultant weight; the co-ordinates of P, referred
to AB and AD as axes, being x and y.
Let Pj Pj, P, P^ be the pressures at the angular points,
and let z^ z^ z^ z^ be the altered lengths of the legs.
Then
^^^ = XP, or «=c-XcP,
and so forth, also neglecting quantities of the second order
we shall consider the pressures at the points ABGD to re-
main vertical ;
.-. P^4-P, + P3 + P,= TF (1),
(P, + PJa;-(P, + P,)(2a-a:)=0 (2),
(P, + PJy-(P3 + P,)(26-y)=0 (3),
and since the base and top of the table remain rigid, the
height of the intersection of its diagonals is
H^a + ^J, or i (a, 4- 2,);
.-. 2!,+ 2, = 21 + ^8;
.-. P. + P, = P. + P3 (4).
9—12.] AND RIDERS. 43
By elimination between these four equations, we get
» 4 Wb J'
with similar values for the other pressures.
If P, = 0, the weight must lie in the line
being a line parallel to BI), and bisecting AB and AD.
Hence, when the weight lies in one of four straight lines
parallel to the diagonals of the table, the table is supported
bj three legs only.
3. Find the equations of equilibrium of a perfectly flexi-
ble uniform inextensible string when acted on by any given
forces.
If a small rough heavy bead be strung upon such a string,
and the string be suspended from two points and acted on
by gravity only, write down the equations for determining
within what portion of the string it is possible for the bead
to rest.
Let AP and BP be the two catenaries into which the
string is divided when the weight rests in one of its limiting
positions of equilibrium as at P. See fig. 32.
Let fi be the coefficient of friction, w the weight of the
ring, T and T' the tensions in the two portions of the string,
0 and 6+ <!> the inclinations of the tangents at the point
P to the horizontal.
The weight cannot affect the horizontal tension, and there-
fore the parameter c must be the same in both catenaries.
Our equations are therefore
T' = Tel'*,
or sec (^ + <^) = sec ^e*^ (1),
c{tan^ + tan(^ + <^)| = w (2),
c sec a— c sec {0+<f>) - csecy3 + c8ec^= h (3),
ctana — ctan(^ + ^)4-ctan)S — ctan^= I (4),
44 SENATE-HOUSE PROBLEMS [Jan. 16,
h and I being the difference in height of A and B, and the
length of the string respectively, and a and /Q the inclinations
to the horizon at the points B and A respectively ; we have
also a fifth geometrical equation indicating that the horizontal
distance between A and B is given, and involving no ad-
ditional unknown quantity. These five equations determine
0y <p, a, ^, and c, and then BP and AP are known.
iv. A particle is attached by a rod without »mass, to the
extremity of another rod, n times as long, which revolves in a
given manner about the other extremity, the whole motion
taking place in a horizontal plane. If 6 be the inclination of
the rods, o) the angular velocity of the second rod at the time
t, prove that
<Pd da) fdo) y. 2 • /i\ ^
dt" at \dt J
If a be the length of the rod without weight, na that of the
rod whose angular velocity is w at the time <,
The angular velocity of the particle about the point of
junction
d9
= T^ + ^-
The acceleration of the point of junction is
dco
'"''-di
perpendicular to the rod na and naa>^ towards the fixed point.
The relative acceleration of the particle perpendicular to
the rod a
(d^ d(D
'"'[de'^dt
therefore since the whole acceleration in that direction is zero
d^O dto (da a , 2 ' /i\ n.
V. A bead is capable of free motion on a fine smooth
wire in the form of any plane curve, and is acted on by
9—12.] AND EIDERS. 45
given forces ; compare the pressure on tlie wire with the
weight of the bead.
If the wire be a horizontal circle, radius a, and the bead
be acted on only by the tension of an elastic string the
natural length of which is a, fixed to a point in the plane
of the circle at distance 2a from its centre, find the con-
dition that the bead may just revolve ; and prove that in
this case the pressures at the extremities of the diameter
through the fixed point will be twice and four times the
weight of the bead if that weight be such as to stretch the
string to double its natural length.
Let S be the point to which the string is fixed, C the cen-
tre of the circle, SA CB a straight line meeting the circle in
A, B. See fig. 33.
If W be the weight of the particle, r the length of the
string, a the radius of the circle,
the tension = W. ,
a
and its accelerating efiect = g . .
If V be the velocity of the particle when the length of the
string is r, w, v the velocities at B and A^
v^ — u^ = — 2g I dr ;
.-. vT-v'^g- '- ;
.". u^=4:ag.
u*
Pressure at A is that due to the acceleration —
a
= 4Tr.
Pressure on J? is the tension of the string at B
= 2W.
9. A distant circular window is viewed by a short-
sighted man through his eye-glass, the axis of which passes
46 SENATE-HOUSE PROBLEMS [Jan. 16,
through the centre of the window and is perpendicular to its
plane. Prove that the image of the window formed by-
primary focal lines will be spherical, provided the window be
filled with concentric rings of stained glass, and the refractive
index of the colour throughout any ring be
(/z-l)(2/t+l) r^
fi being the index of the central colour, r the radius of the
ring in question, and d the distance of the window from
the lens.
When a small pencil of parallel rays passes centrically and
with small obliquity {j>) through a lens, the distance of the
primary focal line from the centre of the lens is
where j= {fi-l) (^-- -^ .
In the case supposed, let f^ be the focal length for the
direct ray and x the refractive index for rays which are in-
cident at the angle <^ ; therefore to satisfy the required con-
dition we must have the following relation between <j> and x,
••i-ii + 2ij'''-7-;rrT'
. .•.x-i=/x-i-(M-i)f-^).«',
putting /Lt for x in the coefficient of the small quantity ^*, and
/"* . .
substituting -^ as the obvious equivalent of that quantity.
9—12.] AND RIDERS. 47
11. Find the parallax in right ascension of a heavenly
boclj, in terras of the latitude of the place of observation, and
the hour angle and declination of the body, assuming the
distance of the body from the Earth to be so great that the
sine and circular measure of the parallax may be considered
equal.
Shew that the locus of all the bodies, which on this
assumption have their parallaxes in right ascension for a
given place and time equal to a given quantity, is a right
circular cylinder touching the plane of the meridian of the
place along the axis of the heavens.
The parallax in R. A. for bodies satisfying the condition
mentioned in the question is
a cos Z sin A
r ' cos h '
a being the Earth's radius, r the distance of the body, I the
latitude of the place, h the hour angle, and h the declination
of the body.
If this be equal to a given quantity m,
a cos I sin h
cos 8
= m
J 2^ a cos I . ■, 5,
or r cos o = . r sm A cos o.
m
Refer to the plane of the equator as that of xy, the axis
of y being perpendicular to the meridian ;
, , a cos /
proving the proposition.
( 48 )
Monday, Jan. 16. \\ to 4.
Junior Modeeatob. Ai*abic numbers.
Junior Examiner, Roman numbers.
1. Shew how to expand a' in a series of ascending powers
of x.
Prove that the series
^ "^ 172 "^ T7273 "^ 1 . 2 . 3 . 4 "^ • • • " ^^•
It may be easily proved by direct multiplication that the
series
X-
whence, putting a; = 1, we have
2' 3'
An expression may also be found for the general series
2" 3"
'^1.2"*"l.2.3'^
Since
,.=i+,.+j£l+_i'L+
1^ — 4. SENATE-HOUSE PROBLEMS AND RIDERS. 49
expanding each term, the required series is easily seen to be
the coeflScient of f in the expansion of e*'. By Herschel's
Theorem this is equal to e^^^.O". Hence the required expres-
sion is
vi. Find the value oi p, in order that the straight line
represented by the equation a; cos^+y sin^=p may touch
the ellipse
8 2
a b
Prove that the locus of the vertices of an equilateral
triangle described about the ellipse -^ + ^ = 1 is given by
the equation
4 (JV+oy-a'J*) =3 (a;«+/-a»-57.
Let X, y be the co-ordinates of one of the vertices of an
equilateral triangle described about the ellipse, 6 the incli-
nation to the axis of y, of a tangent drawn through it. We
have then, for the determination of 6, the equation
a; cos ^ + y sin ^ = {a^cos^d + S^sin'^)*,
which, rationalized, gives
(y*- b') tan'^ + 2xy tan $ + {a?- a") = 0.
If 6^, 6^ be the two values of 0 given by this equation, we
must have, in order that the two tangents represented by it
TT
may be inclined at an angle — ,
TT
that is,
Now
tan(^,-^J=tanJ = V3,
tan 6, — tan 6^ _
1 +tan^,.tan^.
tan^,+ tan^,= -^^,,
50 SENATE-HOUSE PROBLEMS [Jan. 16,
2xyV j^-a'
^'•^-^^3;
. 4a;y-4(a;'-a')(y-y)_
or 3(a^+y-a'^-&y = 4(6V+ay-a''J*),
a relation between x and y, which gives the required locus*.
8. Prove that, if a straight line be drawn from the origin
to cut the straight line — j— = '^- = at right angles,
" I m n
its equations will be
X
, y_
a — It h — mt c — nt^
, rt? + hm + en
where t = ^u— — 2~, — 2 •
Let the given line be
X — a _y — h _z — c _
I m n '
and the required line
X _y _z
\ fi v' -^
Then, since these must intersect and also be at right angles,
we have
a—lt_b — 'mt_c — 'nt \
Vk + myi. + nv = 0, J
* It will be observed that this curve consists of two closed i>ortion8, one
wholly within the other. The outer one alone satisfies the condition of the
question. The ianer is the locus of the intersection of tangents, inclined to
one another at an angle -^ .
1^—4.] AND EIDERS. 51
, ^ al+ bm + en
whence t = -« » § ,
and the ratios \ : fi : v are found.
9. If a, /S, 7 "be the distances of a point from the three
faces of a tetrahedron which meet in the vertex, prove that
the equation of the plane passing through the vertex, and
through the centres of the circles inscribed in and circum-
scribed about the base, is
(cos B — cos C) PjCl + (cos G — cos A) p^^
+ (cos A — cos B) 'p^ = 0,
where A, B, C are the angles of the base, and p^,^^-,^^ the
perpendiculars from the vertex on the sides of the base.
Let the equation to the required plane be
ia + if^ + iVV = 0.
Let r = radius of inscribed circles and let (a8), (/8S), (7S) be
the angles made by the three faces 0/87 with the third face 3,
tlien a = r sin (aS) , /8 = r sin (j88) , 7 = r sin (7S) must satisfy
this equation ;
/. L sin (aS) + if sin (/38) + iVsin (78) = 0.
But the volume of the pyramid = Jj? sin (78), area of base 8,
with two similar expressions. Therefore substituting for
sin (78), 8in(/38), 8in(aS), we get
L M N ^
p p p
Again, if ^ = radius of circumscribing circle, then
a = B cos A sin (aS),
and two similar expressions for /S and 7 must satisfy tlie
equation ;
Leon A McosB NcoaC_
P P P
E2
52 SENATE-HOUSE PROBLEMS AND EIDERS. [Jan, 16.
Hence by cross multiplication we find
L ^ M N
(cos B — cos C)p (cos G — cos A) p (cos A — cos B) p" '
10. Find the equation of the sphere, passing through a
given point and through the circle in which the polar plane
of that point with respect to a given sphere cuts that sphere.
Let the equation to the given sphere be
os' + f + z' = a!',
and the co-ordinates of the given point g, h, k; then the
equation to the plane of contact is
and the equation to every sphere passing through the inter-
section of these two is included in
a^ + y" + z'' - a" -\ {ffx + hi/ + kz -a") = 0.
But by the question, this goes through x=g, y = h, z= k.
Hence by substitution \= 1, and the required sphere is
a? + y' + z^ = gx + hi/ + kz.
11. If a sphere touch an ellipsoid and also cut it, the
common section cannot be a plane curve unless the point of
contact be one of four fixed points on the ellipsoid.
When two surfaces of the second degree intersect, if one
intersection be a plane curve, the other is plane also. Hence;
as all the plane sections of a sphere are circles if a sphere cut
an ellipsoid in a plane curve, both that curve and the other
intersection must be circular sections. Hence in the limit
when the sphere touches the ellipsoid, it must touch it at the
four umbilici or the points which are the evanescent circular
sections.
( 53 )
Tuesday, Jan. 17. 9 to 12.
Senior Modkbatob. Arabic numbers.
JONIOB MoOEBATOB. Boman numbers.
i. Find a point the distances of which from three given
points, not in the same straight line, are proportional to p^
q, and r respectively, the four points being in the same plane.
Let A, B and G be the three points.
Divide AB in D so that AD : DB :'. p : q, and in AB pro-
duced take a point E such that
AE ', BE '.: p '. q)
upon BE, as diameter, describe a circle.
Every point upon this circle has its distances from A and B
proportional to^ and q respectively.
Describe a similar circle relative to A and C.
The points of intersection of these circles, when intersection
is possible, satisfy the required condition.
2. If rP, TQ be two tangents drawn from any point
T to touch a conic in P and Q^ and if S and H be the
foci, then
SF.SQ~HF.BQ'
64 SENATE-HOUSE PROBLEMS [Jan. 17,
The construction of the figure being indicated by the ques-
tion, we have, see fig. 34,
SP_ sin STP SQ_ sin STQ
ST sin SPT' ST~ sin SQT'
SP.SQ sin STP. sin STQ
Similarly
• ST' ~ sin SPT. sin SQT'
HP. HQ _ sin HTP. sin HTQ
HT' ~ sin HPT. sin HQT'
Now these angles STP, HTQ are known to be equal ; and
the angles SPT, HPT axe supplements, and also the angles
SQT, HQT. Hence the above two expressions are equal,
and hence
SP.SQHP.HQ
ST* - Jljr2 .
If the conic become a parabola, these expressions become
each equal to unity.
iii. A polygon is inscribed in an ellipse so that each side
subtends the same angle at one of the foci. Prove that, if
the alternate sides be produced to meet, their points of in-
tersection will lie on a conic section having the same focus
and directrix as the original ellipse, and that the chords
joining the consecutive points of intersection all subtend the
same constant angle at the focus as the sides of the original
polygon.
The polar equation of a chord of an ellipse, the focus being
the pole and the line to the nearer vertex the prime radius, is '
- = e cos ^ + sec /3 cos (a — 6),
r
2^ being the angle subtended by the chord at the focus, and
a — fi, a + fi the angles corresponding to its extremities.
If this be taken as the equation of one of the produced
sides of the polygon in question, then the side which mtersects
it has for its equation
- = c cos ^ + sec ^ cos {a + 4^ — 0);
T
9—12.] AND RIDERS. 55
therefore at the point of intersection
cos (a - ^) = cos (a + 4/3 - ^),
^ = a + 2/3,
a=^-2A
- = e cos ^ + sec i8 cos 2/9,
r
I I e cos 0
sec /3 cos 20 ' r sec /3 cos 2y8
the equation to a conic section having the same focus as the
ellipse, but latus rectum and eccentricity each altered in the
ratio of 1 : sec /8 cos 2/3.
Hence the directrix is the same as before.
Also since
^ = a + 2/9,
it follows that the line joining the focus with the point of in-
tersection of any two alternate sides bisects the angle subtended
by the intermediate side, and therefore the sides of the new
polygon each subtend the angle 2/3 at the focus.
This proposition may be also solved in an obvious manner
by the method of reciprocal polars.
4. Prove that the equiangular spiral is the only curve
such that its radius of curvature is proportional to the re-
ciprocal of the radius of curvature at the corresponding point
of the reciprocal polar.
Let PY be the tangent to one curve, QZ the tangent to the
corresponding point of the reciprocal polar. See fig. 35.
Let OP=r, OY=p, and OQ = r'OZ=p'. Then/r = /c»,
pr = /c*, where k is some constant. Also if pp be the two
radii of curvature at P, Q,
p =r
jS)
66 SENATE-HOUSE PROBLEMS [Jan. 17,
Hence by the question — 3- = constant, but - = sin OPY ;
therefore the angle OFY= constant, or the curve is the equi-
angular spiral.
5. If two plane sections of a right cone be taken, having
the same directrix, the foci corresponding to that directrix lie
on a straight line which passes through the vertex.
Let ABC be the given cone ; and a perpendicular through
L to the plane of the paper the given latus rectum. Fig. 36.
Draw L CB perpendicular to the axis of the cone, and de-
scribe a sphere touching the cone in B and C; draw LP, LQ
tangents to the sphere in the plane of the paper, we know
that P and Q are the foci of the two sections.
Now PQ is the polar line of L, and BCL the polar line of
A. Because the pole of PQ lies on BCL theretore the pole
of PQ, therefore PQ, passes through A.
vi. Find the equation of the envelope of the perpendiculars
to the successive focal radii of a parabola drawn through the
extremities of these radii.
Refer to polar co-ordinates, the focus S being the pole and
AS the prime radius.
If ^ Q be a radius vector to the parabola perpendicular to
which any line is drawn, and if J^ ^ be inclined at the angle..'
<f> to AS, its length being p, we have
p = osec*|,
if 0 and r be the polar co-ordinates of the point where the
line perpendicular to ^^ is intersected by its consecutive,
we have
r =pBec{0 — ^) =a sec" ^ sec {$ — <f)),
or - = cos" ^ cos [B — 6),
r 2
9—12.] AND EIDERS. 57
and 0 = — cos ^ sin ^ cos (^ — </>) + cos' ^ sin {6 — <f>)
^ sin {0 — <f))= sin ^ cos {0 — (f>);
(«-3f) = 0; .-.^^l.e;
= cos -^ sm
sin
^-<^ = 1^;
the equation required.
a .6
r = a sec - ,
o
vii. If two concentric rectangular hyperbolas have a com-
mon tangent, the lines joining their points of intersection to
their respective points of contact with the common tangent
will subtend equal angles at their common centre.
Let 0 be the common centre.
Let OX and 0 F be the asymptotes of one hyperbola,
OX' ... OY' the other.
Let XOX' = 2a, and let the line OC which bisects this
angle be taken as the prime radius.
Let r cos {d — ^) = b be the equation of the common tan-
gent, b being the perpendicular upon it from the centre, and
/S the angle between this perpendicular and the prime radius.
Then if P and P' be the respective points of contact of this
tangent with the hyperbolas, it follows from the geometry of
the hyperbola that
C0P = -{2a + fi),
and COP'= (2a -yS).
Now the equations of the hyperbolas are
r* cos 2 (^ + a) = a\
r* cos 2 (^ - a) = a*y
58 SENATE-HOUSE PROBLEMS [Jan. 17,
and combining tliese with r cos [d — ^) = b, or the equation of
the common tangent, we obtain
OP=^cos2(a+^)=a',
OP'cos'2{a + ^) = b';
therefore cos 2 (a + /S) = -? ,
a
and similarly cos 2 (a — /9) = — , .
a
Also if 0^ be the value of 0 corresponding to Q^ the point
of intersection of the two hyperbolas, we obtain
cos 2 {0^ + g) _ cos 2 (^, - a) ,
a a^
therefore ^^^Ij^ = ^, = 52ii(^ ,
cos 2 (^, - a) a/ cos 2 (a + /9) '
whence we obtain tan 20^ = — tan 2/3),
and hence P0Q = 2ri = P' OQ.
viii. If P be a point on a geodesic line AP, drawn on a
conoidal surface, s the length of AP, or, N, and 0 the projec-
tions of s, P, and the axis on any plane perpendicular to the
axis, and p the projection of ON on the tangent to AP at P,
then
dp _d<T
da ds '
Since the geodesic line on any surface satisfies the condition
that its oscillating plane at any point contains the normal to
the surface at that point, we have for its differential equations
d'x dz d*z _
ds^ dx ds* '
d^y dz d^z _
'd^'^djd?~'
d'x dSf d^z ( dz dz\ ^
9—12.] AND RIDERS. 59
In a conoidal surface if the axis coincide with the axis
of «,
dz dz
d^x d^y ^
""^ ds'\^ds^y ds)-\ds) ^\ds)'
doc duU
But 2; -7- +^ ->- = the quantity denoted by j^ in the question,
and
(dc^\(dy\_(d^\\
\ds] "^ \ds) ~ \dsl '
dp _ fdaV
dp da-
da ds '
9. A string is placed on a smooth plane curve under the
action of a central force F, tending to a point in the same
plane ; prove that, if the curve be such that a particle could
freely describe it under the action of that force, the pressure
of the string on the curve referred to a unit of length will be
jPsin <^ c
where ^ is the angle which the radius vector from the centre
of force makes with the tangent, p is the radius of curvature,
and c is an arbitrary constant.
If the curve be an equiangular spiral with the centre of
force in the pole, and if one end of the string rest freely on
the spiral at a distance a from the pole, then the pressure is
equal to
/A sin <^ /I 1\
60
SENATE-HOUSE PROBLEMS
[Jan. 17,
Let T be the tension at any point, R the pressure referred
to a unit of area, then by the ordinary equations for the equi-
librium of a string
+ F sin <f> = M;
.'. B = F8hi(f} +
fF cos <f)ds
(1).
Let V = velocity of the particle freely describing the curve,
then
;^=-2i?C0S«^,
— = F8m6:
P
' 0 - -^^^^ ^ JFcos <^ — c
2 p
(2).
Subtracting (2) from (1),
2 p
— ^ normal force + - .
The quantity c depends on the tightness with which the
string is tied. If one end be free it is to be determined from'
the condition that jr=0 at that end.
If the curve be an equiangular spiral, ^ is constant and
jP=^. Also from (2)
^ F»m<t> T-c
therefore when r — a, and p sin <^ = r, we have
_fi sin <f> c sin <f>
9—12.] AND RIDERS. 61
2r \r or J
10. If a string, the particles of which repel each other
with a force varying as tne distance, be in equilibrium when
fastened to two fixed points, prove that the tension at any
point varies as the square root of the radius of curvature.
The law of attraction being as the distance, the attraction
of the whole arc on any particle is the same as if the whole
mass was collected at its centre of gravity. Take this point
for origin, and let T be the tension at any point whose radius
vector is r, let p be the perpendicular on the tangent and p
the radius vector. Then the equations of equilibrium will be
dT=-tirdr,\
" p'"" 2 '
1 dr
'''-f^-^'dp^
1
'^ p
.'. roc - oc vp.
p
11. If any uniform arc of an equiangular spiral attract a
particle, placed at the pole with a force varying inversely as
the square of the distance, prove that the resultant attraction
acts along the line joining the pole with the intersection of
the tangents at the extremities of the arc.
Prove also that, if any other given curve possess this same
property, the law of attraction must be F= ^ ->- > where ^ is
the perpendicular drawn from the attracted particle on the
tangent at the point of which the radius vector is r.
It may be easily proved, if a string be in equilibrium in
the form of an equiangular spiral under the action of a force
62 SENATE-HOUSE PROBLEMS [Jan. 17,
in the pole, that the force must vary as -jt-- — r«. And the
^ '' (dist.)
resultant repulsion of the centre of force on the string, being
in equilibrium with the tensions at the extremities, must pass
through the intersection of the tangents at the extremities of
the arc. Now let the arc become a rigid wire attracting the
pole, then all the forces are unaltered, and therefore their
resultant, just as before, acts along the line joining the pole
with the intersection of the tangents at the extremities of
the arc.
In the same way, if any other curve possesses this pro-
perty, the law of attraction must be such that a string may
rest in equilibrium under a centre of force whose law of
repulsion is the same. By writing down the equations of
equilibrium it is easily seen that this must be the law
p' dr '
xii. A material particle is acted on by a force the direc-
tion of which always meets an infinite straight line AB at
right angles, and the intensity of which is inversely propor-
tional to the cube of the distance of the particle from the
line. The particle is projected with the velocity from infinity
from a point P at a distance a from the nearest point 0 of
the line in a direction perpendicular to OF^ and inclined at
the angle a to the plane AOP. Prove that the particle is
always on the sphere of which 0 is the centre, that it meets
every meridian line through AB at the angle a, and that it
reaches the line AB in the time
V/* cos a '
fi being the absolute force.
Suppose first that the particle is constrained to move in
contact with the smooth surface of the sphere mentioned in
the question. Then if r be the distance of the particle from
the attracting line at any instant, and </> the azimuth of the
plane containing r, we have, by taking moments round the
axis,
4^=a
9—12.] AND RIDERS. 63
Also if V be the corresponding velocity of the particle, we
have
J r T T
Since the velocity is that from infinity and therefore C"=0,
dt s/fi '
But r-^ is the resolved part of the whole velocity per-
pendicular to the meridian. Hence it follows that the path
on the sphere is always inclined to the meridian at an angle
C
whose sine is -r- ; therefore if 6 be the angular distance of
the particle from the pole of the attracting line,
dd Ju
a ,- = — v cos a = — cos a
dt r
dt ar a^ sin 6
dd \Jfjb . cos a Va*" • ^^^ ^ '
.-. <= , "' [\xx,ede= , ^' .
V/A cos a j Q V /* cos <*
Also the normal pressure = difierence betweeu the force
veh}'
resolved along the radius and — '—
° a
= ^8in^--^ = 0,
r ar
whence it follows that the particle describes the path men-
tioned in the question.
13. If a particle slide along a smooth curve which tunis
with uniform angular velocity g> about a fixed point 0, then
the velocity of the particle relatively to the moving curve
is given by the equation
64 SENATE-HOUSE PBOBLEMS [Jan. 17,
v' = c' + a>V',
where r is the distance of the particle from the point 0 ; and
the pressure on the curve will be given by the formula
— = — h (op + 2g)v,
m p
where m is the mass of the particle, and p the perpendicular
from 0 on the tangent.
Take ox oy axes moving with the curve, and let R be
the reaction at any point. Then we have the equations
d^x „ 1 d . o. R di/ 1
df ydt^ ^ ' mds ' I
d'y , If?, Rdx I
-j^— eery H 7- (war) = .
If CO be constant, as given in the question, these equations
become
d'x 2 ^ (R , ^ \dy
where v= velocity of the particle relative to the curve.
From these equations we infer that the motion relative t^
the curve is the same as if the curve was fixed, and a re-
pulsive force ©V acted from 0. Hence, resolving along the
tangent,
dv_ , dr ^
ds~ ds*
.'. v* = c + CO v.
R
Also the pressure will be — + 2(»t;. Resolving along the
normal, we get
9—12.] AND RIDERS. 65
— = — tor sin © — — I- 2(ov :
p ^ \m / '
H V*
.'. = — h ta'o + 2a>v.
in p
If o) be not constant, the equation for v cannot be inte-
grated, but the expression for E is
=-+(o'p + 2(ov + -j- . V(»^-»*).
m p ^ at ^ ^ '
14. A string is laid on a smooth table in the form of a
catenary, and an impulse is communicated to one extremity
in the direction of the tangent, prove (1) that the initial
velocity of any point, resolved parallel to the directrix, is
proportional to the inverse square of the distance of that
point from the directrix, and (2) that the velocity of the
centre of gravity of any arc, resolved in the same direction,
is proportionally to the angle between the tangents at ex-
tremities of the arc directly, and to the length of the arc
inversely.
The equation to determine the tension at any point is
da* p'~^'
c' + s'
or since in the catenary p = , this becomes
*LL ? T=0
ds' (c» + «")"
One integral of this equation we know must be
Hence, according to rule, assume
substituting, we get
d'u 2s du
d^^~^W~d8'
()fi SENATE-HOUSE PROBLEMS [Jan. 17,
S
u = A tan"* - + B,
where A and B arc arbitrary constants. Substituting and
reducing, we get
T=i/{A^ + B),
where <f) is the angle the tangent makes with the directrix,
so that tan 6 = - .
^ c
The velocity resolved in any direction is given by
1h
d .,
■■~ds^-
rcos<f))
d I,
~ ds\
r'-]=Ac
d(f>
ds
Ac
P
Ac'
Again, the motion of the centre of gravity of any arc is
the same as if all the forces acted directly at that point
parallel to their original directions. Hence, if S be the length
of any arc, TT' the tensions at the extremities, (jx}}' the
angles those tangents make with the directrix, then the
required velocity of the centre of gravity will be given by
S.v= T ' cos <f>'-T COS <f>,
^CA{cf>' -<!>), ^
whence the proposition follows at once.
XV. A right circular cone floats with its axis horizontal
in a fluid, the specific gravity of which is double that of
the cone, the vertex of the cone being attached to a fixed
point in the surface of the fluid. Prove that for stability
of equilibrium the semi-vertical angle of the cone must be
less than 60".
If A be the area of the plane of floatation, h the length
of the axis, and a the radius of the base, then if BO be the
angle through which the cone is displaced about an axis
9—12.] AND EIDERS. 67
through the vertex, perpendicular to the vertical plane through
the centre of gravity and the vertex ; it follows from Gul-
dinus's properties that the moment tending to turn the body
in the opposite direction to the displacement is increased by
A^hBe.\h, or — Be,
and diminished by
TT O
( V being the volume of fluid displaced).
Hence the test of stability becomes
a* A*
6 ^2 '
or y < v3,
tan a < V3,
a<60%-
a being
the
semi
-vertical angle of the
cone.
xvi. A ribbon of very small uniform thickness h is
coiled up tightly into a cylindrical form, and placed with its
curved surface in contact with a perfectly rough plane in-
clined to the horizon at an angle a, the axis of the cylinder
being parallel to the intersection of the plane with the
horizon. Prove that the time in which the whole will be
unrolled is very approximately — ./(—r-^: — j , where d is
the diameter of the original coil.
We shall consider the section of the coiled up portion of
ribbon made by the plane of motion as circular, and neglect
the motion of the centre of gravity perpendicular to the in-
clined plane. The velocity given by these assumptions at
any instant is less than the true velocity, and the time thus
arrived at for the uncoiling of the whole is greater than the
true time, but the error is evanescent except when a finite
number of revolutions remain to be made, and the time of
F2
6S SENATE-HOUSE PROBLEMS [Jan. 17,
unrolling will therefore be very nearly that arrived at on
these assumptions, and cannot be greater than such time.
Let r be the radius of the coil at the time <, 6 being the
angle remaining to be gone through ;
3 ^d'B
Now
. ^ B-
d^B 47r^sina 1
" de~ 3A 'B'
fdB\^ Sttq sin a , /o
0 being
the initial value of B.
Let log Qj = a^;
d0\ .92 -2s2 fdx\
e-^.
* ' dx V \27r^ sin a/
V V27r^ sm a/ j o
= «V7r // 3A \
2 V V27r<7 sin a/ *
But f = ^a;
2 27r '
4 V \5'^ s"^ ^
9—12.] AND KIDERS. 69
17. If three beads, the masses of which are tw, m, w",
slide along the sides of a smooth triangle ABC, and attract
each other with forces which vary as the distance, find the
position of equilibrium. Prove also that, if they be slightly
disturbed, the displacement of each will be given by a series
of three terms of the form
isin (w^ + X),
where L and \ are arbitrary constants, and the values of n
are the three positive roots of the equation
(n»-a) (n'-/8) (w''-7)-cosMwW(n'-a) -cos«5w"w(n'-/3)
— cos' Cmm (w* — 7) — 2 cos A cos B cos CvmCm — 0,
where a, /9, 7 represent
m" + m', m + w", in + ?n,
respectively.
Let ABC be the triangle, BEF the positions of the three
particles. It is obvious that when they are in equilibrium,
the perpendiculars to the three sides of the triangle at DEF,
must meet in the centre of gravity G of the three masses.
Let a, /3, 7 be the perpendicular distances of G from the
three sides. Then taking moments about the perpendicular
a, we have
7n'/3 sin C = w"7 sin B\
w'/8 _ w"7 _ ma
' ' h c a '
Thus the ratios of a, /S, 7 are found. Draw BG bo that
sin^^G^ , X- 7 1 1 ^^ xT. ^sin^C^ 13
-. — 7TW^= known ratio - and draw CG so that -: — ttttt^ = - ,
sm CBG a sm BOG a
their intersection G is the required centre of gravity. By
drawing the three perpendiculars GD^ GE, GFy the positions
of m, ni, m" are determined.
To find the time of a small oscillation.
70 SENATE-HOUSE PROBLEMS [Jan. 17,
Let 05, y, 2 be the displacements of m, m, m" from their
positions of rest. The attraction of m on a unit of mass of
m, is equivalent to m ED which is equivalent to m'E' C along
E' C, and m CD' along CD' , where E' and D' are supposed to be
the positions of mm at the moment under consideration.
Thus the whole attraction of m' on a unit of mass of m when
resolved along BG is
m'{CD-x)+m' {EC- y) cos (7.
By treating m" in the same way we get the equation
d X
-7-; + (w' + m") X + m'y cos C+m"z cos B = terms independent
o(x,y,z.
d*x
But when a;, y, z all vanish, -p = 0, therefore the right-
hand side of this equation is zero.
We have also the similar equations
-A + (m" + m)y-\- m"z cos A + mxcoa C=Of
d'z
-j-a + {ni + m') z + mx cos B + m'y cos ^ = 0.
To solve these, put x = Levant, y = Ms,\rint, z = Nsm.nty
and we get -''
{m' + m" -n^)L+ m' cos C . M+ m" cos B.N=0,
and two other similar equations. Whence by cross-multi-
plication the ratios of L, M, N are found to be as the three
quantities
mm" cos A cos C — m" cos B (m" + w — w*),
m"m cos B cos C— m" gosA {m + m" — n*),
{m + m" — n') (m" + »i — n*) — »i'»i cos* C.
9—12.] AND RIDERS. 71
Substituting in the last equation we get
(„« _ a) {n' - /8) (7i' - 7) - cosM mm" {n* - o)
— cos' B m'm (n' — yS)
— cos* (7 w w' (n* — 7) — 2 cos A cos B cos G m m'm' = 0,
where a, yS, 7 stand for m' + m", m" + m, m + m' respectively.
xviii. The bore of a gun-barrel is formed by the motion
of an ellipse of which the centre is in the axis of the barrel,
and the plane is perpendicular to that axis, the centre moving
along the axis, and the ellipse revolving in its own plane
with an angular velocity always bearing the same ratio to
the linear velocity of its centre. A spheroidal ball fitting
the barrel is fired from the gun. If V be the velocity with
which the ball would have emerged from the barrel had there
been no twist ; prove that the velocity of rotation with which
it actually emerges in the case supposed is -j-p — , g.^. ,
the number of revolutions of the ellipse corresponding to the
whole length I of the barrel being n, and k being the radius of
gyration of the ball about the axis coinciding with the axis
of the barrel, and the gun being supposed to be immovable.
The ball passes along the barrel under the action of a force
which is a function of the distance from the breech. If then
we assume that there is no friction between the ball and the
barrel, the vis viva must be the same whether or not there
be a twist. If therefore to and v be the velocities of rotation
and translation at emergence in the case supposed, and V the
velocity with no twist, we have
_, CO 27m I
But ~ = -i—> •*•*'= 7; — ^l
•■•"'(*'+iJ^) = '">
2TrnV
^{I' + ^Tt'ii'k*)'
72 SENATE-HOUSE PROBLEMS [Jan. 17,
xix. An elastic ring of length I, mass m, and elasticity E
is placed over the vertex of a smooth cone, the semi-vertical
angle of which is a, and stretched upon it to any size. Sup-
posing it then set free, prove that the time before it leaves
the cone is
1 //ml\
the action of gravity being neglected.
Let A be the vertex of the cone, 0 the centre of the ring
at any instant ; let AO = x, and a be the value of x when
the string is unstretched. Fig. 37.
Then an element of the ring whose unstretched length is
T
ds and stretched length da. is acted on by a force -— . da
*' "^ a: tan a
along OP, and a force Rda along the normal to the cone.
cc ^" a
Now T=E. , therefore if a be the mass of a unit of
a
length of the unstretched string, we have, resolving along
and perpendicular to AP,
H^^j^i — ~) = 1 .d<T.Bma (1),
'^ dr Vcos a/ x tan a ^
Bd<r= — .cosa (2):
X tan a
d'x E x—a d<T . >
/. -j^ = . .-7-. cos' a.
dr fi ax as
But by similar figures -j- = -t
d^x . Ecoa*af v ^
therefore time before the ring regains its natural length is
_ !!" /[ ^' ^
~2VUcos»a;
9—12.] AND RIDERS. 73
_1 /ffiA'n^a*\
"2VV ^ )
cosec a
cosec a
=V(^0
cosec a ;
and it appears from equation (2) that B = 0 when T= 0, that
is, when x = a or the string has regained its natural length.
Therefore the ring leaves the cone in the time
wc
„ , cosec a.
( 74 )
Tuesday, Jan. 17. \\ to 4.
Seniob Examutbb. Arabic numbers.
Junior Exauinbb. Eoman numbers.
2. If jpg- be the image of PQ^ placed perpendicular to the
axis QCq^ of a lens or mirror CB,, QBq the course of a ray
from Q to q, shew that PQ -. pqv. RqC : RQC.
Hence prove that, with all combinations of lenses for eye-
pieces, the magnifying power of a telescope, arranged for
parallel or diverging emergent pencils, is the ratio of the
diameter of the object-glass or mirror to that of its image
formed on emergence from the eye-piece.
If a be the breadth of the object-glass or mirror,
a^ that of its first image,
flj, ttg ... a^j those of the 2nd ... n—i\ images,
h those of the last,
e the angle made with the axis by the axis of a pencil
incident centrically on the object-glass or mirror,
Vii Vi"-Vn t^iG angles made by the axis after refraction
or reflection at the successive lenses or mirrors, we have, by
the proposition,
^ = 1«
1^4.] SENATE-HOUSE PEOBLEMS AND EIDERS. 75
g«-t_ Vn_. *
.*. Y^ = — = magnifying power.
5. Prove that the locus of a point, through which one of
the principal axes is in a given direction, is a rectangular
hyperbola in the plane of which the centre of gravity lies, and
of which one of the asymptotes is in the given direction ;
tmless the given direction be that of one of the principal axes
through the centre of gravity.
Let the origin be the centre of gravity and the axis of x
the given direction.
(f , 7], ^) any position of the point P,
{x, y, z) that of any particle.
Since the line parallel to Ox through P is a principal axis,
tm {x - ^) (i/ - rj) = 0,
and Xm {x — ^){z—^=0;
and since 2 {mx) = 0, % {my) = 0, and S [mz) = 0,
^_t{'nucy)
yy_%{'mXZ)
If the given direction be that of one of the principal axes
through the centre of gravity, the point P lies in the plane
of yz or in Ox.
In other cases the locus is a rectangular hyperbola, one of
whose asymptotes is in the given direction, and the plane of
which has for its equation
V ^ K ^
X[mxy) % {mxz) '
( 76 )
Wednesday, Jan. 18. 9 to 12.
Skniob Modebatob.
1. A PARABOLA touches one side of a triangle in its
middle point and the other two sides produced. Prove that
the perpendiculars drawn from the angles of the triangle upon
any tangent to the parabola are in harmonical progression.
Let ABC (see fig. 38) be the triangle, and DFE the para-
bola touching the side BC in its middle point F, and the
other two sides produced in B and E respectively.
Then it follows from the geometry of the parabola that AD
and AE are bisected in B and C.
Hence if the sides of the triangle opposite B and C be 5
and c respectively, the equation to the parabola referred to
AE and AB as axes is
^/(^^^/(l) = ■.
and the equation to the tangent at any point x y is
The perpendiculars upon this tangent from the points A^
B, C, the co-ordinates of which are 0,0, 0,c and 6,0 respec-
tively, are therefore proportional to
'■iV(^)' ^V(^)'
9—12.] SENATE-HOUSE PROBLEMS AND RIDERS. 77
and 2 2 2 W jhif') - ^ (ex')]
/cxY nhy\ - ^{by')->sj{cx') ^'
i-v^y; vUv
whence the proposition is proved.
There is another method of proving this proposition.
By tangential co-ordinates if a^y be the perpendicular from
ABC upon the tangent to an inscribed conic section, then
I m n ^
Z, m, and n being any constants, and the condition that this
conic be a parabola is l-\-in + n = 0. (Salmon, Higher Plane
Curves, p. 8, Art. 7.)
If then ?=m = l, and n = — 2 we shall have an inscribed
parabola, and since in this case
1 12
the perpendiculars from the angles upon any tangent are in
harmonical progression. Also the form of the tangential
equation shews that the parabola bisects one side internally
and the other two sides externally in its points of contact witn
these sides.
2. Find the length of the longest straight line which can
be drawn in the interval between two similar, similarly
situated and concentric ellipsoids ; and, if a line shorter than
the line so determined be moved about in the interval, prove
that its point of contact with the interior ellipsoid can never
lie within the cone represented by the equation
*' y* 1 *' _ A
a'[a*(\-m^-7^] '*' h'[V{l-m^-r'] ^ c'[c\l-",n*)-i*] " "'
78 SENATE-HOUSE PROBLEMS [Jan. 18,
a, b, c being the semi-axes of the outer ellipsoid, m the ratio
of the linear magnitudes of the inner and outer ellipsoid, and
2r the length of the line in question, which is assumed greater
than 26 V(l -»»*).
What is the meaning of the boundary so determined when
2r is less than 2b \/(l — fn^) aud greater than 20^(1 — "0 ^
The longest line must be a tangent to the inner ellipsoid.
Let any tangent be drawn to this ellipsoid, and let r be the
parallel central radius of the outer ellipsoid, and let a diameter
of length 2a' be drawn to the outer ellipsoid passing through
the point of contact of the line in question. The segments of
this diameter made by the point of contact are a (1 — m) and
a (1 + w), and therefore if x be the semi- length of the touch-
ing line
a;' = -^ a' (1 - w) a' (1 + w) = r' ( 1 - wi») ;
therefore any tangent to the inner ellipsoid = 2r ^f {I — m') ^
where r is the parallel central radius to the outer ellipsoid,
and therefore the longest tangent or the line sought
= 2aV(l-m').
Again, if a central section be made of the outer ellipsoid
by a plane the direction-cosines of the normal to which are
/, m, n, the equation for determining the semi-axes of this sec-
tion is
_ra' m'b' nV _^ ,.
a«_p«+^,«_p« + ?3^«-^ ^^^ -
p being one of these semi-axes, if then we take p such that
r = />V(l-w'),
with the restriction p > b, we shall determine the limiting re-
lation between the direction-cosines of the normal to a plane
touching the inner ellipsoid consistent with the major axes of
the section of the outer ellipsoid made by this plane not being
less than 2r, and therefore the cone formed by lines drawn to
the points of contact of such planes will form a boundaiy
within which no point of contact can be situated.
9-12.]
AND RIDERS.
If X, y, z
be a point on such
a cone
X
y _
..(2),
eliminating
between (2)
and
9
(1), and
remembering
that
''"l-m"'
we obtain
a» [a* (1 - m') -r^\^ b^ [V (1 - m^) - r«}
If 2r be < 2b V(l — w') and > 2c \/(l — wi*)> the cone de-
termines a boundary within which if the point of contact be
situated the line cannot reach the outer ellipsoid in both
extremities, for within this cone the minor axes of the sections
made by planes of contact will be less than 2r.
3. If, in a rigid body moving in any manner about a
fixed point, a series of points be taken along any straight
line in the body, and through these points straight lines be
drawn in the direction of the instantaneous motion of the
points, prove that the locus of these straight lines is an hy-
perbolic paraboloid.
Let AB\)& the line along which the points are taken at any
instant, and let A'B' be consecutive positions of the line AB
after a very small interval of time dt.
Since the lines the locus of which is required are drawn
through the several points of AB in the direction of the
respective instantaneous motions of these points, each line
must in the limit pass through A'B' as A'B' is brought indefi-
nitely near to AB. Each line also must be parallel to the
plane to which the instantaneous axis is perpendicular.
The locus in question is therefore that traced out by a line
moving so as to be always parallel to a fixed plane and
always to pass through two straight lines.
This locus is the hyperbolic paraboloid.
80 SENATE-HOUSE PROBLEMS [Jan. 18,
4. If/(x, y, z) =0 be the equation to a surface, and r be
a straight line drawn through the point x, y, z of which the
magnitude and direction are any given functions of oj, y, z,
state what is the relation between the original surface and
that whose equation is n '^''/{x, y, z) = 0, supposing that in
the latter equation x, y, and z have been expressed in terms
of r and any two other variables independent of r, and that
n is a given numerical quantity, and prove that if the two
surfaces coincide for all values of w, the line r must lie
altogether in either of them.
Apply this to find the partial differential equations of
conical and conoidal surfaces respectively when referred to
any system of rectangular axes.
£
Generally the operation a ^f{x) is equivalent to writing
ax for X in f{x). See Carmichael's Calc. Op. ch. in. sec. 1.
If then mf{x, y, z) we replace the variables x, y, z by three
others, r, s and <, and then perform the operation
n'-fir, s, t),
this is equivalent to writing nr for r in the expression
f{r, 8, t).
Hence the surface determined by the equation
is such that any line drawn according to the same law
as (r) is drawn in the surface
f{r, s, t) = 0,
is n times as large as in this latter surface.
If these two surfaces be identical for all values of n, it
would clearly follow that the line drawn according to this
law must lie in either of them, since by keeping s and t
constant, but varying r in any manner, we still remain on
either surface.
9—12.] AND RIDERS. 81
If a, P, 7 be the co-ordinates of the vertex of a conical
surface, and r the length of the generating line to any point
X, y, z on the cone, we have
and since by what has just been proved, if /(a;, y, a) =0 be
the equation to a conical surface, this is identical with the
surface
«''V(^,y,2)=0
for all values of (n), it follows that
Bat r|=(.-„)|^ + (y-y3)^ + (.-7)|.
Hence by substitution we get the partial differential equa-
tions of conical surfaces, or
(«-<.)|+(y-«f+(-7)f = 0.
Again, in conoidal surfaces. If a, /9, 7 be any point on the
axis, /, 7n, n the direction-cosines of the axis, and r the length
of the portion of a generating line drawn from the point
x,y,z to the axis ; then
-{l{x-a.) + m{i/-^)+n{z-y) }';
and since by reasoning similar to the above
'•^(«>y»2!)=0,
and
'•;^= [(^-a) -l[l{x-a) + m {y -^) +n{z-y)]]j^
+ [(!/- ^-m{l{x-a) + m{y-fi) + n{z-y)}]-^
+ [{^-y)-n[l{x-a)+m{y-fi)+n{z-y)]]-^,
82 SENATE-HOUSE PROBLEMS [Jan. 18,
we get, putting p for I {x —a) + m {1/ — ^) + n (z — 7),
5. From a flexible envelope in the form of an oblate
spheroid, of which the eccentricity of the generating ellipse
is e, the part between two meridians, the planes of which are
inclined to each other at the angle 27r (1 — e), is cut away
and the edges are then sewed together; prove that the
meridian curve of the new envelope will be the curve of
sines.
Let B0 be the angle between two consecutive meridians
in the original envelope. Let 1/ be the distance of any
point P in one of these meridians from the axis of revolution,
and 8 the arc measured up to P from some fixed j)oint on
this meridian.
In the new envelope it is clear that if B6' be the new
value of Bd, then BO' = — . Hence if y' be the new value
of y, 1/' = ey and s is unaltered. When these conditions are
satisfied the element at P has the same value in both en-
velopes, since
et/ . — ds-=y dO ds,
and the old envelope may be developed into the new en-
velope.
Now in the meridian curve of the oblate spheroid
\dy)
<dyJ b'-f
ds^ _ y-cV
Hence, putting ey for ?/, we have in the new meridian
curve
U) ~
^y) -t'b*-f
9—12.] AND RIDERS. 83
and therefore ^_j =_,^_-, ;
.'. x = b V(l — e^) sin"* -^ ,
which proves the proposition.
6. If an uniform inextensihle and fle^cible string be
stretched over a smooth surface of revolution, prove that
the following equations hold :
l(^4:)+'^'-=<'' ■■(').
where ds is the element of the string at any point, dx
and di/ are coiTesponding elements of the arc of the circle
tln-Qugh that point perpendicular to the meridian, and of the
meridian respectively, X and Y are the resolved parts of the
impressed forces along these directions, and r is the distance
from the axis, the mass of an unit of length of the string
being taken as unity. Hence prove that, if such a string
be acted upon by a force at all points perpendicular to the
axis of revolution, and inversely proportional to the square
of the distance from that axis, the string will, if properly
sus^jended, cut every meridian in the same angle.
In the figure (fig. 39), let PQ be an element ds of tlie
string, PN and MQ being the elements dx and dy.
Draw tangents PT and QT to the meridians AP and AQ,
meeting the axis produced in 0, and let a be the elementary
angle between them.
Resolve the forces acting upon the element PQ of the
string in the directions PN and PO, by whicli means the
reaction of the smooth surface does not enter into the equa-
tions of equilibrium.
02
84 SENATE-HOUSE PROBLEMS [Jan. 18,
The resolved part of the tension at P in direction NP is
T ^ and in direction OP is T^.
as as
The resolved part of the tension at Q in direction PN is
as \ as J as
and in direction PO it is
also the resolved part of the impressed force on the element
PQ in direction PN= Xds, in direction PO it is Yds.
II»- l(^l)-.T^I + ^ = « (•)■
i(T^) + ^-_T'^.+ Y=0 (2).
ds \ dsj ds ds
, dx dx
also „=_ =
-••i
Substituting, we obtain
ds \ ds) r \dsj dy
whence, since
\ds)~^ \ds
we obtain
T
T
9—12.] AND EIDERS. 85
IKf)-^S+^^=« (^).
with the law of force assumed in the latter part of the ques-
tion
d^c dti
multiplying (1) by ^ j (2) l>7 ^ , and adding, we get
as r as r
If the string be so suspended that (7=0,
r=^, or Tr=:fjL;
dec
and therefore from equation (1) ~r- = constant, proving the
proposition in the question. N.B. Equation (1) might be
lound at once by taking moments about the axis.
7. A string is wound round a vertical cylinder of radius
a in the form of a given helix, the inclination to the horizon
being i. The upper end is attached to a fixed point in the
cylinder, and the lower, a portion of the string of length
I sec « having been unwound, has a material particle attached
to it which is also in contact with a rough horizontal plane,
the coefficient of friction being fi. Supposing a liorizontal
velocity V perpendicular to the free portion of the string
to be applied to the particle so as to tend to wind the
string on the cylinder, determine the motion ; and prove that
the particle will leave the plane after the projection of the
unwound portion of string upon the plane haa described the
angle of which the circular measure is
log
ga
2fi tan t ° 2fjiV* tan* t — 2fiffl tan t+ga'
86 SENATE-HOUSE PROBLEMS [Jan. 18,
In this case, fig. 40, the string coiled round the cylinder
remains the same helix throughout the motion, and the
particle in contact with the horizontal plane describes the
involute of the circular base.
Let PQ be the projection of the unwrapped string at any-
time t, P being the particle of mass unity, suppose. Let
6 be the angle through which PQ has removed from the
commencement of motion, and let v be the velocity of P,
the tension of the string being T and pressure on the plane E.
Resolving vertically,
E=g— T sin i;
horizontally and tangentially,
horizontally and normally,
- = Tcoat\ PQ = p;
dv tiv' ,
.*. V -7-= — aq + - — tan i,
as "^ p
dv v^ ^ .
• or v-j — ft — tan i = — fj^.
But p = l-a0, and ds= {I — aO) dO ;
ti dv "v^ ^ '
•• l-ad'dd ^l-ad
.'. -j^ — 2/jb tan i.v^ = — 2fiff il—ad) ;
ad
tan t 2fi tsur i
7« = (7 + -^ "^
tan I 2/i.tau't*
9—12.] AND RIDERS. 87
when the particle leaves the plane
g{l-ae)
V =
g^tan i-9 _- _
tani
1 ^9 _ ^ L
(7-2^tan^- 2/.tanS-- y^-gUoU+'^S. ^oi^i
e = —^—..\oi
ag
2/i tan i ' ° 2fj, tan'' * F'"* — 2figl tan i + ag *
8. A particle is acted on by two centres of force residing
in the same point, one attractive, the other repulsive, and
varying inversely as the square and cube of the distance
respectively. Two consecutive equal apsidal distances are
drawn and the portion of the plane of motion included be-
tween them is rolled into a right circular cone. Prove that
the trajectory described under the circumstances mentioned
above becomes a plane curve on the surface of the cone, and
that it will be an ellipse, parabola, or hyperbola, according
as the velocity in the trajectory was less than, equal to, or
greater than that from infinity.
The diflferential equation of the trajectory on the plane is
^ + u(l+-%] = ^, (1),
hV h'
fi and /x' being the absolute intensities of the attractive and
repulsive forces respectively.
Its polar equation is therefore
u = ;^, + Aco3{ne + B) (2).
Ifn'=l + ^;,
The angle between two consecutive equal apsidal distances
27J-
is — , and therefore the equation to the projection of the
trajectory on the cone made on the plane perpendicular to the
cone's axis is
n nh
8$ SENATE-HOUSE PROBLEMS [Jan. 18,
or w = -^ + nA cos {(f> + B);
since «=co8ec a where a is the semi- vertical angle of the
cone, and therefore the cosine of the angle between each gene-
rating line and the base of the cone is - .
° n
This being the equation to a conic section it follows that the
trajectory on the cone must be a plane curve, since none but
a plane curve on the surface of tlie cone can be projected into
a conic section on the plane perpendicular to the axis. Also
the projection, and therefore the original curve, is an ellipse,
parabola, or hyperbola, according as
Now by (1)
and by (2)
nh
v'=h'\ (^^y + u'\ = 2fiu - fi'u' + C,
= AV + w*AM»sin«(«^ + j5)
= n'h'A^ - (n' - 1) h'u' -h 2fiu - 4r,
^ n n
= wViM' - fi'u' + 2fiu - ^, .
Equating these values of v', we obtain
It, h'A Tr% = ^»
n h
Hence A* < = > -^ , according as
n n
(7< = >0,
9—12.] AND RIDERS. 89
or according as the velocity in the original trajectory was less
than, equal to, or greater than that from infinity.
9. A particle is describing an orbit round a centre of
force which is any function of the distance, and is acted upon
by a disturbing force which is always perpendicular to the
plane of the instantaneous orbit and inversely proportional
to the distance of the body from the original centre of force.
Prove that the plane of the instantaneous orbit revolves uni-
formly round its instantaneous axis.
Refer the motion of the particle to the following three
moveable axes, viz :
The radius vector to the particle.
The tangent to the particle's path,
and The perpendicular to the plane of the instantaneous
orbit.
Let dd be the elementary angle described by the radius
vector in the plane of the orbit in the time dt, and let d<^ be
the elementary angle through which the normal to the instan-
taneous plane has revolved in the same time.
The velocities in the above three directions are respectively
du dd
di' ''It' '
dr .
also in the time dt, the direction in which — is measured haa
dO
moved towards the direction in which r -y- is measured through
the angle ~r- dt , and the direction in which r -7- is measured
° at at
dr .
has moved from the direction in which -. is measured through
the same angle and towards the normal to the instantaneous
plane through the angle ~^dt ,
90 SENATE-HOUSE PROBLEMS [Jon. 18,
If then F be the central force and N the disturbing force,
^_ fde
de ^
m-- «.
i4(4^'' (^).
'f-f=^ (^)-
But iVr= ^ , and from (2) r» ^ = A ;
r ' ^ ' dt ' .
.'. -^ = T- is constant.
10. A die in the form of a parallelopiped the edges of
which are 2a, 2J, and 2c, is loaded in such a manner that the
centre of gravity remains coincident with the centre of figure,
but the principal moments of inertia about the centre of gra-
vity become equal ; if it then fall from any height and with-
out rotation upon a horizontal plane composed of adhesive
material so that no point which has once come in contact with
the plane can separate from it, prove that the chance of one
of the faces bounded by the edges 2&, 2c coming uppermost is
2- . _. be
— sm
^ V{(«''+6'0(a'+c*)r
Since the principal moments of inertia about the centre of
gravity of the die are equal and there is no initial rotation,
the die will by the impact of one of its corners upon the
horizontal plane acquire a velocity of rotation about an axis
perpendicular to the vertical plane through the comer and
the centre of gravity. This will continue until an edge
through the comer meets the plane. When this takes place,
since by the adhesiveness of the plane no point can separate
from it, the die must begin to rotate round this edge until a
face meets the plane, and on this face the die will rest. Now
it follows from the foregoing that the face on which the die
will rest is that which was met by the vertical through the
9—12.] AND KIDERS. 91
centre of gravity when the die first began to descend. To
find therefore tlie chance of the face 2b, 2c, lying uppermost
we must construct the pyramid, having for vertex the centre
of the die and for base the face 2b, 2c; and producing its
inclined faces to the surface of the sphere of radius unity
described about the centre of the die, we must find the area
of the spherical quadrilateral whose comers are A, B, C, D,
the points in whicli the lines of intersection of the inclined
fiices meet the spherical surface. Now the angles of this
spherical quadrilateral are equal, and therefore by dividing
it into two spherical triangles we obtain for its area
AA - 27r,
(-
TT
or ::_ 2
where A is the mutual inclination of the inclined faces of the
pyramid.
Now taking the line through the centre of the die perpen-
dicular to the face 2b, 2c, for axis of x, we see that the equa-
tions to the inclined faces are
hx — az = 0,
cx — ay = 0,
respectively ; and therefore the cosine of the angle between
them is
- V(a' + h') (a* + c') '
and taking the negative sign, since the angle sought is obtuse,
we have
-4 = 7r — cos ,- - :
^{a* + b') (a» + c*)
and therefore the area sought is
. (tt _! be )
4-^— — cos ' - - h ,
or 4 sin , =r ,
V(a* + 6») (a' + c')/
the whole surface of the sphere = 47r.
92 SENATE-HOUSE PROBLEMS [Jan. 18,
Hence the chance of either of the required faces lying
uppermost is
2 . _, be
— sin
11. A uniform sphere is placed in contact with the exterior
surface of a perfectly rough cone. Its centre is acted on by
a force, the direction of which always meets the axis of the
cone at right angles, and the intensity of which varies in-
versely as the cube of the distance from that axis. Prove
that, if the sphere be properly started, the path described by
its centre will meet every generating line of the cone on which
it lies in the same angle.
The centre of the sphere is always situated on a cone co-
axial with and similar to the original cone, the vertex of
which is situated in the produced axis of the original cone
and at a distance below the vertex of this cone equal to
a cosec a [a being the radius of the sphere). Take this point
as origin and refer the position of the sphere's centre to the
polar co-ordinates r, <f>, where ^ is the angle at which the
plane containing the line r and the axis is inclined to a cer-
tain fixed plane. Let F be the impressed force on the sphere's
centre resolved along this line. It is clear that this is the
only part of the impressed force which is not counteracted
by the reaction of the cone, and that it = — g .
Let X and Y be the forces of friction along and perperL-
dicular to a generating line, we at once obtain these equations
for the motion of the centre of the sphere, assuming its mass
to be unity,
§-,sin.«(f)'=^-^ W.
s(''™4t)=^' (^)-
Also for the motion of the sphere about its centre let m^, co^, «,
be the instantaneous angular velocities about the three prin-
cipal axes normal to the cone, parallel to the direction of r,
9—12.] AND RIDEKS. 93
and perpendicular to this direction on the surface of the cone
respectively.
Now a)j approaches o>j with the angular velocity -^ cos a,
d6 .
*"« '"» "i-^^^"'
«), recedes from Wj with angular velocity ~- cos a, and
from Wj with the angular velocity -^ sin a.
Hence if A be the moment of inertia of the sphere round
a diameter,
^w-^^'f '"'"=•' t^)'
^t-^'''f^'-=-^'' w-
A ~Y^ + Aa).-— sin a + ^cd, -^ cos a = Xa ... (5).
dt dt ^ dt ^ '
Also ^+"®« = ^ (^)'
ram a-£ — aw2 = 0 (7).
From (4) and (7)
A d I . d<f>\ A dr d<f> . _ „
a ' dt\ dt ) a' dt' dt '
which combined with (2) gives F= 0, and
r^8ina§=a;
<?<^_ C
dt "'r'sina'
94 SENATE-HOUSE PROBLEMS [Jan. 18,
therefore from (3)
d(o^ C dr ^ ^ C
-jr + -J- -77 cot a = 0, or w, = — cot a + (7 .
at r a at ^ ra
C
Let C = 0, then <», = — cot a :
^ ra
, , rs'mad^ C ^ ,.
and also ©, = 7- = — irom (7).
' a at ra ^ '
But the equation of vis viva gives us
(1)'+ '^ ^'■'' " (f ) + ^ (< + < + -/) = - 2 / ^. . * + C".
Substituting for co^ and Wjj and remembering that
and that r' sm' a ( ^ j = -y ,
we obtain
If now the circumstances of projection be such that C" =-0
as well as C, we obtain
dr 1 . dd)
But -y and r sin a -^ are proportional to the velocities of
the sphere's centre, parallel and perpendicular respectively to
the generating line, drawn to its centre on the cone in which
its centre is always situated ; hence the proposition is proved.
12. A small rigid vertical cylinder, containing air, is
rigidly closed at tlie bottom, and covered at the top by a disk
of very small weight which fits it air-tight. Supposing the
9 — 12.] AND RIDERS. 95
air in the cylinder to be set in vibration, prove that the period
27J-
of a vibration is — , «i being a root of the equation
ml kBYI
m tan — = ;
a fia
where I is the length of the tube, a the velocity of sound in
air, /x the mass, k the area of the disk, p cc p (1 + ^s) the re-
lation between the pressure and density when the latter is
suddenly changed from p to p (1 + s), and 11 the pressure of
the air on the cylinder before motion commences.
When the disk is first of all placed in the cylinder it sinks
through a space which is very small in consequence of the
small weight of the disk, and comes to rest when the pressure
n of the air in the cylinder satisfies the condition
K
P being the pressure of the external air.
The diiFerential equation of the disturbance is
^_ ,d^
df'"" da?'
For which we may assume the integral
^ = ^ sin m {at + x) + A' sin m {at — x).
Now measuring x from the bottom of the cylinder we must
have the following relations :
(1) When a: = 0 -,^ = 0 for all values of t, whence we
readily obtain A + A' = 0.
[2] When x = l,
because the condensation s at any point is — -,^ ;
96 SENATE-HOUSE PROBLEMS [Jan. 18,
.'. — fi ^IwV {sin m {at+l) — sin m {at — I)}
= — U^KAm (cos m{at + I)+ cos m {at -T)],
2 cos mat sin W6 = 2 2 cos mat cos w6,
7 W/e.
or wi tan »it = — « :
therefore the time of a vibration is
— , where m satisfies the equation
ma
m tan wt = j- ,
or writins: m for ma, and therefore — for m, this time be-
° a
27r
comes — , where m satisfies the equation
m
ml k/311
mtan — = .
a fia
13. A circular drumhead of uniform thickness is stretched
with a tension of uniform magnitude at all points in its cir-
cumference, and is then set in vibration by a small disturbance
commencing at the centre. Prove (1) that if ^ be the trans-
versal disturbance at the time < of a point the initial distance
of which from the centre was r, then
and (2) that the general primitive of this differential equa-
tion is
z=\ (^ (a< + r cos B) de + j'yfr {at + r cos 0) log (r sin*^) dO,
Jo J a
9—12.] AND RIDEKS. 97
(j> and yfr being arbitrary functions, and a a constant depend-
ing upon the tension and constitution of the drumhead.
The disturbance will be arranged symmetrically about the
centre, and we may consider the motion of an elementary
annulus, the undisturbed radius of which was r, and its
breadth dr.
Let T be the tension of the drumhead, and fi the mass of an
unit of area.
The mass of the annulus just referred to is therefore
/i27rr dr.
The tension along the inner circumference resolved ver-
tically is
(neglecting the longitudinal displacements) and that along
the outer circumference is
dr dr \ drj
Hence our equation of disturbance becomes
f.2'^dr.^ = ^^{2'^T^^dr;
d^_T d f dz\
•*• '' de~ fi'dr V dr) '
T being constant throughout the drumhead ;
. dy_Tndz d^
' ' df fM \r dr dr*
To solve the equation
d^z _ ,ndz d^\
df'"" \rd?'^ drV'
Assume z= \ ^ (ai + r cos 6) dd ;
dz f'
.'. -J- = \ ^' {at + r cos 6) cos 0 dd ;
98 SENATE-HOUSE PROBLEMS [Jan. 18,
and ^ = «•/ '<#>" («< + »• cos 6) de ;
... ^ _ a« ^ = a' f V' («< + r cos ^ (1 - cos'^) rf^
Civ Cv* J Q
= o* ['<^" (a< + r cos 6) 8m^0dd
Jo
= ^^ r(sin e<f>' {at + r cos ^)}
+ - [V (a< +r cos ^ cos ^(7^
= 0+--
because sin ^ = 0 at both limits.
Again assume
z=l -^ {at + r cos 6) log (r sin' &) dO,
■/ 0
^ = f V (a< + »• cos 6) log (r sin' &) cos ^t/^
+ - {'-^ {at + r cos &) d0,
^^ =/ V" («< + »• cos ^) log (r sin' ^) cos' ^rf^
+ - i '^' (a< + r cos 6) cos ^(7^
- -a I V^ (a< + r cos ^) <f^,
** •'o
also
T^ = a' I -(/r (a« + r cos 0) log (r sin' 0) d0 ;
9—12.] AND RIDERS. 99
.-. ^ ~ a' -^ = aM ^" {at + r cos 0) log (r sin' 0) sin' 0
-2- r^' {at + r cos ^) cos 0d0
+ -J ->^ (a< + r cos ff) d0
'"' •'o
a' f'
= I i/r' (a< 4- r cos ^) log (r sin*^) sin 0
a* C
+ - I '«^' (a< + r cos ^) log (r sin'^) cos ^<?^
a' f
+ 2- -i/r' (a< + r cos ^) cos ^J^
' J 0
a' r'
- 2 - / ^fr' {at + r cos ^) cos <7^
** •'o
and since the first term vanishes this expression reduces
itself to
- 1 -i/r' (a< + r cos 6fy log (r sin'^) cos 0d0
*" •'0
+
and therefore to
dz
' Jr'
Hence both these values of z satisfy the differential equation,
which is therefore also satisfied bj their sum, i. e. by the
expression given in the question, and since this expression
involves two arbitrary functions it is tlie general primitive of
the differential equation.
h2
( 100 )
Wednesday, Jan. 18. 1^ to 4.
Sbniob Exauiker. Roman numbers.
JOKIOB ExAMiMEB. Arabic numbers.
4. Find the position of the point, the sum of the squares
on the distances ot which from the three sides of a triangle is
the least possible ; and prove that the angles, which the sides
respectively subtend at this point, exceed the supplements of
those which they subtend at the centre of gravity of the
triangle by the respective angles of the triangle.
Let a, ^8, 7 be the distances of the point required from the
three sides of the triangle, a, J, c the lengths of those sides,
K the area of the triangle. Then we have to make
r'= a*^ + /S"^ 4- 7*, a maximum,
a, /3, 7 being subject to the relation
aa + J/3 + 07 = 2Z".
Hence, by the method of indeterminate multipliers,
a b c'
The position of the point is thus determined. Now, fig. 41,
let AJWhe the triangle, F the required point, G the centre
of gi-avity of the triangle. Then
sin FA F _y _c
^7AC~^~b'
1^—4.] SENATE-HOUSE PROBLEMS AND RIDERS. 101
and
sin QAB_2ire2L of triangle GAB GA.AG AG _h
sin GA G ~ area of triangle GA G ' GA.AB ~ ~AB ~ c '
whence,
sin PAB : sin P^O :: sin GAG : sin GAB,
and PAB + PAG= GAG+ GAB;
.-. PAB = GAG, PAG = GAB;
similarly,
PBG= GBA, PBA = GBG, PGA = GGB, PGB= GCA.
Hence PGB=G-GGB,
similarly, PBG=B-GBG;
.'. PCB+PBG=B+G-{GGB-i-GBG)
^B+G-TT + BGG
= BGG-A;
.-. ir-BPG=BGG-A,
or BPG = 'ir-BGG+A,
similarly, GPA = tt - GGA + B,
APB = 7r-AGB+G.
Hence the angles which the sides respectively subtend at P
exceed the supplements of those which they respectively sub-
tend at G, by A, B, G, respectively.
vi. Trace the curves represented by the equations
{a^-Aa')y*-12a'x{a-i/)=0 (1),
sin y — m sin a; = 0 (2) .
In (1) explain the circumstance that the asymptotes parallel
to the axis of y appear to contradict the statement of (v). In
(2) distinguish between the cases in which w> = or < 1.
102 SENATE-HOUSE PROBLEMS [Jan. 18,
First. The equation
(jc» - 4a*) y» - 12a* a; (a - ^) =s 0
may be written in the form
{a? + 3aa; - Aa')i/'-Zax (y - 2a)' = 0,
or {x + 4a) {x—a)i^— Sax {y — 2a)* = 0,
when X is positive, it cannot be < a,
and when a; is negative, >4a.
When a; = 0, y = 0,
x = a, y = 2a, two values,
a; = 2a, y=«>> or <»>
x = — 2a, y =Qo , or a,
a; = — 4a, y = 2a, two values,
near the origin y' + 3aa; = 0.
The curve touches a; = a, and — 4a.
The axis of x and a;' — 4a* = 0 are asymptotes.
The curve lies both above and below the axis of a; at an
infinite distance.
The form of the curve is as in fig. (42).
The asymptote x = 2a is met in two points at an infinite
distance by the branches to which it is itself an asymptote^
and one more point where it meets the other parallel
asymptote. Similarly for the asymptote x = — 2a.
Secondli/,
ami/ = mBmx. Let m < 1 and = sin 7.
For every value of a;, if y = ^, the equation will be satis-
fied by
2n7r + y8, and (2n+ 1) 7r-/8,
x = 0, y = 0,
X increases, y increases,
1^ — 4.] AND RIDERS. 103
IT
X increases, y diminishes,
a; = TT, y = 0,
and the shape is the same on the opposite side from a; = tt
to a; = 27r, when it recurs, and similarly for x negative.
The shape is the same, see fig. 43, on the lines
^ = 0, y = ±27r, y= + 47r,...
and inverted on
y = + 7r, y = + 37r....
If w > 1, the equation is the same interchanging x and y,
and therefore the figure is the same, as if the above were
turned through 90'.
K m = 1, y = 2mr + x, or (2n + 1) tt — a;. See fig. 44.
10. One circle rolls within another ; apply the above for-
mula to find the area of the curve traced out by a given point
within the rolling circle.
Let C be the centre of the fixed, 0 of the moving circle,
P their point of contact at any time, fig. 45, V the point
which traces out the required curve ; let 0 F" produced meet
the circumference of the moving circle in Q, and let A be
the point of the fixed circle with which Q originally coin-
cided. Let GP=a, OP=b, PCA^d, so that FOQ = ^0,
0
let 0 V= c, and let x, y be the co-ordinates of V. Then
a; = (a - J) cos ^ + c cos , ^,
y = [a — h) 9>\xi 6 — c sin —7— 6 ;
.-. xdy - ydx = {a- hydd - c' (^)' dd
. ,. a- 2b - ,^ (a -b)* a- 2b - ,^
+ {a-b)c cos — T — dad - - — r— ^ c cos — , — add,
104 SENATE-HOUSE PROBLEMS [Jan. 18,
and, integrating this between the limits 0 and 27r, we get for
tlie area of the curve,
(a _ J)' ^1 _ _j 2^ _ __ i, sm -J- 2,.
13. Define a developable surface ; and, from your definition,
deduce the partial differential equation of such surfaces.
Find the equation of the developable surface generated by
the plane which moves in such a manner as to be always in
contact with the surfaces
^s + ja + ^2-1,
x' / z'
a—r 0 — r c — r
Let Ix + niJ/ + nz = l (1)
be the equation of a plane touching the ellipsoid
^.+2^. +1=1.
We then have to find the locus of the ultimate intersections
of (1), subject to the conditions
?a'+»nV + nV=l (2),
P +m' +n' =0 (3). .
[The equation (3) is impossible, but the form of the equation
of the required surface may nevertheless be found. A similar
method, moreover, may be applied to find the developable
surface circumscribed about any two concentric and similarly
situated surfaces of the second degree.]
Multiply (2) by X, (3) by //., and add to (1) and differentiate,
and we get
x+ Xla' + fil = 0 (4),
1/ +\mb'+fji.m=0 (5),
z +\nc' + fin=-0 (6),
1^—4.] A\D RIDERS. 105
(4)Z+(5)?»+ (6) n gives
1 + X = 0.
Hence x—la* + fil = Of
X
.'. l = -
a — fi
z
Similarly m = f^ — , n =
Hence (1) becomes
^ -li* ^
x" y" s'
a* — fi b* — fi c^ — fi
And (3) gives
(a>^/x)'+(J'-M)'^(c--M)"
The latter of these equations is the differential of the former,
hence the required result will be the same as that of elimi-
nating fi between the equation
{fM-a:'){fi-b^{f,-c')+x'{fi-b'){jM-c')
+ y' (ji -c') (ji-a") + z' i^L-a') (jM-b^) =0 (7),
and that obtained by differentiating it.
Now, writing (7) under the form
its differential is
3/jJ'-2Pfi+ Q = 0.
The result of the elimination of fi between these is
4 (F'-SQ) iO'-SPM) = {OR-PQ)\
^ The equation of the required surface is therefore given, by
putting in the foregoing,
R = a'6V - J'cV - (fay - a*b'z\
The required surface is, therefore, of the eighth degree.
106 SENATE-HOUSE PROBLEMS AND RIDERS. [Jan. 18.
14. Explain what is meant hy A'.O"'; and prove that,
if/(e') be expanded in a series proceeding by ascending
powers of t. the coeflScient of f h"^ — .
* 1.2 m
Prove that, if m be less than r,
{l + log(l+A)}\0'" = r (r-1) (r-2) ... {r-m+ 1).
By the theorem enunciated in the former pert of the question,
{l + log(l + A)|^o"'
1.2 ... w
will be the coefficient of T in (1 + loge')', that is, in (1 + 1)'.
Hence
{l4-log(l-4-A)}''0'"^r(r-l)...(r-m + l)^
1.2...m ~ 1.2... 7» '
.-. {l+log(l + A)|'0'" = r(r-l)...(r-.m4-l).
( 107 )
Thursday, Jan, 19. 9 to 12.
JCHIOB MODSBATOB.
1. If at the extremities P, Q of any \:7f(i diameters CP,
(7^ of an ellipse, two tangents i^, Qq^ be drawn cutting each
other in T and the diameters produced in p and g-, then the
areas of the triangles TQp^ TPq are equal.
Project the ellipse, orthogonally, into its auxiliary circle ;
then the areas of any two triangles TQp, TPq, fig. 46, in
the primitive are in the ratio of their projections. But in the
auxiliary circle these areas are equal by symmetry. Hence
also they are equal in the primitive ellipse.
2. If a straight line CN be drawn from the centre to
bisect that chord of the circle of curvature at any point P of
an ellipse which is common to the ellipse and circle, and if it
be produced to cut the ellipse in Q and the tangent in T,
prove that CP= CQ, and that each is a mean proportional
between CN and CT,
If two diameters be drawn in any ellipse, making equal
angles with the major axis, then their conjugates will also
make equal angles with the same axis. This is obvious from
the consideration that the conjugate of any diameter is parallel
to the tangents at the extremities of that diameter.
Now CP, fig. 47, by construction is the conjugate of the
diameter parallel to PT, and CQ the conjugate of that parallel
to Pr, Also, by a known proposition in Conies, PjT and PV
108 SENATE-HOUSE PROBLEMS [Jan.l9,
make equal angles with the major axis. Therefore CP, CQ
make equal angles with the axis and are consequently
equal.
Also by Goodwin's Conies, CN. CT= CCt-
3. If a, J, c be the sides of a triangle, and r the radius
of the inscribed circle, then the distances of the radical
centre of the three escribed circles from the sides of the
triangle will be respectively
5+c c+a a+i
''"2^' ''^T' '*~2r'
Let ABC be the triangle, fig. 48, and let the side AB
touch the two escribed circles in B and E, Now, by defi-
nition, the radical axis of the two circles bisects the common
tangent BE, and is perpendicular to the straight line joining
the centres of the circles. Also it is evident that the straight
line joining the centres passes through C and bisects the
angle exterior to A CB. Again, it is proved in most treatises
on Trigonometry, that BA = BE, so that the middle points
of BE and of the side AB are the same. Therefore the
radical axis bisects the side AB and is parallel to the bisector
of the angle A CB.
If a = 0, y9 = 0, 7=0 be the equations to the sides of the
triangle, the equation to any straight line parallel to the
bisector of the angle C is
a — ^ = constant,
but since this passes through the middle point of AB, it
c c
must be satisfied by a = -sxnB, fi=- sin a.
Hence the equation to the radical axis is
a — ^ = - (sin -B — sin -4).
Similarly, another radical axis will be
/9 — 7= - (sin (7— sin B),
9--12.] AND RIDERS. 109
But aa + J|9 + C7 = 2A,
where A is the area of the triangle. Solving these equations
to determine /9, remembering that
. ac sin B
we get
and the values of 7 and a may be written down from sym-
metry.
Cor. If the radical centre coincides with the centre of
the inscribed circle, the triangle must be equilateral.
4. Two equal heavy particles are connected by a string
which passes thi-ough a small smooth ring. Prove that the
equation to the plane vertical curve on which the particles
will rest in all positions is
rc08^ = a + -\/r (r) — i/r (? — r),
where 9 is the angle the radius vector makes with the vertical,
I is the length of the string, -v/r an arbitrary function, and a an
arbitrary constant.
Take the smooth fixed ring as origin, and the axis of a?
vertical. Let x, r, x', r be the co-ordinates of the two
particles. Then by virtual velocities we have
dx + dx = 0 ;
also
.*. x + x' = c,\
r + r =1.)
Let x=<f){r) be the equation to the curve, then x =^ (r^,
and we have the functional equation
<f>{r)+<l>(l-r)=c.
Solving this in the manner exhibited in Ilerschel's ex-
amples, we get
<t> {r) = a + ^lr {r) + yjr [l - r).
110 SENATE-HOUSE PROBLEMS [Jan. 19,
5. If four equal particles, attracting each other with forces
which vary as the distance, slide along the arc of a smooth
ellipse, they cannot generally be in equilibrium unless placed at
the extremities of the axes ; but if a fifth equal particle be fixed
at any point and attract the other four according to the same
law, there will be equilibrium if the distances of the four
particles from the semi-axis major be the roots of the
equation
where p and q are the distances of the fifth particle from the
axis minor and axis major respectively.
If the four particles be placed on an arc of an ellipse in
equilibrium, the resultant attraction on any particle must
be normal to the curve. Hence, by Todhunter's Statics,
Art. 220, the four normals at the four particles must meet in
their centre of gravity.
Let X, y be the co-ordinates of any particle, then the
equation to the normal is
y-y=-^ (»''-«) (!)•
Let Ilk be the co-ordinates of the point in which the four
normals meet, then
h-y = ^{h-x) (2),
also i]Vil'=l (3).
a I b
Eliminating x we get
(a« - hjy' + 2h'k (a* -1')^+ (JV + a'A' - (a* - hj] by
-26*^♦(a'-t')3^-5•^' = 0 (4),
Since k is the ordinate of the centre of gravity, k is one-
fourth of the sum of the roots of this equation ;
•• ** (a'-b') ^^'-
9 — 12.] AND RIDERS. Ill
This equation can only be satisfied by
a'-J» = 0, or a' + 7&*=0,
or by ^ = 0. Taking the latter supposition, the equation (4)
reduces to
hence two of the particles must be situated at the extremities
of the major axis. To find the positions of the other two,
write for y^, its value obtained from (3), and we get
but since h is the abscissa of the centre of gravity, we have
A = - : hence this equation can only be satisfied, first by
a; = 0, and then the four particles are at the extremities of
the two axes: secondly by e= — , and in this particular
ellipse there will be equilibrium if two of the particles are at
the extremities of the major axis, and the other two are at
the extremities of any ordinate.
The case a' + 7 J' = 0 is impossible. In the case a* = 6', the
ellipse becomes a circle, and equation (4) reduces to
3^ =
*" + <''(lj'
But since h and h in this case must both vanish because
all the normals pass through the centre, this expression may
have any value. Hence there will be equilibrium in a circle
if the four particles are at the extremities of any two
diameters.
If we have five particles, it is necessary that the point
{hk) should coincide with the centre of gravity of the five
masses. Hence equation (4) becomes
112
SENATE-HOUSE PROBLEMS
\Jan. 19,
bk =
Wk
a*- 6"^ + ^'
and similarly
bh =
2a'A
substituting these in equations (2) and (3) we get
the required result.
6. A heavy string is placed in equilibrium on a smootli
sphere ; prove that, if 6 be the length of the spherical arc
drawn from the highest point of the sphere perpendicular to
tlie great circle touching the string at any point P, then
sin^ =
where z is the perpendicular from P on any horizontal plane,
and a, b are constants.
Shew that the form of the string can be a circle only when
its plane is vertical or horizontal.
Let z be the highest point of the sphere, fig. 49, AB the
string, and PQN the great circle toucning it along the ele-
ment PQ.
Let T be the tension at P, then resolving the forces on tlie
element PQ along its arc, we get
dT=gdz',
.-. T=gz + c.
Again, take moments about the vertical through z. Re-
solving T perpendicular to the axis, we get Tsin zPN, and
the moment is therefore
hence
T sin zPN . sin zP^T sin 6,
9—12.] AND RIDERS. 113
sin a =
z + b'
where a and b are arbitrary.
The curve of the string could not be a circle, for the alti-
tude of the centre of gravity must be a maximum or minimum.
Now unless the plane of the circle be vertical or horizontal,
a slight motion, without change of form, will clearly elevate
or depress the centre.
7. If three particles of masses m, m, m" attracting each
other start from rest, shew that if at any instant parallels to
their directions of motion be drawn so as to form a triangle
the momenta of the several particles are as the sides of that
triangle.
Let V, V, vi" be the velocities of the particles. Since the
three particles start from rest, the area conserved round any
point is zero. Now the area conserved by any particle of
mass w moving with velocity v is mvj), where p is the length
of the perpendicular from the origin on the direction of motion.
Hence
mv .p-\-mv.p-\-mv.p =0.
Therefore if three forces represented by mv, m'v\ m"v"
were to act along the directions of motion, the sum of their
moments about every point would be zero. Therefore these
forces are in equilibrium, and if a triangle be constructed by
drawing lines parallel to their directions, the forces will be
proportional to the sides of that triangle.
Hence also the three directions of motion being produced
meet always in one point 0.
Let Fy Fy F" be the resultant forces on the three particles
each due to the attraction of the other two. Then, these
being the resultants, two and two, of all the internal forces
of the system, must balance each other. Therefore the three
forces F, F', F' meet in a point 0', and are proportional to
the sides of a triangle formed by drawing parallels to the
straight lines joining 0' to the particles.
I
114 SENATE-HOUSE PROBLEMS [Ja«.19,
The points 0, 0' are not in general the same, nor are they
fixed in space. If the law of attraction be directly as the
distance, they both coincide with the centre of gravity of
the system.
8. If from any point on a surface a number of geodesic
lines be drawn in all directions, shew (1) that those which
have the greatest and least curvature of torsion bisect the
angles between the principal sections, and (2) that the radius
of torsion of any line, making an angle 6 with a principal sec-
tion, is given by the equation
R \p, pj
sin 6 cos 0.
where p^ , p^ are the radii of curvature of the principal sec-
tions.
Take the given point as origin 0, and the normal as the
axis of z, and let the equation to the surface be
22 = <^ {x, y)
Let OP be any geodesic line and ON the projection of OP
on the plane of xy.
The osculating plane of any geodesic line contains the nor-
mal to the surface on which it is drawn. Hence NOZ is the
osculating plane at 0, and also the osculating plane at P con-
tains the normal to the surface at P.
Let de be the angle between two consecutive normal planch
to the curve, du the angle between two consecutive oscillating
planes. Then clearly tlie normal OZ is turned into the con-
secutive normal to the surface at P, by turning it, first through
the angle dc, then through du ; and the planes of these angles
are at right angles.
The equation to the normal at P is
^-x ^-n-y ^};-z
ax by — 1 '
.•• cos du . cos de = -77—- — ■, , , . ^ „. j
9—12.] AND RIDERS. Ill
-•• 4 + -» = «' cos' e + h' sin' 0.
But by Euler*'s tlieorem since a = — , b = — ,
Pi p.
- = a cos' 0 + b sin* 6 ;
P
therefore substituting, we get
~ = {a — h) sin 0 cos ^
= (— — — ] sin ^ cos 0.
\Pi pJ
This is a maximum or minimum when 0 = t, hence the
4
tangents, the geodesic lines of greatest and least torsion, bisect
the angles between the principal sections.
If B^ be the least radius of torsion, and B the radius of a
geodesic line making an angle <j> with it, then the above ex-
pression becomes
cos 2^ *
The expression for B may also be put into the form
B~ 2d0[pJ'
9. If du and de be the angles of torsion and contingence
of any curve of double curvature, and if sin (f> be the ratio of
the radius of circular curvature to the radius of spherical
curvature, prove that the square of the angle of contingence
of the locus of the centres of circular curvature is
d<f> + duY + cos' ^eT.
Let CC\ fig. 50, be an element of the locus of the centres
of circular curvature corresponding to a point A on the original
curve. This element ultimately lies in the normal plane at^,
I 2
116 SENATE-HOUSE PROBLEMS [Jan. 19,
let it make an angle <f} with the principal normal A C pro-
duced. Now any element CC is brought into the position of
the next by turning it, first, through an angle du round the
tangent AA' to the curve, secondly, through an angle de
round a perpendicular CO at C to the osculating plane CAA';
the element has thus been brought into the consecutive nor-
mal plane to the original curve, and we have therefore,
thirdly, only to increase (f> by 2^.
The change may therefore be effected by turning CC in
two planes at right angles to each other through the two
angles C CP=du + d(\>, and PCQ = de cos ^.
Therefore if d-^ be the angle between the old and new
positions of CC,
cos c?i/r = cos {du + d^) . cos {de cos ^),
or d^\ = {du + d<^y + ^T^ cos'' <^*.
10. A particle is projected with velocity V along an in-
finitely thin ellipsoidal shell attracting according to the law of
nature; prove that when it leaves the ellipsoid the perpen-
dicular from the centre on the tangent plane is
7C-^)'
where R is the radius-vector parallel to the initial direction of
motion, and P the perpendicular on the tangent, /x the attra^*
tion of a mass equivalent to a unit of area of the ellipsoid at a
unit of distance.
First we must find the attraction of the ellipsoidal shell on
the particle. Let P be the position of the particle at any
instant, take a point Q just inside the shell, and situated on
the normal at P. Round P take any very small area A which
may be ultimately considered as plane. Since the point Pis
on the ellipsoid, its distance is infinitely small compared with
the linear dimensions of the area A. Hence the attraction of
A on P or ^ is ultimately the same as that of an infinite
plane on a point at a finite distance from it, and is therefore
normal and equal to 27r/x. The attraction of the whole ellip-
9—12.] AND EIDERS. 117
soidal shell on Q is zero, hence the attraction on Q of the
whole shell except the area A is 277/1.. But it is evident that
the whole shell less the area A exerts equal attractions on P
and Q, because the distance PQ diminishes without limit
compared with the distance of P from the nearest point of the
attractive mass. Therefore the attraction of the whole shell
on P is normal and equal to 47r/Lt.
Now since this attraction is normal to the path of the parti-
cle, its velocity will be always the same and equal to V; and
it will describe a geodesic line on the ellipsoid.
The pressure on the ellipsoid will be 47r/x , where p is
the radius of curvature. Hence when the particle leaves the
ellipsoid, we have
47rit = — .
P
Now, because the path is a geodesic line, p = -5 , where r
is the radius vector of the ellipsoid parallel to the direction of
motion, and c^ is a constant = P^. (See Hymers' Solid
Geometry^ Problems on Sect. x). Hence we have
V(
V\PR\
If p be the perpendicular from the centre on the tangent
plane at the point where the particle leaves the ellipsoid, we
also have
^Vi — v^~)'
Now the shell has been supposed bounded by similar ellip-
soids, hence /x is really variable and proportional to the tiiick-
ness h of the shell. Let fi = fi^h. Also by similar figures this
thickness is proportional to^; let n be the infinitely small
ratio of the thickness of the shell at the extremity of any
axis to that semi-axis. Then h = np; therefore fi = ii^n.p,
and let m = fi^n. Tlien substituting in the above expression,
we get
118 SENATE-HOUSE PROBLEMS [Jan. 19,
11. An infinitely thin ellipsoidal shell attracting accord-
ing to the law of nature is bounded by two similar and simi-
larly situated ellipsoids. A very small piece is cut out of
the shell and replaced in its original position. Shew that
the force necessary to hold the piece in equilibrium is propor-
tional to the square of the thickness of the shell.
Let dB be any element of the small piece of area B cut
out of the ellipsoid. Round dB describe a small area A
which may be ultimately considered plane and with respect
to which dB is infinitely small. Then the attraction oi A
on this element dB of itself is clearly zero. Let / be the
attraction of the remainder of the shell on a unit of mass
supposed collected at dB. Then, since the shell is infinitely
thin, we may consider y to be the same throughout the thick-
ness h of dB, and therefore the force necessary to hold dB in
equilibrium is /. dB. h. But we have proved that/= lirfi.h,
hence the force = 27rfMdB . h^. Hence the force on the whole
very small area B is 'iiriiB . h'.
12. A sphere of radius a is suspended from a fixed point
by a string of length I and is made to rotate about a vertical
axis with an angular velocity w. Prove that, if the string
make small oscillations about its mean position, the motion
of the centre of gravity will be represented by a scries of
terms of the form
Zcos {Kt + M),
where the several values of k are the roots of the equation
Let G be the centre of gravity of the sphere, BGC the
diameter to the extremity of which the string is tied. Take
the fixed extremity of the string as the origin, and fixed axes
in space, so that g acts parallel to the positive direction of
the axis of z. Let Wj, a^, o), be the angular velocities of the
9—12.] AND KIDERS. 119
sphere about diameters parallel to the axes. Let P, Q be
the direction-cosines oi GC and F Q' of the string referred to
the axes x and y.
The squares of small quantities being neglected according
to the usual rule, it is also obvious that the tension of the
string will be the weight of the sphere.
The equations of motion are therefore
d(o, moment of forces
rfr+'""'= A "
^®2 ^,, -^9 (P< p\ I
where A is the square of the radius of gyration of the sphere
2
about a diameter = - a^.
5
Also cDj = rate at which GO approaches
Similarly
' d . _i p. dP
.. = -^^(cosP) = ^.
dQ
* dt
If XT/ be the co-ordinates of the centre of gravity, we
have
d*x „ ^
df=-^^' I
d'y
Also x = lF-\-aP, y = lQ' + aQ.
Substituting for x, y, u)^, ©^ their values in terms of P, Q,
P\ Q\ we get
12U
SENATE-HOUSE TROBLEilS
de ^ dt A^^ ^ ^'
, d'P' ^P ^
[Jan, ly,
de
de
To solve these put P = Z cos {kI + M) , Q=M sin {kI + 3/ ) ,
P' = L' cos {Kt + M), Q = M' sin (/c« + J/), we get
L'{g-l^)=aieL,
M'{g-l^) = a^M,
M{"-i-K^-jM' = -nLK.
Eliminating the ratios of L, M, L', M\ we get
2
when we put ^ =- a', this reduces to the result given in the
o
question.
13. A string is in equilibrium in the form of a circle about
a centre of force in the centre. If the string be now cut at
any point A, prove that the tension at any point P is instan-
taneously changed in the ratio of 1 —
cT — •
4. £-(»-•)
: 1, where
e' + 6-'
0 is the angle subtended at tlie centre by the arc AP.
This is a particular case of a more general proposition.
Suppose a string to be in equilibrium in any curve in one
plane under the action of any forces. Let Pds, Qda be the
resolved parts of these along the tangent and normal to any
element ds. In order to refer the motion to moving axes,
9—12.] ASD SIDEKS. 121
let v, F be the relodties of the element along the tangent and
nonnaL Then the equations are
(1),
dm
dt
d^
dt^
■0*;^
(2),
where T is the tension, p the radius of carratme, and ^ die
angle the tangent makes with any fixed straight line.
In the beginning of the motion just after the string is cot,
wem.y «j«ct the 8q»«« of muU qnantitk., hence rf and
H^ may be neglected.
The geometrical equations are to be found from the con-
dition that any element PQ = ds of the string is inextensible.
The tangmtaai and normal velocities of P and Q are respec-
tively «, o and u+dm, v+dc. Hence the velocity ciaepanr
tioa of P and Q aloB|^ the tangent is dm — wd^, which must
be werOj and the veloaty of lotatioa of Q round Pis dc-\- ud^
which most be d* . ^ . Henoe we hare the two equations
s-r« •• <')•
s-rf '■*'■
Differentiating (3) we get
d*u 1 dv
dsdt ~pdi^
since the anuJl tenn r jT may be neglected in the begin-
nii^of the
122 SENATE-HOUSE PROBLEMS [Jan. 19,
Substituting from equations (1) and (2)
dP oTT^Q r,
ds ds^ p p^ '
d^T_T^_dP Q
ds* p^ ds p'
This is the general equation to determine the tension of
the string at the instant after the string is cut.
If the string be in the form of a circle, as in the question,
^ = — i*^ is independent of s and 5=0, and p = a the radius ;
d^T T ^-F
ds* a* a"
Now s = ad, hence we get
T= Fa + A^ + Be-''.
To determine the arbitrary constants we observe that r= 0
when ^ = 0, and 0 = 2Tr;
But just before the string was cut we have
T=Fa\
Hence the result given in the question follows at once.
If the string be a catenary under the action of gravity;
we have
P=-g ,, i g> and Q = -
9<^
V(s' + c') «^^ ^- Vl^' + C)
The equation becomes
which has been integrated in a previous question.
, dP Q
whence -7-5 = —
as p
d'T T_
9—12.] AND RIDERS. 123
If the string be in the form of an equiangular spiral under
the action of a central repulsive force in the pole varying
inversely as the cube of tne distance, the resulting equation
can be easily integrated.
14. An inelastic string is suspended from two fixed points
so that it hangs in the form of a catenary of which the para-
meter is c. Suppose it to make small oscillations in a ver-
tical plane, prove the equation
where a is the angle the tangent at any point makes with the
horizon when the string is at rest, and a + ^ is the value of
the same angle at the time t.
Shew that there are sufficient data to determine all the
arbitrary functions.
Let M, V be the velocities of any element ds of the string
resolved along the tangent and normal. Then the general
equations of motion of the string are
du d(f> . , ^, dT
dv d<b / ,N T'dd>
where T' is the tension. Now the tension when the string
is at rest is gy= — — . Let T'—-^- yT. Substitute and
^^ cos a cos a
remember that in small oscillations we may neglect the squares
of the small quantities m, v, ^, we get
du , . co»* a dT ,.
- = -gcosa.4,+-^-^ (1).
dv . , , d(t> cos" a ^ ,_.
-^=gBma.<f>+gcoBa.^ + -j-T (2).
124 SENATE-HOUSE PROBLEMS [Jan. 19,
We have also the two general geometrical equations
du V _
ds p '
dv u _ d<f>
ds
dt'
where - = -j- + -^ is the reciprocal of the radius of curvature.
p as as '^
Changing the independent variable and neglecting the squares
of small quantities these reduce to
d^u c d(b
du'
COS'* a dt
.(3).
Ti 1 1 rt , . , , ,„ p du dv dj^4>
!• or the sake oi orevitj put uvfp ^^^ '1} ^ ~Ji ^ ~7^ ^^'
spectively.
In order to eliminate T from equations (1) and (2) differen-
tiate the second, we get
cPu , d'd) cos* a dT 2 cos a sin a rwi
d'u , fd^4> ^ 2(6^ _ 2 cos g sin g ^
Eliminating Tfrom this by means of (2), we get
fd^u \ ^ , . du' , . , d^6i
cos g f -j-i + w j + 2 (sm g -^ u cos a) = y cos g -r\
+ 2^ sm g cos g -— + 2^<p
(4).
But by (3)
j% + M = — — (f> ;
aoL cos g ^
du , /"sing,,,
.*. sm g -J u cos a = c / — r— © ,
og j cos* g ^ '
*J— 12.j AND RIDEKS. 125
substitute these in (4), we get
^ _»" . ft Tsina ,„ , _ d^<f> ^ . d(h ^ ,,
A + 2c / — 5- 9 =<7 (cos a-T-^ + 2sina cos a -7^ + 26).
cos a ^ j cos'* a ^ "^ ^ d(i da ^'
Differentiate again, we have
c d6" 3c sin a .„ » /cPrf) , dd>\
-f-+ 2— <^ =^cos'a -T-5+4-/ ;
cos a aa cos a V^^i da)
, «<f> „ « • ,ti
cos a ^— + 3 cos a sm ad) , , , ,^
da ^ ff {^<f> .d^\
integrating both sides, we have
_£_ 2^0 )
cos^a cl^a^^^^"^ J'
which is the same as the result given in the question.
An expression for the tension may be found as follows.
Differentiating (2) and adding the result to (1), we obviously
get
c d'd} d'cb ^cos'a dT Tdco&'a
cos a dtr ^ da c da c da
or cos a -1 sma . 2 = - ' - - '
da ' 2 Vcos'a clt'' c d(^ )
.-. cosa.r= |/[4<^+/w}rfa;
cV<'> a
2 cos a •'^
( 126 )
Thursday, Jan. 19. 1^ to 4.
Skniob Modebatoe. Koman numbers.
Junior Modebatob. Arabic numbers.
1. Find a superior limit to the numerical values of x
consistent with the convergency of the series
2 V 3 V w" . ic"
3+ -+1:1
(n + 1)"^'. a;"
1.2 1.2.d 1.2...n
Here w«^, =
[_«+_
11' . iC"
tt_ w" w + 1 V w
a;
=(-D-
?<„=» -r^ = ?«„=» f 1 + ^1 a; = ex,
?t« \ n
and the superior limit to the values of x is therefore - .
2. If the sides of a spherical triangle be small compared
with the radius of the sphere, then each angle of tlie spherical
triangle exceeds by one-third of the spherical excess the cor-
responding angle of the plane triangle, the sides of which are
of tiic same lengtli as the sides of the spherical triangle.
If the sides of a right-angled plane triangle of given area be
bent so as to form a spherical triangle on a given sphere of
great radius, the alteration of area in the triangle is very
nearly proportional to the square of the hypothenuse.
1|— 4.] SENATE-HOUSE PROBLEMS AND EIDERS. 127
It is proved in Todhunter's Spherical Trigonometry, Art.
109, that the area of any spherical triangle whose sides
are a,'97 exceeds that of the plane triangle by the fraction
„ of the latter, where r is the radius of the sphere.
If the plane triangle be right-angled Oi^ + 0'=Y, and it maybe
proved in exactly the same way, that the alteration in area is
= r^ of the plane triangle. This expression varies as y*, be-
cause by the question the area of the plane triangle is
constant.
3. Two tangents OA, OB are drawn to a conic, and are
cut in P and <^ by a variable tangent ; prove that the locus
of the centres of all circles described about the triangle OFQ
is an hyperbola.
Taking OA, OB as axes, let the equation to the tangent
PQ be
-+|-1=0..
a p
and the equation to the conic
(1),
^-«(s+f-0*
(2).
Then the equation to the circle is
x' + 1/^ + 2xy cos CO = ax + ^y
Then since (1) touches (2) the roots of the equation
.(3).
must be equal ;
" \a a)\fi b) 4/c
To determine the centre of (3), we have
.(4).
a+y cosQ) = - ,
a
y + a; cos 0) = - .
128 SENATE-HOUSE PROBLEMS [Jan. 19,
Substituting in (4) the equation to the required locus ia
{x + 1/ cos fo) {y + X cos (o)
_ 2h (x + y cos &)) + 2a (;/ 4- x C03 w) — ah
\ Tb '
4
which is evidently an hyperbola.
If the given conic be a parabola, we must have 4/c = ah^
hence we get for the locus
x-\-y cos G) y -\-x cos « _ 1
'~a ^ h *'
which is a straight line. This is evident, a priori, for it is a
known property that all these circles pass through the focus.
iv. If w be a function of three independent variables
x, y, z, which are connected by three equations with three
new independent variables ^, ?/, ^, shew how to express the
partial differential coefficients of w, to the first and second
orders respectively, with respect to x. y, z, in terms of the
con'esponding partial differential coefficients with respect to
Apply this method to prove that, if at a certain point in a
surface r = t and s = 0 when the axes of x and y are taken
parallel to a particular pair of lines, at right angles to each
other, in the tangent plane at that point, the axis of z being
normal, then the following relations will hold at that point
whatever be the direction of the co-ordinate axes provided
they be rectangular, viz.
r s t
TT/? ^M ~ l+<?' '
where p, q, r, s, t, denote
dz dz d*z d'z d'z
di' dj/' dx'' d^' dy"
respectively.
Let tlie equation to the surface referred to the original
axes, whereof tliat of z is normal and those of x and y are
at right angles to each and tangential, be
u = z-f{x,y) = 0,
1^—4.] AND RIDERS. 129
let ^, T], ^ be the new axes, and let the relations between the
old and new variables be
du du du du dz dz
' dx dy
d. ., , - du , du
bimilarly for -7- and —rr, ;
again,
d^u ,d^z ^ d'^z ^d^z ,, ^
d^z _ <^*z _ , d^z _
oar ay^ <mry
-,..,, f, d'^u , J'm
Similarly tor ^ and ^ ,
also
d'xi , d*z yd'z . , T N <^« r
Similarly for ^^f^^^^^^'
Hence differentiating the equation
du da d^ _
d^'^d^d^^'^'
with regard to f and substituting we obtain, remembering
that
K
130 SENATE-HOUSE PROBLEMS [JuJl. 19,
Similarly, we should find
whence the proposition is proved.
V. If the differential equations of the first order
give rise to the same differential equation of the second order,
shew how the general solution of an equation of the fonn
.^{,(.,„|),t(.,..|)l = o,
may be found without integration.
Apply this or any other method to the discovery of the
general solution of the equation
Here 2a;y — — - x^yp = a* ;
l^ut
x\x- -^j = c, and y (y -px) = Ik
V
give rise to the same differential equation of the 2nd order.
Now p = - (y --] = -^ ;
ic V yJ X - c
1^ 4.] AND RIDERS. 131
ha? + cy' = he,
hx^ + T 2^' = «*•
6. Enunciate and explain D'Alembert's principle. Apply
it to determine the small oscillations in space of a uniform
heavy rod of length 2a, suspended from a fixed point by an
inextensible string of length I fastened to one extremity.
Prove that, if x be one of the horizontal co-ordinates of that
extremity of the rod to which the string is fastened,
fc = u4 sin (Wj< + a) + i5 sin (n/ + /9),
where n^, n^ are the two positive roots of the equation,
aln* - (4a + 3?) gn^ + 3/ = 0,
and A, B, a, /8, are arbitrary constants.
Take 0 the point of suspension of the string for origin, and
the axis of z vertically downwards. Let p, q, p', q' be the
cosines of the angles made by the string and rod respectively
with the axes of x and y, and let u be the distance of any ele-
ment du of the rod from that extremity to which the string is
attached. Then the co-ordinates of this element will be
x = lp + up',^
y = lq + uq',\ (1).
z= I + u, \
Then the equations of motion will be
,^^«^ = .Zi, (2),
dr dr m
\
K '1
132
SENATE-HOUSE PROBLEMS
[Jan. 19,
where T is the tension of the string and m the mass of the
rod. By D'Alembert's principle, the equation of moments
round x will be
which becomes by (1)
^ jfjd'q , d'q'\ ^j ,d^q Sa' dV „ ,, . ,,
or -2al{lJ + u^)-2la^-^—^--^ = 2aff{k + aq),
which by equation (2) reduces to
92
z^f+^^'=_
de'^l''~d(
Therefore the four equations of motion are
and two similar equations for g', g'.
To solve these put
p= A sin {nt + a), q = Bsm [nt + a),
we get
In^A-if an^B=gAA
MA-\--an^B = gB'A
n —
3
4a^+3Z
al
and the values of n are found.
vii. A rigid bodv is rotating about an axis through its
centre of gravity when a certain point of the body becomes
1^—4.] AND RIDERS. 133
suddenly'fixed, the axis being simultaneously set free ; find the
equations of the new instantaneous axis ; and prove that, if it
be parallel to the originally fixed axis, the point must lie in
the line represented by the equations
d^hc + l^my + <^nz = 0,
(j._e')|+(c'-a»)|4-(a'-5')^ = 0;
the principal axes through the centre of gravity being taken as
axes of co-ordinates, a, b, c the radii of gyration about these
lines, and I, m, n the direction-cosines of the originally fixed
axis referred to them.
In order that the new instantaneous axis may be parallel to
the originally fixed axis the plane passing through the im-
pulse at the fixed point and the centre of gravity must be
diametral to the originally fixed axis. Hence the point
must lie in the plane
o*ic 4- J'wy + c'w2; = 0 (1).
Again, in order that rotation round the original axis
through the centre of gravity combined with a velocity of
translation parallel to the blow may reduce the point to rest,
the line joining the point with the centre of gravity must co-
incide with the projection of the axis upon the diametral
plane; let X, /*, v be the direction-cosines of the normal to the
plane passing through the axis and the normal to the diametral
plane, then
TK + mfi + wv = 0,
c^Vk -f- h^miJk -H <?nv = 0,
Ik (c' — a*) -I- mfi (c* — &*) = 0, and so on ;
therefore the equation to the plane is
(t« _ c») ^ + (c» - a») ^ + (a' - J») - = 0 (2);
(1) and (2) determine the line in which the point must be
situated.
134 SKNATE-HOUSE PROBLEMS [Jan. 19,
ix. Prove tlie following relation between the perturbations
of a planet in longitude and radius vector :
h being twice the sectorial area described in a unit of time by
the undisturbed planet round the Sun ; and find the corre-
sponding relation whatever be the law of force, provided it be
central and a function of the distance only, and provided
such a function as R can be found.
Let F be the central force.
Our equations of motion give us
whence, proceeding as in Airy's Tracts, we get
put r 4- Sr, 6 + Zd, for r and 6 respectively, r and 6 being the
functions of t given in the undisturbed motion, and we obtain
dr dhr ,r,d'r ^ d\Br „ . fdOV ^ ,d0 dhO
'dt'-dt+'^'de^^'-^r-^'^'^dJ-^'df-dT
= eFBr + AFBr + Ar^-^Br-6[^dt-^r'^^,
dr J dt dr
which, as in Airy's Tracts, is easily reduced to
dr
dR
dr'
d
dt\
\ dt dt J dt dt
„ -_ , dF ^ „ rd(R) , , dR
= 8FBr+Ar -r-Br-6 —j-! dt-ir-j^,
dr J dt dr
■U — ^-l ^ND RIDERS. 135
Put h for r' -j- , and we get
+ 3 -\-^dt + 2r -J AF8r - 2r ^- SrV .
j at dr dr )
This is reduced to the equation in the question when F is
such a function of r that
10. If the ohject-glass of a telescope be covered over by
a diaphragm, pierced in the centre by a small hole, the form
of which is a rectangle, state generally the nature of the
spectra formed about the image of a star on a screen placed at
the focus.
If the hole be circular and the screen be pushed towards
the lens, prove that, when the light is homogeneous, the centre
is alternately bright and dark. Trace also the order of the
colours seen if the light be not homogeneous.
This rider is obviously the same as the problem solved by
Airy in Art. 79 of his Iract on Light. For the introduction
of the lens is merely making the incident pencil convergent
instead of divergent, that is, the a in Airy's investigation is
to be made negative. The intensity of the illumination
will be
4\Vi» . ,/27ra-&
rsm
{a-hf
J^ira-b ^
The interpretation of this result is nearly the same as that
given by Airy, except that we now begin at the central
spot.
136 SENATE-HOUSE PROBLEMS AND RIDERS. [JaJl. 19.
When b=a, the intensity becomes ttV; when b—a = ± — g-,
c
the sine is unity, and intensity is measured by Ac*; when
h — a = ± — — , the intensity is zero, and as J — a continues to
increase, we have alternately brightness and darkness.
If the light be not homogeneous, we have, when h = a, a.
white spot, and as J ~ a increases ; the violet disappears first,
leaving a red spot, which gives place to the other colours in
order.
( 137 )
Friday, Jan. 20. 9 to 12.
SxinOB MoDKBATOB. Romaa numbers.
Sbniob Exahineb. Arabic numbers.
1. OA, OB are common tangents to two conies having a
common focus S, CA, CB are tangents at one of their points
of intersection, BD, AE tangents intersecting CA, CB in
D, E. Prove that SDE is a straight line.
Let the conies be reciprocated into two circles within both
of wliich S lies. Fig. 51.
oa, o5, their points of intersection, correspond to OA, OB;
ca, cb, the points of contact of a common tangent, to CA, CB.
The straight line b joining cb and ob meets one circle in hd,
a ca ... oa the other in ae,
bd, ae correspond to BD, AE,
and d, e which join Id, ca and ae, cb to D and E.
It is required to shew that d, e are parallel.
The angle between d and a = that between b and o = that
between a and e.
ii. Define the terra potential of a mass, the particles of
which attract according to tlie law of nature ; and prove that,
if a body moveable about a fixed axis be subject to the ac-
tion of an attracting mass of which the potential is V, then
138 SENATE-HOUSE PEOBLEMS [Jan. 20,
///
dV
-jrz dm is the moment which must be impressed upon the
body about that axis in order to produce equilibrium, where
6 is the inclination of the plane through the fixed axis and
the particle of which the mass is din to a fixed plane.
A uniform straight line, the particles of which attract ac-
cording to the law of the inverse square, acts upon a rigid
uniform circular arc in the same plane with the line, of which
the radius is equal to the line, and which is moveable about
an axis through its centre perpendicular to its plane, the axis
being coincident with one extremity of the line. Prove that
the moment necessary to produce equilibrium when the bound-
ing radii are inclined at the angles a and yS to the line pro-
duced is proportional to
sec - 4- 1
log— ;g— .
sec 1+1
Let AB be the straight line, P a point at a distance r
from AB, Fig. 52,
Q a point in the line at dist. x from B,
P(^= cc* + r' + Ixr cos 6 ;
ioVI(a; +
dx
0 ^[{x + r cos^plVihi^
_. V2 V(l 4- cos g) + (1 + cos &)
~ ^^ 1+cos^
a a 6
2 cos- + 2 cos*- 1+cos- Q
= log ^ ^ ^ = log -^ = log (sec- + 1 ) .
2 cos* - cos -
2 *t
Hence since in this case dm = rdO, the moment on
crrdv sec 1+1
sec — + 1
9— 12.J AND KIDEIJS. 139
4. If an elastic string, whose natural length is that of a
uniform rod, be attached to the rod at both ends and sus-
pended by the middle point, prove hy means of Vis Viva
that the rod will sink until the strings are inclined to the
horizon at an angle $, which satisfies the equation
a Q
cot' - - cot - - 2w = 0,
where the tension of the string, when stretched to double its
length, is n times the weight.
If the string be suspended by a point, not in the middle,
write down the equation of Vis Viva.
C is the point of suspension. Fig. 53.
ABC= 6 at time t.
Let AB = 2a, BC= r = a sin 0,
m the mass of the rod.
The moving effect of the tension of CB
r — a
= rvrng . .
By Vis Viva, at the time t,
f T ~~ d
«M?*= 2mga tan ^ — 4 \nmg dr
= ^Tfiga tan v — Inmg . -^ ,
since v = 0 when ^ = 0, and r=a\
therefore the rod comes to rest when
tan^-w(sec^-l)*=0,
. /, /, .4^
or gm B cos ^ — 4w sm — = 0,
6 6
or cot' - — cot - — 2n = 0.
If the fixed point divide tlie string into the portions a — c^
a + c, and these be inclined at angles 6, & to the horizon, and
140 SENATE-HOUSE PROBLEMS [Jan. 20,
he of length r, r at the time «, ^ the inclination of the rod,
the equation of Vis Viva gives
/dfrV 72/#V c {{^r-a-^cf {r'-a-c)\
where a = ^ (r sin ^ + r' sin 6') ,
and 2a cos <^ = r cos 6 -\- r cos ^',
2a sin ^ = — r sin d+r cos ^',
which with two more dynamical equations are sufficient to
determine the problem of the motion.
5. If an oblate spheroid be moveable about its centre,
and B be the inclination of its equator to a fixed plane, i/r the
inclination of the line of intersection of its equator with this
plane to a fixed line in the plane, A and C the respective
moments of inertia about the axis of figure and a line in the
equator respectively, L and M the moments of impressed
couples about the line of intersection of the equator with the
fixed plane, and a line in the equator perpendicular to this
latter line respectively, to the angular velocity about the axis
of figure, prove that
0^-0 f-^ j sin ^ cos ^ 4- A(o smd-^ = i,
ci ('^4:,ine] + c^i^. cos e-Aa,^ = M,
dt\dt J dt dt dt
hence deduce the precessional and nutational velocity of the
Earth's axis, assuming the effect of the Sun's action to be a
couple of winch the moment is m sin A cos A about an axis
in the equator 90° distant from the Sun, m being a very small
quantity, A and C very nearly equal, and the Sun's motion
in declination and right ascension being neglected.
The angular velocities about the axis of figure, the line of
nodes, and the axis in the equator 90° distant from the line of
nodes respectively, are
d0 , dyft . ^
9—12.] AND EIDERS. 141
and the line of nodes recedes from the axis of figure with the
angular velocity -j- sin 6, and approaches the perpendicular
to the line of nodes in the equator with the angular velocity
-T- cos 6. Similarly the perpendicular to the line of nodes in
the equator approaches the axis with the angular velocity -j- ,
and recedes from the line of nodes with the velocity -— cos 6.
Hence by the formulae for accelerations of angular momenta
referred to moving axes,
= acceleration of momentum round the line of nodes,
^dt\-i''''^)^I'''^'^dt-Tt'^''
= acceleration of momentum round the axis perpendicular
to the line of nodes,
whence the equations in the question.
In the case of the Sun's action upon the Earth
X = m sin A cos A sin a,
M= — m sin A cos A cos a,
and the squares of the very small quantities -— and -j- are
to be neglected as well as the differencee between A and C.
Hence the equations become
d^d . ^ d-^ m . . . .
-rf-^ + (o.9m0 — =-77 sm A cos A sm a,
d(^ dt C *
d ( . ^d-^lA dd m . . .
^, . sni ^ -p- — Q) -J- = — -T7 sm A cos A cos a,
dt \ dt) dt C
142 SENATE-HOUSE PROBLEMS [Jan. 20,
differentiate the first and subtract the second multiplied by o),
remembering that A and a are to be considered constant, and
we have
d'O .dd m
-To +o> -jT = Ty ® sm A cos A cos a,
dr at C '
d' fde\ ^dd m
Similarly,
dO m . .
. -r = — 7y sm A cos A cos a.
dt 0)6
. ^d-dr m . . . .
ama -~-= — ^ sin A cos A sm a,
dt a>C
neglecting the arbitrary parts of the respective integrals.
6. If a solid of revolution be immersed in a heavy homoge-
neous fluid with its axis vertical, prove that, when the total
normal pressure on the surface is a minimum, its form must
be such that the numerical value of the diameter of curvature
of the meridian at any point is a harmonic mean between the
segments of the normal to the surface at that point intercepted
between the point and the surface of the fluid and between
the point and the axis, respectively.
In this case Jxi/ds is to be a minimum.
Hence we must substitute icy for fi in the formula,
1 1 fdu, dii
- = /-cosa + -7-
p fM \dx dy
giving us
1 cos a cos ^
(l^^^«+f ^°^^)'
+
p X y
2 1 1
2/3 a; sec a y sec fi '
proving the proposition.
viii. Explain the phenomenon of external conical refrac-
tion where a small pencil of light passes through a biaxal
crystal ; and describe an experiment by Aviiich this pheno-
menon may be manifested.
9—12.] AND RIDERS. 143
If the crystal be bounded by planes perpendicular to the
line bisecting the acute angles between the optic axes, write
down equations Avhence the equation of the cone of emerging
rays may be obtained.
Let Z, m, n be the direction-cosines of the perpendicular
to any wave-front incident upon the second surface of the
crystal ; the axes of reference being the axes of elasticity ;
let X, fi, V be the direction-cosines of the perpendicular to the
corresponding wave-front after emergence. Then \, fi, v are
tlie direction-cosines to the corresponding emergent ray, and
if the point of the second surface at which the light emerges
be taken as origin, the equation of the cone will be deter-
mined by eliminating \ /*, v between the equations
X-fi-p ^^^'
and an equation between X, fi, v which remains to be found.
(1) The emergent ray, the normal at the point of emer-
gence and the perpendicular to the front of the incident wave,
lie in the same plane, whence
^ = ^ (B).
V n
(2) The sines of the angles of emergence and incidence
are to each other as w : v, u being taken for the velocity of
light in air, and v for the velocity with which the wave-front
under consideration was propagated through the crystal,
whence
'-^='-^ (C).
(3) Also the value of v in terras of I, m, n is to be found by
substituting the values corresponding to the multiple point in
the equations a (page 18 of Griffin's tract on Double Refrac-
tion), whence we obtain the following relations,
v^ — a' _ - ^^
S 8
V —c _ vn
(D).
144 SENATE-HOUSE PROBLEMS AND RIDERS. [Jan. 20.
Between the four equations of B, G, and D, and the addi-
tional equation
we may eliminate I, m, n, and v, and obtain the relation
sought between X, /x, and v.
The final relation between x, y, and z gives a cone of the
fourth degree.
( 145 )
Friday, Jan. 20. 1^ to 4.
JcmoB MoDEBATOB. Boman numbers.
JUNIOB EXAHINKB. Arabic numbers.
1. If a, /3, 7 be the respective distances of a straight line
from the three angular points of a triangle ABC, these dis-
tances being reckoned positive or negative according as their
directions fall within the angles of the triangle itself or their
supplements, investigate the following relation,
(a sin Af+ (/3 sin J5)*+ (7 sin C)'- 2 cos ^ sin B sin Cfiy
— 2 cos -B sin (7 sin A<ya — 2 cos Osin -4 sin Bafi
= 4i?'sinM sin'^^siu'C,
where R is the radius of the circumscribed circle.
Referring to fig. 54, we have
AP=a, BQ^-fi, CR = -'i.
Hence, if 50 = a, CA==h, AB = c, &ndBAP=~-0,
-47 = a + iS = c cos f Y - ^j »
similarly a + 7 = i cos f— +^) ;
143 SENATE-HOUSE PROBLEMS [Jail, 20,
.'. & (a + /8) + c (a + 7) = 2bc cos — cos 0,
J (a + /3) + c (a + 7) = 2lc sin —sin Q.
A A
^Multiply these by sin — , cos — , respectively, and add
squares, then, since he sin A = — ^ ,
i'^(a4-ySr+cMa + 7r-2Jccos^(a + /3)(a+7)=:^,
or, since 2bc cos A — }?-\-(? — <^,
aV + 1'/3* + cV - 25c cos A.^'y-2ca cos B. ya
a'bV
— 2ah cos C. a^ =
Now,
4i^ •
sin A sin ^ sin C 1
a b c 2B'
.: (a sin Ay+ (yS sin By+ (7 sin Cy - 2 cos ^ sin i?sin C. ^y
— 2 cos jB sin Csin -4 . 7a — 2 cos Csin ^ sin 5 . o^
= 4i2sin''^sin'^sin'C,
the required result.
2. State the positive and negative characteristics of d
singular solution of a differential equation ; and shew how it
is deduced from the complete primitive. Shew also how the
singular solution of a differential equation of the first order
is obtained from the equation itself.
Obtain the singular solution of the equation of which
y cos*m = 2 cos (a; — 2 m)
is the complete primitive ; and find the singular solution of
the equation
1| — i.] AND RIDERS. J4T
(a) y coa^m = 2 cos {x — 2m).
This may be put into the form
{y — cos x) cos 2m — sin x sin 2ni = ^ .
Differentiating with respect to m, we get
(y — cos x) sin 2m + sin x cos 2m = 0,
whence, adding squares and reducing,
-y — 2v COS a; + 1 = 0,
4
the required singular solution.
(/8) This may be written under the form
The condition for a singular solution is
dp
which, in this case, gives
Z{x + yyp'-2{x'-f)p = Q,
2 x — y
Substituting this value for p in the original equation,
we get
^iKx+y) 9 VaJ + y/
or 4(ic-^)'' = 27 (a; + 3/)^
the required singular solution.
iii. Prove that, in any curve of double curvature, the
locus of the centres of spherical curvature is the edge of n -
gression of the envelope of the normal planes. Prove also
that this locus cannot be an evolute.
The normal plane to the locus of the centres of circular
curvature bisects the radius of spherical curvature.
148 SENATE-HOUSE PROBLEMS [/an. 20,
If two consecutive normal planes be drawn to a curve, their
intersection is a generator of th? envelope of the planes, or, as
it is usually called, of the polar surface. The envelope of
these generators is known to be the edge of regression, tliat
is, any two consecutive generators intersect on the edge of
regression.
The intersection of two consecutive normal planes is a
straight line through the centre of circular curvature, and it
is clearly such that if any point be taken on it, that point is
equally distant from three consecutive points on the curve.
Therefore the intersection of two consecutive generators is
equally distant from four consecutive points on the curve,
i. e. it is the centre of spherical curvature.
Hence the proposition follows.
It is also clear that the edge of regression cannot be an
evolute, because its tangents, which are the generators of the
above polar surface, do not pass through the original curve.
Let A, A', A" be three consecutive points on a curve, and
let the plane of the paper be the normal plane at A', Let CO
be the intersection of the normal planes at A, A'; CO the
intersection of those at A', A". Let the plane that passes
through the three points A, A\ A" cut CO, C 0 in C and C.
Then C, C are ultimately two consecutive centres of circular
curvature, and 0 is the corresponding centre of spherical
curvature. Fig. 55.
Now A' CO, A' CO, are two right angles in one plane, and
therefore a circle described on A' 0 as diameter will pass
through 0 and C. And CC is ultimately a tangent to the
circle, hence a normal to CC bisects A'O the diameter. But
CC is also ultimately a tangent to the locus of C, whence
the normal plane to the locus of C bisects A'O the radius of
spherical curvature.
4. Determine the class of cui^ves which possess the pro-
perty that the locus of the extremity of the polar subtangent
of any one is similar to the curve itself.
Shew that r^e"* = a is the equation of such a curve.
U — 4.] AND RIDERS. 149
It may be shewn that, if - =f{6) be the equation of a
given curve, that of the locus of the extremity of its polar
subtangent is - =/' ( ^ — 9 ) •
Now, if /OT=^
/■Kf) =
» '"-!)+' .(.-f)
6
a
=hH^-iy
,./(o
Hence, the equation of the locus of the extremities of the
polar subtangents of the curve r^e"^ = a, is
\ 2 m/
l^.-('-^i) = la,
representing a curve of similar form to the given one, but
with its dimensions varied in the ratio e : m, and turned
through an angle g ~ ~ •
V. If a homogeneous sphere roll on a perfectly rough
plane under the action of any forces whatever, of which the
resultant passes through the centre of the sphere, the motion
of the centre of gravity will be the same as if the plane were
smooth and all the forces were reduced in a certam constant
ratio ; and the plane is the only surface which possesses this
property.
. Take the plane as the plane of xi/, and take axes fixed in
space. Let w,, o),, 0)3 be the angular velocities about diameters
parallel to the axes. Let v^, v^, v, be the velocities of the
centre, X, Y, Z the impressed forces, and F^ G the frictions
resolved parallel to the axes. Let a — radius of the sphere.
Then the equations of motion will be
150
SENATE-HOUSE PROBLEMS
[Jan. 20,
idea „
^ dt ^'
(1)
dt
= x+i?;
= r+G^,
(2)
and since the point of contact is at rest we have the geo-
metrical equations
V, — awj = 0, "1
r„ + awj :i= 0. J
(3)
By differentiating (3) and substituting from (2) we have
~ d' dt '
r, 1^ dvy
a at
Hence the equations of motion of the centre are
dv, _ a*
X,
dt a' + A^
These are the very equations we should have had if the
plane had been smooth and forces had been reduced in the
a'
ratio -= — 5,
a +/C
A rough plane is tlie only surface which possesses the
property enunciated in the question.
li-4.]
AND RIDERS.
151
Let I, m, n be the direction-cosines of the normal at any
point of a surface, X', Y', Z' the resolved parts of the friction,
and v^, i\, V, of the velocities of the centre of gravity parallel
to the axes. Then by the question
Also the equations of motion are
^ -j-^ = {mZ'—n F')a,
K^^={nX' -lZ')a,
and the geometrical equations are
v^ = {a>,m - o>„n) a, j
V, = {(oj, - (ojn) a ;
K^ dm. dv. 7 dv.
• — — ; = nm. — = — at —^
X at at
dt
dn ,, »
- ^ a {Ua^ + mWy-^-rm^ ;
dn ,j ^ , \ ^'"«
" dt
where /i is a constant, also
dl n , , V dm^
-^ {Ita^ + mtOy + na),) = fi -^ ,
dm ,, . da>y
dt ^ ' ■*■ ^^"^ + »»«.)= /^ ^ >
152 SENATE-HOUSE PROBLEMS [Jan. 20,
Multiply these hy q)„ cd^, &>„ and add, we get, since
where D, is the resultant of the angular velocities tw^, tw,,, <»,.
Now Iv^ + mVy + wu, = 0 ;
dl dm dn
^ J dl dm dn
by cross-multiplication, we get
.. j^ = F{v„n-v,m)
= Pa'{(i)^-l {1(0^ + mwy + nwi)]
.'. = Ql say ;
dl ^, dm ^ dn ^
•••jr^^' ^=^'"' Tr^^
whence it easily follows that Z, m, n are constants.
vii. If the Earth be completely covered by a sea of small
depth, prove that the depth in latitude I is very nearly equal
to i/(l— esin^Z) wliere // is the depth at the equator, and
e the ellipticity of the Earth.
The surface of the Earth and the surface of the water rest-
ing on the Earth will both be surfaces of equilibrium, and
therefore will be similar spheroids. Draw two parallel tan-
gent planes to thfe Earth and to the surface of the sea, the
1| — 4.] AND RIDERS. 153
distance li between these planes is the depth of the sea at the
point at which the tangent planes were drawn. Let p be the
perpendicular from the centre on either of these planes, then
by similar figures, the ratio - is constant. Let I be the lati-
tude of the place, a, h the semi-axes of the spheroid, then
/ = a''cos7+J'8in7
= a'(l-2e8in»Z);
.'. p = a{l — e sin'^) ;
where H is some constant. But putting ? = 0, we get h = H;
therefore H is the depth of the sea at the equator.
ix. The base of an infinite cylinder is the space contained
between an equilateral hyperbola and its asymptotes. A
plane is drawn perpendicular to the base, and cutting it in
a straight line parallel to an asymptote, and the portion of
the cylinder between this plane and its parallel asymptote is
filled with homogeneous fluid, under the action of no im-
pressed forces. The plane being suddenly removed, deter-
mine the motion ; and prove that the free surface of the fluid
will remain plane, and advance with a uniform velocity
proportional to »Jisr, where ct is the pressure at an infinite
distance, which is supposed to remain constant tliroughout
the motion.
Since the fluid starts from rest, the function <^ exists, and
we have
d'<f> d'6 ^
— — + = 0.
dx^ dy^
Transforming to polar co-ordinates
dr\ dr]^ rdS"'
To solve this, assume
0 = ylr"€"^;
154 SENATE-HOUSE PROBLEMS [Jan. 20,
or generally (j) can he expressed in a series whose general
term is
<^= (^r" + -sj cosn^ (1).
The number of terms to be taken and the values of n
clearly depend on the geometrical conditions of the bounding
surfaces.
Now we know that the curve ^ = constant cuts all the
lines of motion at right angles, hence this curve must also
cut at right angles the sides of the containing vessel. Let
r'$' be the co-ordinates of any point of the hyperbola or
of its asymptotes, then we must have
The hyperbola and Its asymptotes may be included in the
single equation r'* sin 1& = 2a*, where a has the two values
a = a and a = 0.
Hence -r^, = —rcot20.
Again, from the value of <f> we have
Hence equation (2) becomes
X (^r"- ^ cosn^ . tan 2^ = S (^r-+ ^ sinw^ ... (3).
This equation will evidently be satisfied if we take n = 2
and B = 0, hence we have
</) = ^r*cos2^ (4).
1^ 4.] AND RIDERS. 155
This value of <^ determines the motion, and we shall know-
that it is the true value if the other conditions of the pro-
blem are satisfied. These conditions are that the fluid starts
from rest, and that along the free surface to be determined
from equation (4) the pressure should equal zero.
First, to determine the motion from (4) ; we have
hence the velocity of any particle distant r from the centre
is 2Ar ; and all the particles move along hyperbolas having
the axes for asymptotes. Take any particle whose co-
ordinates are x^, y^ at time ^ = 0, its co-ordinates at any
other time are
AOt
X— XJ£
y=y^^
hence, if two particles have the same abscissae at the time
f=0, they always have the same abscissae, and therefore the
free boundary of the fluid being originally a straight line,
it will be always a straight line.
Secondly, to determine the pressure at any point ; we have
^-=^-1'-! (^)-
Let ^, 7} be the co-ordinates of any point in the free surface
of the fluid ; then
tl A
or
H'^-^'^M'^'-wh'-
This by hypothesis is a straight line parallel to the axis of rj ;
156 SENATE-HOUSE PROBLEMS AND RIDERS. [Jan. 20.
where A^ is an arbitrary constant ;
The pressure at infinity ia p = xr, and throughout tlie fluid
we have
p= G-4.A^j? (6);
.-. tzr = (7,
hence along the free surface, equating the two values of C,
= fo + V<»«>
where f^ is the abscissa of the free surface at the time < = 0.
Hence the boundary moves uniformly with a velocity V®.
SENATE-HOUSE PROBLEMS AND RIDERS
FOR THE YEAR EIGHTEEN HUNDRED AND SIXTY.
MODEBATOBS :
Henbt William Watson, M.A. Trinity College.
Edwaed John Rooth, M.A. St Peter's College.
EXAUINEBS :
Pkboival Fbost, M.A. St John's College.
NOBMAN MACLEOD Febbebs, M.A. Gonville and Caius College.
Tuesday, January 3. 9 to 12.
In the amsioers to tlie first six questions the symbol — muM not be
used. The only ahhreviation admitted for tlie square described
on AB is sq. on AB, and for the rectangle contained by AB and
CD, the rect. AB, CD.
1. Define parallel straight lines.
Parallelograms upon the same base, and between the same
parallels, are equal to one another.
If a straight line DME be drawn through the middle point M
of the base BC of a triangle ABC, so as to cut ofi" equal parts AD,
AE from the sides AB, AC, produced if necessary respectively,
then shall BD be equal to CE.
2. Describe a square that shall be equal to a given rectilineal
figura
Shew how to construct a rectangle which shall be equal to a
given square; (1) when the sxun, and (2) when the difference of
two adjacent sides is given.
3. If, from any point without a circle, two straight lines be
drawn, one of which cuts the circle, and the other touches it, the
rectangle contaiued by the whole line which cuts the circle, and
158 SENATE-HOUSE PROBLEMS [Jan. 3,
the part of it without the circle, shall be equal to the square on
the line which touches it.
If two chords AB, AC be drawn from any point ^ of a circle,
and be produced to J) and U, so that the rectangle AC, AE is
equal to the rectangle AB, AD, then, if C? be the centre of the
circle, AO i& perpendicular to DE.
iv. Describe an isosceles triangle, having each of the angles at
the base double of the third angle.
If -4 be the vertex, and BD the base of the constructed triangle,
D being one of the points of intersection of the two circles em-
ployed in the construction, and E the other, and AE be drawn
meeting BD produced in F, prove that FAB is another isosceles
triangle of the same kind.
V. If the outward angle of a triangle, made by producing one
of its sides, be divided into two equal angles by a straight line
which also cuts the base produced ; the segments between the
dividing line and the extremities of the base have the same ratio
which the other sides of the triangle have to one another.
If the two sides, containing the angle through which the bisect-
ing line is drawn, be equal, interpret the result of the proposition.
Prove, from this proposition and the preceding, that the straight
lines, bisecting one angle of a triangle internally and the other two
externally pass through the same point.
vi. If two straight lines be cut by parallel planes, they shall
be cut in the same ratio.
If three straight lines, which do not all lie in one plane, be cut
in the same ratio by three planes, two of which are parallel, shew /
that the third will be parallel to the other two, if its intersections
with the three straight lines are not all in one straight line.
vii. Define a parabola ; and prove, from the definition, that it
cannot be cut by a straight line in more than two points.
Prove that, if the tangent at P meet the directrix in D, DSP
is a right angle.
viii. If /* be a point in an ellipse of which the foci are «S^and H,
the straight line, which bisects the angle between SP produced and
IIP, meets the ellipse in no point but P.
P, Q are points in two confocal ellipses, at which the line join-
ing the common foci subtends equal angles ; prove that the tangents
9 — 12.] AND RIDERS. 159
at P, Q are inclined at an angle which is equal to the angle sub-
tended by PQ at either focus.
ix. Assundng the property of the tangent to an ellipse enun-
ciated in 8, prove that *ST. UZ=BC.
If a circle, passing through Y and Z, touch the major axis in
Q, and that diameter of the circle, which passes through Q^ meet
the tangent in P, then PQ = BG.
10. Prove that, if in any ellipse any diameter CD be drawn
parallel to the tangent at the extremity of any other diameter CP,
then CP will also be parallel to the tangent at the extremity of CD.
If PG, the normal at -P, cut the major axis in G, and if DR^
PN be the ordinates of D and P, prove that the triangles PGN,
DEC are similar; and thence deduce that PG bears a constant
ratio to CD.
11. Define an asymptote to an hyperbola; and prove that, if
from any point in the curve straight lines be drawn parallel to and
terminated by the asymptotes, their rectangle is invariable.
In an hyperbola, supposing the two asymptotes and one point
of the curve to be given in position, shew how to construct the
curve ; and find the position of the foci.
12. If a right cone be cut by a plane which is not parallel to
a line in the surface, and which meets only one sheet of the cone,
the section will be an ellipse.
Given a right cone and a point within it, there are but two
sections which have this point for focus ; and the planes of these
sections make equal angles with the straight line joining the given
point and the vertex of the cone.
Tuesday, January 3. 1^ to 4.
1. The sum of £177 is to be divided among 15 men, 20 women,
and 30 children, in such a manner that a man and a child may
together receive tis much as two women, and all the women may
together receive £60; what will they respectively receive 1
2. A wine merchant buys 12 dozen of port at 84». |)er dozen,
and 60 dozen more at 48«. per dozen ; he mixes them, and sells the
mixture at 72». per dozen; what profit per'cent. does he realize on
his original outlay I
1 60 SENATE-HOUSE PROBLEMS [Jan. 3,
3. Assuming that a" is defined by the equation a", a" = a"*"
for all values of m and n, and that a' = a, interpret the expression
o", when m is any commensurable quantity, positive or negative.
Extend your mode of interpretation, so as to assign a meaning to
such a symbol as a^'.
4. Solve the equations,
x+2 x+3
-3-^-2-='' <^^
2x + {x*-ay 2x-(x*-ay _ 3
2x-{x'-ay^2x + (x*-ay~2 ^ ''
x' — yz = a', i/'-zx = b'', s?-xy = <^ (3).
5. Explain the terms permutation and combination; and
find the number of permutations of n things taken r together.
If ^P represent the number of permutations of n things taken
r together, and a^, a^, a^--- be the successive terms of a descending
arithmetical progression, of which the common difierence is d,
prove that
P P P= P
6. Prove the Binomial Theorem for a positive integral value
of the index.
Prove that 2"-^2"-' + '^^— ^^ 2"-»- ... + (- 1)"= 1.
vii. Define a logarithm ; and prove that log„iV^= ^og Jb . log^iV;
and, given that log,^2 = -30103, find log„50.
viii. Define the principal trigonometrical i-atios ; and trace the
, . . ,. sin (tt cos 6) /, . -.
changes m sign 01 ; - . .' , as ^ vanes from 0 to tt.
^ ^ cos (tt sin d) *
ix. Prove the formula sin {A + B) = sin A cos B + cos A sin j5,
A and B being each less than a right angle ; and assuming its
truth when the values of A and B are unlimited, deduce the ex-
pression for cos {A — B).
X. Prove that
. . . J, ^ . A+B A-B
sin .4 + sm jd = 2 sin — _— cos — ^ — ,
^ 2
and sinS (ii - 16") = 4 cos (.4 - 46")co8 (.4 + 15*) sin (.4 - 15"),
1^ — 4.] AND RIDERS. 161
and find sin A and sin B from the equations
a sin^A + b s,\n°B = c,
a sin 2A —b sin 2B = 0.
xL Prove, a priori, that sin A, when expressed in terms of
A
sin — , has two equal values of opposite signs ; and that cos A,
A
when expressed in terms of cos -^ , has only one value ; and give a
geometrical illustration-
xiL Prove that, when Q is less than ^ , sin B, 6, and tan 6 are
in order of magnitude, and that they vanish in a ratio of equality.
A railway passenger seated in one comer of the carriage looks
out of the windows at the further end and observes that a star
near the horizon is tx-aversing these windows in the direction of
the train's motion and that it is obscured by the partition between
the comer window on his own side of the carriage and the middle
window while the ti'ain is moving through the seventh part of a
mile. Shew that the train is on a curve the concavity of which is
directed towards the star, and which, if it be circular, has a I'adius
of nearly three miles ; the length of the carriage being seven feet
and the breadth of the partition four inches.
xiii If a, 6, and B be given, shew under what circumstances
there will be two triangles satisiying the conditions of the problem.
Prove that the circles circumscribing both triangles are equal
in magnitude, and that the distance between their centres is
^(i'cosec'^-a*).
"Wednesday, January 4. 9 to 12.
1. Enunciate the proposition of the parallelogram of forces;
and, assuming its truth for the magnitude, prove it also for the
direction, of the resultant.
2. When three forces acting at a point are in equilibrium, each
force is proportional to the sine of the angle between the other two.
Two equal particles, eawih attracting with a force varying di-
rectly as the distance, are situated at the opposite extremities of a
diameter of a horizontal circle, on whose cii-cumference a small
M
162 SENATE-HOUSE PROBLEMS [/an. 4,
smooth ring is capable of sliding; pi'ove that the ring will be kept
at rest in any position under the attraction of the particles.
3. "When three forces, acting in one plane on a i-igid body
produce equilibrium, the algebraical sum of the moments of either
pair about any point in the line of action of the third is zero.
Two equal heavy j)articles are situated at the extremities of the
latus rectum of a parabolic arc without weight, which is placed
with its vertex in contact with that of an equal parabola, whose
axis is vertical and concavity downwards ; prove that the parabolic
arc may be turned through any angle without disturbing its equi-
librium, provided no sliding be possible between the curves.
4. Find the position of equilibrium when a common balance
is loaded with given unequal weights.
If the tongue of the balance be very slightly out of adjustment,
prove that the true weight of a body is the arithmetic mean of its
apparent weights, when weighed in the opposite scales.
5. Prove that every rigid body has one and only one centre
of gi'avity.
In the figure of Euclid, Book i. Prop. 47, if the perimeters of
the squares be regarded as physical lines uniform throughout, prove
that the figure will balance about the middle point of the hypo-
thenuse with that line horizontal, the lines of construction having
no weight.
6. Enunciate the principal laws of statical friction.
A uniform heavy rod, having one extremity attached to a fixed
point, about which it is free to move in all directions, passes over
the circumference of a rough ring whose centre is at the fixed point
and whose plane is inclined at a given angle to the horizon ; find
the limiting position of equilibrium.
vii. Explain how uniform velocity and uniform acceleration are
measured.
A point, moving with a unifonn acceleration, describes 20 feet
in the half-second which elapses after the first second of its motion ;
compare its acceleration witli that of a falling heavy particle; and
give its numerical measure, taking a minute as the unit of time,
and a mile as that of space.
viiL Describe any experiment by which it is shewn, that a force
acting on a given i)article, produces an acceleration, proportional
9—12.] AND RIDERS. 163
to the statical measure of the force. Hence deduce a definition of
mass.
ix. A heavy particle slides down a smooth inclined plane of
given height; prove that the time of its descent varies as the
secant of the inclination of the plane to the vertical.
X. Prove that the path of a projectile in a vacuum is a para-
bola.
A heavy particle is projected from a given point with a given
velocity so as to pass through another given point ; prove that, in
general, there will be two parabolic paths which the particle may
describe; and give a geometrical construction to determine their
foci. Also find the locus of the second point in order that there
may be only one parabolic path.
xi. Two imperfectly elastic balls of given masses, moving in
the same directions with given velocities, impinge directly on one
another ; determine their velocities after impact.
A series of perfectly elastic balls are arranged in the same
straight line, one of them impinges directly on the next, and so
on ; prove that, if their masses form a geometrical progression of
which the common ratio is 2, their velocities after impact will form
a geometrical progi'ession of which the common ratio is |.
xii. Define the cycloid ; and prove that, if a particle oscillate
in a cycloid, the time of an oscillation will be independent of the
arc of vibration.
Wednesday, January 4. 1| to 4.
1 . Explain what is meant by " the pressure of a fluid referred
to a unit of area." Prove that the pressure at any depth z below
the surface of a homogeneous fluid of density p, contained in a
vessel of any form, may be found from the formula p - gpz + U,
where 11 is the pressure of the atmosphere.
A uniform tube is bent into the form of a parabola, and placed
with its vertex downwards and axis vertical : supposing any
quantities of two fluids of densities p, p to be poured into it, and
r, r to be the distances of the two free surfaces respectively, from
the focus, then the distance of the common surface from the focus
wiiibe^^^::^:^.
p-p
m2
164 SENATE-HOUSE PROBLEMS [Jan. 4,
2. The whole fluid pressure on a surface immersed in a fluid
is equal to the weight of a column of fluid, having for base the
area of the surface immersed and for height the depth of the centre
of gravity of the surface below the surface of the fluid. In what
case will this give the resultant pressure 1
A parallelogram is immersed in a fluid with one side in the
surface ; shew how to draw a line from one extremity of this side
dividing the parallelogram into two parts on which the pressui-es
are equal.
3. A heavy homogeneous body being wholly immersed in a
fluid; shew how to find the magnitude and line of action of the
force required to keep it in any given position.
A heavy hollow right cone, closed by a base without weight, is
immersed in a fluid, find the force that will sustain it with its axis
horizontal.
4. State the law that connects the temperature, density, and
elastic force of any gas.
If a quantity of heavy elastic fluid of uniform temperature be
placed in a vessel, prove that, if it be divided into indefinitely thin
horizontal strata of equal thickness, the densities of the strata will
be in geometrical progression.
A given weight of heavy elastic fluid of uniform temperature is
confined in a smooth vertical cylinder by a piston of given weight;
shew how to find the volume of the fluid.
5. Describe the action of the Fire-engine; and explain the
Tise of the air vessel.
If yi be the area of the section of each pump, I the length of tlie
stroke, n the number of strokes per minute, £ the area of the hose,
find the mean velocity with which the water rushes out.
6. Explain the terms specific gravity and density ; and shew
how to compare the specific gravities of two fluids by weighing the
same body in each.
Supposing some light material, whose density is p, to be
weighed by means of weights of density p, the density of the
atmosphere when the barometer stands at 30 inches being unity ;
shew that, if the mercury in the barometer fall one inch, the ma-
terial will appear to be altered by ; ^,„n, — sttt of its former
(p-l)(30p-29)
■weight. Will it appear to weigh more or less 1
1^ — 4.] AND RIDERS. 166
vii. A small convergent pencil of light is incident directly on
a concave spherical mirror; investigate the relation between the
distances of the conjugate foci from the surface.
If the convergence be measured by the angle of the cone of
rays, prove that the convergence of the reflected is greater than
that of the incident pencil by a constant quantity.
viiL Find the number of images of a bright point, which can
be formed by reflections at two plane miiTors inclined at an angle
■which is contained an exact number of times in two right angles.
ix. Find the deviation of a ray refracted tkrough a prism in
a plane perpendicular to its edge.
A bright point is at the bottom of still water, and an eye is
vertically above it, at the same distance from the surface ; if a
small isosceles prism, of which the refi-active angle i is nearly two
right angles, be interposed so as to have its base in contact with
the water, prove that the angular distance between the images of
the point in the two faces is , - (tt — t), /x', fi being the refractive
indices for water and for the prism, respectively.
X. Investigate the position of the geometrical focus of a pencil
of i-ays directly refracted through a concave lens of focal length/!
Prove that, as the focus of an incident convergent pencil moves
from the lens, the distance between the conjugate foci always in-
creases, except when the focus of incident rays passes between the
distances y and 2/" from the lens.
xi Describe the eye, regarded as an optical instrument ; illus-
trating the description by di-awing pencils from any point of an
object not in the axis, when seen (1) distinctly and (2) indistinctly.
If the focal length of a convex lens be 3 inches, and the shortest
distance of distinct vision be 6 inches, prove that, when the eye is
always placed so as to see distinctly under the greatest possible
angle, the lens magnifies when within 6 inches of the object, and
diminishes at greater distances.
xii. Trace the course of an oblique pencil of rays from a star
to the eye through the common Astronomical Telescope; and cal-
culate the magnifying power, when the telescope is adjusted for
vision by rays diverging from a given distance from the eye-glass.
If the object-glass be divided, so as to form two semicircular
lenses, and these be displaced along the line of division, what must
1G6 SENATE-HOUSE PROBLEMS [JuJl. 4,
be the displacement of the centres in order that a double star may
appear as three stars t
Thursday, January 5. 9 to 12.
L Thrke concentric circles are drawn in the same plane.
Draw a straight line, such that one of its segments between the
inner and outer circumference may be bisected at one of the points
in which the line meets the middle circumference.
ii. A quadrilateral circximscribes an ellipse. Prove that either
pair of opposite sides subtends supplementary angles at either focus.
iii. A polygon of a given number of sides circumscribes an
ellipse. Prove that, when its area is a minimum, any side is par-
allel to the line joining the points of contact of the two adjacent
sides.
4. If the tangent at any point P of an hyperbola cut an asymp-
tote in Tf and if SF cut the same asymptote in Q, then aS'^ = QT.
5. Prove that the sum of the products of the first n natural
numbers taken two and two together is
{n-l)n{n + l){Sn + 2)
24 •
6. The centres of the escribed circles of a triangle must lie
without the circumscribing circle, and cannot be equidistant from
it unless the triangle be equilateral.
vii. If perpendiculars be drawn from the angles of an equila-
teral triangle upon any tangent to the inscribed circle, prove that
the sum of the reciprocals of those perpendiculars which fall upon
the same side of the tangent is equal to the reciprocal of that per-
pendicular which falls upon the opposite side.
viii. Four equal particles are mutually repulsive, the law of
force being that of the inverse distance. If they be joined together
by four inextensible strings of given length so as to form a quadri-
lateral, prove that, when there is equilibrium, the four particles
lie in a circle.
9. A heavy rod is placed in any manner resting on two points
of a rough horizontal curve, and a string attached to the middle
point C of the chord is pulled in any direction so that the rod is
on the point of motion. Prove that the locus of the intersection
9 — 12.] AND RIDERS. 167
of the string with the directions of the frictions at the points of
support is an arc of a circle and a part of a sti-aight line.
Find also how the force must be applied that its intersections
with the frictions may trace out the remainder of the circle.
X. A. rigid wire without appreciable mass is formed into an
arc of an equiangular spiral and carries a small heavy particle fixed
in its pole. If the convexity of the wire be placed in contact with
a perfectly rough horizontal plane, prove that the point of contact
with the plane will move with uniform acceleration, and find tliis
acceleration.
11. If two parabolas be placed with their axes vertical, vertices
downwards, and foci coincident, prove that there are three chords
down wliich the time of descent of a particle under the action of
gravity from one curve to the other is a minimum, and that one of
these is the principal diameter and the other two make an angle of
60° with it on either side.
12. If a particle slide along a chord of a circle under the
action of a centre of force varying as the distance, the time will
be the same for all chords provided they terminate at either ex-
tremity of the diameter through the centre of force.
13. A hollow cone floats with its vertex downwards in a
cylindrical vessel containing watei\ Determine the equal quan-
tities of water that may be poured into the cone and into the
cylinder that the position of the cone in space may be unaltered.
xiv. A hemispherical bowl is filled to the brim with fluid, and
a rod specifically heavier than the fluid rests with one end in con-
tact with the concave surface of the bowl and passes over the rim
of the bowl, find an equation for determining the position of equi-
librium.
XV. A ray of light passes through a medium of which the
refractive index at any point is inversely proportional to the dis-
tance of that point from a cei*tain plane. Prove that the path of
the ray is a circular arc of which the centre is in the above-
mentioned plane.
16. A small bead is projected with any velocity along a cir-
cular wire under the action of a force varying inversely as the
fifth power of the distance from a centre of force situated in the
circumference. Prove that the pressure on the wire is constant.
17. A bi-ight spot of white light is viewed through a right
168 SENATE-HOUSE PROBLEMS [Jan. 5,
cone of glass the vertex of which is pointed directly towards the
spot. Describe the appearances seen ; and prove that, if a section
of the locus of the images corresponding to different values of the
refractive index be made by a plane through the axis of the cone,
it will be a rectangular hyperbola.
xviii. An elastic string passes through a smooth straight tube
whose length is the natural length of the string. It is then pulled
out equally at both ends until its length is increased by v2 times
its original length. Two equal perfectly elastic balls are attached
to the extremities and projected with equal velocities at right
angles to the string and so as to impinge upon each other. Prove
that the time of impact is independent of the velocity of projection,
and that after impact each ball will move in a straight line, as-
suming that the tension of the string is proportional to the exten-
sion throughout the motion.
xix. A particle is projected along a chord of an ellipse from
any point in the curve, and when it again meets the ellipse has a
certain impulse towards the centre of the ellipse impressed upon
it, causing it again to describe a chord, and so on for any number
of times. Prove that, if after a given number of such impulses,
the pai-ticle pass through another given point on the circumference
of the curve, the polygonal area so described about the centre is
a maximum, when the successive chords are described in equal
times.
Thursday, January 5. 1 to 4.
1. Enukciate and prove Newton's second Lemma. /."
Hence prove that two quantities may vanish in an infinite
ratio to one another; and explain accurately what is meant by
this phrase.
2. Enunciate and prove Newton's tenth Lemma.
If the curve employed in the proof of this lemma be an arc of
a parabola, the axis of which is perpendicular to the straight line
on which time is measured, prove that the accelerating effect of
the force will vary as the distance from the axis of the parabola,
3. If particles describe different circles with uniform velocity,
their accelerations tend to the centres of the circles ; and are to
each other as the squares of arcs described in the same time,
divided by the radii of the circles.
1 — 4.] AND RIDERS. 169
One circle rolls uniformly within another of twice its radius ;
prove that the resultant acceleration of a particle situated on the
circumference of the rolling circle tends to the centre of the fixed
circle, and varies as the distance from that centre.
iv. Prove that the accelerating effect of a force, under the
action of which a body moves in a central orbit, is measured by
OR
the ultimate value of 2 - ^ , QR being the subtense, parallel to
the direction of the force at P, of the aitj PQ described in the
timer.
Deduce the equation F* = ^F . PV.
Prove that, when a body moves along a smooth tube under
the action of any force tending to a point and varying as the dis-
tance from the point, the difference of the squares of the velocities
at the beginning and end of an arc varies as the difference of the
squares of the distances of the extremities of the arc from the
fixed point.
V. Find the law of force tending to the centre of an ellipse,
under the action of which a body can describe the ellipse.
A body is revolving in an ellipse under the action of such a
force, and when it ari'ives at the extremity of the major axis, the
force ceases to act until the body has moved through a distance
equal to the semi-minor axis, it then acts for a quainter of the
periodic time in the ellipse ; prove that, if it again ceases to act
for the same time as before, the body will have arrived at the other
extremity of the major axis.
vi. When a body revolves in an ellipse under the action of a
force tending to the focus, find the velocity at any point of its
orbit, and the [leriodic time.
If on arriving at the extremity of the minor axis, the force
has its law changed, so that it varies as the distance, the magni-
tude at that point remaining the same, the periodic time will be
unaltered, and the sum of the new axes is to their difference as
the sum of the old axes to the distance between the foci.
vii. Explain the changes in the length of days in the north
temperate zone, during the passage of the Earth from Libra to
Aries.
Describe the position of the Earth in its orbit to-day, and our
position on it at three o'clock this afternoon with reference to the
ecliptic and the position of the Sun.
170 SENATE-HOUSE PROBLEMS [Jaw. 5,
viii. Describe the apparent path of the Moon with reference to
the Earth and the Sun, shewing by a figure the direction of the
curvature of its absolute path; and shew how many Lunar Eclipses
can occur in a year.
What distance of the Moon would, with the same inclination
of the orbit, have ensured an eclipse at every opposition 1
ix. Explain the use of the reading microscope in the mural
circle ; and prove that, when a pair of microscopes is used, the
error arising from want of perfect coincidence in the centres of
rotation and gi-aduation, will be eliminated if the axes of the mi-
croscopes be coincident.
Shew how a double observation is made with the mural circle.
10. Explain the origin of the tides; and prove that, suppos-
ing the Earth to be accurately a sphere covered with water, when
it is high water at a given point on the surface of the glf>be, it is
also high water at the antipodes of that point. Prove that the
highest spring tides will take place at the time of an eclipse.
1 1 . Define a tropical, a sidereal, and an anomalistic year ;
stating to which of the three the average length of a civil year
is adjusted, and why.
Explain the statement, that the perihelion of the Earth's orbit
completes a tropical revolution in about 20,000 years, and a side-
real revolution in about 100,000.
12. Define Parallax ; stating how the position of a heavenly
body is afiected by it. In what positions of a star are its right
ascension and declination respectively unafiected by it? State
also which of the heavenly bodies is most affected by it, and in
what position it is so.
Monday, January 16. 9 to 12.
1. Investigate the conditions necessary and sufficient for the
equilibrium of a rigid body, acted on by any number of forces in
any directions in space.
A uniform heavy ellipsoid has a given point in contact with a
smooth horizontal plane. Find the plane of the couple necessaiy
to keep it at rest in this j)osition ; and investigate its equation
referred to the principal axes of the ellipsoid.
2. If a heavy rigid body rest upon more than three immove-
able points of support, prove that the pressure at each jioint is
indeterminate.
9 — 12.] AND RIDERS. 171
An oblong table has the legs at the four comers alike in all
respects and slightly compressible. Supposing the floor and top
of the table to be perfectly rigid, find the pressures on the legs,
when the table is loaded in any given manner, supposing the com-
pression to be proportional to the pressure ; and prove that, when
the resultant weight lies in one of ftmr straight lines on the sur-
face of the table, the table is supported by three legs only.
3. Find the equations of eqiulibrium of a perfectly flexible
uniform inextensible string when acted on by any given forces.
If a small rough heavy bead be strung upon such a string, and
the string be suspended from two points and acted on by gravity
only, write down the equations for determining within what por-
tion of the string it is possible for the bead to rest.
iv. Prove that, when any number of particles Pj, P^,...P^ are
moving in any manner, the acceleration of P^ is the resultant of
the accelerations of P^ relative to -?*„_,, of P_^_, relative to P,_^,...
of /*j relative to /*„ and of P^.
A particle is attached by a rod without mass, to the extremity
of another rod, n times as long, which revolves in a given manner
about the other extremity, the whole motion taking place in a
horizontal plane. If 6 be the inclination of the rods, w the an-
gular velocity of the second rod at the time t, prove that
d 6 d<3i (dm - 9 . \ ~.
-r-i + -y-+W(-y-C0S^ + O) Sm 5 ) = 0.
dt dt \dt )
V. A bead is capable of free motion on a fine smooth wire in
the form of any plane curve, and is acted on by given forces ; com-
pare the pressure on the wire with the weight of the bead.
If the wire be a horizontal circle, radius o, and the bead be
acted on only by the tension of an elastic string, the natural length
of which is a, fixed to a point in the plane of the circle at distance
2a from its centre, find the condition that the bead may just
revolve ; and prove that in this case the pressures at the extremi-
ties of the diameter through the fixed point will be twice and
four times the weight of the bead if that weight be such as to
stretch the string to double its natural length.
vi. Find the nature of the orbit, when a particle moves under
the action of a centitd force which varies inversely as the cube of
the distance.
If a particle, acted on by a central force, and moving in a
resisting medium in which the retardation = k (vel.)* describe an
172 SENATE-HOUSE PROBLEMS [Jan. 16,
equiangular spiral, the pole of which is the centre of force ; prove
that the central force
1 ?5L
oc -je~<^*,
where a is the angle of the spiral.
vii. If an incompressible fluid be in equilibrium under the
action of any forces, prove that the direction of the resultant force
at any point is perpendicular to the surface of equal pressure at
that point.
If the particles of a mass of fluid rotating uniformly about a
fixed axis, attract one another according to such a law that the
surfaces of equal pressure are similar coaxial oblate spheroids,
pi'ove that the resultant attraction of a spheroid, the particles of
which atti-act according to the 8:ime law, is the resultant of two
forces perpendicular to the equator and the axis of revolution
respectively, and varying as the distance of the attracted point
from them.
viii. Prove that, when the density of a mass of air is suddenly
changed from p to p', the pressure is altered in the ratio ( — ) »
where X is the ratio of the specific heats of air, on the supposi-
tions of the pressure and volume remaining constant respec-
tively.
9. A small pencil of light, diverging from a given point,
passes centrically and with small obliquity through a lens ; deter-
mine the position of the primary focal line.
A distant circular window is viewed by a short-sighted man
through his eye-glass, the axis of which passes through the centre
of the window and is perpendicular to its plane. Prove that the
image of the window formed by primary focal lines will be spher-
ical, provided the window be filled with concentric rings of
stained gla.ss, and the refractive index of the colour throughout
any ring be
(;.-l)(2^-fl) r'
fj. being the index of the central colour, r the radius of the ring in
question, and d the distance of the window from the lens.
x- Prove that, in order to determine the time at a given
place by a single altitude of a star, the most favourable stars to
observe are those near the prime vertical
9 — 12.] AND RIDERS. 173
1 1 . Find the parallax in right ascension of a heavenly body,
in terms of the latitude of the place of observation, and the hour
anglt; and declination of the body, assuming the distance of the
body from the Earth to be so great that the sine and circular mea-
sure of the parallax may be considered equal.
Shew that the locus of all the bodies, which on this assumption
have their parallaxes in right ascension for a given place and time
equal to a given quantity, is a right circular cylinder touching the
plane of the meridian of the place along the axis of the heavens.
Monday, January 16. 1^ to 4.
1. Shew how to expand a* in a series of ascending powers
of X.
Prove that the series
2' 3' 4^
1 ^ 172 ■" ITTTS -^ 07374 ^• = ^^-
IL Prove de Moivre's theorem ; and thence prove that, what-
ever be the unit of angular measure, if cos 1 + ^(— 1) sin \=k,
COS 5 = — g — , 8m&=—
V(-i)"
Prove also that the limit of —^ , as ^ is indefinitely dimi-
nished, is
iii. Give Cardan's solution of a cubic equation : and prove
that, when the roots are all real, they will be exhibited under an
imaginaiy form.
Solve the equation, a;'— 6a; — 9 = 0.
iv. Enunciate Sturm's Theorem; and apply it to find the
number and position of the real roots of the equation,
a'+6x* + 4 = 0.
V. Find the area of the triangle, the co-ordinates of the an-
gular points of which are (Aj, k^, (A,, A;J, (A» k^.
174 SENATE-HOUSE PROBLEMS [Jan. 16,
Hence deduce the equation of a straight line passing through
two given points.
vi. Find the value of p, in order that the straight line repre-
sented by the equation, x cos 6 + i/sind =p, may touch the ellipse
1- — = 1.
a 0
Prove that the locus of the vertices of an equilateral triangle
a;* ?/" . .
described about the ellipse, -^ + 't; = 1, is given by the equation
Ct 0
4 (6V+ ay- a'b') = 3{x'+ f- a'- bj.
vii. Investigate the criterion by which it is determined whe-
ther the equation, ax'+ 2hxy + cy'+ 2dx+ 2ey +1 = 0, represents an
ellipse, parabola, or hyperbola.
Prove that, however rectangular co-ordinate axes be shifted,
the ratio of the quantities 6*— ac, (a + c)', will remain unaltered.
What is the geometrical meaning of this ratio 1
8. Investigate the condition that the straight lines
x — a _y—h _ z—c
I m w '
x — a_y — h'_z — c'
V m n *
may be at right angles to one another.
Prove that, if a straight line be drawn from the origin to
cut the first of the above straight lines at right angles, its equa-
tions will be
x _ y _ z
a— It b — mt c -nt'
. al + bm + en
where t = «i,,,^» •
t +171 +n
9. Find the equation of a plane in the form
Ix + my + nz —p = 0,
where I, m, n, are its direction-cosines.
If a, )3, y be the distances of a point from the three faces of
a tetrahedron which meet in the vertex, prove that the equation
1| — 4.] AND RIDERS. 175
of the plane passing through the vertex, and through the centres
of the circles inscribed in and circumscribed about the base, is
(cos5-cos(7);)ia + (cosC- cosA)p^ + (cos A - cos B) p^y= 0,
where A, B, C are the angles of the base, and p^J p^ p^ the per-
pendiculars from the vertex on the sides of the base.
10. Define the polar plane of a given point with respect to
a given sphere ; and find its equation, referred to the centre of
the sphere as origin.
Find the equation of the sphere, passing through a given
point and through the circle in which the polar plane of that
point with respect to a given sphere cuts that sphere.
11. Shew how to find the real circular sections of the s\ir£aM»
of which the equation is
Ax'+By'+Cz'^l;
and describe their positions relative to the difi*erent classes of sur-
faces represented by the above equation.
If a sphere touch an ellipsoid and also cut it, the common sec-
tion cannot be a plane curve unless the point of contact be one of
four fixed points on the ellipsoid.
Tuesday, Jantmry 17. 9 to 12.
L FiNT> a point the distances of which from three given points,
not in the same straight line, are proportional to p, q and r respec-
tively, the four points being in the same plane.
2. If TP, TQ be two tangents drawn from any point T to
touch a conic in P and Q, and il" S and U be the foci, then
iii. A polygon is inscribed in an ellipse so that each side sub-
tends the same angle at one of the foci. Prove that, if the alter-
nate sides be produced to meet, their points of intersection will lie
on a conic section having the same focus and directrix as the ori-
ginal ellipse, and that the chords joining the consecutive points of
intci-section all subtend the same constant angle at the focus as
the sides of the original polygon.
176 SENATE-HOUSE PROBLEMS [Jan. 17,
4. Prove that the equiangular spiral is the only curve such
that its radius of curvature is proportional to the reciprocal of the
radius of curvature at the corresponding point of the reciprocal
polar.
5. If two plane sections of a right cone be taken, having the
same directrix, the foci corresponding to that directrix lie on a
straight line which passes through the vertex.
vi. Find the equation of the envelope of the perpendiculars to
the successive focal radii of a parabola drawn through the extremi-
ties of these radii.
vii. If two concentric rectangular hyperbolas have a common
tangent, the lines joining their points of intersection to their re-
spective points of contact with the common tangent will subtend
equal angles at their common centre.
viii. If P be a point on a geodesic line AP, drawn on a con-
oidal surface, s the length ofAF, a; iV, and 0 the projections of «,
P, and the axis on any plane perpendicular to the axis, and p the
projection of OiV on the tangent io AP a.t P, then
dp d(r
da- ds'
9. A string is placed on a smooth plane curve under the
action of a central force F, tending to a point in the same plane ;
prove that, if the curve be such that a particle could freely de-
scribe it under the action of that force, the pressure of the string
on the curve referred to a unit of length will be = — ^— + -,
where <}> is the angle which the radius vector from the centre of
force makes with the tangent, p is the radius of curvature, and c
is an arbitrajy constant.
If the curve be an equiangular spiral with the centre of force
in the pole, and if one end of the string rest freely on the spiral at
a distance a from the pole, then the pressure is equal to
2r\r^ a') '
10. If a string, the particles of which repel each other with
a force varying as the distance, be in equilibrium when fastened to
two fixed points, ])rove that the tension at any point varies as the
square root of the radius of curvature.
9 — 12.] AND RIDERS. 177
11. If any uniform arc of an equiangular spiral attract a
particle, placed at the pole with a force varying inversely as the
square of the distance, prove that the resultant attraction acts along
the line joining the jiole %vith the intersection of the tangents at
the extremities of the ara
Prove also that, if any other given curve possess this same
property, the law of attraction must be F=—g -J-, where p is the
perpendicvdar drawn from the attracted particle on the tangent at
the point of which the radius vector is r.
xii. A material particle is acted on by a force the direction of
which always meets an infinite straight line AB at right angles,
and the intensity of which is invei-sely proportional to the cube of
the distance of the particle from the line. The particle is pro-
jected with the velocity from infinity from a point /* at a distance
a from the nearest point 0 of the line in a dii-ection perpendicular
to OP, and inclined at the angle a to the plane AOP. Prove that
the particle is always on the sphere of which 0 is the centre, that
it meets every meridian line through AB at the angle a, and that
it reaches the line ^^ in the time
^(/i.) cos a '
/x being the absolute force.
13. If a particle slide along a smooth curve which turns
with uniform angular velocity to about a fixed point 0, then the
velocity of the pai-ticle relatively to the moving curve is given by
the equation
t>* = c* + to) V,
where r is the distance of the jjarticle from the point 0 ; and the
pressure on the curve will be given by the formula
— = — + to» n + 2(i>r,
m p *^
where m is the mass of the particle, and p the perpendicular from
0 on the tangent.
14. A string is laid on a smooth table in the form of a cate-
nary, and an impulse is communicated to_ one extremity in the di-
rection of the tangent, prove (1) that the initial velocity of any
point, resolved parallel to the directrix, is proportional to the in-
verse square of the distance of that pouit from the directrix, and
N
178 SENATE-HOUSE PROBLEMS [Jan. 17,
(2) that the velocity of the centre of gmvity of any arc, resolved in
the same direction, is proportional to the angle between tlie tan-
gents at extremities of the arc directly, and to the length of the
arc inversely.
XV. A right circular cone floats with its axis horizontal in
a fluid, the specific gravity of which is double that of the cone, the
vertex of the cone being attached to a fixed point in the surface of
the fluid. Prove that for stability of equilibrium the semi-vertical
angle of the cone must be less than 60°.
xvi. A ribbon of very small uniform thickness h is coiled up
tightly into a cylindrical form, and placed with its curved s ivfnce
in contact with a perfectly rough plane inclined to the horizon at
an angle a, the axis of the cylinder being parallel to the inter-
section of the plane with the horizon. Prove that the time in
which the whole will be unrolled is very approximately
4 V \(/h sin a) '
where d is the diameter of the original coil.
17. If three beads, the masses of which are m, ni, m", slide
along the sides of a smooth triangle ABC, and attract each other
with forces which vary as the distance, find the position of equi-
librium. Prove also that, if they be slightly disturbed, the dis-
placement of each will be given by a series of three terms of the
form
L sin {nt + A),
where L and X are arbitrary constants, and the values of n are t^e
three positive roots of the equation
(n* - a) {n' - /3) {n' -y)- cos' A m'm" (n' - a) - cos'^m"w (n' - /3)
— cos'Cmm (n' — y) — 2 cos A cos B cos Cmm'm" = 0,
where a, /8, y represent m" + m\ m + m", m+m respectively.
xviiL The boi-e of a gun barrel is formed by the motion of an
ellipse of which the centre is in the axis of the barrel, and the
plane is perpendicular to that axis, the centre moving along the
axis, and the ellipse revolving in its own plane with an angular
velocity always bearing the same ratio to the linear velocity of its
centre. A spheroidal ball fitting the barrel is fii-ed from the gun.
9—12.] AND EIDERS. 179
If V be the velocity with which the ball would have emerged from
the barrel had there been no twist; prove that the velocity of
rotatioa with which it actually emerges in the case supposed is
2Trrjv
J{1' + A-r'n'k') '
the number of revolutions of the ellipse corresponding to the whole
length I of the barrel being n, and k being the radius of gyration
of the ball about the axis coinciding with the axis of the barrel,
and the gun being supposed to be immovable.
xix. An elastic ring of length I, mass m, and elasticity U is
placed over the vertex of a smooth cone, the semi-vertical angle
of which is a, and stretched upon it to any size. Supposing it
then set free, prove that the time before it leaves the cone is
V(x)
cosec a,
the action of gravity being neglected.
Tuesday, Jan. 17. 1| to 4.
1. Investigate the condition of achromatism which is required
in Huyghens' eye-piece ; and find the magnifying power of Gregory's
Telescope with this eye-piece. Draw a figure representing the
course of the pencil.
2. If pq be the image of PQ, placed perpendicular to the axis
QCq of a lens or mirror C2i, Qliq the course of a ray from Q to q,
shew that PQ : pq :: RqC : BQC.
Hence prove that, with all combinations of lenses for eye-
pieces, the magnifying power of a telescope, airanged for parallel
or diverging emergent pencils, is the ratio of tlie diameter of the
object-glass or mirror to that of its image formed on emergence
from the eye-piece.
iiL Define the equation of time; state the causes to which it is
due, and prove that it vanishes four times a year. Find also roughly
when it attains its maxima and minima values^ assuming the longi-
tude of perihelion to be 100".
N2
180 SENATE-HOUSE PROBLEMS [Jan. 17,
iv. Two particles move under the influence of gravity, and of
their mutual attractions; prove that their centre of gravity will
describe a parabola, and that each particle will describe, relatively
to that point, areas proportional to the times.
5. Define a principal axis through any point of a rigid body ;
and, having given one principal axis through a point, find the posi-
tions of the other two.
Prove that the locus of a point, thi'ough which one of the
principal axes is in a given direction, is a rectangular hyperbola
in the plane of which the centre of gravity lies, and of which
one of the asymptotes is in the given direction; unless the given
direction be that of one of the principal axes through the centoe of
gravity.
vi. Investigate the efiect of the central disturbing force on the
position of the apsides of the Moon's orbit, supposing the line of
apsides near syzygy.
By what two causes is the excess of progression over regres-
sion, during a synodic revolution of the Sun and line of apsides
increased 1
vii. Ifp be the pressure, and p the density, at any point, x, y, z
of a mass of fluid in motion, w, v, w the component velocities of the
fluid at that point, X, F, Z the component accelerations due to the
forces acting on the fluid parallel to the co-ordinate axes, investi-
gate the equations.
\ dp _ y. du du du du
p dx dt dx dy dz* y
1 dp „ dv dv dv dv
p dy dt dx dy dz
1 dp _ „ dio din dw dw
p dz dt dx dy dz'
State the hypothesis of steady motion ; and point out the modi-
fication which will be introduced into the above equation.s, if the
motion be steady.
8. Give a full account of the methods of ap]>roximation adopted
in the Lunar Theoiy; and state of what circumstances advantage
is taken in order to conduct the approximation to the solution of
the equations in the Planetary Theory.
1| i.] AND EIDERS. 181
9. Assuming the equations
d^u „ ((Pu \ cTdB T du P .
3/ /3
r = 2^ sin 2 (^-^,
investigate that term in u the argument of which is
(2 - 2m - 2c) ^ - 2^ + 2a.
Webnesday, Jan. 18. 9 to 12,
1. A PARABOLA touches one side of a triangle in its middle
point and the other two sides produced. Prove that the perpendi-
culai-s drawn from the angles of the ti'iangle upon any tangent to
the parabola are in harmonical progression.
2. Find the length of the longest straight line which can be
drawn in the interval between two similar similarly situated and
concentric ellipsoids ; and, if a line shorter than the line so deter-
mined be moved about in the interval, prove that its point of
contact with the interior ellipsoid can never lie within the cone
represented by the equation
a* {a* (1 - 7J»*) - r'} "^ h' [b' (1 - m^) - r'} "^ c* {c» (1 - m«) - r*} ~ '
a, b, c being the semi-axes of the outer ellipsoid, m the ratio of
the linear magnitudes of the inner and outer ellipsoid, and 2r the
length of the line in question, which is assumed greater than
2bJ(l-m').
What is the meaning of the boundary so determined when 2r
is less than 26 ^^(1 — m*) and greater than 2c ^(1 — w»*)1
3. If, in a rigid body moving in any manner about a fixed
point, a series of points be taken along any straight line in the
body, and through these points straight lines be drawn in the
direction of the instantaneous motion of the points, prove that the
locus of these stiuight lines is an hyperbolic paraboloid.
182 SENATE-HOUSE PROBLEMS [Jan. 18,
4. If / («, y, 2) = 0 be the equation to a surface, and r be a
straight line drawn through the point x, y, z of which the mag-
nitude and direction are any given functions of x, y, z, state
what is the relation between the original surface and that whose
equation is n '"'/ {x, y, z) = 0, supposing that in the latter equa-
tion X, y, and z have been expressed in terms of r and any two
other variables independent of r, and that n is a given numerical
quantity, and prove that if the two surfaces coincide for all values
of w, the line r must lie altogether in either of them.
Apply this to find the partial differential equations of conical
and conoidal surfaces respectively when referred to any system of
rectangular axes.
5. From a flexible envelope in the form of an oblate sphe-
roid, of which the eccentricity of the generating ellipse is e, the
part between two meridians, the planes of which ai'e inclined to
each other at the angle 27r (1 — e), is cut away and the edges are
then sewed together ; prove that the meridian curve of the new
envelope will be the curve of sines.
6. If an uniform inextensible and flexible string be stretched
over a smooth surface of revolution, prove that the following
equations hold :
s(^4^)-^|^^'=«-
where ds is the element of the string at any point, dx and dy are
corresponding elements of the arc of the circle through that poii^t
j)eq>endicular to the meridian, and of the meridian respectively,
X and Y are the resolved parts of the impressed forces along these
directions, and r is the distance from the axis, the mass of an unit
of length of the string being taken as unity. Hence prove that, if
such a string be acted upon by a force at all points perpendicular
to the axis of revolution, and inversely proportional to the square
of the distance from that axis, the string will, if properly suspended,
cut every meridian in the same angle.
7. A string is wound round a vertical cylinder of radius a
in the form of a given helix, the inclination to the horizon being
i. The upper end is attached to a fixed point in the cylinder,
and the lower, a portion of the string of length I sec i having
been unwound, has a material particle attached to it which is also
9—12.] AND RIDERS. 183
in contact with a rough horizontal plane, the coefficient of friction
being fji. Supposing a horizontal velocity V perpendicular to tlie
free jjortion of the string to be applied to the particle so as to
tend to wind the string on the cylinder, determine the motion ;
and prove that the particle will leave the plane after the projec-
tion of the unwound portion of string upon the plane has described
the angle of which the circular measure is
2fi tan { ' " 2fxtSLn'i.V' —2figl tani + ag *
8. A particle is acted on by two centres of force residing in
the same point, one attractive, the other repulsive, and varying
inversely as the square and cube of the distance respectively. Two
consecutive equal apsidal distances are drawn and the portion of
the plane of motion included between them is rolled into a right
circular cone. Prove that the trajectory described under the cir-
cumstances mentioned above becomes a plane curve on the surface
of the cone, and that it will be an ellipse, parabola, or hyperbola,
according as the velocity in the trajectory was less than, equal to,
or greater than that from infinity.
9. A particle is describing an orbit round a centre of force
which is any function of the distance, and is acted upon by a dis-
turbing force which is always perpendicular to the plane of the
instantaneous orbit and inversely proportional to the distance of
the body from the original centre of force. Prove that the plane
of the instantaneous orbit revolves uniformly round its instanta-
neous axis.
10. A die in the form of a pai-allelopiped the edges of which
are 2a, 2b, and 2c, is loaded in such a manner that the centre of
gravity remains coincident with the centre of figure, but the prin-
cipal moments of inertia about the centre of gravity become equal ;
if it then fall from any height and without rotation upon a hori-
zontal plane composed of adhesive material so that no point which
has once come in contact with the plane can separate from it, prove
that the chance of one of the faces bounded by the edges 26, 2c
coming uppermost is
2 . _, be
-sm
T Jiia' + b*) {a* + €')}'
11. A uniform sphere is placed in contact with the exterior
Biu-face of a perfectly rough cone. Its centre is acted on by a force,
184 SENATE-HOUSE PROBLEMS [Jan. 18,
the direction of which always meets the axis of the cone at right
angles, and the intensity of which varies inversely as the cube of
the distance from that axis. Prove that, if the sphere be properly
started, the path described by its centre will meet every generating
line of the cone on which it lies in the same angle.
12. A small rigid vertical cylinder, containing air, is rigidly
closed at the bottom, and covered at the top by a disk of very
small weight which fits it air-tight. Supposing the air in the
cylinder to be set in vibration, prove that the period of a vibra-
27r
tion is — , m being a root of the equation
m
, ml kBU
m tan — = — — :
a fia
where I is the length of the tube, a the velocity of sound in air,
fjL the mass, k the area of the disk, pec p (I + ^s) the relation be-
tween the pressure and density when the latter is suddenly
changed from p to p (1 + «), and 11 the pressure of the air in the
cylinder before motion commences.
13. A circular drumhead of uniform thickness is stretched
with a tension of uniform magnitude at all points in its circum-
ference, and is then set in vibration by a small disturbance com-
mencing at the centre. Prove (1) that if z be the transversal
disturbance at the time t of a. point the initial distance of which
from the centre was r, then
<fg_ , /Idz d'z\
^~" [rdr'^d?)'
and (2) that the general primitive of this differential equation is
z= j <l> {at + r cos 6) d6 + I \p{at + r cos &) log (r sin*^) dO,
<f> and i/r being arbitrary functions, and a a constant depending
upon the tension and constitution of the drumhead.
1^ — 4.] AND RIDERS. 185
Wedkesday, Jan. 18. 1^ to 4.
1. Define the terms Limit, Independent and Dependent
Variable, and Differential Coefficient.
If X represent the time which has elapsed since a given epoch,
and y the space which a moving point has described in that time,
what will -J- represent 1
ii. Prove thaf^-4-T =.,,,, [ , where 6 lies between 0 and 1,
<l){x) <f> (Bx)
with certain limitations : and deduce Stirling's Theorem.
3. Change the independent variables in
<rV d'r j/c/F dV\
"^ dx'^'^dy'^'^\di^'d^r
from X and y to m and v, having given x-k- y = u, y = uv.
4. Explain the use of the method of indeterminate multipliers
in determining the values of a function of several variables con-
nected by given equations, which are maxima or minima.
Find the position of the point, the sum of the squares on the
distances of which from the three sides of a triangle is the least
possible; and prove that the angles, which the sides respectively
subtend at this point, exceed the supplements of those which they
subtend at the centre of gravity of the triangle by the respective
angles of the triangle.
V. Prove that, if a curve represented by an algebraical equa-
tion have an asymptote, and the curve lies on the same side at both
ends, there will be an odd number of points at an infinite distance
in which the asymptote meets the curve.
vi Trace the curves represented by the equations
(a:'-4a*)y'-12a'a;(a-y) = 0 (1),
sin y — m, sin a; = 0 (2).
In (1) explain the circumstance that the asymptotes parallel to
the axis of y appear to contradic^ the statement of (v.). In (2)
distinguish between the cases in ^hich m > = or < 1.
186 SENATE-HOUSE PKOBLEMS [Jan. 18,
vii. If / (x, y) = 0 be the equation of a ciirve in a rational
integi-al form, in which (Xg, y^ is a multiple point through which
n branches pass ; shew that the directions of the branches are
determined by the equation
<t>^{0) = 0, where <f> (r) =/ (x^ + Ir, y„ + mr).
Find the form of the curve represented by the equation
{y-by-{x'-2axy-a' = 0,
at the multiple point.
viii. Find the co-ordinates of the centre of curvature at any
point of a plane curve.
The equation of a cii'cle is x' + y' = c', prove that the equation
of the directrix of a parabola of which the axis is parallel to the
axis of x and which has the closest possible contact at the point
{x^ yj is 2x = 3x^.
9. Integrate the following diffei-entials :
dx dO dx
£c»(A + a;)»' (m + wcos^)'' (x* - 2aa; + a" + c«) (a; - 6)"
and evaluate I * log sin Odd.
10. Prove that the area of a curve, represented by the equa-
tion y (a;, y) = 0, will be given by the formula ^j{xdy — ydx), the
integrals being taken within proper limits.
One circle rolls within another; apply the above formula to
find the area of the curve traced out by a given point within the
rolling circle.
xi. Prove that, if a plane intersect a surface, it will generally
be a tangent plane to the surface at every multiple point of the
curve of intersection. What exceptions are there ?
Prove that a tangent plane to a developable surface meets the
surface in a straight liue and a curve touching that straight line.
xii. Explain generally the principle of Variation of Parame-
ters, as applied to the solution of a diiferential equation ; and illus-
1^—4.] AND RIDERS. 187
trate this method of solution by assuming the result in the form of
that obtained fi-om the fii^st three terms f
g-2m|H-m'3,4(|-'»3')=0.
13. Define a developable surface ; and, from your definition,
deduce the pai'tial differential equation of such surfaces.
Find the equation of the developable surface generated by the
plane which moves in such a manner as to be always in contact
with the surfaces
<v* »•* ^'
X y z ^
—T+— 1 =1
a' b'
c
X' y' z'
— -I — 1
-» ^ JvH ^ ^ „a
14. Explain what is meant by A". O" ; and prove that, if
/{e') be expanded in a series proceeding by ascending powers of
t, the coefficient of r isV^—t-^i— .
1. 2 m
Prove that, if m be less than r,
{l+log(l + A)}'.0'" = r(r-l)(r-2)...(r-m + l).
Thursday, Jan. 19. 9 to 12.
1. If at the extremities P, Q of any two diameters CF, CQ of
an ellipse, two tangents Pp, Qq be drawn cutting each other in
T and the diameters produced in p and q, then the areas of the
triangles TQp, TPq are equal
2. If a straight line CN be drawn from the centre to bisect
that chord of the circle of curvature at any point P of an ellipse
which is common to the ellipse and circle, and if it be prodticed to
cut the ellipse in Q and the tangent in T^ prove that CP = CQ, and
that each is a mean proportional between CN and CT,
3. If a, 5, c be the sides of a triangle, and r the radius of
the inscribed circle, then the distances of the radical centre of
188 SENATE-HOUSE PROBLEMS [Jan. 19,
the three escribed circles from the sides of the triangle will be
respectively
b + c c + a a + b
^~2^' ^~2b~' '^~W
4. Two equal hea\'y particles are connected by a string
which passes through a small smooth fixed ring. Prove that the
equation to the plane vertical curve on which the particles will
rest in all positions is,
r cos 0 = a+ \j/{r) — \p{l — r),
where 6 is the angle the radius vector makes with the vertical, I is
the length of the string, ij/ an arbitrary function and a an arbitrary
constant.
5. If four equal particles, attracting each other with forces
which vary as the distance, slide along the arc of a smooth ellipse,
they cannot generally be in equilibrium unless placed at the extremi-
ties of the axes; but if a fifth eqiial particle be fixed at any pf»int
and attract the other four according to the same law, there will be
equilibrium if the distances of the four particles from the semi-axis
major be the roots of the equation
where p and q are the distances of the fifth particle from the axis
minor and axis major respectively.
6. A heavy string is placed in eqxiilibrium on a smooth
sphere ; prove that, if 0 be the length of the spherical arc drawn
from the highest point of the sphere perpc ndicular to the great
circle touching the string at any point P, then
sm 0 = J- ,
z+ b
where z is the perpendicular from F on any horizontal plane, and
a, b are constants.
Shew that the form of the string can be a circle only when its
plane is vertical or horizontal
7. If three particles of masses m, rn, m", attracting each other
start from rest, shew that if at any instant parallels to their direc-
tions of motion be drawn so as to form a triangle the momenta of
the several particles are proportional to the sides of that triangle.
9 — 12.] AND EIDERS. 189
8. If from any point on a surface a number of geodesic lines
be drawn in all directions, shew, (1) that those which have the
greatest and least cui^vature of toi-sion bisect the angles between
the principal sections, and (2) that the radius of torsion of any
line, making an angle 9 with a principal section, is given by the
equation
ji U p)
sin 6 cos 6,
where pj, p^ axe the radii of curvature of the principal sections.
9. If t/it and de be the angles of torsion and contingence of
any curve of double curvature, and if sin <^ be the ratio of the
radius of circular curvature to the radius of spherical curvature,
prove that the square of the angle of contingence of the locus of
the centres of circular curvature is
d(f> + du [' + cos*^>^*.
10. A particle is projected with 'velocity V along an infi-
nitely thin ellipsoidal shell attracting according to the law of
nature ; prove that, when it leaves the ellipsoid the perpendicular
from the centre on the tangent plane is y [ — ^-^ — J where
R is the radius-vector parallel to the initial direction of motion,
and P the perpendicular on the tangent, /x the attraction of a
mass equivalent to a unit of area of the ellipsoid at a unit of dis-
tance.
11. An infinitely thin ellipsoidal shell attracting according to
the law of nature is bounded by two similar and similarly situated
ellipsoids. A very small pie«e is cut out of the shell and replaced
in its original position. Shew that the force necessary to hold the
piece in equilibrium is proportional to the square of the thickness
of the shell.
12. A sphere of radius a is suspended from a fixed point by a
string of length I and is made to rotate about a vertical axis with
an angular velocity <d. Prove that, if the string make small
oscillations about its mean jwsition, the motion of the centre of
gravity will be represented by a series of terms of the form
L cos {kI + M),
190 SENATE-HOUSE PROBLEMS [Jan. 19,
where the several values of k are the roots of the equation
13. A string is in equilibrium in the form of a circle about
a centre of force in the centre. If the string be now cut at any
point A, prove that the tension at any point P is instantaneously
changed in the i-atio of 1 : 1, where 6 is the angle
subtended at the centre by the arc AP.
14. An inelastic string is siispended from two fixed points
so that it hangs in the form of a catenary of which the parameter
is c. Suppose it to make small oscillations in a vertical plane,
prove the equation
§=f-HS-**-/(')}.
where a is the angle the tangent at any point makes with the
horizon when the string is at rest, and a + <^ is the value of the
same angle at the time t.
Shew that there are sufficient data to determine all the arbi-
trary functions.
Thursday, Jan. 19. 1| to 4.
L Shew that the series w, + w, + ... + w„+ . .. will be convot-
gent> if the fraction — =-^ converge to a limit less than unity as n
is indefinitely increased, and divergent, if, supposing all the terms
to be of the same sign, this limit be equal to or greatt-r than
unity.
Find a superior limit to the numerical values of x consistent
with the convergency of the series
2 V 3V m". x"
''■' iT2 "-TTira "" ■••'"1.2...^^ •••
2. If the sides of a sj)herical triangle be small compared with
the radius of the sphere, then each angle of the spherical triangle
1^ — 4.] AND RIDERS. 191
exceeds by one-third of the spherical excess the corresponding
angle of the plane triangle, the sides of which ai-e of the same
length as the sides of the spherical triangle.
If the sides of a right-angled plane triangle of given area be
bent so as to form a spherical triangle on a given sphere of great
radius, the altemtion of area in the triangle is very nearly propor-
tional to the square of the hypothenuse.
3. If a, j8, y, be the distances of a point from three given
straight lines, determine the position of the conic, ayS = ky' ; and
prove that the equation of the tangent at any point may be put
into the form,
Two tangents OA, OB ai'e drawn to a conic, and are cut in
P and ^ by a variable tangent; prove that the locus of the
centres of all circles described about the triangle OFQ is an
hyperbola.
iv. If ?* be a function of three independent variables x, y, z,
which are connected by three equations with three new inde-
pendent variables ^, -q, ^ shew how to express the partial differen-
tial coefficients of w, to- the firet and second ordera respectively,
with respect to x, y, z, in terms of the corresponding partial
differential coefficients with respect to ^, t;, ^.
Apply this method to prove that, if at a certain point in a
surface r = t and « = 0 when the axes of a; and y are taken parallel
to a particular pair of lines, at right angles to each other, in the
tangent plane at that jwint, then the following relations will hold
at that point whatever be the direction of the co-ordinate axes
provided they be rectangular, viz.
1 +;>* pq \-\-<f*
, , , dz dz d*z d'z d*z . ,
where p,(?,r,,,<, denote^, ^, ^, ^^, ^, respectively.
V. If the differential equations of the first order
192 SENATE-HOUSE PROBLEMS [Jan. 19,
^ve rise to the same differential equation of the second order,
shew how the general solution of an equation of the form
may be found without integration.
Apply this or any other method to the discovery of the general
solution of the equation
6. Enunciate and explain d'Alembert's principle. Apply it
to determine the small oscillations in space of a uniform heavy rod
of length 2a, suspended from a fixed point by an inextensible
string of length I fastened to one extremity. Prove that, if x be
one of the horizontal co-ordinates of that extremity of the rod to
which the string is fastened,
£c = j1 sin (n^t + a) + £ sin {njt + fi),
where n^, n^ are the two positive roots of the equation,
ahi*- {ia + dl) (jn'+ 3g'= 0,
and A, B, o, /3, are arbitrary constants.
vii. A rigid body is rotating about an axis through its centre
of gravity, when a certain point of the body becomes suddeyJy
fixed, the axis being simultsmeously set free ; find the equations of
the new instantaneous axis ; and prove that, if it be parallel to the
originally fixed axis, the point must lie in the line represented by
the equations,
a'lx + b'rny + c'nz = 0,
(6« - c») y + (c' -a')^+ (a'- 6') - = 0 ;
^ ' I ^ ^ m ^ ' n
the principal axes through the centre of gravity being taken as
axes of co-ordinates, a, h, c the radii of gyration about these lines,
and I, m, n the direction-cosines of the originally fixed axis referred
to them.
1| — 4.] AND RIDERS. 193
8. Explain the physical meaning of the term
15
-^ inea cos {(2 - 2m - c) 6 - 2/3 + a}
in the expression for the reciprocal of the Moon's radius vector.
Calculate roughly the proportionate alteration in the Moon's
mean distance produced by this term, and its period. Why is this
term usually taken in combination with the elliptic inequality]
ix. Prove the following relation between the perturbations of
a planet in longitude and radius vector
h\ dt r dt JJ dt J dr j'
h being twice the sectorial area described in a iinit of time by the
undisturbed planet round the Sun: and find the corresponding
relation whatever be the law of force, provided it be centi*al and
a function of the distance only, and provided such a function as li
can be found,
10. If the object-glass of a telescope be covered over by a
diaphragm, pierced iu the centre by a small hole, the form of
which is a rectangle, state generally the nature of the spectra
formed about the image of a star on a screen placed at the
focus.
If the hole be circular and the screen be pushed towards the
lens, prove that, when the light is homogeneous, the centre is
alternately bright and dark. Trace also the order of the coloui'S
seen if the light be not homogeneous.
Friday, Jan. 20. 9 to 12.
1. When the reciprocal of a circle ia taken with respect to
another circle, investigate the nature of the reciprocal conic, and
the polars of its centre and further focus.
OA, OB are common tangents to two conies having a common
focxis S, CAy CB are tangents at one of their points of intersection,
BD, AE tangents intersecting CA, CB iu D, E. Prove that SDE
is a straight line.
194 SENATE-HOUSE PROBLEMS [Jan. 20,
ii. Define the term potential of a mass, the particles of which
attract according to the law of nature ; and prove that, if a body
moveable about a fixed axis be subject to the action of an at-
tracting mass of which the potential is V, then /// — - dm is the
moment which must be impressed upon the botly about that axis
in order to produce equilibrium, where 6 is the inclination of the
plane through the fixed axis and the particle of which the mass
is dm to a fixed plane,
A uniform straight line, the particles of which attract accord-
ing to this law, acts upon a rigid uniform circular arc ia the same
plane with the line, of which the radius is equal to the line, and
which is moveable about an axis through its centre perpendicular
to its plane, the axis being coincident with one extremity of the
line. Prove that the moment necessary to produce equilibrium
when the bounding radii are inclined at the angles a and /3 to the
line produced is proportional to
sec n + 1
log
sec ^ + 1
3. Define lines of curvature on a surface; and find their dif-
ferential equation. Prove that one line of curvature at any point
very near an umbilicus passes through that umbilicus.
4. State and prove the principle of Vis Viva. If an elastic
string, whose natural length is that of a uniform rod, be attached
to the rod at both ends and suspended by the middle point, prove
by means of Vis Viva that the rod will sink until the strings
are inclined to the horizon at an angle ^, which satisfies the
equation
6 6
cot»2-cot^-2n=0,
where the tension of the string, when stretched to double its length
ia n times the weight.
If the string be suspended by a point, not in the middle, write
down the equation of Vis Viva.
V. If a spheroid of revolution be moveable about its centre
which is fixed, and 6 be the inclination of its equator to a fixed plane,
9 — 12.] AND RIDERS. 195
^ the inclination of the line of intersection of its eqnator with this
plane to a fixed line in the plane, A and C the respective moments
of ineitia about the axis of figure and a line in the equator respec-
tively, L and J/ the moments of impressed couples about the line
of intersection of the equator with the fixed plane, and a line in
the eq\iator perpendicular to this latter line respectively, w the
angular velocity about the axis of figure, prove that
C -nr-0 1 -r) sm 6 cos 9 + Aw sm 6 ~- = L,
df \dtj dt '
„ d /dib . A ^d^ dO a A ^^ nr
C -ri-f- sin 6 + C-^ -^ cos B - Aw -^ = M,
dt \dt ) dt dt dt '
hence deduce the processional and nutational velocity of the
Eiirth's axis, assuming the effect of the Sun's action to be a couple
of which the moment is m, sin A cos A about an axis in the equator
90' distant from the Sun, m being a very small quantity, A and C
very nearly equal, and the Sun's motion in declination and right-
ascension being neglected.
vi. If /t be a given function of the co-ordinates x and y of
any point in a plane curve, prove that if the curve be so deter-
mined as to render the integraJ jiids between given limits a maxi-
mum or minimum, then
1 1 /dui da
— = — \-j- cos a + — cos
p II \dx ay
.).
p being the radius of curvature at any point, and a, ^ the acute
angles which the normal at that point makes with the axes of x
and y respectively.
If a solid of revolution be immersed in a heavy homogeneous
fluid with its axis vertical, prove that, when the total normal
pressure on the surface is a minimum, its form must be such
that the numerical value of the diameter of curvature of the
meridian at any point is a harmonic mean between the segments
of the normal to the surface at that point intercepted between the
point and the surface of the fluid and between the point and the
axis, respectively.
7. Supposing the orbits of a disturbing and disturbed planet
to be in the same plane, prove that the rate of change of the
19(5 SENATE-HOUSE PROBLEMS [/an. 20,
longitude of perihelion of the instantaneous ellipse of the disturbed
planet is
In what respects do the theories of the motion of the apsides in
lunar and planetary disturbances present themselves respectively
in simpler aspects?
viii. Explain the phenomenon of external conical refraction
where a small ])encil of light passes through a biaxal ciystal ;
and describe an experiment by which this phenomenon may be
manifested.
If the ciystal be bounded by planes peqiendicular to the line
bisecting the acute angles between the optic axes, write down
equations whence the equation of the cone of emei-ging i-ays may
be obtained.
Friday, Jan. 20. 1| <o 4.
1. If a, /?, y be the respective distances of a straight line
from the three angular points of a triangle ABC, these distances
]>eing reckoned positive or negative according as their directions
fall within the angles of the triangle itself or their supplements,
investigate the following relation,
(asin ^)»4- OSsin 5)* + (y sin (7)' - 2 cos ^ sin 5 sin C^y - 2 cos 5
sin G sin Aya — 2 cos C sin A sin Ba^ = 4^* sinM sin*^ sin'C,
where R is the radius of the circunMcribed circle.
2. State the positive and negative characteristics of a singular
solution of a differential equation; and shew how it is deduced
from the complete primitive. Shew also how the singular solution
of a differential equation of the fir.st order is obtained from the
equation itself.
Obtain the singular solution of the equation of which
y cos*m = 2 cos {x — 2m)
is the complete primitive ; and find the singular solution of the
equation
\
1^ — 4.] AND RIDERS. 197
iii. Prove that, in any curve of double curvature, the locus of
the centres of spherical curvature is the edge of regression of the
envelope of the normal planes. Prove also that this locus cannot
be an evolute.
The normal plane to the locus of the centres of circular curva-
ture bisects the i-adius of spherical curvature.
4. Determine the class of curves which possess the property
that the locus of the extremity of the polar subtangent of any one
is similar to the curve itself
Shew that r^c" = a is the equation of such a curve.
v. Investigate the general equations of motion of a sphere
under the action of any forces.
If a homogeneous sphere roll on a perfectly rough plane under
the action of any forces whatever, of which the resultant passes
through the centre of the sphere, the motion of the centre of gi"avity
will be the same as if the plane were smooth and all the forces
were reduced in a certain constant ratio; and the plane is the only
surface which possesses this property.
6. Assuming the following equations for the rate of variation
of the inclination and of the longitude of the line of nodes of a
planet m disturbed by a planet m\
do, na dR
dt /A (1 — e*)^ sin i di '
di _ na C 1 dR i /dR
di^~ /x(l-e^)» (^i^ dQ'^ 2\d^
dR dR\^
investigate general expressions for i and il, so far as they are
affected by the effect of the following terms in the expansion
of R,
- m' [2D^ee cos (ct' -zt) + \aa'D^ {tan' i-2 tan i tani'coa (Q - W)\\
vii. Prove that, if the Earth be considered as a homogeneous
mass of fluid in the form of a spheroid, revolving *s^ith a uniform
angiilar velocity about its axis, gi-avity at any point acts along
the normal, and is proportional to the part of the normal inter-
cepted between the point of contact and the plane of the equator.
198 SENATE-HOUSE PROBLEMS AND RIDERS. [Jan. 20.
If the Earth be completely covered by a sea of small depth,
prove that the depth in latitude I is very nearly equal to
H{l—esin'l) where // is the depth at the equator, and c the
ellipticity of the Earth.
8. A thin plate of Iceland spar, cut perpendicTilarly to its
axis, is interposed between two tourmalines, used as a polarizing
and analyzing plate, and a pencil of parallel rays is transmitted
through the crj'stals. Assuming that the difference of retardation
of the ordinary and extraordinary rays, when a ray is transmitted
through a plate of Iceland spar in a direction inclined to its axis
at a small angle i varies as Tsin'i, investigate an expression for
the illumination at any point of the field of view, the axes of the
tourmalines being parallel ; and hence deduce a description of the
phenomena observed, supposing the light (1) to be homogeneous,
(2) to be white.
ix. State under what circumstances in the motion of a fluid
we may assume udx + vdy + wdz to be a perfect differential of some
function.
Assuming this function to be <^, and the fluid to be homoge-
neous, investigate the equation
fi + ^ + ^^o.
dx' dxf di?
The base of an infinite cylinder is the space contained between
an equilateral hyperbola and its asymptotes. A plane is drawn
perpendicular to the base, and cutting it in a straight line parallel
tx) an asymptote, and the portion of the cylinder between this plague
and its parallel asymptote is filled with homogeneous fluid, vrnder
the action of no impressed forces. The plane being suddenly
removed, determine the motion; and prove that the free surface
of the fluid will remain plane, and advance with a uniform
velocity proportional to Jts, where -jn" is the pressure at an infi-
nite distance, which is supposed to remain constant throughout
the motion.
CAMBRIDGE: PRINTED BY C. J. CLAY. M-A. AT THb UNIVERSITY PRESS.
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