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JOEL LEWIS: Hi.

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Welcome back to recitation.

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In lecture, you've learned
about gradients and about

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directional derivatives.

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So I have a problem here
to test your understanding

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of those objects and give you
some practice computing them.

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So I have three functions here.

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And for each function,
I've given you

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a point and a direction.

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So what I'd like
you to do is compute

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the gradients of
the functions, then

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evaluate their gradients
at the point given,

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and then compute the directional
derivative of the function

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at the point in the direction
of the vector given.

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So here, for example,
in part a, I've

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given you the function f of x,
y equal to x squared y plus x

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y squared, at the point x
equal minus 1, y equal 2,

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and in the direction
of the vector [3, 4].

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And then we've got two more
examples here. g of x, y, z

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is equal to the square root
of x squared plus y squared

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plus z squared, P is
the point 2, 6, minus 3,

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and v is the
direction [1, 1,  1].

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And our third example, h is a
function of four variables, w,

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x, y, and z.

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It's given by w*x plus w*y plus
w*z plus x*y plus x*z plus y*z,

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at the point 2, 0, minus 1,
minus 1, in the direction 1,

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minus 1, 1, minus 1.

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So why don't you pause
the video, take some time

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to work out these three
problems, come back,

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and we can work
them out together.

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I hope you had some luck
working out these problems.

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Let's get started
on the first one.

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So on the first one,
our function f of x, y

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is equal to x squared
y plus x y squared.

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So to compute the
gradient, I just

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have to compute the first
partials of our function,

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and then put them
into a single vector.

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So that's what the gradient is.

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So we have that the
gradient of f at (x,

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y) is equal to--
it's the vector,

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and its first component is the
first partial of f with respect

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to x, so that's going to
be 2x*y plus y squared.

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And its second component is the
partial of f with respect to y,

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so that's going to be
x squared plus 2x*y.

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So this is the gradient
of our function f.

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Now, the question asked you
to compute this gradient

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at the particular
point minus 1, 2.

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So we have gradient of
f at the point minus 1,

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2 is equal to-- well, we
just put in x equal minus 1,

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y equal 2 into this formula.

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So at x equal minus 1, y
equals 2, this is 2 times

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minus 1 times 2 plus 2 squared.

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So that's minus 4 plus
4, so that's just 0.

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And the second term is
minus 1 squared plus 2 times

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minus 1 times 2.

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So that's 1 plus 4--
nope-- 1 plus negative 4,

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so that's negative 3.

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All right.

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So this is the gradient
of our function f.

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This is the gradient at
the point minus 1, 2.

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Now we are asked for the
directional derivative.

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So we were asked for the
directional derivative

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of f in a direction.

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So we're given-- I didn't give
you-- so I gave you a vector v.

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But when we take directional
derivatives, what we want

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is a unit vector.

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Right?

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So the vector v that I gave
you-- was right over here--

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it was this vector, [3, 4].

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And so we need to find the unit
vector u in the direction of v

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in order to apply
our usual formula.

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So u is just v divided
by the length of

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v. So in this case,
the vector [3, 4],

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by the Pythagorean
theorem, it has length 5.

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So this is equal to the
vector 3/5 comma 4/5.

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And so our
directional derivative

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in the direction of v--
which is this vector

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u-- it's just equal to
the gradient at that point

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dotted with the
direction vector.

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So this is equal to gradient
of f at that point dot u.

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Which in our case
is equal to-- well,

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since we're interested at the
point P-- it's equal to 0,

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minus 3, dot our vector
u, which is [3/5, 4/5].

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And that dot product
is negative 12/5.

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So in this direction,
the function

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is decreasing at about this
rate at the given point.

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All right.

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That's part a.

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Let's go on to part b.

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So, you know, you've probably
noticed that these questions

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are all fairly similar.

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The only real complication
is that I've been increasing

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the number of variables.

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But of course, you
know how to take

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a gradient for a function
of three variables as well.

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So in this case, our
function g of x, y, z

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is given by the square
root of x squared

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plus y squared plus z squared.

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So, if you like,
this is the function

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that measures the distance
of a point from the origin.

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And so we're asked first for the
gradient, so the gradient of g.

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What do I do?

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Well, again, I just take
these partial derivatives.

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This is going to be a tiny
bit messy, I'm afraid.

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Because I take a
partial derivative

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of this expression
with respect to x,

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I've got to apply the
chain rule, right?

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Because this is a composition
of functions here.

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So OK.

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So, what do I get?

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I get the derivative of the
inside times the derivative

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of the out.

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So the outside function
is a square root,

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so that's to the
one halfth power.

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So I get 1/2 times the thing
inside to the minus one halfth

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power.

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So it's over 2 times
the square root

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of x squared plus y
squared plus z squared.

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And then I need to multiply by
the derivative of the inside,

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which is times 2x.

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OK.

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So, and then the 2's
cancel, and that's

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x over square root of x squared
plus y squared plus z squared.

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And similarly, this is a
nice symmetric function,

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and, you know, if I changed
x and y, it's the same.

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So the other partial
derivatives look very similar.

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So they're going to be y over
the square root of x squared

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plus y squared plus z squared.

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And z over the square root of
x squared plus y squared plus z

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squared.

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Sorry.

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That's a little bit of
a long formula there.

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But there we have it.

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That's the derivative-- or,
sorry, not the derivative.

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That's the gradient.

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The vector of
partial derivatives.

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So now you were asked also
to compute this gradient

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at a particular point.

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So the point in question-- I
have to look back over here.

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And the point in question was
this point, 2, 6, minus 3.

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So at the point
2, 6, minus 3, we

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want to compute the gradient.

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So we just take these
numbers back over

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to our formula over there, and
we're going to put them in.

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So we take the gradient
of g at 2, 6, minus 3.

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Well, OK, so this square root of
x squared plus y squared plus z

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squared appears in all terms.

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So let's compute that first.

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So x squared is 4, y squared
is 36, and z squared is 9.

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So I add those numbers
together, I get 49.

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And then I take a square
root of that and I get 7.

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So these denominators are all
going to be 7, and then up

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top I just have x, y, and z.

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So this is going to
be 2 over 7 comma

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6 over 7 comma minus 3 over 7.

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All right.

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So just what I get by plugging
the values at our point

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into this formula.

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So this is the
gradient at that point.

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And now once again,
I want to compute

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a particular directional
derivative of this function.

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So in order to do that, I just
need the right unit vector,

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and I need to dot it.

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So the direction
that I asked you

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was the direction v with
coordinates 1, 1, 1.

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So of course again, this
isn't a unit vector.

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So we need to divide it by
its length to find the unit

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vector that we're going to use.

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So the length of this vector
is the square root of 1

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squared plus 1 squared
plus 1 squared,

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so that's the square root of 3.

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So u is equal to--
well, I'm just

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going to write it as 1
over the square root of 3

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times the vector [1, 1, 1].

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And so the directional
derivative dg/ds

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in the direction u hat is--
again, it's what I get,

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I dot the gradient with this
u-- it's gradient dot u.

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Which in our case, so that
1 over the square root of 3

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lives out front.

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And now I dot 2/7, 6/7,
minus 3/7 with [1, 1, 1].

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And that's just going to give
me 2/7 plus 6/7 minus 3/7.

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And OK, so we can finish
off the arithmetic there,

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and I think that's 5
over 7 square root of 3

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if I didn't make any mistake.

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Now, the last one,
very, very similar.

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Now we have a function
of four variables.

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There's really nothing
new there at all.

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So let me rewrite.

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So for part c, we
have h of w, x, y,

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z is equal to w*x plus w*y plus
w*z plus x*y plus x*z plus y*z.

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So, for the gradient
of h, I just

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take the four
partial derivatives.

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Now here, w is first.

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So I take the partial
derivative with respect

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to w as the first coordinate.

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So that's going to
be x plus y plus z.

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And then I take the partial
derivative with respect

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to x as the second coordinate.

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So that's going to
be w plus y plus z.

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And similarly, the last two are
w plus x plus z, and w plus--

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that's a plus-- x plus y.

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Whew.

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OK.

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So that's my gradient.

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00:11:18,760 --> 00:11:20,260
Those are my
partial derivatives.

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00:11:20,260 --> 00:11:23,620
Now, I asked you again for the
gradient at a particular point.

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So that's grad h at the
point 2, 0, minus 1, minus 1.

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00:11:30,690 --> 00:11:32,200
So we can just plug that in.

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00:11:32,200 --> 00:11:36,330
Here, it's 0 plus minus 1
plus minus 1 is minus 2.

215
00:11:36,330 --> 00:11:41,290
Here it's 2 plus minus 1
plus minus 1, that's 0.

216
00:11:41,290 --> 00:11:45,660
Here, it's 2 plus 0 plus
minus 1 so that's 1.

217
00:11:45,660 --> 00:11:51,560
And lastly, it's 2 plus
0 plus minus 1 is 1.

218
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So that's the gradient of this
function h at this point P.

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And now I gave you again, I
gave you a vector v. So this

220
00:12:02,800 --> 00:12:07,900
was the vector 1,
minus 1, 1, minus 1.

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And I asked you to find
the directional derivative

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at this point in that direction.

223
00:12:11,930 --> 00:12:13,770
Again, we need a unit vector.

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00:12:13,770 --> 00:12:15,260
This isn't a unit vector.

225
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So u is equal to v over the
length of v. Well, what is

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the length of v in this case?

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It's over the square root of--
well, we square the coordinates

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and add them.

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So that's 1 plus
1 plus 1 plus 1.

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00:12:33,305 --> 00:12:35,520
So that's 4; square rooted is 2.

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00:12:35,520 --> 00:12:36,970
So that's v over 2.

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00:12:36,970 --> 00:12:40,330


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And finally, I get that the
directional derivative dh/ds

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in this direction
u, is what I get

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00:12:50,980 --> 00:12:54,550
when I dot this
gradient together

236
00:12:54,550 --> 00:12:57,340
with the direction
in which I'm heading.

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So this is minus 2, 0, 1, 1.

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00:13:02,160 --> 00:13:07,330
The gradient dotted with the
unit vector of the direction.

239
00:13:07,330 --> 00:13:12,060


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00:13:12,060 --> 00:13:13,500
Which we just computed.

241
00:13:13,500 --> 00:13:15,710
And so finally, you could
expand this out yourself.

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00:13:15,710 --> 00:13:17,800
I won't write down
the final answer,

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but it's just a
dot product, right?

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All right.

245
00:13:22,180 --> 00:13:25,950
So hopefully this has
gotten you totally

246
00:13:25,950 --> 00:13:27,850
comfortable with
computing gradients

247
00:13:27,850 --> 00:13:30,920
and with computing directional
derivatives from gradients.

248
00:13:30,920 --> 00:13:32,580
So I'll end there.

249
00:13:32,580 --> 00:13:33,080