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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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Those are my
partial derivatives.
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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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So we can just plug that in.
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Here, it's 0 plus minus 1
plus minus 1 is minus 2.
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Here it's 2 plus minus 1
plus minus 1, that's 0.
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00:11:41,290 --> 00:11:45,660
Here, it's 2 plus 0 plus
minus 1 so that's 1.
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And lastly, it's 2 plus
0 plus minus 1 is 1.
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So that's the gradient of this
function h at this point P.
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00:11:59,360 --> 00:12:02,800
And now I gave you again, I
gave you a vector v. So this
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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.
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Again, we need a unit vector.
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This isn't a unit vector.
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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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So that's 4; square rooted is 2.
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So that's v over 2.
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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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when I dot this
gradient together
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with the direction
in which I'm heading.
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So this is minus 2, 0, 1, 1.
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The gradient dotted with the
unit vector of the direction.
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Which we just computed.
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And so finally, you could
expand this out yourself.
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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.
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So hopefully this has
gotten you totally
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comfortable with
computing gradients
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and with computing directional
derivatives from gradients.
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So I'll end there.