JOEL LEWIS: Hi.
Welcome back to recitation.
In lecture, you've learned
about gradients and about
directional derivatives.
So I have a problem here
to test your understanding
of those objects and give you
some practice computing them.
So I have three functions here.
And for each function,
I've given you
a point and a direction.
So what I'd like
you to do is compute
the gradients of
the functions, then
evaluate their gradients
at the point given,
and then compute the directional
derivative of the function
at the point in the direction
of the vector given.
So here, for example,
in part a, I've
given you the function f of x,
y equal to x squared y plus x
y squared, at the point x
equal minus 1, y equal 2,
and in the direction
of the vector [3, 4].
And then we've got two more
examples here. g of x, y, z
is equal to the square root
of x squared plus y squared
plus z squared, P is
the point 2, 6, minus 3,
and v is the
direction [1, 1,  1].
And our third example, h is a
function of four variables, w,
x, y, and z.
It's given by w*x plus w*y plus
w*z plus x*y plus x*z plus y*z,
at the point 2, 0, minus 1,
minus 1, in the direction 1,
minus 1, 1, minus 1.
So why don't you pause
the video, take some time
to work out these three
problems, come back,
and we can work
them out together.
I hope you had some luck
working out these problems.
Let's get started
on the first one.
So on the first one,
our function f of x, y
is equal to x squared
y plus x y squared.
So to compute the
gradient, I just
have to compute the first
partials of our function,
and then put them
into a single vector.
So that's what the gradient is.
So we have that the
gradient of f at (x,
y) is equal to--
it's the vector,
and its first component is the
first partial of f with respect
to x, so that's going to
be 2x*y plus y squared.
And its second component is the
partial of f with respect to y,
so that's going to be
x squared plus 2x*y.
So this is the gradient
of our function f.
Now, the question asked you
to compute this gradient
at the particular
point minus 1, 2.
So we have gradient of
f at the point minus 1,
2 is equal to-- well, we
just put in x equal minus 1,
y equal 2 into this formula.
So at x equal minus 1, y
equals 2, this is 2 times
minus 1 times 2 plus 2 squared.
So that's minus 4 plus
4, so that's just 0.
And the second term is
minus 1 squared plus 2 times
minus 1 times 2.
So that's 1 plus 4--
nope-- 1 plus negative 4,
so that's negative 3.
All right.
So this is the gradient
of our function f.
This is the gradient at
the point minus 1, 2.
Now we are asked for the
directional derivative.
So we were asked for the
directional derivative
of f in a direction.
So we're given-- I didn't give
you-- so I gave you a vector v.
But when we take directional
derivatives, what we want
is a unit vector.
Right?
So the vector v that I gave
you-- was right over here--
it was this vector, [3, 4].
And so we need to find the unit
vector u in the direction of v
in order to apply
our usual formula.
So u is just v divided
by the length of
v. So in this case,
the vector [3, 4],
by the Pythagorean
theorem, it has length 5.
So this is equal to the
vector 3/5 comma 4/5.
And so our
directional derivative
in the direction of v--
which is this vector
u-- it's just equal to
the gradient at that point
dotted with the
direction vector.
So this is equal to gradient
of f at that point dot u.
Which in our case
is equal to-- well,
since we're interested at the
point P-- it's equal to 0,
minus 3, dot our vector
u, which is [3/5, 4/5].
And that dot product
is negative 12/5.
So in this direction,
the function
is decreasing at about this
rate at the given point.
All right.
That's part a.
Let's go on to part b.
So, you know, you've probably
noticed that these questions
are all fairly similar.
The only real complication
is that I've been increasing
the number of variables.
But of course, you
know how to take
a gradient for a function
of three variables as well.
So in this case, our
function g of x, y, z
is given by the square
root of x squared
plus y squared plus z squared.
So, if you like,
this is the function
that measures the distance
of a point from the origin.
And so we're asked first for the
gradient, so the gradient of g.
What do I do?
Well, again, I just take
these partial derivatives.
This is going to be a tiny
bit messy, I'm afraid.
Because I take a
partial derivative
of this expression
with respect to x,
I've got to apply the
chain rule, right?
Because this is a composition
of functions here.
So OK.
So, what do I get?
I get the derivative of the
inside times the derivative
of the out.
So the outside function
is a square root,
so that's to the
one halfth power.
So I get 1/2 times the thing
inside to the minus one halfth
power.
So it's over 2 times
the square root
of x squared plus y
squared plus z squared.
And then I need to multiply by
the derivative of the inside,
which is times 2x.
OK.
So, and then the 2's
cancel, and that's
x over square root of x squared
plus y squared plus z squared.
And similarly, this is a
nice symmetric function,
and, you know, if I changed
x and y, it's the same.
So the other partial
derivatives look very similar.
So they're going to be y over
the square root of x squared
plus y squared plus z squared.
And z over the square root of
x squared plus y squared plus z
squared.
Sorry.
That's a little bit of
a long formula there.
But there we have it.
That's the derivative-- or,
sorry, not the derivative.
That's the gradient.
The vector of
partial derivatives.
So now you were asked also
to compute this gradient
at a particular point.
So the point in question-- I
have to look back over here.
And the point in question was
this point, 2, 6, minus 3.
So at the point
2, 6, minus 3, we
want to compute the gradient.
So we just take these
numbers back over
to our formula over there, and
we're going to put them in.
So we take the gradient
of g at 2, 6, minus 3.
Well, OK, so this square root of
x squared plus y squared plus z
squared appears in all terms.
So let's compute that first.
So x squared is 4, y squared
is 36, and z squared is 9.
So I add those numbers
together, I get 49.
And then I take a square
root of that and I get 7.
So these denominators are all
going to be 7, and then up
top I just have x, y, and z.
So this is going to
be 2 over 7 comma
6 over 7 comma minus 3 over 7.
All right.
So just what I get by plugging
the values at our point
into this formula.
So this is the
gradient at that point.
And now once again,
I want to compute
a particular directional
derivative of this function.
So in order to do that, I just
need the right unit vector,
and I need to dot it.
So the direction
that I asked you
was the direction v with
coordinates 1, 1, 1.
So of course again, this
isn't a unit vector.
So we need to divide it by
its length to find the unit
vector that we're going to use.
So the length of this vector
is the square root of 1
squared plus 1 squared
plus 1 squared,
so that's the square root of 3.
So u is equal to--
well, I'm just
going to write it as 1
over the square root of 3
times the vector [1, 1, 1].
And so the directional
derivative dg/ds
in the direction u hat is--
again, it's what I get,
I dot the gradient with this
u-- it's gradient dot u.
Which in our case, so that
1 over the square root of 3
lives out front.
And now I dot 2/7, 6/7,
minus 3/7 with [1, 1, 1].
And that's just going to give
me 2/7 plus 6/7 minus 3/7.
And OK, so we can finish
off the arithmetic there,
and I think that's 5
over 7 square root of 3
if I didn't make any mistake.
Now, the last one,
very, very similar.
Now we have a function
of four variables.
There's really nothing
new there at all.
So let me rewrite.
So for part c, we
have h of w, x, y,
z is equal to w*x plus w*y plus
w*z plus x*y plus x*z plus y*z.
So, for the gradient
of h, I just
take the four
partial derivatives.
Now here, w is first.
So I take the partial
derivative with respect
to w as the first coordinate.
So that's going to
be x plus y plus z.
And then I take the partial
derivative with respect
to x as the second coordinate.
So that's going to
be w plus y plus z.
And similarly, the last two are
w plus x plus z, and w plus--
that's a plus-- x plus y.
Whew.
OK.
So that's my gradient.
Those are my
partial derivatives.
Now, I asked you again for the
gradient at a particular point.
So that's grad h at the
point 2, 0, minus 1, minus 1.
So we can just plug that in.
Here, it's 0 plus minus 1
plus minus 1 is minus 2.
Here it's 2 plus minus 1
plus minus 1, that's 0.
Here, it's 2 plus 0 plus
minus 1 so that's 1.
And lastly, it's 2 plus
0 plus minus 1 is 1.
So that's the gradient of this
function h at this point P.
And now I gave you again, I
gave you a vector v. So this
was the vector 1,
minus 1, 1, minus 1.
And I asked you to find
the directional derivative
at this point in that direction.
Again, we need a unit vector.
This isn't a unit vector.
So u is equal to v over the
length of v. Well, what is
the length of v in this case?
It's over the square root of--
well, we square the coordinates
and add them.
So that's 1 plus
1 plus 1 plus 1.
So that's 4; square rooted is 2.
So that's v over 2.
And finally, I get that the
directional derivative dh/ds
in this direction
u, is what I get
when I dot this
gradient together
with the direction
in which I'm heading.
So this is minus 2, 0, 1, 1.
The gradient dotted with the
unit vector of the direction.
Which we just computed.
And so finally, you could
expand this out yourself.
I won't write down
the final answer,
but it's just a
dot product, right?
All right.
So hopefully this has
gotten you totally
comfortable with
computing gradients
and with computing directional
derivatives from gradients.
So I'll end there.
