CHRISTINE BREINER: Welcome back to recitation.
In this video I'd like us to work on the following problem.
What values of b will make this vector field F a gradient
field where F is determined by e to the x plus y times x plus
bi plus xj?
So the e to the x plus y is in both the i
component and the j component.
And then once you've determined what values b will
make that a gradient field--
for this b, or I should've said these b's--
find a potential function f using both
methods from the lecture.
So why don't you pause the video, work on this, and then
when you are ready to look at how I do it bring
the video back up.
OK.
Welcome back.
So I'm going to start off working on the first part of
this problem, which is to find the values of b that will make
this vector field F a gradient field.
And to clarify things for myself, I'm going to write
down what M and what N are based on F. So just to have it
clear, M is equal to e to the x plus y times x plus b and N
is equal to x times e to the x plus y.
So those are our values for M and N.
And now if I want f to be a gradient field, what I have to
do is I have to have M sub y equal N sub x.
So I'm going to determine M sub y and I'm going to
determine N sub x and I'm going to compare them and see
what value of b I get.
So M sub y, fairly straightforward because this
is a constant in y.
And the derivative of this in terms of y is just this back.
Right?
It's an exponential function with the value that it has in
y is linear.
So you get exactly that thing back.
So it actually is just either the x plus y times x plus b.
So the derivative of M sub y is just itself.
The derivative of M with respect to y.
Sorry.
Not the derivative of M sub y.
OK.
That's an x.
Let me just rewrite that.
OK, now N sub x is going to have two parts.
N sub x, the derivative with respect to x of this is 1.
And so I have an e to the x plus y.
And the derivative with respect to the x of e to the x
plus y is just e to the plus y.
For the same reason as the derivative with respect to y
was the same.
So then I'm just going to get a plus x e to the x plus y.
So that means if I factor that out, I get an e to the x plus
y times 1 plus x.
And we see that if F is going to be a gradient field then b
has to equal 1.
Because it can only have one value, and so b
has to equal 1.
To get N sub x to equal M sub y, b has to equal 1.
So now what I'm going to do is I'm going to erase that b, put
in a 1, so that the rest of my calculations refer to that.
So now the second part said for this b find a potential
function f using both methods from the lecture.
So we're going to go through both methods and hopefully we
get the same answer both times.
So let me come back here.
The first method is where I'm integrating along a curve from
(0, 0) to (x1, y1).
So I'm going to do it in the following way.
I'm going to let C1--
so here's (0, 0)--
I'm going to let C1 be the curve from (0,
0) up to (0, y1).
And then C2 be the curve--
so it's parameterized in that direction--
C2 be the curve from (0, y1) to (x1, y1).
OK?
So that's what I'm going to do and I'm going to let C equal
the curve C1 plus C2.
So I'm going to have C be the full curve.
And what I'm interested in doing is finding f of (x1,
y1), which will just equal the integral along C of f dot dr.
So now we need to figure out some important things
about C1 and C2.
What's happening on C1 and what's happening on C2.
And the first thing I want to point out-- actually, before I
do that, let me remind you that this is going to be the
integral on C of M dx plus N dy.
So this will be helpful to refer back to.
That's really what we're also doing here.
So on C1, what do I notice?
On C1, x is 0 and dx is 0.
And y goes between 0 and y1.
And then on C2, y is equal to y1.
So dy is equal to 0 and x is going between 0 and x1.
So those are the values that are important to me.
So if you notice from this fact and this fact, we see
that if we look at the integral just along C1,
there's going to be no M dx term.
And if we look at the integral along C2, there's going to be
no dy term because of that.
So let me write down the terms that do exist, and we'll see
some other things drop out also.
If I look along first just C1, I'm only going to get--
I said the dy term, which--
let me just make sure-- dx is 0.
I'm only going to get the dy term, which is--
well, x is 0 there.
So I'm going to get 0 times e to the 0 plus y.
dy.
From 0 to y1.
Well that's nice and easy to calculate, thank goodness.
That's just 0.
So all I have to do for this one is just leave it at 0.
That's everything that happens along C1.
That's what I'm interested in.
I just get 0 there.
And if I integrate along C2, as I mentioned, dy is 0.
So we don't have any component with N dy.
We just have the component M dx that we're integrating.
OK, so if I integrate along C2, I just have M dx and M is
e to the x plus y times x plus 1.
And y is fixed at y1.
So it's e to the x plus y1 times x plus 1 dx.
And I'm going from 0 to x1.
I'm going to make sure I didn't make any mistakes.
I'm going to check my work here.
Yes, I'm looking good.
OK.
So this one is 0.
So all I have to do is find an antiderivative of this.
And the term-- if I multiply through, I see that here I get
exactly the same thing when I look for an antiderivative.
And here I get, I believe, two terms when I look for an
antiderivative.
But I'm going to get some cancellation.
And ultimately, when I'm all done I'm going to get this.
xe to the x plus y1 evaluated at 0 and x1.
You could do this.
This is really now a single variable problem.
So I'm not going to work out all the details, but you might
want to do an integration by parts on that first part of
it, if that helps.
Or an integration by parts on the whole thing.
That would also do the trick.
So what do I get here?
Then I get x1 e to the x1 plus y1.
And then when I put in 0 for x here, I get 0, so that's it.
So this, plus possibly a constant, is equal to my f.
So I see that in general I get f of xy is equal to xe to the
x plus y plus a constant.
So that's what I get in the first method.
So now let's use the second method.
So I should say f of (x1, y1).
In the second method, what I do is M, remember, is
equal to f sub x.
So f sub x is equal to M which is equal to e to the x plus y
times x plus 1.
So if I want to find an antiderivative--
if I want to find f, I should take an antiderivative, right?
With respect to x.
And so notice I already did that, actually.
If I just put this as y, I already did that here.
And so I should get something that looks like this.
xe to the x plus y plus possibly a function that only
depends on y.
And the reason is when I take a derivative with respect to x
of this, obviously this would be 0.
So it doesn't show up over here.
So this, we make sure that--
oh, that I shouldn't write equals.
Sorry.
That I shouldn't write equals.
OK?
This would imply that this is equal to f.
Sorry about that.
f sub x was equal to M was equal to this.
That implies--
when I take an antiderivative of an x--
that xe to the x plus y plus g of y is equal to f.
So I apologize.
That wouldn't have been an equals because obviously those
two things are not equal.
That would imply, I think--
yeah, that would imply something very bad
mathematically.
So make sure you understand I put an equals sign where I
should not have. This is actually a derivative of that.
So this is antiderivative of this.
So now I have a candidate for f.
And so now I'm going to take the derivative of that.
And what's the derivative of that with respect to y?
So f sub y based on this is going to be equal to xe to the
x plus y plus g prime of y.
So the prime here indicates it's in a derivative in y.
And now that f sub y should also equal N. And N equals xe
to the x plus y.
So what do I get here?
I see xe to the x plus y has to equal xe to the x plus y
plus g prime of y.
Which means g prime of y is equal to 0.
Which means when I take an antiderivative of that I just
get a constant.
That means g of y was a constant.
So that implies that this boxed expression right here is
f of xy if g of y is just a constant.
So let me go through that logic one more time.
I had f sub x.
I took an antiderivative to get f but it involved a
constant in x that was a function of y.
I take a derivative of that in y.
I compare that to what I know the derivative is in y.
That gives me that this is 0.
So its antiderivative, which is g of y, is just a constant.
And so all together this implies that f of xy is equal
to xe to the x plus y plus a constant.
Which is exactly what I got before.
Fortunately, I got two answers that are the same.
So that's it.
I'll stop there.