Question

finding the potential function for a conservative vector field.

I have xy dx and (x+y)dy that I'm trying to find a vector valued function of but I can't get the two integrals to work together where the partial derivatives equal those two equations. Any help?

Answer 1

Don't you mean a potential function for a conservative vector field?

Answer 2

yes, sorry. my mind is a bit frazzled

Answer 3

Okay, let's just try some stuff.

Answer 4

First integration the partial with respect to \(x\).\[
\int xy~dx = \frac{x^2}2y+g(y)
\]

Answer 5

yes thats what I got. and xy+y^2/2 for the other part

Answer 6

Now we differentiate with respect to \(y\):\[
\frac{x^2}{2}+g'(y)
\]Set it equal to partial with respect to \(y\): \[
x+y = \frac{x^2}2+g'(y) \implies g'(y) = x-\frac{x^2}{2}+y
\]

Answer 7

This means it cannot be a potential function, since \(g'\) and \(g\), are not constant with respect to \(x\).

Answer 8

So there's no solution?

Answer 9

There is no potential function.

Answer 10

You can still integrate over the field though.

Answer 11

Hm... I'm trying to evaluate a line integral and this is the only way my instructor pointed us toward. (She doesn't teach :/ ) I'll look around for anyother way to do it. thanks!

Answer 12

Another way to test is to use the fact that \(f_{xy}=f_{yx}\).
In this case were were getting \(f_{xy}=y\) and \(f_{yx}=1\).

Answer 13

Do you know what the path is?

Answer 14

y=x^2

Answer 15

What are the start and end points?

Answer 16

(-1,1) and (2,4)

Answer 17

We can do this.

Answer 18

Is it the parameterization way? I saw that in Dawkin's notes but I couldn't understand it

Answer 19

Yes

Answer 20

We will parametrize this curve by letting \(x=t\).

Answer 21

Then our curve \(\mathbf r\) becomes: \[
\mathbf r(t) = (x,y) = (x,x^2) = (t,t^2)
\]

Answer 22

Our \(d\mathbf r\) becomes: \[
d\mathbf r = \mathbf r'(t)~dt = (1,2t)~dt
\]

Answer 23

Our vector field \(\mathbf F\) becomes: \[
\mathbf F = (xy,x+y) = (tt^2,t+t^2) = (t^3,t+t^2)
\]

Answer 24

wait I was good up until the last one

Answer 25

oh, thats cuz of x=t. But why is it only the y part of the coordinates?

Answer 26

All together: \[
\int_C\mathbf F\cdot d\mathbf r = \int_{-1}^2\mathbf F\cdot \mathbf r'(t)~dt = \int_{-1}^2 (t^3,t+t^2)\cdot (t,t^2)~dt
\]

Answer 27

Can you finish it?

Answer 28

wait, the bounds just changed. Am I supposed to use the X-coord or the Y-coord?

Answer 29

We could have not done the whole \(x=t\) thing and still arrived at: \[
\int_C\mathbf F\cdot d\mathbf r = \int_{-1}^2\mathbf F\cdot \mathbf r'(x)~dx = \int_{-1}^2 (x^3,x+x^2)\cdot (x,x^2)~dx
\]It would be the same.

Answer 30

and I thought it was F dot ||r'(t)||?

Answer 31

if that's it then yes, I can finish it

Answer 32

This is a vector field, not a scalar function.

Answer 33

You can't dot with a scalar anyway.

Answer 34

yeah you're right. Alright I think Ive got it

Answer 35

wait should the last part be (1,2t) instead of (t,t^2)

Answer 36

?

Answer 37

Yes

Answer 38

okay, so does 15.9167 or 191/12 sound somewhat okay?

Answer 39

Well, anyway Thanks so much for your help!!!