# Show F is conservative

1. Dec 11, 2013

### jaydnul

1. The problem statement, all variables and given/known data
Show $F=<3x^2y-y^2,x^3-2xy>$ is conservative. Find a scalar potential f. Evaluate $∫FdR$ where C connects (0,0) to (2,1).

2. Relevant equations
Conservative if $P_y=Q_x$

3. The attempt at a solution
So it is conservative, but I don't know where to go from here. Thanks

2. Dec 11, 2013

### vela

Staff Emeritus
What's a scalar potential?

3. Dec 11, 2013

### jaydnul

Where the gradient of f is equal to F, right? I just have no clue how to do this.

4. Dec 11, 2013

### vela

Staff Emeritus
I'm sure there are examples you could consult in your textbook. In any case, you're right. If f is the scalar potential, then $\nabla f = F$, so that means
\begin{align*}
\frac{\partial f}{\partial x} &= P \\
\frac{\partial f}{\partial y} &= Q
\end{align*} Plug in the function you have for P for this problem, and integrate the first equation. What do you get?

5. Dec 11, 2013

### jaydnul

$x^3y-xy^2+c$

My book is so awful

6. Dec 11, 2013

### jaydnul

Never mind I think I got it. Sometimes when I post it jogs my memory. I remember my teacher doing something with a function of y in the place of c. Thanks anyways vela!!

7. Dec 11, 2013

### D H

Staff Emeritus
That's correct. You integrated 3x2y-y2 with respect to x to yield x3y-xy2+c. That constant c can be any function of y because what you integrated was a partial derivative with respect to x.

Consider $\vec F = 2x\hat x + 2y\hat y$. This is obviously conservative, but now when you do the integrations you get $U(x,y)=x^2+c$ on one hand versus $U(x,y)=y^2+c$ on the other. The way around this apparent problem to realize that the first c is a function of y, the second a function of x. With that, $U(x,y)=x^2+f(y)=y^2+g(x)$, so $U(x,y)=x^2+y^2+c$.

8. Dec 11, 2013

### vanhees71

The only question left is the sign. Usually one defines the potential such that it adds positively to the total energy, which implies that
$$\vec{F}(\vec{x})=-\vec{\nabla} V(\vec{x}).$$

9. Dec 11, 2013

### D H

Staff Emeritus
That's the convention used in physics. Mathematicians often use the opposite convention. This question was asked in the mathematics section, so it's probably more apropos to use the non-negated convention.

10. Dec 11, 2013

### vanhees71

Oh dear, that's very confusing for a physicist