Is F conservative and what is its scalar potential?

[jaydnul]

Homework Statement

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).

Homework Equations

Conservative if P_y=Q_x

The Attempt at a Solution

So it is conservative, but I don't know where to go from here. Thanks

 

[vela]

What's a scalar potential?

 

[jaydnul]

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

 

[vela]

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?

 

[jaydnul]

x^3y-xy^2+c

My book is so awful

 

[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!

 

[D H]

I remember my teacher doing something with a function of y in the place of c.

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 $U(x,y)=x^2+y^2+c$.

 

[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}).

 

[D H]

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}).

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.

 

[vanhees71]

Oh dear, that's very confusing for a physicist