Tricky question involving exact differentials

• Jan 23rd 2011, 12:26 PM
Runty
Tricky question involving exact differentials
I've been asking around on this question for a bit, and though I have an answer for the first half of it, I'm a little bit iffy on the solidarity of the answer I have. Primarily, however, I could use a hand with the second half of this, since it isn't worded that well.

Show that $F=\left(\frac{x}{x^2+y^2}, \frac{y}{x^2+y^2}\right)$ and $G=\left(\frac{y}{x^2+y^2}, \frac{x}{x^2+y^2}\right)$ are such that one of $F.dP$, $G.dP$ is exact but the other is not $(dP=(dx,dy))$.

Let $C$ be the shorter arc of $x^2+y^1=1$ from $(1/\sqrt{2}, -1/\sqrt{2})$ to $(1/\sqrt{2}, 1/\sqrt{2})$.
Evaluate $\int_C F.dP$ and $\int_C G.dP$.

As you can see, this one is actually quite tough. If you could provide any assistance on it.
• Feb 3rd 2011, 08:50 AM
TheEmptySet
Quote:

Originally Posted by Runty
I've been asking around on this question for a bit, and though I have an answer for the first half of it, I'm a little bit iffy on the solidarity of the answer I have. Primarily, however, I could use a hand with the second half of this, since it isn't worded that well.

Show that $F=\left(\frac{x}{x^2+y^2}, \frac{y}{x^2+y^2}\right)$ and $G=\left(\frac{y}{x^2+y^2}, \frac{x}{x^2+y^2}\right)$ are such that one of $F.dP$, $G.dP$ is exact but the other is not $(dP=(dx,dy))$.

Let $C$ be the shorter arc of $x^2+y^1=1$ from $(1/\sqrt{2}, -1/\sqrt{2})$ to $(1/\sqrt{2}, 1/\sqrt{2})$.
Evaluate $\int_C F.dP$ and $\int_C G.dP$.

As you can see, this one is actually quite tough. If you could provide any assistance on it.

An equation is exact if it has a potential function e.g there is a scalar function $f(x,y)$ such that with $r=x\vec{i}+y\vec{j}$

$\displaystyle \nabla f \cdot d\vec{r}=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy$

Now for $F$ this implies

$\displaystyle \frac{\partial f}{\partial x}=\frac{x}{x^2+y^2}$

Taking the partial integral of both sides with respect to x gives

$\displaystyle f(x,y)=\frac{1}{2} \ln (x^2+y^2)+g(y)$

Since we know what the partial with respect to $y$ is we can determine the function $g(y)$

$\displaystyle \frac{\partial f}{\partial y}=\frac{y}{x^2+y^2}+g'(y)$

This implies that $g'(y)=0 \implies g(y)=c$

So the potential function is

$\displaystyle f(x,y)=\frac{1}{2}\ln(x^2+y^2)$

For Part B:
Since $F$ has a potential function the fundamental theorem of lines integrals applies and the integral is

$\displaystyle \int_{C}\vec{F}\cdot d\vec{P}=\int_{C}\nabla f \cdot d\vec{P}=f\left( \frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right) - f\left( \frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right)=0$

Now for $G$ if you try to mimic the process you will find that you cannot find a potential function (the arbitary function of integration will is a function of one variable, but you will find that it is not!

Because of this the linear integral must be evaluated directly. Since it is a circular arc parametrize it with polar coordinates.
• Feb 3rd 2011, 07:33 PM
Abu-Khalil
Also if $\vec F\in\mathrm C^1$ and its domain is simply conected you may check if

$\displaystyle\frac{\partial Q}{\partial x}=\frac{\partial P}{\partial y}$

and then conclude that there exists a potential. (but in this case doesn't apply 'cause $\mathbb{R}^2\setminus\{\vec 0\}$ is not simply connected)

On the other hand, if $\vec F=\vec \nabla f$ and $f\in\mathrm C^2$ then $\displaystyle\frac{\partial Q}{\partial x}=\frac{\partial P}{\partial y}$ so u may use the contrapositive for $\vec G$.