Function being path independent?

**cad2blender** · December 5th 2009, 02:54 PM

I have a question about this problem. I know that I have to parametrize the circle using polar coordinates but after that I am completely lost. My book says that being path independent also means the same as being a conservative field correct? So does that mean that I have to take the partials of those terms? What am I missing here? Thanks!

**mr fantastic** · December 5th 2009, 03:14 PM

Originally Posted by cad2blender

I have a question about this problem. I know that I have to parametrize the circle using polar coordinates but after that I am completely lost.

My book says that being path independent also means the same as being a conservative field correct? Mr F says: Correct.

So does that mean that I have to take the partials of those terms? What am I missing here? Thanks!

If a vector field is conservative then $\nabla \times F = 0$ . Does this happen here?

**TheEmptySet** · December 5th 2009, 03:18 PM

Originally Posted by cad2blender

I have a question about this problem. I know that I have to parametrize the circle using polar coordinates but after that I am completely lost. My book says that being path independent also means the same as being a conservative field correct? So does that mean that I have to take the partials of those terms? What am I missing here? Thanks!

So the parametric equation for the circle is

$\vec{r}(t)=<4\cos(t),4\sin(t)> \implies \vec{r'}(t)=<-4\sin(t),4\cos(t)>dt$

Then the line integral is

$\int_{0}^{2\pi}\vec{F}(r(t))\cdot r'(t)dt=$
$\int_{0}^{2\pi}<3(4\cos(t))^2,(4\sin(t))>\cdot <-4\sin(t),4\cos(t)>dt=$
$\int_{0}^{2\pi}-192\sin(t)\cos^2(t)+16\sin(t)\cos(t)dt$

$=64\cos^3(t)+8\sin^2(t) \bigg|_{0}^{2\pi}=0$

You can see the vector field is conservative.

$\frac{\partial f}{\partial x}=3x^2 \implies f(x,y)=x^3+g(y)$

Now taking the partial with respect to y gives

$\frac{\partial f}{\partial y}=g'(y)=y \implies g(y)=\frac{1}{2}y^2$

So the potential function is

$f(x,y)=x^3-\frac{1}{2}y^2$ and notice that

$\nabla f =\vec{F}$

So the vector field is conservative.

**cad2blender** · December 5th 2009, 03:21 PM

hmmm ok then, so the question was simply asking me if it was conservative just in different wording? That's what threw me off. Thanks a ton guys

Thread: Function being path independent?

LinkBack

Thread Tools

Display

Function being path independent?

Similar Math Help Forum Discussions

Function, using two-path test, finding limits

degree of a path, path homotopy.

integrating a vector function along a path

conservative and independent of path - line integral

Independent of path; conservative vector field

Search Tags