# Vector analysis

1. Nov 19, 2009

### Denver Dang

1. The problem statement, all variables and given/known data
Let the curve C be paramatized into polar coordinates given by:

$$$r\left( t \right)=\left( r\left( t \right)\cos \theta \left( t \right),\,\,\,\,\,r\left( t \right)\sin \theta \left( t \right) \right),\,\,\,\,\,a\le t\le b$$$
where r and theta is continuous derivatives and r(t) > 0.

Show that:

$$$F\left( r\left( t \right) \right)\cdot r'\left( t \right)=\theta '\left( t \right)\,\,\,\,\,and\,\,\,\,\,\int_{C}{F\cdot dr=\theta \left( b \right)}-\theta \left( a \right)$$$

2. Relevant equations

F is given by:

$$$F\left( x,y \right)=\frac{-y\,i+x\,j}{{{x}^{2}}+{{y}^{2}}}$$$

3. The attempt at a solution

No idea... I get stupid when it changes into polar coords.
So anyone with a little hint maybe ?

Regards

2. Nov 19, 2009

### clamtrox

Start by calculating what F is in polar coordinates: x = r cos theta, y = r sin theta.

3. Nov 19, 2009

### Denver Dang

Then I get:

$$F\left( r,\theta \right)=\frac{\cos \left( \theta \right)j-\sin \left( \theta \right)i}{r}$$

Last edited: Nov 19, 2009
4. Nov 19, 2009

### Dick

So far, so good. Now what's R'(t) where R(t)=r(t)cos(theta(t))i+r(t)sin(theta(t))j? Try not to confuse the scalar r(t) with the vector R(t). It's R'(t) you want to dot with F.

5. Nov 19, 2009

### Denver Dang

Well, that's what confuses me :)
How do I differentiate a function with functions in it ? And with respect to what ? r or theta ? If I do it with respect to both, then I get to different R'(t), dont I ?

6. Nov 19, 2009

### Dick

Differentiate with respect to t. You'll need to use the product rule.

7. Nov 19, 2009

### Denver Dang

Hmmm...
But there is no t in R(t) ? :S

8. Nov 19, 2009

### Dick

There are two functions r(t) and theta(t)!!

9. Nov 19, 2009

### Denver Dang

Ehhh, I feel pretty stupid right now :S
Do I know the functions r(t) and theta(t) ? :S

10. Nov 19, 2009

### Dick

No, but you know dr(t)/dt=r'(t) and dtheta(t)/dt=theta'(t). Just write out R(t)' in terms of them.