# Simple question related to polar curves

1. Jun 16, 2013

### Saitama

1. The problem statement, all variables and given/known data
For a curve in Cartesian form, show that
$$\tan \phi = \frac{xy'-y}{x+yy'}$$

2. Relevant equations

3. The attempt at a solution
According to the book notation, $\phi$ is the angle between the radius vector and tangent at any point of the curve. I know that $\tan \phi=rd\theta/dr$ but how am I supposed to convert this to Cartesian form. I am currently studying polar curves by myself. I thought I can do simple stuff before studies begin so any help is appreciated. Thanks! :)

2. Jun 16, 2013

### ehild

Stay with Cartesian coordinates and express tan(theta) with them.

ehild

3. Jun 16, 2013

### Saitama

And how should I do that?

I tried it again and I get the answer now. In Cartesian coordinates, $x=r\cos \theta$(i) and $y=r\sin \theta$(ii).
Differentiating (i) w.r.t $\theta$
$$\frac{dx}{d\theta}=\cos \theta \frac{dr}{d\theta}-r\sin\theta$$
$$\frac{dx}{d\theta}=\frac{x}{r}\frac{dr}{d\theta}-y$$
Let $rd\theta/dr=k$
$$\Rightarrow \frac{dx}{d\theta}=\frac{x}{k}-y$$
Similarly,
$$\frac{dy}{d\theta}=\frac{y}{k}+x$$
Dividing the above two equations and solving for $k$,
$$k=\frac{xy'-y}{x+yy'}$$
And since $k=\tan \phi$, hence proved.

Looks good?

4. Jun 16, 2013

### ehild

Why to mix polar coordinates in? Φ is the angle between the radius vector r=xi+yj and the tangent vector 1i+y'j. Express cosΦ with their dot product and sinΦ with the cross product.

ehild

5. Jun 16, 2013

### Saitama

Thanks ehild! Your method gives the answer in a few steps. It is really a nice way to solve the problem contrary to mine which took more time.