Orthogonal trajectories in polar coordinates

[patric44]

Homework Statement

in the polar coordinates find the set of curves that intersects with a right angle with the set of curves describes by r^{2} = a^{2}cos(theta)

Relevant Equations

r^{2} = a^{2}cos(θ)

there is a problem in a book that asks to find the orthogonal trajectories to the curves described by the equation :
$$r^{2} = a^{2}\cos(\theta)$$
the attempt of a solution is as following :
1- i defferntiate with respect to $\theta$ :
$$2r \frac{dr}{d\theta} = -a^{2}\;\sin(\theta)$$
2- i eliminated "a" from the two equations and get :
$$\frac{dr}{d\theta} = -\frac{1}{2}tan(\theta)$$
then the book said to set $\frac{dr}{d\theta} = -r^{2}\frac{d\theta}{dr}$ ! , this step i don't get ? why would i do that , it doesn't seem to be equal ?!

 

[pasmith]

Homework Statement:: in the polar coordinates find the set of curves that intersects with a right angle with the set of curves describes by r^{2} = a^{2}cos(theta)
Relevant Equations:: r^{2} = a^{2}cos(θ)

there is a problem in a book that asks to find the orthogonal trajectories to the curves described by the equation :
$$r^{2} = a^{2}\cos(\theta)$$

the attempt of a solution is as following :
1- i defferntiate with respect to $\theta$ :
$$2r \frac{dr}{d\theta} = -a^{2}\;\sin(\theta)$$

2- i eliminated "a" from the two equations

$$\frac{dr}{d\theta} = -\frac12 \tan\theta$$

I wouldn't descrive that as "eliminating $a$", so much as "eliminating $r$".

then the book said to set $\frac{dr}{d\theta} = -r^{2}\frac{d\theta}{dr}$ ! , this step i don't get ? why would i do that , it doesn't seem to be equal ?!

There is an abuse of notation here.

A curve $r = f(\theta)$ in plane polar coordinates is given by $\mathbf{r}(\theta) = f(\theta)\mathbf{e}_r(\theta)$. The tangent vector is, by the product rule, given by $$\frac{d\mathbf{r}}{d\theta} = \frac{df}{d\theta}\mathbf{e}_r + f(\theta) \mathbf{e}_\theta$$. If another curve $r = g(\theta)$ is orthogonal to this, then we must have $$\frac{df}{d\theta} \frac{dg}{d\theta} + fg = 0,$$ but because the curves intersect we have $f(\theta) = g(\theta)$ at this point, and so $$\frac{df}{d\theta} = - g(\theta)^2 \frac{d\theta}{dg}.$$ By an abuse of notation which suppress the fact that$r$ is given by a different function of $\theta$ on each side, this could be written $$\frac{dr}{d\theta} = -r^2 \frac{d\theta}{dr}.$$

 

[patric44]

thank you so much, it make sense now