Family of orthogonal trajectories using an ODE in polar coordinates? I am trying to find the family of functions orthogonal to the family described by the differential equation,
$$\frac{dr}{d \theta} = -\frac{r\sin{\theta}}{\cos{\theta}}$$
If this equation were expressed in rectangular coordinates, then I could swap $-\frac{dx}{dy}$ in for $\frac{dy}{dx}$ in the LHS to obtain the differential equation for the corresponding orthogonal family. Unfortunately, since $\frac{dr}{d \theta}$ does not correspond to the slope of the tangent line of the curve at $(r, \theta)$, this can't be solved this way.
Instead, let $\psi = \gamma - \theta$ denote the angle between the tangent, $\gamma$, and the radius, $\theta$. We have that $\tan{\psi} = r \frac{d \theta}{dr}$. I am trying to use this fact to relate $r$ and $\frac{dr}{d \theta}$ with the slope of the tangent. Since the slope of the tangent line to the curve is $\tan{\gamma}$ we can write,
\begin{align*}
\tan{\gamma} &= \tan{\psi + \theta} = \frac{\tan{\psi} + \tan{\theta}}{1 - \tan{\psi}\tan{\theta}} \\
\end{align*}
but I'm not sure where to go from here.
The book I am using says that since $\tan{\psi} = r\frac{d \theta}{dr}$, we can replace $r\frac{d \theta}{dr}$ with its negative reciprocal $-\frac{1}{r}\frac{dr}{d \theta}$ to get,
$$r \frac{d \theta}{dr} = \frac{\sin \theta}{\cos \theta}$$ 
However, this seems wrong to me since the slope of the tangent line at $(r, \theta)$ is $\tan{\gamma}$, not $\tan{\psi}$; why does it seem like the book is suggesting that the slope is $\tan{\psi}$? What am I missing here?
Thanks!
 A: Look  at the differentials triangles in cartesian coordinates the infinitesimal (differential) sides are parallel to $x,y$ axes.
$$ \tan \phi= \dfrac{dy}{dx} $$ 
Negative reciprocal is taken for orthogonal trajectory with angle reckoned anti-clockwise.
Exactly the same happens for polar coordinates as seen in the picture (rough, hand drawn). Except that the infinitesimal triangle sides are parallel to polar radius drawn from origin and its normal along its circumference at point under consideration.
$$ \tan \psi= \dfrac{r d\theta}{dr} $$
and negative reciprocal is taken for orthogonal trajectory in same sense for new complement angle.

For DE of blue circles marked "1"
$$- \tan \theta= \dfrac{r d\theta}{dr} $$
Integrate involving logs etc.
$$ r = C_1 \cos \theta $$
For DE of red circles marked "2"
$$ \cot \theta= \dfrac{r d\theta}{dr} $$
Integrate involving logs etc.
$$ r = C_2 \sin \theta. $$
A: Tangent in polar coordinate is given by $(1/r)(dr/d\theta)$ replace this by negative reciprocal, i.e $-r(d\theta /dr)$(this will be slope of normal) so you have $$-r\dfrac{d\theta}{dr} = -\dfrac{\sin \theta}{\cos \theta}$$
This is variable separable and solvable.
What is written in the book is equivalent. Since the given differential equation doesn't refer to tangent or the normal specifically. So replacing tangent with normal or normal with tangent is the same. 
A: It can be seen that the angle from the radius to the orthogonal trajectory is $\psi + \frac{\pi}{2}$. We can first rewrite the stated differential equation as,
$$\tan{\psi} = r\frac{d \theta}{d r} = -\frac{\cos \theta}{\sin \theta}$$
Since,
$$\tan \left({\psi + \frac{\pi}{2}} \right) = -\frac{1}{\tan{\psi}} = -\frac{1}{r}\frac{d r}{d \theta}$$
to find the orthogonal family, we instead look for solutions to the equation,
\begin{align*}
&-\frac{1}{r}\frac{d r}{d \theta} = -\frac{\cos \theta}{\sin \theta} \\
\Rightarrow & \int \frac{r'}{r} d\theta = \int \frac{\cos \theta}{ \sin \theta} d \theta \\
\Rightarrow & \log|r| = \log|\sin \theta| + C \\
\Rightarrow & r = C \sin \theta.
\end{align*}
A: Actually there is another way of looking at all orthogonal trajectories passing through the singular point the origin in this particular case, whose DE is given.
$$\frac{dr}{d \theta} = -\frac{r\sin{\theta}}{\cos{\theta}};\quad\frac{dr}{r} = -\frac{\sin{\theta} \;d \theta}{\cos{\theta}}$$
Integrating we obtain all circles through origin with an arbitrary diameter $d$.
$$ r= d \cos (\theta+ \gamma) $$
where $ (d,\gamma) $ are arbitrary integration constants.
$$\frac{dr/ds}{d\theta/ds} = -r \tan \theta$$
From differential relations $$ \frac{dr}{ds}=\cos \psi;r d \theta  = ds \sin \psi,$$
after integration
$$ \psi +\theta = k \pi , \; \psi +\theta = \pi. $$
(took principal $k$ value),it is seen that these are all (red) semi-circles by the Thales theorem. We pick up only the orthogonal trajectories.
Actually it serves to answer when intersection angle is constant, but we do not complicate the question in hand.

