1. ## orthogonal trajectaries

Find the orthogonal trajectory of the system $r^2cos^2 2\theta=c^2 sin 2 \theta$.

ans: $r^4(3-cos \theta)=k$.

2. Originally Posted by dimuk
Find the orthogonal trajectory of the system $r^2\cos^2 2\theta=c^2 \sin 2 \theta$.

ans: $r^4(3-\cos \theta)=k$. I'm getting $\color{red}r^4(3-\cos4\theta)=k$.
In polar coordinates, the quantity $\tfrac1r\tfrac{dr}{d\theta}$ represents the slope of the tangent of a curve with respect to the radial direction. (Informally, a point on the curve moves an infinitesimal distance dr in the radial direction and rdθ in the transverse direction.) So the orthogonal system should have "radial slope" $-r/\tfrac{dr}{d\theta}$ at that point.

Write $r^2\cos^2 2\theta=c^2 \sin 2 \theta$ as $r^2 = c^2\sec2\theta\tan2\theta$, and differentiate with respect to θ: $2rr' = 2c^2(\sec2\theta\tan^22\theta + \sec^32\theta)$, where $r' = \tfrac{dr}{d\theta}$. Substitute $c^2 = \frac{r^2\cos^22\theta}{\sin2\theta}$ and simplify a bit, to get $\frac{r'}r = \frac{1+\sin^22\theta}{\sin2\theta\cos2\theta}$.

The orthogonal trajectory will therefore have $\frac{r'}r =-\frac{\sin2\theta\cos2\theta}{1+\sin^22\theta}$. Integrate this: $\int\frac{dr}r = -\int\frac{\sin2\theta\cos2\theta}{1+\sin^22\theta} d\theta$.

To find the integral on the right-hand side, substitute $u = \sin2\theta$, and it becomes $-\int\frac{u}{2(1+u^2)}du = -\tfrac14\ln(1+u^2)$.

That gives the orthogonal trajectory as $\ln r = -\tfrac14\ln(1+\sin^22\theta)\; +$ const., or $r^4(1+\sin^22\theta) =$ const.

Finally, the trig identity $\sin^2\phi = \tfrac12(1-\cos2\phi)$ can be used to write the answer as $r^4(3-\cos4\theta)=$ const.