Answer to Question #343730 in Differential Equations for Vishu

Question

Find the orthogonal trajectories of the following curves (where a is a parameter). r = a(1 + sin θ)

Answer 1

Find the orthogonal trajectories of the following curves:

$r=a\left(1+\sin{\theta}\right)$ ,

Solution:

$r\prime=a\cos{\theta}$ $\Longrightarrow a=\frac{r\prime}{\cos{\theta}}$

then $r=\frac{r\prime}{\cos{\theta}}\left(1+\sin{\theta}\right)$ .

For orthogonal trajectories in polar coordinates substitution is used: $r\prime\rightarrow-\frac{r^2}{r\prime}$

Then $r=-\frac{r^2}{r\prime}\frac{1+\sin{\theta}}{\cos{\theta}}$ ,

$\frac{dr}{d\theta}=-r\frac{1+\sin{\theta}}{\cos{\theta}}$ ,

$\int\frac{dr}{r}=-\int{\frac{1+\sin{\theta}}{\cos{\theta}}d\theta}$ ,

$\ln{r}=-\int{\frac{1+\sin{\theta}}{\cos{\theta}}d\theta}$ ,

$\int{\frac{1+\sin{\theta}}{\cos{\theta}}d\theta}=\int{\frac{\left(1+\sin{\theta}\right)\cos{\theta}}{{cos}^2\theta}d\theta}=\int\frac{\left(1+\sin{\theta}\right)d\left(\sin{x}\right)}{1-{sin}^2\theta}=\int\frac{d\left(\sin{x}\right)}{1-\sin{x}}$

$-\int\frac{d\left(1-\sin{x}\right)}{1-\sin{x}}¬ =-\ln{\left(1-\sin{x}\right)}+\ln{C}$ .

Answer:

$r=С(1-sinθ)$