Answer to Question #343730 in Differential Equations for Vishu

Question #343730

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

1
Expert's answer
2022-05-24T16:01:03-0400

Find the orthogonal trajectories of the following curves:

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

Solution:

r=acosθr\prime=a\cos{\theta} a=rcosθ\Longrightarrow a=\frac{r\prime}{\cos{\theta}}

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

For orthogonal trajectories in polar coordinates substitution is used: rr2rr\prime\rightarrow-\frac{r^2}{r\prime}

Then r=r2r1+sinθcosθr=-\frac{r^2}{r\prime}\frac{1+\sin{\theta}}{\cos{\theta}} ,

drdθ=r1+sinθcosθ\frac{dr}{d\theta}=-r\frac{1+\sin{\theta}}{\cos{\theta}} ,

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

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

1+sinθcosθdθ=(1+sinθ)cosθcos2θdθ=(1+sinθ)d(sinx)1sin2θ=d(sinx)1sinx\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}}

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

Answer:

r=С(1sinθ)r=С(1-sinθ)


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