UPSC 2013 Maths Optional Paper 1 Q5b — Step-by-Step Solution

10 marks · Section B

Orthogonal trajectories (cartesian and polar) · ODEs · asked 7× in 14 yrs · Read the full method →

Question

Obtain the equation of the orthogonal trajectory of the family of curves represented by $r^{n}=a\sin n\theta$ , $(r,\theta)$ being the plane polar coordinates.

Technique

Eliminate the parameter from the family’s ODE; replace $\dfrac{1}{r}\dfrac{dr}{d\theta}\to-r\dfrac{d\theta}{dr}$ ; re-integrate.

Solution

Strategy. Polar orthogonal trajectories: derive the differential equation of the given family (eliminating $a$ ), then replace $\dfrac{1}{r}\dfrac{dr}{d\theta}$ by $-r\dfrac{d\theta}{dr}$ — equivalently $\tan\psi\to-\cot\psi$ — and re-integrate.

Step 1 — ODE of the given family

Differentiate $r^{n}=a\sin n\theta$ with respect to $\theta$ :

n\,r^{n-1}\frac{dr}{d\theta}=a\,n\cos n\theta\;\Longrightarrow\;r^{n-1}\frac{dr}{d\theta}=a\cos n\theta.

Eliminate $a$ by dividing this into the original ( $r^{n}=a\sin n\theta$ ):

\frac{r^{n-1}\,dr/d\theta}{r^{n}}=\frac{a\cos n\theta}{a\sin n\theta}\;\Longrightarrow\;\frac{1}{r}\frac{dr}{d\theta}=\cot n\theta.

Step 2 — Orthogonality substitution

Two polar curves intersect orthogonally iff their values of $\dfrac{1}{r}\dfrac{dr}{d\theta}$ at the intersection are negative reciprocals. Hence replace

\frac{1}{r}\frac{dr}{d\theta}\;\longrightarrow\;-r\,\frac{d\theta}{dr},

in the family’s ODE to get the orthogonal family’s ODE:

-r\frac{d\theta}{dr}=\cot n\theta\;\Longrightarrow\;\tan n\theta\,d\theta=-\frac{dr}{r}.

Step 3 — Integrate

\int\tan n\theta\,d\theta=-\int\frac{dr}{r}\;\Longrightarrow\;-\frac{1}{n}\ln|\cos n\theta|=-\ln|r|+\text{const}.

Rearrange: $\ln|r|=\dfrac{1}{n}\ln|\cos n\theta|+\text{const}$ , i.e. $r^{n}=b\,\cos n\theta$ for a new constant $b$ .

Answer

\boxed{\;r^{n}=b\,\cos n\theta.\;}