← 2024 Paper 1

UPSC 2024 Maths Optional Paper 1 Q5a — Step-by-Step Solution

10 marks · Section B

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

Question

Find the orthogonal trajectories of the family of curves r=c(sec⁡θ+tan⁡θ)r=c(\sec\theta+\tan\theta), where cc is a parameter.

Technique

Differentiate to eliminate cc; apply the polar orthogonality rule dr/dθ→−r2 dθ/drdr/d\theta\to -r^2\,d\theta/dr; integrate.

Solution

Step 1 — Differential equation of the given family.

Differentiating r=c(sec⁡θ+tan⁡θ)r=c(\sec\theta+\tan\theta):

drdθ=c(sec⁡θtan⁡θ+sec⁡2θ)=csec⁡θ(tan⁡θ+sec⁡θ)=sec⁡θ⋅[c(sec⁡θ+tan⁡θ)]=rsec⁡θ.\frac{dr}{d\theta}=c(\sec\theta\tan\theta+\sec^2\theta)=c\sec\theta(\tan\theta+\sec\theta)=\sec\theta\cdot[c(\sec\theta+\tan\theta)]=r\sec\theta.

Step 2 — Orthogonal-trajectory ODE.

Replace dr/dθdr/d\theta by −r2 dθ/dr-r^2\,d\theta/dr (the polar orthogonality rule):

−r2dθdr=rsec⁡θ  ⇒  drdθ=−rcos⁡θ.-r^2\frac{d\theta}{dr}=r\sec\theta\;\Rightarrow\;\frac{dr}{d\theta}=-r\cos\theta.

Step 3 — Solve.

drr=−cos⁡θ dθ  ⇒  ln⁡∣r∣=−sin⁡θ+const  ⇒  r=k e−sin⁡θ.\frac{dr}{r}=-\cos\theta\,d\theta\;\Rightarrow\;\ln|r|=-\sin\theta+\text{const}\;\Rightarrow\;r=k\,e^{-\sin\theta}.

Answer

  r=k e−sin⁡θ.  \boxed{\;r=k\,e^{-\sin\theta}.\;}
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