Question

Find the orthogonal trajectories of the family r=a(1+sin θ).

(a) r=k(sin θ)

(b) r^2=k(cos θ)^2

(c) r=k(1-cos θ)

(d) r=k(1-sin θ)

This question was addressed to me in examination.

My question is based upon Orthogonal Trajectories in section Ordinary Differential Equations – First Order & First Degree of Engineering Mathematics

Answer 1

Correct choice is (d) r=k(1-sin θ)

The explanation: r=a(1+sin θ) taking log we get log r=log a+log(1+sin θ)…differentiating

w.r.t θ we have, \( \frac{1}{r} \frac{dr}{dθ} = \frac{cos⁡θ}{1+sin⁡θ}\) since given equation is polar slope of tangent is given by \( \frac{1}{r} \frac{dr}{dθ}\) and  perpendicular line has a slope of  \(-r\frac{dθ}{dr} \, replacing\,  \frac{1}{r}  \frac{dr}{dθ} \,by\, -r\frac{dθ}{dr}\)

we have \(-r\frac{dθ}{dr} = \frac{cos⁡θ}{1+sin⁡θ}\) …..separating the variables and integrating it

\(\int \frac{dr}{r} + \int \frac{1+sin⁡θ}{cos⁡θ} \,dθ = c \rightarrow log r + \int sec⁡θ \,dθ + \int tan⁡θ \, dθ  = c\)

log r + log(sec⁡θ + tan⁡θ) + log(sec⁡θ) = c = log k

\(\rightarrow log(r(sec⁡θ + tan⁡θ) (sec⁡θ)) = log k \rightarrow r\left(\frac{1}{cos⁡θ} + \frac{sin⁡θ}{cos⁡θ}\right)  \frac{1}{cos⁡θ} = k\)

\(\frac{r(1+sin⁡θ)}{cos^2 θ} = \frac{r(1+sin⁡θ)}{1-sin^2 θ} \rightarrow r = k(1-sin θ)\) is the required orthogonal trajectory.