Question

find the orthogonal trajectories of the family curves r=a theta, where a' is the parameter .r=a theta​

Answer 1

$\large\underline{\sf{Solution-}}$

Given polar curve is

$\red{\rm :\longmapsto\:r \: = \: a \: \theta - - - (1)}$

$\rm :\longmapsto\:Differentiating \: both \: sides \: w.r.t. \: \theta , \: we \: get$

${\rm :\longmapsto\:\dfrac{d}{d\theta } r \: = \dfrac{d}{d\theta } \: a \: \theta }$

${\rm :\longmapsto\:\dfrac{dr}{d\theta } \: = \: a \: \dfrac{d}{d\theta }\: \theta }$

${\rm :\longmapsto\:\dfrac{dr}{d\theta } \: = \: a \:}$

Substituting the value of 'a' in equation (1), we get

$\rm :\longmapsto\:r = \theta \dfrac{dr}{d\theta }$

Now, for Orthogonal trajectories, Replace

$\red{\rm :\longmapsto\:\dfrac{dr}{d\theta } \: = \: - \: {r}^{2}\dfrac{d\theta }{dr }}$

So, we get

$\rm :\longmapsto\:r = - {r}^{2} \theta \dfrac{d\theta }{dr}$

$\rm :\longmapsto\:1 = - {r}^{} \theta \dfrac{d\theta }{dr}$

On separate the variables, we get

$\rm :\longmapsto\:\dfrac{dr}{r} = - \theta \: d\theta$

On integrating both sides, we get

$\rm :\longmapsto\:\displaystyle\int \rm \dfrac{dr}{r} = - \displaystyle\int \rm \theta \: d\theta$

We know,

$\boxed{ \rm{ \displaystyle\int \rm \frac{1}{x} \: dx \: = \: logx + c}}$

and

$\boxed{ \rm{ \displaystyle\int \rm {x}^{n} \: dx \: = \: \frac{ {x}^{n + 1} }{ n+ 1} + c}}$

So, using this, we get

$\rm :\longmapsto\: log(r) = - \dfrac{ {\theta }^{2} }{2} + c$

$\rm :\longmapsto\: 2log(r) = - {\theta }^{2}+ 2c$

$\rm :\longmapsto\: log( {r}^{2} ) + {\theta }^{2} = d$

Answer 2

Answer:

We have, r

2

=a

2

cos4θ=a

2

(1−2sin

2

2θ)...(1)

Differentiating w.r.t. θ, we get

2r

dθ

dr

=−4a

2

sin4θ...(2)

Eliminating a from (2) using (1), we get

r

2

dθ

dr

=−

cos4θ

4sin4θ

...(3)

Replacing

dθ

dr

with −r

2

dr

dθ

in (3), we get

2r

dr

dθ

=

cosθ

4sin4θ

⟹

r

2

dr=

sinθ

cosθ

dθ

Integrating, we get

2logr=

4

1

logsin4θ+2logc

⟹r

8

=c

8

sin4θ