Question

show that angle between the normal at any prove that (r,theta) on the curve r^n=a^ncosntheta and the initial line is (n+1)theta​

Answer 1

Answer:

Step-by-step explanation:

The given curve is,

$\rm{{r}^{n}={a}^{n}\cdot\,cos(n\theta)}$

Now, the angle between radius vector and tangent is φ, the angle between the initial line and the radius vector is θ and the angle between the tangent and the initial line is Ψ.

So, the angle between the initial line and normal will be  $\rm{\psi-\dfrac{\pi}{2}}$

Now,

$\rm{\ln\left({r}^{n}\right)=\ln\left({a}^{n}\cdot\,cos(n\theta)\right)}$

$\rm{\implies\,n\ln(r)=n\ln(a)+\ln(cos(n\theta))}$

Differentiating both sides w.r.t θ,

$\rm{\implies\,\dfrac{n}{r}\cdot\dfrac{dr}{d\theta}=0+n\cdot\dfrac{-sin(n\theta)}{cos(n\theta)}}$

$\rm{\implies\,\dfrac{1}{r}\cdot\dfrac{dr}{d\theta}=-\,tan(n\theta)}$

$\rm{\implies\,r\cdot\dfrac{d\theta}{dr}=-\,cot(n\theta)}$

We know,

$\boxed{\sf{tan(\phi)=r\cdot\dfrac{d\theta}{dr}}}$

So,

$\rm{\implies\,tan(\phi)=-\,cot(n\theta)}$

$\rm{\implies\,tan(\phi)=tan\left(\dfrac{\pi}{2}+n\theta\right)}$

$\rm{\implies\,\phi=\dfrac{\pi}{2}+n\theta\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(1)}$

Also, we know,

$\boxed{\bf{\psi=\theta+\phi}}$

So,

$\rm{\implies\,\psi=\theta+\dfrac{\pi}{2}+n\theta}$

$\rm{\implies\,\psi-\dfrac{\pi}{2}=\theta+n\theta}$

$\rm{\implies\,\psi-\dfrac{\pi}{2}=(n+1)\theta}$

Hence, proved

Answer 2

Answer:

Step-by-step explanation: