Question

find the angle between radius vector and tangent of the curve
$r {}^{n } = {a}^{n} sec(ntheta + \alpha )$
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Answer 1

Answer:

Step-by-step explanation:

The given curve is,

$\rm{{r}^{n}={a}^{n}\cdot\,sec\left(n\theta+\alpha\right)}$

The angle φ between the radius vector and tangent is given by,

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

So,

taking log both sides of the given equation,

$\rm{\implies\,n\,\ln\left(r\right)=n\,\ln(a)+\ln\left(sec\left(n\theta+\alpha\right)\right)}$

Differentiating both sides w.r.t θ,

$\rm{\implies\,n\cdot\dfrac{1}{r}\dfrac{dr}{d\theta}=0+\dfrac{1}{sec\left(n\theta+\alpha\right)}\cdot\dfrac{d}{d\theta}\big\{sec\left(n\theta+\alpha\right)\big\}}$

$\rm{\implies\,n\cdot\dfrac{1}{r}\dfrac{dr}{d\theta}=\dfrac{1}{sec\left(n\theta+\alpha\right)}\cdot\,n\cdot\,sec\left(n\theta+\alpha\right)\cdot\,tan\left(n\theta+\alpha\right)}$

$\rm{\implies\dfrac{1}{r}\dfrac{dr}{d\theta}=tan\left(n\theta+\alpha\right)}$

$\rm{\implies\,r\cdot\dfrac{d\theta}{dr}=\dfrac{1}{tan\left(n\theta+\alpha\right)}}$

$\rm{\implies\,tan(\phi)=cot\left(n\theta+\alpha\right)}$

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

$\rm{\implies\,\phi=\dfrac{\pi}{2}-n\theta-\alpha}$