Question

Find ${\displaystyle \frac{d^{2}y}{{dx}^2}}$ of the following : x = a cos θ, y = b sin θ at θ = ${\displaystyle \frac{\pi }{4}}$.

Answer 1

x = a cos θ, y = b sin θ

Differentiating x and y w.r.t. θ, we get,

${\displaystyle \frac{dx}{d\theta }=\frac{d}{d\theta }\left(a\cos \theta \right)}$

${\displaystyle \frac{dx}{d\theta }=a\frac{d}{d\theta }\left(\cos \theta \right)}$

${\displaystyle \frac{dx}{d\theta }=a(–\sin \theta )}$

${\displaystyle \frac{dx}{d\theta }=– a\sin \theta }$                          ...(1)

and

${\displaystyle \frac{dy}{d\theta }=\frac{d}{d\theta }\left(b\sin \theta \right)}$

${\displaystyle \frac{dy}{d\theta }=b\frac{d}{d\theta }\left(\sin \theta \right)}$
${\displaystyle \frac{dy}{d\theta }=b\cos \theta }$

∴ ${\displaystyle \frac{dy}{dx}=\frac{\left(\frac{dy}{d\theta }\right)}{\left(\frac{dx}{d\theta }\right)}}$

${\displaystyle \frac{dy}{dx}=\frac{b\cos \theta }{–a\sin \theta }}$

${\displaystyle \frac{dy}{dx}=(–\frac{b}{a})\cot \theta }$

and

${\displaystyle \frac{d^{2}y}{{dx}^2}=\frac{d}{dx}\left[(–\frac{b}{a})\cot \theta \right]}$

${\displaystyle \frac{d^{2}y}{{dx}^2}=–\frac{b}{a}.\frac{d}{d\theta }\left(\cot \theta \right)\times \frac{d\theta }{dx}}$

${\displaystyle \frac{d^{2}y}{{dx}^2}=(–\frac{b}{a})(–\cos e{c}^2\theta )\times \frac{1}{\left(\frac{dx}{d\theta }\right)}}$

${\displaystyle \frac{d^{2}y}{{dx}^2}=\left(\frac{b}{a}\right)\cos e{c}^2\theta \times \frac{1}{–a\sin \theta }}$      ..[By (1)]

${\displaystyle \frac{d^{2}y}{{dx}^2}=(–\frac{b}{a^2})\cos e{c}^3\theta }$

∴ ${\displaystyle {\left(\frac{d^{2}y}{{dx}^2}\right)}_{\text{at}\theta =\frac{\pi }{4}}=(–\frac{b}{a^2})\cos e{c}^3\frac{\pi }{4}}$

= ${\displaystyle \frac{–b}{a^2}\times {\left(\sqrt{2}\right)}^3}$

= ${\displaystyle –\frac{2\sqrt{2}b}{a^2}}$