Question

If x=asinthita and y=bcosthita, then d^2y_dx^2 is equal to

Answer 1

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

Given that,

$\rm :\longmapsto\:x = a \: sin \theta$

and

$\rm :\longmapsto\:y \: = \: b \: cos\theta$

So,

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

We know,

$\boxed{ \tt{ \: \dfrac{d}{dx}sinx = cosx \: }}$

So, we get

$\rm \implies\:\boxed{ \tt{ \: \dfrac{dx}{d\theta} = a \: cos\theta \: }}$

Also,

$\rm :\longmapsto\:\dfrac{dy}{d\theta} = \dfrac{d}{d\theta}b \: cos\theta$

We know,

$\boxed{ \tt{ \: \dfrac{d}{dx}cosx = - \: sinx \: }}$

So, we get

$\rm \implies\:\boxed{ \tt{ \: \dfrac{dy}{d\theta} = - \: b \: sin\theta \: }}$

Therefore,

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{dy}{d\theta} \div \dfrac{dx}{d\theta}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = - \: b \: sin\theta \: \div \: a \: cos\theta$

$\bf :\longmapsto\:\dfrac{dy}{dx} = - \: \dfrac{b}{a}tan\theta$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{ {d}^{2}y }{d {x}^{2} } = - \: \dfrac{b}{a} \: {sec}^{2}\theta \: \dfrac{d\theta}{dx}$

$\rm :\longmapsto\:\dfrac{ {d}^{2}y }{d {x}^{2} } = - \: \dfrac{b}{a} \: \times \dfrac{1}{ {cos}^{2} \theta} \times \dfrac{1}{a \: cos\theta}$

$\rm :\longmapsto\:\dfrac{ {d}^{2}y }{d {x}^{2} } = - \: \dfrac{b}{ {a}^{2} \: {cos}^{3}\theta} \:$

As it is given that,

$\rm :\longmapsto\:y = b \: cos \theta\rm \implies\:cos\theta = \dfrac{y}{b}$

So, on substituting the value, we get

$\rm :\longmapsto\:\dfrac{ {d}^{2} y}{d {x}^{2} } = - \: \dfrac{b}{ {a}^{2} \times {\bigg[\dfrac{y}{b} \bigg]}^{3} }$

$\rm :\longmapsto\:\dfrac{ {d}^{2} y}{d {x}^{2} } = - \: \dfrac{b \times {b}^{3} }{ {a}^{2} \times {y}^{3} }$

$\rm :\longmapsto\:\dfrac{ {d}^{2} y}{d {x}^{2} } = - \: \dfrac{{b}^{4} }{ {a}^{2} \times {y}^{3} }$

$\red{\rm \implies\:\boxed{ \tt{ \: \:\dfrac{ {d}^{2} y}{d {x}^{2} } \: = \: - \: \dfrac{{b}^{4} }{ {a}^{2} {y}^{3} } }}}$

More to know :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$