Question

Find d^2y/dx^2 if,x= acos³ ϴ, y = b sin³ ϴat ϴ= pi/4
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Answer 1

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

Given that,

$\rm :\longmapsto\:x = a \: {cos}^{3}\theta$

So,

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

$\rm :\longmapsto\:\dfrac{dx}{d\theta } = a \: \dfrac{d}{d\theta }{(cos\theta)}^{3}$

$\rm :\longmapsto\:\dfrac{dx}{d\theta } = 3a \: {cos}^{2}\theta \dfrac{d}{d\theta }{(cos\theta)}$

$\rm :\longmapsto\:\dfrac{dx}{d\theta } = 3a \: {cos}^{2}\theta( - sin\theta )$

$\rm :\longmapsto\:\dfrac{dx}{d\theta } = - 3a {cos}^{2}\theta sin\theta - - - (1)$

Further, given that

$\rm :\longmapsto\:y = b {sin}^{3}\theta$

$\rm :\longmapsto\:\dfrac{d}{d\theta }y = b \dfrac{d}{d\theta }{sin}^{3}\theta$

$\rm :\longmapsto\:\dfrac{dy}{d\theta } = b \dfrac{d}{d\theta }{(sin\theta )}^{3}$

$\rm :\longmapsto\:\dfrac{dy}{d\theta } = 3b {sin}^{2}\theta \dfrac{d}{d\theta }{(sin\theta )}$

$\rm :\longmapsto\:\dfrac{dy}{d\theta } = 3b {sin}^{2}\theta \: cos\theta$

Therefore,

$\rm :\longmapsto\:\dfrac{dy}{dx}$

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

$\rm \:  =  \: \dfrac{3b {sin}^{2} \theta cos\theta }{ - 3a {cos}^{2} \theta sin\theta }$

$\rm \:  =  \: - \dfrac{b}{a}tan\theta$

$\rm\implies \:\dfrac{dy}{dx} \: =  \: - \: \dfrac{b}{a} \: tan\theta$

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

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

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

$\rm \:  =  \: - \dfrac{b}{a} {sec}^{2}\theta \: \times \dfrac{1}{ - 3a {cos}^{2} \theta sin\theta }$

$\rm \:  =  \: \dfrac{b}{ 3{a}^{2} } \: {sec}^{4}\theta \: cosec\theta$

Thus,

$\rm\implies \:\dfrac{ {d}^{2} y}{d {x}^{2} } =  \: \dfrac{b}{ 3{a}^{2} } \: {sec}^{4}\theta \: cosec\theta$

So,

$\rm\implies \:\bigg(\dfrac{ {d}^{2} y}{d {x}^{2} }\bigg)_{\dfrac{\pi}{4} }=  \: \dfrac{b}{ 3{a}^{2} } \: {sec}^{4}\dfrac{\pi}{4} \: cosec\dfrac{\pi}{4}$

$\rm\implies \:\bigg(\dfrac{ {d}^{2} y}{d {x}^{2} }\bigg)_{\dfrac{\pi}{4} }=  \: \dfrac{b}{ 3{a}^{2} } \: {( \sqrt{2} )}^{4} \times \sqrt{2}$

$\rm\implies \:\bigg(\dfrac{ {d}^{2} y}{d {x}^{2} }\bigg)_{\dfrac{\pi}{4} }=  \: \dfrac{4 \sqrt{2} b}{ 3{a}^{2} }$

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$\begin{gathered}\large{\sf{{\underline{Formula \: Used - }}}}  \end{gathered}$

$\boxed{\tt{ \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1}}}$

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

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

$\boxed{\tt{ \dfrac{d}{dx}tanx = {sec}^{2} x}}$

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Additional Information

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