Question

x=acostheta, y=bsintheta then find d^2y/dx^2=​

Answer 1

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

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

and

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

$\large\underline{\sf{To\:Find - }}$

$\rm :\longmapsto\:\dfrac{ {d}^{2} y}{ {dx}^{2} }$

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}}  \end{gathered}$

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

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

$\boxed{ \sf{ \:\dfrac{d}{dx}k \: f(x) = \: k \: \dfrac{d}{dx} \: f(x)}}$

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

Given that

$\rm :\longmapsto\:x = a \: cos\theta - - - (1)$

and

$\rm :\longmapsto\:y = b \: sin\theta - - - (2)$

Now, Differentiate (1), w. r. t. θ, we get

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

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

Now, Differentiate (2) w. r. t. θ, we get

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

$\rm :\longmapsto\:\dfrac{dy}{d\theta } = \: b \: cos\theta - - - - (4)$

So,

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

On substituting the values from equation (3) and (4), we get

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

$\rm :\longmapsto\:\dfrac{dy}{dx} = - \: \dfrac{ b \: }{\: a \: } cot\theta$

Now, Differentiate both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}\dfrac{dy}{dx} = - \: \dfrac{d}{dx}\dfrac{ b \: }{\: a \: } cot\theta$

$\rm :\longmapsto\:\dfrac{ { \: d}^{2} y \: }{ {dx}^{2} } = - \dfrac{b}{a}( - {cosec}^{2}\theta ) \: \dfrac{d\theta }{dx}$

$\rm :\longmapsto\:\dfrac{ { \: d}^{2} y \: }{ {dx}^{2} } = \dfrac{b}{a {sin}^{2} \theta } \: \: \dfrac{ - 1}{asin\theta }$

$\red{\bigg \{ \because \:\dfrac{dx}{d\theta } = \: - a \: sin\theta \bigg \}}$

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

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

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

Answer 2

Answer:

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

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

and

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

$\large\underline{\sf{To\:Find - }}$

$\rm :\longmapsto\:\dfrac{ {d}^{2} y}{ {dx}^{2} }$

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}}  \end{gathered}$

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

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

$\boxed{ \sf{ \:\dfrac{d}{dx}k \: f(x) = \: k \: \dfrac{d}{dx} \: f(x)}}$

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

Given that

$\rm :\longmapsto\:x = a \: cos\theta - - - (1)$

and

$\rm :\longmapsto\:y = b \: sin\theta - - - (2)$

Now, Differentiate (1), w. r. t. θ, we get

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

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

Now, Differentiate (2) w. r. t. θ, we get

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

$\rm :\longmapsto\:\dfrac{dy}{d\theta } = \: b \: cos\theta - - - - (4)$

So,

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

On substituting the values from equation (3) and (4), we get

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

$\rm :\longmapsto\:\dfrac{dy}{dx} = - \: \dfrac{ b \: }{\: a \: } cot\theta$

Now, Differentiate both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}\dfrac{dy}{dx} = - \: \dfrac{d}{dx}\dfrac{ b \: }{\: a \: } cot\theta$

$\rm :\longmapsto\:\dfrac{ { \: d}^{2} y \: }{ {dx}^{2} } = - \dfrac{b}{a}( - {cosec}^{2}\theta ) \: \dfrac{d\theta }{dx}$

$\rm :\longmapsto\:\dfrac{ { \: d}^{2} y \: }{ {dx}^{2} } = \dfrac{b}{a {sin}^{2} \theta } \: \: \dfrac{ - 1}{asin\theta }$

$\red{\bigg \{ \because \:\dfrac{dx}{d\theta } = \: - a \: sin\theta \bigg \}}$

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

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

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