Question

If x=a sin pt and y=b cos pt ,find the value of d²y/dx² at t=0 . Also show that (a²-x²)y d²y/dx² + b²=0​

Answer 1

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

Given that

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

and

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

On differentiating equation (1) w. r. t. t, we get

$\rm :\longmapsto\:\dfrac{d}{dt}x = \dfrac{d}{dt}a \: sinpt$

$\rm :\longmapsto\:\dfrac{dx}{dt} =a \dfrac{d}{dt} \: sinpt$

$\rm :\longmapsto\:\dfrac{dx}{dt} =a \: cospt \dfrac{d}{dt} \: pt$

$\rm :\longmapsto\:\dfrac{dx}{dt} =a \: p \: cospt - - - (3)$

Now,

On differentiating equation (2), w. r. t. t, we get

$\rm :\longmapsto\:\dfrac{d}{dt}y = \dfrac{d}{dt}b \: cospt$

$\rm :\longmapsto\:\dfrac{dy}{dt} =b \dfrac{d}{dt} \: cospt$

$\rm :\longmapsto\:\dfrac{dy}{dt} =b( - sinpt) \dfrac{d}{dt} \:pt$

$\rm :\longmapsto\:\dfrac{dy}{dt} =-b \: p \: sinpt - - - (4)$

Now,

We know,

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

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

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{ - \: bp \: sinpt}{ \: \: \: ap \: cospt \: \: \: }$

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

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

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

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

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

$\rm :\longmapsto\:\dfrac{ {d}^{2} y}{d {x}^{2} } = - \: \dfrac{bp}{a} \: {sec}^{2}pt \: \dfrac{dt}{dx}\:$

$\rm :\longmapsto\:\dfrac{ {d}^{2} y}{d {x}^{2} } = - \: \dfrac{bp}{a} \: {sec}^{2}pt \: \: \dfrac{1}{ap \: cospt}\:$

$\rm :\longmapsto\:\dfrac{ {d}^{2} y}{d {x}^{2} } = - \: \dfrac{b}{ {a}^{2} } \: {sec}^{2}pt \: \: \dfrac{1}{cospt}\:$

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

[ using (2) ]

$\rm :\longmapsto\:\dfrac{ {d}^{2} y}{d {x}^{2} } = - \: \dfrac{ {b}^{2} }{ {a}^{2} } \: \: \: \dfrac{1}{(1 - sin^{2} pt)}\:\dfrac{1}{y}$

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

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

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

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

$\rm :\longmapsto\:( {a}^{2} - {x}^{2})y \dfrac{ {d}^{2} y}{d {x}^{2} } + {b}^{2} = 0$

Hence, Proved.

Now,

When t = 0,

$\rm :\longmapsto\:x = a \: sinpt = a \times 0 = 0$

and

$\rm :\longmapsto\:y = b \: cospt = b \times 1 = b$

Now, Substitute these values of x and y in

$\rm :\longmapsto\:( {a}^{2} - {x}^{2})y \dfrac{ {d}^{2} y}{d {x}^{2} } + {b}^{2} = 0$

we get

$\rm :\longmapsto\:( {a}^{2} - {0}^{2})b \dfrac{ {d}^{2} y}{d {x}^{2} } + {b}^{2} = 0$

$\rm :\longmapsto\:b{a}^{2}\dfrac{ {d}^{2} y}{d {x}^{2} } + {b}^{2} = 0$

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

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

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