Question

Topic Differentiation

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If x = a cost and y = b sint, prove that d^2y/dx^2 = - b^4/a^2y^3​

Answer 1

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

Given that,

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

and

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

So,

Differentiating (1), w. r. t. t, we get

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

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

We know,

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

So, using this, we get

$\rm :\longmapsto\:\dfrac{dx}{dt}\: = - \: a \: sint$

Now,

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

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

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

We know,

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

So, using this, we get

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

Now,

We know that,

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

$\rm \:  =  \:  \: \dfrac{b \: cost}{ - a \: sint}$

$\rm \:  =  \:  \: - \: \dfrac{b }{ \: a \: } \: cot \: t$

$\bf\implies \:\dfrac{dy}{dx} =  \:  \: - \: \dfrac{b }{ \: a \: } \: cot \: t$

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

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

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

$\rm :\longmapsto\:\dfrac{ {d}^{2}y }{ {dx}^{2} } = \: \dfrac{b}{a} {cosec}^{2}t\times \dfrac{1}{ - a \: sint}$

$\rm :\longmapsto\:\dfrac{ {d}^{2}y }{ {dx}^{2} } = - \: \dfrac{b}{ {a}^{2} } \times \dfrac{1}{ {sin}^{2} t} \times \dfrac{1}{\: sint}$

$\rm :\longmapsto\:\dfrac{ {d}^{2}y }{ {dx}^{2} } = - \: \dfrac{b}{ {a}^{2} } \times \dfrac{1}{ {sin}^{3} t}$

As it is given that,

$\red{\rm :\longmapsto\:y \: = \: b \: sint \: \bf\implies \:sint = \dfrac{y}{b} }$

So, on substituting this value, we get

$\rm :\longmapsto\:\dfrac{ {d}^{2}y }{ {dx}^{2} } = - \: \dfrac{b}{ {a}^{2} } \times \dfrac{1}{ {\bigg(\dfrac{y}{b} \bigg) }^{3} }$

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

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

Hence, Proved

Additional Information :-

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

$\boxed{ \bf{ \: \dfrac{d}{dx}cotx = { - \: cosec}^{2}x}}$

$\boxed{ \bf{ \: \dfrac{d}{dx}cosecx = - \: cosecx \: cotx}}$

$\boxed{ \bf{ \: \dfrac{d}{dx}secx = \: secx \: tanx}}$

$\boxed{ \bf{ \: \dfrac{d}{dx}k = 0}}$

$\boxed{ \bf{ \: \dfrac{d}{dx}x = 1}}$

$\boxed{ \bf{ \: \dfrac{d}{dx} \sqrt{x} = \frac{1}{2 \sqrt{x} } }}$

$\boxed{ \bf{ \: \dfrac{d}{dx}logx = \frac{1}{x}}}$

Answer 2

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

Given that,

$\rm :\longmapsto\:x \: = \: a \: cost \: - - - (1) \\ \\ and \\ \\ \rm :\longmapsto\:y \: = \: b \: sint \: - - - (2)$

So,

Differentiating (1), w. r. t. t, we get

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

We know,

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

So, using this, we get

$\rm :\longmapsto\:\dfrac{dx}{dt}\: = - \: a \: sint$

Now,

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

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

We know,

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

So, using this, we get

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

Now,

We know that,

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{dy}{dt} \div \dfrac{dx}{dt} \\ \\ \rm \: = \: \: \dfrac{b \: cost}{ - a \: sint} \\ \\ \rm \: = \: \: - \: \dfrac{b }{ \: a \: } \: cot \: t \\ \\ \bf\implies \:\dfrac{dy}{dx} = \: \: - \: \dfrac{b }{ \: a \: } \: cot \: t$

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

$\rm :\longmapsto\:\dfrac{d}{dx} \: \dfrac{dy}{dx} = \: \: - \: \dfrac{b }{ \: a \: } \dfrac{d}{dx} \: cot \: t \\ \\ \rm :\longmapsto\:\dfrac{ {d}^{2}y }{ {dx}^{2} } = - \: \dfrac{b}{a} {( - cosec}^{2}t)\dfrac{dt}{dx} \\ \\ \rm :\longmapsto\:\dfrac{ {d}^{2}y }{ {dx}^{2} } = \: \dfrac{b}{a} {cosec}^{2}t\times \dfrac{1}{ - a \: sint} \\ \\ \rm :\longmapsto\:\dfrac{ {d}^{2}y }{ {dx}^{2} } = - \: \dfrac{b}{ {a}^{2} } \times \dfrac{1}{ {sin}^{2} t} \times \dfrac{1}{\: sint} \\ \\ \rm :\longmapsto\:\dfrac{ {d}^{2}y }{ {dx}^{2} } = - \: \dfrac{b}{ {a}^{2} } \times \dfrac{1}{ {sin}^{3} t}$

As it is given that,

$\red{\rm :\longmapsto\:y \: = \: b \: sint \: \bf\implies \:sint = \dfrac{y}{b} }$

So, on substituting this value, we get

$\rm :\longmapsto\:\dfrac{ {d}^{2}y }{ {dx}^{2} } = - \: \dfrac{b}{ {a}^{2} } \times \dfrac{1}{ {\bigg(\dfrac{y}{b} \bigg) }^{3} } \\ \\ \rm :\longmapsto\:\dfrac{ {d}^{2}y }{ {dx}^{2} } = - \: \dfrac{b}{ {a}^{2} } \times \dfrac{ {b}^{3} }{ {y}^{3} } \\ \\ \rm :\longmapsto\:\dfrac{ {d}^{2}y }{ {dx}^{2} } = - \: \dfrac{ {b}^{4} }{ \: \: {a}^{2} \: \: {y}^{3} \: \: }$

Hence, Proved .