NIMCET 2013 Question Paper with solutions

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Question : 38 of 120

Marks: +1, -0

If x = a cos t, y = b sint t, then

{d^2y}/{dx^2}

is :

- $b^4/{a^2y^3}$
$b^4/{a^2y^3}$
$b/{ay^4}$
- $a^4/{bx^3}$

Solution:

Given,

x=a\cos t

and

y=b\sin t

∴{dy}/{dx}={dy{/}dt}/{dx{/}dt}={-b}/{a}{\cot}t

Let

u={dy}/{dx}=\cot t

∴{d^{2}y}/{dx^{2}}={du}/{dx}={du}/{dt}×{dt}/{dx}=({b}/{a}{\cosec}^{2}t)×({-1}/{a\sin t})

={-b}/{a^{2}\sin^{3}t}={-b^{4}}/{a^{2}y^{3}}

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