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MHT CET 2020 Math Solved Paper

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Question : 49 of 50

Marks: +1, -0

If

x=a(\cos t+t \sin t)

y=a(\sin t-t \cos t)

\; \frac{d^{2} y}{d x^{2}}

is

$\sec^{3} t$
$a t \sec^{3} t$
$\; \frac{\sec^{3} t}{a t}$
$\sec^{2} t$

Solution:

It is given that

x=a(\cos t+t \sin t)

and

y=(\sin t-t \cos t) .

Therefore,

\; \frac{d x}{d t} = a[-\sin t + \sin t + t \cos t] = a t \cos t

\; \frac{d y}{d t} = a[\cos t - \{\cos t - t \sin t\}] = a t \sin t

\therefore \; \frac{d y}{d x} = \; \frac{ \left( \frac{d y}{d t} \right) }{ \left( \; \frac{d x}{d t} \right) } = \; \frac{a t \sin t}{a t \cos t} = \tan t

Then,

\; \frac{d^{2} y}{d x^{2}} = \; \frac{d}{d x} \left( \; \frac{d y}{d x} \right)

= \; \frac{d}{d x} \left( \tan t \right)

= \; \frac{d}{d t} \left( \tan t \right) \; \frac{d t}{d x}

= \sec^{2} t \cdot \; \frac{d t}{d x} = \sec^{2} t \cdot \; \frac{1}{a t \cos t}

= \; \frac{\sec^{3} t}{a t}

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