Question

If x=a(t-sint),y=b(1-cost) find d2y/dx2 at t=π/2

Answer 1

Question :-

$\sf{if \: x \: = a \: (t \: - \sin \: t) \: } \\ \: \: \: \: \sf{ y \: = b(1 - \cos \: t)} \\ \\ \sf{then \: find \: \frac{ {d}^{2} y}{ {d}^{2} \: x } \: at \: t \: = \frac{ \pi}{2} }$

Answer :-

$\boxed{ \pink{\sf{ \frac{ {d}^{2} y}{ {d}^{2} x} } \: }\: at \: \: t \: = \frac{ \pi}{2} \: = 0}$

Explanation :-

We have ,

$\rightarrow \sf{ x \: = a(t \: - \sin \: t)} \\ \\ \sf{ differentiate \: with \: respect \: to \: t} \\ \\ \rightarrow \: \bf{ \frac{dx}{dt} = a(1 - \cos \: t)} \: \: \: \\ \\ \pink{\sf{ again \: differentiate \: with \: respect \: to \: t}} \\ \\ \rightarrow \: \sf{\frac{ {d}^{2}x }{d {t}^{2} } = a \: (0 + \sin \: t)} \: ......(1)$

and,

$\rightarrow \: \sf{y \: = b \: (1 - \cos \: t)} \\ \\ \green{\sf{differentiate \: with \: respect \: to \: t}} \\ \\ \rightarrow \: \sf{ \frac{dy}{dt} = b \: (0 + \sin \: t)} \\ \\ \rightarrow \: \sf{ \frac{dy}{dt} = b \: \sin \: t} \: \: \: \\ \\ \red{\sf{ again \: differentiate \: with \: respect \: to \: t}} \\ \\ \rightarrow \: \sf{\frac{ {d}^{2} y}{d {x}^{2} } = b \: \cos \: t} \: \: \: .......(2)$

Now ,divided (2) by (1)

$\rightarrow \: \sf{ \frac{ \frac{ {d}^{2}y }{d {t}^{2} } }{ \frac{ {d}^{2} x}{d {t}^{2} } } = \frac{b \: \cos \: t }{a \: \sin \: t } } \\ \\ \: \sf{ \frac{ {d}^{2} y}{ {d}^{2} x} \: \: \: \: at \: \: t \: = \frac{ \pi}{2}} \: \\ \\ \rightarrow \: \sf{ \frac{b \cos( \frac{ \pi}{2} ) }{a \: \sin( \frac{ \pi}{2} ) }} \\ \\ \rightarrow \: 0$

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Answer 2

Answer:

if x=1-sint, y=1-cost, value of d2y/dx2 at(π/2) will be