Question

3. If x = sint, y = cos 2t, then the value of d^2y/dx^2 when x=1/root3 is
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Answer 1

Given :

$\pink\star$ x=sint , y=cos2t

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To find :

$\star\sf{\dfrac{d^{2} y}{dx^2} =? \: \: \: at \: x =\dfrac{1}{\sqrt{3}} }$

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Differentiation :

$\\ \star \sf \frac{dy}{dx} sinx = cosx \\ \\ \star \sf \: \frac{dy}{dx} cosx = - sinx$

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Solution :

$\: \quad \bf \: x = sint \: ....(1) \\ \\ \quad \bf \: y = cos2t....(2)$

➡ Double Differentiating equation (1) with respect to t

$\implies \sf \: x = sint \\ \\ \implies \sf \: \frac{dx}{dt} = \frac{d}{dt} sint \\ \\ \implies \sf \: \frac{dx}{dt} = cost$

$\\ \implies \sf \: \frac{dx}{dt} = cost \\ \\ \implies \sf \: \frac{ {d}^{2}x }{ {dt}^{2} } = \frac{d}{dt} cost = - sint \\ \\ \implies \sf \: \frac{ {d}^{2} x}{ {dt}^{2} } = - sint \\ \\ \implies \ \boxed{ \sf{ \frac{ {d}^{2}x }{ {dt}^{2} } = - sint}}...(1)$

Double Differentiating equation (2) with respect to t

$\implies \sf \: y = cos2t \\ \\ \implies \sf \: \frac{dy}{dt} = \frac{d}{dt} cos2t \\ \\ \implies \sf \: \frac{dy}{dt} = - sin2t \: \times 2 \\ \\ \implies \sf \: \frac{dy}{dt} = - 2sin2t \\ \\ \implies \sf \: \frac{ {d}^{2}y }{ {dt}^{2} } = - 2 \frac{d}{dt} sin2t \\ \\ \implies \sf \: \frac{ {d}^{2}y }{ {dt}^{2} } = - 2cos2t \times 2 \\ \\ \implies \sf \: \frac{ {d}^{2}y }{ {dt}^{2} } = - 4cos2t \\ \\ \implies \boxed{ \sf{ \frac{ {d}^{2}y }{ {dt}^{2} } = - 4cos2t}}......(2)$

Dividing both equations (1) and (2) we get

$\\ \implies \sf \: \frac{ \frac{ {d}^{2}y }{ {dt}^{2} } }{ \frac{ {d}^{2} x}{ {dt}^{2} } } = \frac{ - 4cos2t}{ - sint}$

$\\ \implies \sf \: \frac{ \frac{ {d}^{2} y}{ \cancel{ {dt}^{2}} } }{ \frac{ {d}^{2} x}{ \cancel{ {dt}}^{2} } } = \frac{ \cancel - 4cos2t}{ \cancel - sint}$

$\\ \implies \sf \: \frac{ {d}^{2}y }{ {dx}^{2} } = \frac{ 4(1 - 2 {sin}^{2} \frac{2t}{2} ) }{sint}$

$\\ \implies \sf \: \frac{ {d}^{2} y}{ {dx}^{2} } = \frac{ 4(1 - 2 {sin}^{2}t)}{sint}$

$\\ \implies \sf \: \frac{ {d}^{2} y}{ {dx}^{2} } = 4 \bigg( \frac{1}{sint} - 2 \frac{ {sin}^{2} t}{sint} \bigg)$

$\\ \implies \sf \: \frac{ {d}^{2}y }{ {dx }^{2} } = 4(cosect - 2sint)$

$\\ \implies \boxed{ \sf{ \frac{ {d}^{2} y}{ {dx}^{2} } = 4(cosect - 2sint)}}$

Put the given values of x in equation we get

$\\ \implies \sf \: \frac{ {d}^{2} y}{ {dx}^{2} } = 4 \bigg(cosec \frac{1}{ \sqrt{3} } - 2sin \frac{1}{ \sqrt{3} } \bigg)$

$\\ \bigstar \boxed{ \sf{ \therefore \: \frac{ {d}^{2} y}{ {dx}^{2} } = 4 \bigg(cosec \frac{1}{ \sqrt{3} } - 2sin \frac{1}{ \sqrt{3} } \bigg) }} \: is \: the \: answer$

Additional information :

Formulas of cosx:

$\\ \bf{(1) cos2x=cos^{2}x-sin^{2}x}$

$\\ \bf{(2) cos2x=1-2sin^{2}x}$

$\\ \bf{(3) cos2x=2cos^{2}x-1}$

$\\ \bf{(4) cos2x=\dfrac{1-tan^{2}x}{1+tan^{2}x}}$