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

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$\text{\large\underline{\red{Given:-}}}$

• x = sint, y = cos2t

$\text{\large\underline{\orange{To Find:-}}}$

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

Differentiation:-

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

$\text{\large\underline{\pink{Solution:-}}}$

$\begin{gathered} \: \quad \bf \: x = sint \: ....(1) \\ \\ \quad \bf \: y = cos2t....(2)\end{gathered}$

Double Differentiating equation (1) with respect to t

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

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

Double Differentiating equation (2) with respect to t

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

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

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

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

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

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

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

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

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

Put the given values of x in equation we get

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

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

$\text{\large\underline{\red{Additional information:-}}}$

Formulas of cosx:-

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

• $\begin{gathered}\\ \bf{(2) cos2x=1-2sin^{2}x}\end{gathered}$

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

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

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