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Solution for If X = 3 C O T − 2 Cos 3 T , Y = 3 Sin T − 2 Sin 3 T , Find D 2 Y D X 2 ? - CBSE (Science) Class 12 - Mathematics

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Question

If \[x = 3 \ cot - 2 \cos^3 t, y = 3\sin t - 2 \sin^3 t,\] find \[\frac{d^2 y}{d x^2} \] ?

Solution

We have,

\[x = 3\cos t - 2 \cos^3 t\]

\[ \Rightarrow \frac{dx}{dt} = 3\left( - \sin t \right) - 6 \cos^2 t\left( - \sin t \right)\]

\[ = - 3\sin t + 6\sin t \cos^2 t\]

Also,

\[y = 3\sin t - 2 \sin^3 t\]

\[ \Rightarrow \frac{dy}{dt} = 3\cos t - 6 \sin^2 t \cos t\]

Now,

\[\frac{dy}{dx} = \frac{\left( \frac{dy}{dt} \right)}{\left( \frac{dx}{dt} \right)}\]

\[ = \frac{3\cos t - 6 \sin^2 t \cos t}{- 3\sin t + 6\sin t \cos^2 t}\]

\[ = \frac{3\cos t\left( 1 - 2 \sin^2 t \right)}{3\sin t\left( - 1 + 2 \cos^2 t \right)}\]

\[ = \frac{\cot t\left( \cos2t \right)}{\left( \cos2t \right)}\]

\[ = \cot t\]

\[So, \frac{d^2 y}{d x^2} = \frac{d}{dx}\left( \frac{dy}{dx} \right)\]

\[ = \frac{d}{dx}\left( \cot t \right)\]

\[ = - {cosec}^2 t \frac{dt}{dx}\]

\[ = \frac{- {cosec}^2 t}{\left( \frac{dx}{dt} \right)}\]

\[ = \frac{- {cosec}^2 t}{- 3\sin t + 6\sin t \cos^2 t}\]

\[ = \frac{- {cosec}^2 t}{- \sin t\left( 1 - 2 \cos^2 t \right)}\]

\[ = \frac{{cosec}^3 t}{\left( - \cos 2t \right)}\]

\[ = \frac{- {cosec}^3 t}{\cos 2t}\]

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Solution for question: If X = 3 C O T − 2 Cos 3 T , Y = 3 Sin T − 2 Sin 3 T , Find D 2 Y D X 2 ? concept: Simple Problems on Applications of Derivatives. For the courses CBSE (Science), CBSE (Commerce), PUC Karnataka Science, CBSE (Arts)
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