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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

ConceptSimple Problems on Applications of Derivatives

#### 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 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.
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