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

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Question

If  \[x = \frac{\sin^3 t}{\sqrt{\cos 2 t}}, y = \frac{\cos^3 t}{\sqrt{\cos t 2 t}}\] , find\[\frac{dy}{dx}\] ?

 

Solution

\[\text { We have, x } = \frac{\sin^3 t}{\sqrt{\cos2t}} \text { and y } = \frac{\cos^3 t}{\sqrt{\cos2t}}\]
\[ \Rightarrow \frac{dx}{dt} = \frac{d}{dt}\left[ \frac{\sin^3 t}{\sqrt{\cos2t}} \right]\]
\[ \Rightarrow \frac{dx}{dt} = \frac{\sqrt{\cos2t}\frac{d}{dt}\left( \sin^3 t \right) - \sin^3 t\frac{d}{dt}\sqrt{\cos2t}}{\cos2t} \left[ \text { Using quotient rule } \right]\]
\[ \Rightarrow \frac{dx}{dt} = \frac{\sqrt{\cos2t}\left( 3 \sin^2 t \right)\frac{d}{dt}\left( \sin t \right) - \sin^3 t \times \frac{1}{2\sqrt{\cos2t}}\frac{d}{dt}\left( \cos 2t \right)}{\cos2t} \]
\[ \Rightarrow \frac{dx}{dt} = \frac{3\sqrt{\cos2t}\left( \sin^2 t \cos t \right) - \frac{\sin^3 t}{2\sqrt{\cos2t}}\left( - 2 \sin2t \right)}{\cos 2t}\]
\[ \Rightarrow \frac{dx}{dt} = \frac{3\cos2t \sin^2 t \cos t + \sin^3 t \sin2t}{\cos2t\sqrt{\cos2t}}\]
\[Now, \frac{dy}{dt} = \frac{d}{dt}\left[ \frac{\cos^3 t}{\sqrt{\cos2t}} \right]\]
\[ \Rightarrow \frac{dy}{dt} = \frac{\sqrt{\cos2t}\frac{d}{dt}\left( \cos^3 t \right) - \cos^3 t\frac{d}{dt}\sqrt{\cos2t}}{\cos2t} \left[ \text { Using quotient rule } \right]\]
\[ \Rightarrow \frac{dy}{dt} = \frac{\sqrt{\cos2t}\left( 3 \cos^2 t \right)\frac{d}{dt}\left( \cos t \right) - \cos^3 t \times \frac{1}{2\sqrt{\cos2t}}\frac{d}{dt}\left( \cos 2t \right)}{\cos2t} \]
\[ \Rightarrow \frac{dy}{dt} = \frac{3\sqrt{\cos2t} \cos^2 t \left( - \sin t \right) - \frac{\cos^3 t}{2\sqrt{\cos2t}}\left( - 2 \sin2t \right)}{\cos 2t}\]
\[ \Rightarrow \frac{dy}{dt} = \frac{- 3\cos2t \cos^2 t \sin t + \cos^3 t \sin2t}{\cos2t\sqrt{\cos2t}}\]
\[ \therefore \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{- 3\cos2t \cos^2 t \sin t + \cos^3 t \sin2t}{\cos2t\sqrt{\cos2t}} \times \frac{\cos2t\sqrt{\cos2t}}{3\cos2t \sin^2 t \cos t + \sin^3 t \sin2t}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{- 3\cos2t \cos^2 t \sin t + \cos^3 t \sin2t}{3\cos2t \sin^2 t \cos t + \sin^3 t \sin2t}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{\sin t \cos t\left[ - 3\cos2t \cos t + 2 \cos^3 t \right]}{\sin t \cos t\left[ 3\cos2t \sin t + 2 \sin^3 t \right]}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{\left[ - 3\left( 2 \cos^2 t - 1 \right)\cos t + 2 \cos^3 t \right]}{\left[ 3\left( 1 - 2 \sin^2 t \right)\sin t + 2 \sin^3 t \right]} \binom{\cos2t = 2 \cos^2 t - 1}{\cos2t = 1 - 2 \sin^2 t}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{- 4 \cos^3 t + 3\cos t}{3\ sint - 4 \sin^3 t}\]

\[ \Rightarrow \frac{dy}{dx} = \frac{- \cos3t}{\sin3t} \binom{\cos3t = 4 \cos^3 t - 3\cos t}{\sin3t = 3\sin t - 4 \sin^3 t}\]
\[ \therefore \frac{dy}{dx} = - \cot3t\]

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