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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
$\text{ If }\text{ x } =\cos\text{t} (3-2\cos^2\text{t)}\ \text{and}\ \text{y}=\sin\text{t}(3-2\sin\text{ t}),\text{ find the value of}\ \frac{\text{dy}}{\text{dx}} = \text{ at t}=\frac{\pi}{\text{4}}.$
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Answer

$x = 3 \cos t – 2 \cos^3t$
$\therefore$$\frac{\text{dy}}{\text{dx}}=-3\sin\text{t}+6\cos^2\text{t}\sin\text{t}$
${\text{y}}=3\sin\text{t}-2\sin^3\text{t}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=-3\cos\text{t}-6\sin^2\text{t}\cos\text{t}$
$\frac{\text{dy}}{\text{dx}} =\frac{3\cos \text{t}(1-2\sin^2\text{t})}{3\sin\text{t}(-1+2\cos^2\text{t})} = \cot\text{t}$
$\text{at t}=\frac{\pi}{4},\frac{\text{dy}}{\text{dx}}=1$

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