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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

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

$\text{x}=\cos\text{t}(3-2\cos^2\text{t})\text{ and }\text{y}\sin\text{t}(3-2\sin^2\text{t})$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=-\sin\text{t}(3-2\cos^2\text{t})+\cos\text{t}(4\cos\text{t}\sin\text{t})$ and $\frac{\text{dy}}{\text{dx}}=\cos\text{t}(3-2\sin^2\text{t})+\sin\text{t}(-4\sin\text{t}\cos\text{t})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\sin\text{t}+6\sin\text{t}\cos^2\text{t}$ and $ \frac{\text{dy}}{\text{dx}}=3\cos\text{t}-6\sin^2\text{t}\cos\text{t}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\sin\text{t}+(1-2\cos^2\text{t})$ and $\frac{\text{dy}}{\text{dt}}=3\cos\text{t}(1-2\sin^2\text{t})$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=3\sin\text{t}\cos2\text{t}\cos2\text{t}$ and $\frac{\text{dy}}{\text{dt}}=3\cos\text{t}(1-2\sin^2\text{t})$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dx}}}{\frac{\text{dx}}{\text{dt}}}=\frac{3\cos\text{t}\cos2\text{t}}{3\sin\text{t}\cos2\text{t}}=\cot\text{t}$
Now, $\Big(\frac{\text{dy}}{\text{dt}}\Big)_{\text{t}=\frac{\pi}{4}}=\cot\frac{\pi}{4}=1$

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