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Differentiation — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDifferentiation5 Marks
Question
If $\text{x}=\text{a}\sin2\text{t}(1+\cos 2\text{t})$ and $\text{y}=\text{b}\cos\text{t}(1-\cos2\text{t}),$ show that at $\text{t}=\frac{\pi}{4},\frac{\text{dy}}{\text{dx}}=\frac{\text{b}}{\text{a}}\text{ t}=\frac{\pi}{4},\frac{\text{dy}}{\text{dx}}=\frac{\text{b}}{\text{a}}$
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Answer

Consider the given functions,
$\text{x}=\text{a}\sin(2\text{t})(1+\cos2\text{t})\text{ and y}=\text{b}\cos2\text{t}(1-\cos2\text{t})$
Rewriting the above function, we have,
$\text{x}=\text{a}\sin2\text{t}+\frac{\text{a}}{2}\sin4\text{t}$
Differentiating the above function w.r.t. 't', we have,
$\frac{\text{dx}}{\text{dx}}=2\text{a}\cos2\text{t}+2\text{a}\cos4\text{t}\ .....(\text{i})$
$\text{y}=\text{b}\cos2\text{t}(1-\cos2\text{t})$
$\text{y}=\text{b}\cos2\text{t}-\text{b}\cos^22\text{t}$
$\frac{\text{dy}}{\text{dt}}=-2\text{b}\sin2\text{t}+2\text{b}\cos2\text{t}\sin2\text{t} \\ =-2\text{b}\sin2\text{t}+\text{b}\sin4\text{t}\ .....(\text{ii})$
From (1) and (2),
$\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{-2\text{b}\sin2\text{t}+\text{b}\sin4\text{t}}{2\text{a}\cos2\text{t}+2\text{a}\cos4\text{t}}$
$\therefore\frac{\text{dy}}{\text{dx}}\Big|_{\frac{\pi}{4}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}\Bigg|_{\text{t}=\frac{\pi}{4}}=\frac{-2\text{b}}{-2\text{a}}=\frac{\text{b}}{\text{a}}$

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