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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
If x and y are connected parametrically by the equations given in Exercise without eliminating the parameter, Find $\frac{\text{dy}}{\text{dx}}.$
$\text{x}=\frac{\sin^3\text{t}}{\sqrt{\cos2\text{t}}},\text{y}\frac{\cos^3\text{t}}{\sqrt{\cos2\text{t}}}$
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Answer

The given equations are $\text{x}=\frac{\sin^3\text{t}}{\sqrt{\cos2\text{t}}}\text{ and y}=\frac{\cos^3\text{t}}{\sqrt{\cos2\text{t}}}$Then, $\frac{\text{dx}}{\text{dt}}= \frac{\text{d}}{\text{dt}}\Big[\frac{\sin^3\text{t}}{\sqrt{\cos2\text{t}}}\Big]$
$=\frac{\sqrt{\cos2\text{t}}.\frac{\text{d}}{\text{dt}}(\sin^3\text{t})-\sin^3\text{t}.\frac{\text{d}}{\text{dt}}\sqrt{\cos2\text{t}}}{\cos2\text{t}}$
$=\frac{\sqrt{\cos2\text{t}}.3\sin^2\text{t}.\frac{\text{d}}{\text{dt}}(\sin\text{t})-\sin^3\text{t}\times\frac{1}{2\sqrt{\cos2\text{t}}}.\frac{\text{d}}{\text{dt}}(\cos2\text{t})}{\cos2\text{t}}$
$=\frac{3\sqrt{\cos2\text{t}}.\sin^2\text{t}\cos\text{t}-\frac{\sin^3\text{t}}{2\sqrt{\cos2\text{t}}}.(-2\sin2\text{t})}{\cos2\text{t}}$
$=\frac{3\cos2\text{t}\sin^2\text{t}\cos\text{t}+\sin^3\text{t}\sin2\text{t}}{\cos2\text{t}\sqrt{\cos2\text{t}}}$
$\frac{\text{dy}}{\text{dt}}=\frac{\text{d}}{\text{dt}}\Big[\frac{\cos^3\text{t}}{\sqrt{\cos2\text{t}}}\Big]$
$=\frac{\sqrt{\cos2\text{t}}.\frac{\text{d}}{\text{dt}}(\cos^3\text{t})-\cos^3\text{t}.\frac{\text{d}}{\text{dt}}(\sqrt{\cos2\text{t}})}{\cos2\text{t}}$
$=\frac{\sqrt{\cos2\text{t}}\cos^2\text{t}.\frac{\text{d}}{\text{dt}}(\cos\text{t})-\cos^3\text{t}.\frac{1}{2\sqrt{\cos2\text{t}}}.\frac{\text{d}}{\text{dt}}(\cos2\text{t})}{\cos2\text{t}}$
$=\frac{3\sqrt{\cos2\text{t}}.\cos^2\text{t}(-\sin\text{t})-\cos^3\text{t}.\frac{1}{2\sqrt{\cos2\text{t}}}.(-2\sin2\text{t})}{\cos2\text{t}}$
$=\frac{-3\cos2\text{t}\cdot\cos^2\text{t}\cdot\sin\text{t}+\cos^3\text{t}.\sin2\text{t}}{\cos2\text{t}\cdot\sqrt{\cos2\text{t}}}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\Big(\frac{\text{dy}}{\text{dt}}\Big)}{\Big(\frac{\text{dx}}{\text{dt}}\Big)}=\frac{-3\cos2\text{t}.\cos^2\text{t}.\sin\text{t}+\cos^3\text{t}\sin2\text{t}}{3\cos2\text{t}\sin^2\text{t}\cos\text{t}+\sin^3\text{t}\sin2\text{t}}$
$=\frac{-3\cos2\text{t}.\cos^2\text{t}.\sin\text{t}+\cos^3\text{t}(2\sin\text{t}\cos\text{t})}{3\cos2\text{t}\sin^2\text{t}\cos\text{t}+\sin^3\text{t}(2\sin\text{t}\cos\text{t})}$
$=\frac{\sin\text{t}\cos\text{t}[-3\cos2\text{t}.\cos\text{t}+2\cos^3\text{t}]}{\sin\text{t}\cos\text{t}[3\cos2\text{t}\sin\text{t}+2\sin^3\text{t}]}$
$=\frac{[-3(2\cos^2\text{t}-1)\cos\text{t}+2\cos^3\text{t}]}{[3(1-2\sin^2\text{t})\sin\text{t}+2\sin^3\text{t}]}$ $\begin{bmatrix}\cos2\text{t}=(2\cos^2\text{t}-1). \\\cos2\text{t}=(1-2\sin^2\text{t}) \end{bmatrix}$
$=\frac{-4\cos^3\text{t}+3\cos\text{t}}{3\sin\text{t}-4\sin^3\text{t}}$
$=\frac{-\cos3\text{t}}{\sin3\text{t}}$ $\begin{bmatrix}\cos3\text{t}=4\cos^3\text{t}-3\cos\text{t}. \\\sin3\text{t}=3\sin\text{t}-4\sin^3\text{t} \end{bmatrix}$
$=-\cot3\text{t}$

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