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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY1 Mark
MCQ
If $\text{x}=\text{a}\cos^3\theta,\text{y}=\text{a}\sin^3,$ then $\sqrt{1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2}=$
  • A
    $\tan^2\theta$
  • B
    $\sec^2\theta$
  • C
    $\sec^2\theta$
  • ✓
    $|\sec\theta|$
✓

Answer

Correct option: D.
$|\sec\theta|$
We have, $\text{x}=\text{a}\cos^2\theta$

$\Rightarrow\frac{\text{dx}}{\text{d}\theta}=\text{a}\frac{\text{d}}{\text{d}\theta}(\cos^2\theta)$

$\Rightarrow\frac{\text{dx}}{\text{d}\theta}=3\text{a}\cos^2\theta\frac{\text{d}}{\text{d}\theta}(\cos\theta)$

$\Rightarrow\frac{\text{dx}}{\text{d}\theta}=-3\text{a}\cos^2\theta\sin\theta\ .....(\text{i})$

And,

$\text{y}=\text{a}\sin^3\theta$

$\Rightarrow\frac{\text{dx}}{\text{d}\theta}=\text{a}\frac{\text{d}}{\text{d}\theta}(\sin^3\theta)$

$\Rightarrow\frac{\text{dx}}{\text{d}\theta}=3\text{a}\sin^2\theta\frac{\text{d}}{\text{d}\theta}(\sin\theta)$

$\Rightarrow\frac{\text{dx}}{\text{d}\theta}=3\text{a}\sin^2\theta\cos\theta\ .....(\text{ii})$

Dividing (ii) by (i), we get,

$\frac{\frac{\text{dy}}{\text{d}\theta}}{\frac{\text{dx}}{\text{d}\theta}}=\frac{3\text{a}\sin^2\theta\cos\theta}{-3\text{a}\cos^2\theta\sin\theta}$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\sin\theta}{-\cos\theta}$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\tan\theta$

Now, $\sqrt{1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2}=\sqrt{1+\tan^2\theta}$

$=\sqrt{\sec^2\theta}=|\sec\theta|$

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