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, $\sqrt{1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2}=\sqrt{1+\tan^2\theta}$
$=\sqrt{\sec^2\theta}=|\sec\theta|$
$\Rightarrow\frac{\text{dx}}{\text{d}\theta}=\text{a}\frac{\text{d}}{\text{d}\theta}(\cos^2\theta)$
$\Rightarrow\frac{\text{dx}}{\text{d}\theta}=3\text{a}\cos^2\theta\frac{\text{d}}{\text{d}\theta}(\cos\theta)$
$\Rightarrow\frac{\text{dx}}{\text{d}\theta}=-3\text{a}\cos^2\theta\sin\theta\ .....(\text{i})$
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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