Question

If $x=sin^{-1}\left(3t-4t^{3}\right)$ and $y=cos^{-1}\left(\sqrt{1-t^{2}}\right)$, then $\frac{dy}{dx}$ is equal to

• A. $\frac{1}{2}$
• B. $\frac{2}{3}$
• C. $\frac{1}{3}$
• D. $\frac{2}{5}$

Answer 1

The Correct Option is C

$x=\sin ^{-1}\left(3 t-4 t^{3}\right) $
and $y=\cos ^{-1}\left(\sqrt{1-t^{2}}\right)$
Put $t=\sin \theta \,\,\,\,\,\,\dots(i)$
Then, $x =\sin ^{-1} \left(3 \sin \theta-4 \sin ^{3} \theta\right)$
$=\sin ^{-1}(\sin 3 \theta)=3 \theta=3 \sin ^{-1} t $
and $ y=\cos ^{-1} \sqrt{1-\sin ^{2} \theta} $
$=\cos ^{-1}(\cos\, \theta)=\theta=\sin ^{-1} t $
Now, $\frac{d x}{d t} =\frac{3}{\sqrt{1-t^{2}}}$
$ \frac{d y}{d t}=\frac{1}{\sqrt{1-t^{2}}} $
$ \Rightarrow \,\frac{d y}{d x} =\frac{d y}{d t} \cdot \frac{d t}{d x} $
$ \frac{d y}{d x} =\frac{1}{\sqrt{1-t^{2}}} \times \frac{\sqrt{1-t^{2}}}{3}$
$=\frac{1}{3} $