Question

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

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

Answer 1

Answer: (C)

$x={\sin }^{-1}\left(3t-4t^3\right)$
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, $\frac{dx}{dt}=\frac{3}{\sqrt{1-t^2}}$
$\frac{dy}{dt}=\frac{1}{\sqrt{1-t^2}}$
$\Rightarrow \frac{dy}{dx}=\frac{dy}{dt}\cdot \frac{dt}{dx}$
$\frac{dy}{dx}=\frac{1}{\sqrt{1-t^2}}\times \frac{\sqrt{1-t^2}}{3}$
$=\frac{1}{3}$