If x = sin^{-1}(3t - 4t^3) and y = cos^{-1}(s | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given $x = \sin^{-1}(3t - 4t^3)$ and $y = \cos^{-1}(\sqrt{1 - t^2})$.For $y = \cos^{-1}(\sqrt{1 - t^2})$,let $t = \sin 	heta$,then $\sqrt{1 - t^2

Vedclass · Accepted Answer

$1/3$

Vedclass · Answer

(D) Given $x = \sin^{-1}(3t - 4t^3)$ and $y = \cos^{-1}(\sqrt{1 - t^2})$.For $y = \cos^{-1}(\sqrt{1 - t^2})$,let $t = \sin 	heta$,then $\sqrt{1 - t^2} = \cos 	heta$.So,$y = \cos^{-1}(\cos 	heta) = 	heta = \sin^{-1} t$.For $x = \sin^{-1}(3t - 4t^3)$,let $t = \sin 	heta$,then $x = \sin^{-1}(3 \sin 	heta - 4 \sin^3 	heta) = \sin^{-1}(\sin 3 	heta) = 3 	heta = 3 \sin^{-1} t$.Now,differentiate $x$ and $y$ with respect to $t$:$\frac{dx}{dt} = 3 \cdot \frac{1}{\sqrt{1 - t^2}}$$\frac{dy}{dt} = 1 \cdot \frac{1}{\sqrt{1 - t^2}}$Therefore,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{1/\sqrt{1 - t^2}}{3/\sqrt{1 - t^2}} = \frac{1}{3}$.

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

Similar Questions

If $x = \sec \theta - \cos \theta$ and $y = \sec^n \theta - \cos^n \theta$,then $(x^2 + 4) \left(\frac{dy}{dx}\right)^2 =$

The rate of change of $\sqrt{x^2+16}$ with respect to $\frac{x}{x-1}$ at $x=5$ is

If $\tan x = \frac{2t}{1-t^2}$ and $\sin y = \frac{2t}{1+t^2}$,then the value of $\frac{dy}{dx}$ is

The $2^{nd}$ derivative of $a \sin^3 t$ with respect to $a \cos^3 t$ at $t = \frac{\pi}{4}$ is

If $x=a\left\{\cos \theta+\log \tan \left(\frac{\theta}{2}\right)\right\}$ and $y=a \sin \theta$, then $\frac{dy}{dx}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module