If x = a cos^3 theta and y = a sin^3 theta,th | vedclass

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

Question

(D) Given $x = a \cos^3 	heta$ and $y = a \sin^3 	heta$. First,find the derivatives with respect to $	heta$: $\frac{dx}{d	heta} = 3a \cos^2 	heta

Vedclass · Accepted Answer

$|\sec 	heta|$

Vedclass · Answer

(D) Given $x = a \cos^3 	heta$ and $y = a \sin^3 	heta$. First,find the derivatives with respect to $	heta$: $\frac{dx}{d	heta} = 3a \cos^2 	heta (-\sin 	heta) = -3a \cos^2 	heta \sin 	heta$. $\frac{dy}{d	heta} = 3a \sin^2 	heta (\cos 	heta) = 3a \sin^2 	heta \cos 	heta$. Now,find $\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{3a \sin^2 	heta \cos 	heta}{-3a \cos^2 	heta \sin 	heta} = -\frac{\sin 	heta}{\cos 	heta} = -	an 	heta$. Then,$\left( \frac{dy}{dx} ight)^2 = (-	an 	heta)^2 = 	an^2 	heta$. Finally,$\sqrt{1 + \left( \frac{dy}{dx} ight)^2} = \sqrt{1 + 	an^2 	heta} = \sqrt{\sec^2 	heta} = |\sec 	heta|$.

If $x = a \cos^3 \theta$ and $y = a \sin^3 \theta$,then $\sqrt{1 + \left( \frac{dy}{dx} \right)^2} = $

Similar Questions

If $x = 2 \cos^3 \theta$ and $y = 3 \sin^2 \theta$,then $\frac{dy}{dx} = $

The derivative of $\sin^2 x$ with respect to $\cos^2 x$ is

If $x = t + \frac{1}{t}$ and $y = t - \frac{1}{t}$,then $\frac{dy}{dx} = $

Let $x = at^2 - 1$, where $a > 0$ and $y = t^3 + 1$. If at $t = 1$, $\frac{d^2y}{dx^2} = \frac{3}{16}$, then the value of $a$ is:

The derivative of $f(\tan x)$ with respect to $g(\sec x)$ at $x = \frac{\pi}{4}$,where $f^{\prime}(1) = 2$ and $g^{\prime}(\sqrt{2}) = 4$,is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module