If x = a cos theta and y = b sin theta,then f | vedclass

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

Question

(C) Given $x = a \cos 	heta$ and $y = b \sin 	heta$. First,find $\frac{dy}{dx}$ using the chain rule: $\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta}

Vedclass · Accepted Answer

$-\frac{3b}{a^3} \csc^4 	heta \cot 	heta$

Vedclass · Answer

(C) Given $x = a \cos 	heta$ and $y = b \sin 	heta$. First,find $\frac{dy}{dx}$ using the chain rule: $\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{b \cos 	heta}{-a \sin 	heta} = -\frac{b}{a} \cot 	heta$. Next,find the second derivative $\frac{d^2y}{dx^2}$: $\frac{d^2y}{dx^2} = \frac{d}{dx} \left( -\frac{b}{a} \cot 	heta ight) = -\frac{b}{a} (-\csc^2 	heta) \frac{d	heta}{dx} = \frac{b}{a} \csc^2 	heta \left( \frac{1}{-a \sin 	heta} ight) = -\frac{b}{a^2} \csc^3 	heta$. Finally,find the third derivative $\frac{d^3y}{dx^3}$: $\frac{d^3y}{dx^3} = \frac{d}{dx} \left( -\frac{b}{a^2} \csc^3 	heta ight) = -\frac{b}{a^2} (3 \csc^2 	heta \cdot -\csc 	heta \cot 	heta) \frac{d	heta}{dx} = \frac{3b}{a^2} \csc^3 	heta \cot 	heta \left( \frac{1}{-a \sin 	heta} ight) = -\frac{3b}{a^3} \csc^4 	heta \cot 	heta$.

If $x = a \cos \theta$ and $y = b \sin \theta$,then $\frac{d^3y}{dx^3}$ is equal to

Similar Questions

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:

If $x=a(\cos t+t \sin t)$ and $y=a(\sin t-t \cos t),$ find $\frac{d^{2} y}{d x^{2}}.$

What is the slope of the tangent to the curve $x = t^2 + 3t - 8$ and $y = 2t^2 - 2t - 5$ at the point $(2, -1)$?

If $x = \frac{9t^2}{1+t^4}$ and $y = \frac{16t^2}{1-t^4}$, then $\frac{dy}{dx} = $

For $a > 0, t \in \left( 0, \frac{\pi}{2} \right)$,let $x = \sqrt{a^{\sin^{-1} t}}$ and $y = \sqrt{a^{\cos^{-1} t}}$. Then,$1 + \left( \frac{dy}{dx} \right)^2$ equals

Vedclass Test Series

Exam Paper Generator

Online Exam Module