The 2^{nd} derivative of a sin^3 t with respe | vedclass

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

Question

(A) Let $y = a \sin^3 t$ and $x = a \cos^3 t$. First,find the derivatives with respect to $t$: $\frac{dy}{dt} = 3a \sin^2 t \cos t$ $\frac{dx}{dt} = 3

Vedclass · Accepted Answer

$\frac{4\sqrt{2}}{3a}$

Vedclass · Answer

(A) Let $y = a \sin^3 t$ and $x = a \cos^3 t$. First,find the derivatives with respect to $t$: $\frac{dy}{dt} = 3a \sin^2 t \cos t$ $\frac{dx}{dt} = 3a \cos^2 t (-\sin t) = -3a \cos^2 t \sin t$ Now,find $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3a \sin^2 t \cos t}{-3a \cos^2 t \sin t} = -	an t$ Next,differentiate $\frac{dy}{dx}$ with respect to $x$: $\frac{d^2y}{dx^2} = \frac{d}{dx}(-	an t) = -\sec^2 t \cdot \frac{dt}{dx}$ Since $\frac{dt}{dx} = \frac{1}{dx/dt} = \frac{1}{-3a \cos^2 t \sin t}$,we have: $\frac{d^2y}{dx^2} = -\sec^2 t \cdot \frac{1}{-3a \cos^2 t \sin t} = \frac{\sec^2 t}{3a \cos^2 t \sin t} = \frac{1}{3a} \cdot \frac{\sec^4 t}{\sin t}$ At $t = \frac{\pi}{4}$,$\sec(\frac{\pi}{4}) = \sqrt{2}$ and $\sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$: $\left(\frac{d^2y}{dx^2}ight)_{t = \pi/4} = \frac{1}{3a} \cdot \frac{(\sqrt{2})^4}{1/\sqrt{2}} = \frac{1}{3a} \cdot \frac{4}{1/\sqrt{2}} = \frac{4\sqrt{2}}{3a}$

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

Similar Questions

If $x = a \sin^3 t$ and $y = a \cos^3 t$, then the value of $\frac{d^2y}{dx^2}$ at $t = \frac{\pi}{3}$ is equal to

If $x = a \cos \theta$ and $y = a \sin \theta$,then $\frac{d^2 y}{d x^2} =$ . . . . . . . (where $a \neq 0$ and $\theta \neq k \pi, k \in Z$)

For $x \neq -1, y \neq -1$, if $x = \frac{1 - \sqrt[3]{y}}{1 + \sqrt[3]{y}}$, then $\frac{dx}{dy} =$

If the parametric equations of a curve are given by $x = \cos \theta + \log \tan \frac{\theta}{2}$ and $y = \sin \theta$,then the points for which $\frac{dy}{dx} = 0$ are given by

If $x$ and $y$ are connected parametrically by the equations,without eliminating the parameter,find $\frac{dy}{dx}$ for $x = a \sec \theta$ and $y = b \tan \theta$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module