The second order derivative of a sin^3 t with | vedclass

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

Question

(C) Let $y = a \sin^3 t$ and $x = a \cos^3 t$. First, we find the first derivative $\frac{dy}{dx}$ using the chain rule: $\frac{dy}{dt} = 3a \sin^2 t

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $y = a \sin^3 t$ and $x = a \cos^3 t$. First, we find the first derivative $\frac{dy}{dx}$ using the chain rule: $\frac{dy}{dt} = 3a \sin^2 t \cos t$ $\frac{dx}{dt} = -3a \cos^2 t \sin t$ $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3a \sin^2 t \cos t}{-3a \cos^2 t \sin t} = -	an t$. Now, we find the second derivative $\frac{d^2 y}{dx^2}$ with respect to $x$: $\frac{d^2 y}{dx^2} = \frac{d}{dx}(-	an t) = \frac{d}{dt}(-	an t) \cdot \frac{dt}{dx} = (-\sec^2 t) \cdot \frac{1}{-3a \cos^2 t \sin t} = \frac{1}{3a \cos^4 t \sin t}$. Evaluating at $t = \frac{\pi}{4}$: $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$ and $\sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$. $\left. \frac{d^2 y}{dx^2} ight|_{t=\pi/4} = \frac{1}{3a (\frac{1}{\sqrt{2}})^4 (\frac{1}{\sqrt{2}})} = \frac{1}{3a (\frac{1}{4}) (\frac{1}{\sqrt{2}})} = \frac{4\sqrt{2}}{3a}$.

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

Similar Questions

If $x = \sqrt{10^{\sin^{-1} t}}$ and $y = \sqrt{10^{\cos^{-1} t}}$,then $\frac{dy}{dx} = $ . . . . . .

If $x=\log _e\left(\frac{\cos \frac{y}{2}-\sin \frac{y}{2}}{\cos \frac{y}{2}+\sin \frac{y}{2}}\right)$ and $\tan \frac{y}{2}=\sqrt{\frac{1-t}{1+t}}$,then the value of $\left(\frac{dy}{dx}\right)_{t=\frac{1}{2}}$ is

If the tangent to the curve given by $x=t^{2}-1$ and $y=t^{2}-t$ is parallel to the $X$-axis,then the value of $t$ is

Find the rate of change of $\sqrt{x^2 + 16}$ with respect to $\frac{x}{x - 1}$ at $x = 3$.

If $x = 2 \sqrt{2} \sqrt{\cos 2 \theta}$ and $y = 2 \sqrt{2} \sqrt{\sin 2 \theta}$, $0 < \theta < \frac{\pi}{4}$, then the value of $\frac{dy}{dx}$ at $\theta = 22 \frac{1}{2}^{\circ}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module