If x=3 tan t and y=3 sec t,then the value of | vedclass

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

Question

(B) Given $x=3 	an t$ and $y=3 \sec t$. First,find the derivatives with respect to $t$: $\frac{dx}{dt} = 3 \sec^2 t$ $\frac{dy}{dt} = 3 \sec t 	an t

Vedclass · Accepted Answer

$\frac{1}{6 \sqrt{2}}$

Vedclass · Answer

(B) Given $x=3 	an t$ and $y=3 \sec t$. First,find the derivatives with respect to $t$: $\frac{dx}{dt} = 3 \sec^2 t$ $\frac{dy}{dt} = 3 \sec t 	an t$ Now,find the first derivative $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3 \sec t 	an t}{3 \sec^2 t} = \frac{	an t}{\sec t} = \sin t$ Now,find the second derivative $\frac{d^2y}{dx^2}$: $\frac{d^2y}{dx^2} = \frac{d}{dx}(\sin t) = \frac{d}{dt}(\sin t) \cdot \frac{dt}{dx} = \cos t \cdot \frac{1}{3 \sec^2 t} = \frac{\cos t}{3 \sec^2 t} = \frac{\cos^3 t}{3}$ Finally,evaluate at $t = \frac{\pi}{4}$: $\left(\frac{d^2y}{dx^2}ight)_{t=\frac{\pi}{4}} = \frac{(\cos(\pi/4))^3}{3} = \frac{(1/\sqrt{2})^3}{3} = \frac{1/(2\sqrt{2})}{3} = \frac{1}{6\sqrt{2}}$

If $x=3 \tan t$ and $y=3 \sec t$,then the value of $\frac{d^2 y}{d x^2}$ at $t=\frac{\pi}{4}$ is

Similar Questions

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

If $t$ is a parameter and $x=t+\frac{1}{t}, y=t-\frac{1}{t}$, then $\frac{d^2 y}{d x^2}=$

If $x = a(t + \sin t)$ and $y = a(1 - \cos t)$,then $\frac{dy}{dx}$ equals

If $x=\cos \theta$ and $y=\sin 5 \theta$,then $\left(1-x^2\right) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}$ is equal to (in $y$)

The angle made by the tangent of the curve $x = a(t + \sin t \cos t)$; $y = a(1 + \sin t)^2$ with the $x$-axis at any point on it is

Vedclass Test Series

Exam Paper Generator

Online Exam Module