If x = at^2 and y = 2at,then find y_2 (where | vedclass

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

Question

(A) Given $x = at^2$ and $y = 2at$. First,find the first derivative $\frac{dy}{dx}$: $\frac{dx}{dt} = 2at$ $\frac{dy}{dt} = 2a$ $\frac{dy}{dx} = \frac

Vedclass · Accepted Answer

$-\frac{1}{2at^3}$

Vedclass · Answer

(A) Given $x = at^2$ and $y = 2at$. First,find the first derivative $\frac{dy}{dx}$: $\frac{dx}{dt} = 2at$ $\frac{dy}{dt} = 2a$ $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2a}{2at} = \frac{1}{t} = t^{-1}$. Now,find the second derivative $y_2 = \frac{d^2y}{dx^2}$: $y_2 = \frac{d}{dx}(\frac{dy}{dx}) = \frac{d}{dt}(\frac{dy}{dx}) \cdot \frac{dt}{dx}$ $y_2 = \frac{d}{dt}(t^{-1}) \cdot \frac{1}{2at}$ $y_2 = (-t^{-2}) \cdot \frac{1}{2at} = -\frac{1}{t^2} \cdot \frac{1}{2at} = -\frac{1}{2at^3}$. Thus,the correct option is $A$.

If $x = at^2$ and $y = 2at$,then find $y_2$ (where $t \neq 0$).

Similar Questions

If $\tan x = \frac{2t}{1-t^2}$ and $\sin y = \frac{2t}{1+t^2}$,then the value of $\frac{dy}{dx}$ is

If $y=t^2+t^3$ and $x=t-t^4$,then $\frac{d^2 y}{d x^2}$ at $t=1$ is

If $\frac{dy}{dx} = 4$ and $\frac{d^2y}{dx^2} = -3$ at a point $P$ on the curve $y = f(x)$,then $\left(\frac{d^2x}{dy^2}\right)_P = $

The angle made by the tangent at $\theta = \pi / 3$ on the curve $x = a(\theta + \sin \theta)$,$y = a(1 - \cos \theta)$ with the $X$-axis is

Differentiate $\sin ^2 x$ with respect to $\cos ^2 x$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module