If x = t^2 and y = t^3,then frac{d^2y}{dx^2} | vedclass

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

Question

(B) Given $x = t^2$ and $y = t^3$.First,find the first derivative $\frac{dy}{dx}$ using the chain rule for parametric equations:$\frac{dx}{dt} = 2t$ a

Vedclass · Accepted Answer

$\frac{3}{4t}$

Vedclass · Answer

(B) Given $x = t^2$ and $y = t^3$.First,find the first derivative $\frac{dy}{dx}$ using the chain rule for parametric equations:$\frac{dx}{dt} = 2t$ and $\frac{dy}{dt} = 3t^2$.$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3t^2}{2t} = \frac{3}{2}t$.Now,differentiate $\frac{dy}{dx}$ with respect to $x$ to find $\frac{d^2y}{dx^2}$:$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{3}{2}t\right) = \frac{d}{dt}\left(\frac{3}{2}t\right) \cdot \frac{dt}{dx}$.Since $\frac{dt}{dx} = \frac{1}{dx/dt} = \frac{1}{2t}$,we have:$\frac{d^2y}{dx^2} = \frac{3}{2} \cdot \frac{1}{2t} = \frac{3}{4t}$.

If $x = t^2$ and $y = t^3$,then $\frac{d^2y}{dx^2} =$

Similar Questions

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

Let $x(t) = 2 \sqrt{2} \cos t \sqrt{\sin 2t}$ and $y(t) = 2 \sqrt{2} \sin t \sqrt{\sin 2t}$,$t \in (0, \frac{\pi}{2})$. Then $\frac{1 + (\frac{dy}{dx})^2}{\frac{d^2y}{dx^2}}$ at $t = \frac{\pi}{4}$ is equal to:

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

The derivative of $\sin x$ with respect to $\log x$ is

If $\tan u=\sqrt{\frac{1-x}{1+x}}$ and $\cos v=4 x^{3}-3 x$,then $\frac{d u}{d v}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module