If y=t^2+t^3 and x=t-t^4,then frac{d^2 y}{d x | vedclass

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

Question

(B) Given $y=t^2+t^3$ and $x=t-t^4$. First,find the derivatives with respect to $t$: $\frac{dy}{dt} = 2t + 3t^2$ $\frac{dx}{dt} = 1 - 4t^3$ Then,$\fra

Vedclass · Accepted Answer

$-\frac{4}{3}$

Vedclass · Answer

(B) Given $y=t^2+t^3$ and $x=t-t^4$. First,find the derivatives with respect to $t$: $\frac{dy}{dt} = 2t + 3t^2$ $\frac{dx}{dt} = 1 - 4t^3$ Then,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2t + 3t^2}{1 - 4t^3}$. Now,differentiate $\frac{dy}{dx}$ with respect to $x$ using the chain rule: $\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right) \cdot \frac{dt}{dx} = \frac{d}{dt}\left(\frac{2t + 3t^2}{1 - 4t^3}\right) \cdot \frac{1}{1 - 4t^3}$. Using the quotient rule: $\frac{d}{dt}\left(\frac{2t + 3t^2}{1 - 4t^3}\right) = \frac{(1 - 4t^3)(2 + 6t) - (2t + 3t^2)(-12t^2)}{(1 - 4t^3)^2}$. Thus,$\frac{d^2y}{dx^2} = \frac{(1 - 4t^3)(2 + 6t) + 12t^2(2t + 3t^2)}{(1 - 4t^3)^3}$. At $t=1$: $\left.\frac{d^2y}{dx^2}\right|_{t=1} = \frac{(1 - 4)(2 + 6) + 12(1)^2(2(1) + 3(1)^2)}{(1 - 4)^3} = \frac{(-3)(8) + 12(5)}{(-3)^3} = \frac{-24 + 60}{-27} = \frac{36}{-27} = -\frac{4}{3}$.

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

Similar Questions

If $\cos x = \frac{1}{\sqrt{1 + t^2}}$ and $\sin y = \frac{t}{\sqrt{1 + t^2}}$,then $\frac{dy}{dx} = $

If $x$ and $y$ are connected parametrically by the equations,without eliminating the parameter,find $\frac{dy}{dx}$ for $x=a\left(\cos t+\log \tan \frac{t}{2}\right), y=a \sin t$.

If $x=\sec \theta-\cos \theta$ and $y=\sec ^n \theta-\cos ^n \theta$,then $\frac{d y}{d x}=$

Find the slope of the normal to the curve $x = 1 - a \sin \theta$,$y = b \cos^2 \theta$ at $\theta = \pi / 2$.

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module