If x = t + frac{1}{t} and y = t - frac{1}{t}, | vedclass

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

Question

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

Vedclass · Accepted Answer

$-4t^3(t^2 - 1)^{-3}$

Vedclass · Answer

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

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

Similar Questions

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

The position of a point at time $t$ is given by $x = a + bt - ct^2$ and $y = at + bt^2$. Its acceleration at time $t$ is:

At any two points of the curve represented parametrically by $x = a(2 \cos t - \cos 2t)$ and $y = a(2 \sin t - \sin 2t)$,the tangents are parallel to the $x$-axis. The values of the parameter $t$ corresponding to these points differ from each other by:

If $f^{\prime}(x)=\sqrt{2 x^2-1}$ and $y=f(x^3)$,then find the value of $\frac{dy}{dx}$ at $x=1$.

If $x = \frac{3at}{1 + t^3}$ and $y = \frac{3at^2}{1 + t^3}$,then $\frac{dy}{dx} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module