If t = frac{v^2}{2},then left( - frac{df}{dt} | vedclass

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

Question

(B) Given $t = \frac{v^2}{2}$.Differentiating with respect to $t$,we get $\frac{dt}{dt} = \frac{1}{2} 	imes 2v \frac{dv}{dt}$.Since $f = \frac{dv}{dt

Vedclass · Accepted Answer

$f^3$

Vedclass · Answer

(B) Given $t = \frac{v^2}{2}$.Differentiating with respect to $t$,we get $\frac{dt}{dt} = \frac{1}{2} 	imes 2v \frac{dv}{dt}$.Since $f = \frac{dv}{dt}$,we have $1 = v \cdot f$,which implies $f = \frac{1}{v}$ or $v = f^{-1}$.Now,differentiate $f = v^{-1}$ with respect to $t$:$\frac{df}{dt} = -v^{-2} \frac{dv}{dt}$.Substitute $\frac{dv}{dt} = f$ and $v = \frac{1}{f}$:$\frac{df}{dt} = -\left( \frac{1}{f} ight)^{-2} \cdot f = -f^2 \cdot f = -f^3$.Therefore,$-\frac{df}{dt} = -(-f^3) = f^3$.

If $t = \frac{v^2}{2}$,then $\left( - \frac{df}{dt} \right)$ is equal to,(where $f$ is acceleration)

Similar Questions

Find the derivative of the following function: $\frac{\sec x-1}{\sec x+1}$

If $y = |\cos 4x| + |\sin 4x| + |\tan 4x|$,then find $\frac{dy}{dx}$ at $x = \frac{\pi}{6}$.

If $f(x) = 1 + x + x^2 + \ldots + x^{1000}$,then $f^{\prime}(-1) = $ . . . . . .

Find the derivative of the following function: $5 \sec x + 4 \cos x$.

Let $f(x)$ and $g(x)$ be twice differentiable functions satisfying $f''(x) = g''(x)$ for all $x \in R$, $f'(1) = 2g'(1) = 4$ and $g(2) = 3f(2) = 9$. Then $f(25) - g(25)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module