Question

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

Answer 1

Explanation for the correct answer:

Finding the differentiation:

$$\begin{gathered} \mathrm{t}=\frac{{\mathrm{v}}^2}{2} \\ \Rightarrow {\mathrm{v}}^2=2\mathrm{t} \end{gathered}$$

Differentiating both sides with respect to $\mathrm{t}$, we get

$\begin{array}{rcl}2\mathrm{v}\frac{d\mathrm{v}}{d\mathrm{t}}& =& 2\\ & \Rightarrow & \frac{d\mathrm{v}}{d\mathrm{t}}=\frac{1}{\mathrm{v}}...\left(1\right)\end{array}$

We know that acceleration is the rate of change of velocity then

$\frac{\mathrm{dv}}{\mathrm{dt}}=\mathrm{f}...\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$ we get,

$\mathrm{vf}=1$

By differentiating the above equation with respect to $\mathrm{t}$, then

$$\begin{gathered} \mathrm{v}\frac{\mathrm{df}}{\mathrm{dt}}+\mathrm{f}\frac{\mathrm{dv}}{\mathrm{dt}}=0\mathbf{[}\mathbf{\because }\mathbf{}\frac{\mathbf{d}\left(\mathrm{xy}\right)}{\mathbf{dt}}\mathbf{=}\mathbf{x}\frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{+}\mathbf{y}\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{]} \\ \Rightarrow \frac{\mathrm{df}}{\mathrm{dt}}+\frac{\mathrm{dv}}{{\mathrm{v}}^2\mathrm{dt}}=0 \\ \Rightarrow \frac{\mathrm{df}}{\mathrm{dt}}=-\frac{\mathrm{dv}}{{\mathrm{v}}^2\mathrm{dt}} \\ \Rightarrow -\frac{\mathrm{df}}{\mathrm{dt}}=\frac{1}{{\mathrm{v}}^3}\left[\because \mathrm{from}\left(1\right)\right] \\ \Rightarrow -\frac{\mathrm{df}}{\mathrm{dt}}={\mathrm{f}}^3 \end{gathered}$$

Hence, the value of a derivative is ${\mathbf{f}}^{\mathbf{3}}$.