Water is poured into a container that has a small leak. The mass m of the water is given as a function of timetby$\mathbf{m}\mathbf{}\mathbf{=}\mathbf{}\mathbf{5}\mathbf{.}\mathbf{00}{\mathbf{t}}^{\mathbf{0}\mathbf{.}\mathbf{8}\mathbf{}}\mathbf{‐}\mathbf{}\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{t}\mathbf{}\mathbf{+}\mathbf{}\mathbf{20}\mathbf{.}\mathbf{00}$, with$t\ge 0$,min grams, andtin seconds. (a) At what time is the water mass greatest, and (b) what is that greatest mass? In kilograms per minute, what is the rate of mass change at (c)$\mathbf{t}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{s}$and (d)$\mathbf{t}\mathbf{}\mathbf{=}\mathbf{}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{s}$?

The mass of the water as a function of time,

$m\left(t\right)=5.00t^{0.8}-3.00t+20.00$

… (i)

The mass of the water is greatest when the amount of water lost through the leak is smallest. The smallest leak corresponds to the smallest rate of change in the flow with respect to time. Therefore, differentiating the given equation and equating it to zero will give us the time at which the mass of the water is greatest.

Mass of water is greatest when,

$\frac{dm}{dt}=0$ … (ii)

Differentiating equation (i) with respect to $\frac{dm}{dt}$gives,

$\frac{dm}{dt}=4000t^{0.2}-3.00$ … (iii)

Use this value in equation (ii) to calculate time.

$$\begin{gathered} 4000t^{0.2}-3.00=0 \\ t=4.21\mathrm{s} \end{gathered}$$

Therefore, the mass of water is greatest at $t=4.21\mathrm{s}$.

To find the mass of water at$t=4.21\mathrm{s}$, substitute the value of time in equation (i).

$$\begin{gathered} m=5.00{\left(4.21\mathrm{s}\right)}^{0.8}-3.00\left(4.21\mathrm{s}\right)+20.00 \\ =23.2\mathrm{gm} \end{gathered}$$

Thus, the greatest mass of water is$23.2\mathrm{gm}$.

To calculate the rate of change of mass of water at$t=2.00\mathrm{s}$, substitute the value of time in equation (iii).

$$\begin{gathered} \frac{dm}{dt}=\left[4.00{\left(2.00\mathrm{s}\right)}^{‐0.2}-3.00\right]\mathrm{g}/\mathrm{s} \\ 0.48\mathrm{g}/\mathrm{s} \end{gathered}$$

But$1\mathrm{kg}=1000\mathrm{gm}$and$1\mathrm{minute}=60\mathrm{s}$

$$\begin{gathered} \frac{dm}{dt}=0.48\frac{\mathrm{g}}{s}.\left(\frac{1\mathrm{kg}}{1000\mathrm{g}}\right)\left(\frac{60\mathrm{s}}{1\min }\right) \\ =2.89\times {10}^{‐2}\mathrm{kg}/\min \end{gathered}$$

Thus, the rate of change of mass of water is $2.89\times {10}^{‐2}\mathrm{kg}/\min$.

Similarly, to calculate the rate of change of mass of water at$t=5.00\mathrm{s}$, substitute the value of time in equation (iii).

$$\begin{gathered} \frac{dm}{dt}=\left[4.00{\left(5.00\mathrm{s}\right)}^{‐0.2}-3.00\right]\mathrm{g}/\mathrm{s} \\ =-0.101\mathrm{g}/\mathrm{s} \end{gathered}$$

But$1\mathrm{kg}=1000\mathrm{gm}$and$1\mathrm{minute}=60\mathrm{s}$,

$$\begin{gathered} \frac{dm}{dt}=-0.101\mathrm{g}/\mathrm{s}.\left(\frac{1\mathrm{kg}}{1000\mathrm{g}}\right).\left(\frac{60\mathrm{s}}{1\min }\right) \\ =-6.05\times {10}^{‐3}\mathrm{kg}/\min \end{gathered}$$

Thus, the rate of change of mass of water is$-6.05\times {10}^{‐3}\mathrm{kg}/\min$.