Question

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)$\mathbf{t}\mathbf{}\mathbf{=}\mathbf{}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{s}$?

Short answer

(a) The mass of water is greatest at $4.21\mathrm{s}$

(b) The greatest mass of water is $23.2\mathrm{gm}$.

(c) The rate of mass change at $t=2\mathrm{s}$is $2.89\times {10}^{‐2}\mathrm{kg}/\min$

(d) The rate of mass change at $t=5\mathrm{s}$is $-6.05\times {10}^{‐3}\mathrm{kg}/\min$.

Step-by-step solution

Given data

The mass of the water as a function of time,

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

… (i)

Understanding the concept

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.

(a) Determination of the time at which the mass of 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 $t=4.21\mathrm{s}$.

(b) Determination of the greatest mass of water

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$23.2\mathrm{gm}$.

(c) Determination of the rate of change of mass of water at 

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 $2.89\times {10}^{‐2}\mathrm{kg}/\min$.

(d) Determination of the rate of change of mass of water at 

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$-6.05\times {10}^{‐3}\mathrm{kg}/\min$.