I have come across the following question in a past exam paper of a module that I will be teaching this semester.

The volume $V$ of m$^3$ of earth removed from a pit after $t$ hours is given by $$V=10t-\frac{t^2}{3}.$$ Find the rate at which soil is being removed, in cubic metres per minute, after three hours.

Considering the level that these students are at I think the question is actually a typo and it should say minute rather than hour. However I still want to do the problem as stated and I have four equivalent approaches and I was wondering which is superior.

Approaches (A) and (B) seem easy but I am not sure are the better students going to be convinced that we are not 'cheating' somehow. I like approaches (D) but it will possibly confuse some students (especially as they don't do related rates until next semester).

I am tending towards approach (B) but would be interested in any enlightening opinions.

Approach A

If $V$ is the volume of earth removed after $t$ hours then (suppressing the fact that $t$ is measured in hours and hence the 'cheat') then the volume of soil removed after $t$ minutes is given by $$\frac{V}{60}=\frac{1}{6}t-\frac{t^2}{180}.$$ Hence we differentiate to get $$\frac{1}{6}-\frac{t}{90}$$ now put $t=3$ to get the answer $\displaystyle \frac{2}{15}$ m$^3$ min$^{-1}$.

Approach B

To find the rate of change differentiate. We want to evaluate at $t=3$ hours.

$$\frac{dV}{dt}=\left.10-\frac{2}{3}t\right|_{t=3}=8\text{ m}^3\text{ hour}^{-1}.$$

Now we simply convert to minutes via hour = 60 minutes and so hour$^{-1}=\frac{1}{60}\text{ min}^{-1}$ to get $\displaystyle \frac{8}{60}=\frac{2}{15}$ m$^{3}$ min$^{-1}$.

Approach C

A minute is a sixtieth of an arrow so we can find the volume as a function of minutes by writing $$10\left(\frac{t}{60}\right)-\frac{\left(\frac{t}{60}\right)^2}{3}=\frac{1}{6}t-\frac{t^2}{10800}.$$ We find the rate of change by differentiating and evaluate at $t=180$ minutes $$\frac{1}{6}-\left.\frac{t}{5400}\right|_{t=180}=\frac{1}{6}-\frac{180}{5400}=\frac{2}{15} \text{ m}^3\text{ min}^{-1}.$$

Approach D Let $\tau$ be the time measured in minutes. We have $$\frac{dV}{d\tau}=\frac{dV}{dt}\cdot\frac{dt}{d\tau}$$ where $$\frac{dV}{dt}=\left.10-\frac{2}{3}t\right|_{t=3}=8,$$ and and we have $\displaystyle \frac{dt}{d\tau}=\frac{1}{60}$.


The simplest approach here is to convert the inputs to the unit the formula wants the the output to the units you want.

We have:

$$V = 10 t - \frac{t^2}{3}$$

With $t$ in hours and $V$ in $m^3$. Therefore

$$\frac{dV}{dt} = 10 - \frac{2t}{3}$$

Which after 3 hours is

$$10 - \frac{2 \times 3}{3} = 8 \frac{m^3}{hr}$$

Which we convert to the units we want

$$\frac{8 \frac{m^3}{hr}}{60 \frac{min}{hr}} = \frac{8}{60} \frac{m^3}{min} \approx 0.1333 \frac{m^3}{min}$$

Second approach: Convert formula to use the units we want.

We have $t$ in hours but want to use $\tau$ in minutes so replace $t$ with $\frac{\tau}{60}$ in the original equation

$$\begin{align} V & = 10 \cdot \frac{\tau}{60} - \frac{\left( \frac{\tau}{60} \right)^2}{3} \\ & = \frac{\tau}{6} - \frac{\tau^2}{10800} \\ \frac{dV}{d \tau} & = \frac{1}{6} - \frac{1}{5400} \tau \end{align}$$

Now we want the answer after 3 hours which is 180 min

$$\frac{dV}{d \tau} = \frac{1}{6} - \frac{1}{5400} \times 180 \approx 0.1333 \frac{m^3}{min} $$

  • $\begingroup$ I think Approach B is superior too. $\endgroup$ – JP McCarthy Sep 20 '13 at 9:24

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