Question 17
The volume of water in a tank is represented by a function of the form
$V(t)=Ae^{kt}$, where $V$ is in litres and $t$ is in minutes.
Initially, the volume is 100 litres and it is decreasing by 50 litres per minute.
Determine the time at which the volume is decreasing at the rate of $\dfrac{50}{7}$ litres per minute.
Express your answer in the form $\ln(a)$.
[6 marks]The tank slows down as it empties
The rate is the gradient of the volume graph. Watch the tangent flatten as the water drains, until it is falling at only 50/7 litres per minute.
Both facts about the start are needed. The volume finds $A$ and the rate finds $k$. Then the same derivative answers the question.
Decide which equation the words are asking for
The black curve is the volume and the blue curve is its derivative, the rate of change. “Decreasing at the rate of $\frac{50}{7}$ litres per minute” is a statement about one of them. Choose the equation and see where the orange line meets the curve it belongs to.
The dashed line is the tangent to the black curve. Its gradient is the height of the blue dot, and it is always half the volume, with a minus sign.
Choose an equation. Ask yourself two things: is 50/7 a volume or a rate, and does “decreasing” change the sign?
A rate of change is a derivative, and decreasing means negative. The orange line meets the blue curve once, at $t=\ln(49)\approx3.89$. That is the question’s answer, and the volume there is $\frac{100}{7}$ litres.
The volume is falling the whole time, so $V'(t)$ is always negative. The orange line never meets the blue curve. Algebra says the same thing: you would get $e^{-\frac{t}{2}}=-\frac{1}{7}$, and an exponential is never negative. The word “decreasing” is what gives the minus sign.
That treats 50/7 as a volume. The orange line meets the black curve at $t=\ln(196)\approx5.28$, when only 50/7 litres are left. The question gives litres per minute, which is a rate, so it is about $V'(t)$.
Work through the solution one mark at a time
Try each step yourself before you reveal it. This is technology-free, so every value stays exact.
“Initially” means $t=0$. The volume then is 100 litres, and $e^0=1$.
$$\begin{aligned}V(0)&=100\\Ae^{k\times0}&=100\\A&=100\end{aligned}$$
So $V(t)=100e^{kt}$.
Marker: correctly determines $A$
“Decreasing by 50 litres per minute” is a rate, so it is the derivative. Decreasing makes it negative: $V'(0)=-50$.
$$\begin{aligned}V'(t)&=100ke^{kt}\\V'(0)&=100k\\-50&=100k\\k&=-\frac{1}{2}\end{aligned}$$
So $V(t)=100e^{-0.5t}$ and $V'(t)=-50e^{-0.5t}$. If $A$ had a slip, you can still earn this mark by follow-through.
Marker: determines $k$
Now the target rate. Decreasing at $\frac{50}{7}$ litres per minute means $V'(t)=-\frac{50}{7}$. Divide both sides by $-50$.
$$\begin{aligned}-50e^{-0.5t}&=-\frac{50}{7}\\e^{-0.5t}&=\frac{1}{7}\end{aligned}$$
Equivalent forms of the equation are accepted, such as $-\frac{50}{7}=-50e^{-0.5t}$.
Marker: establishes equation in $t$
Take the natural log of both sides to bring the exponent down.
$$\begin{aligned}-0.5t&=\ln\left(\frac{1}{7}\right)\\t&=-2\ln\left(\frac{1}{7}\right)\end{aligned}$$
This mark may be implied by your later working.
Marker: determines $t$
The question wants a single log, $\ln(a)$. Use the power law to move the $-2$ inside.
$$\begin{aligned}t&=\ln\left(\left(\frac{1}{7}\right)^{-2}\right)\\&=\ln\left(7^2\right)\\&=\ln(49)\end{aligned}$$
Any equivalent expression in the form $\ln(a)$ is accepted, for example $t=\ln\left(\left(\frac{1}{7}\right)^{-2}\right)$.
Marker: determines $t$ in required form
The last mark is for how you set it out across the whole response. Use equals signs properly, say where you used each piece of given information (“At $t=0$, $V=100$”), and use function and derivative notation such as $V(0)$ and $V'(t)$.
Marker: shows logical organisation communicating key steps
Each phrase of the question gave you something
There are three conditions, and they sit on two different functions. The volume condition belongs to $V$, and both rate conditions belong to $V'$. You have to build the model from the first two before you can use the third, then finish with log laws to reach the form asked for. It is still familiar, because the model is given and each phrase points at one standard step.
If I had to guess where you might lose marks, it would be the signs. The 2020 subject report asks for more practice using a negative value for a decreasing rate of change, especially in worded questions. Here “decreasing” appears twice, so both rates are negative. Use +50 and $+\frac{50}{7}$ and you get $k=\frac{1}{2}$ and a negative time. Get $k$ right but use $+\frac{50}{7}$ and the equation has no solution. The other easy loss is stopping at $-2\ln\left(\frac{1}{7}\right)$, which is correct but not yet in the form $\ln(a)$.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording and marking-guide steps are from the 2020 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2020, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.
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