Question 17
A chemical is added to the water in a swimming pool at 10:00 am to prevent algae. The amount of chemical absorbed into the water over time $t$ (hours) is represented by
$A=10t^2-4t^3,$$0\le t\le 1\tfrac{2}{3}$
Determine the time of day when the rate of absorption of the chemical is at its maximum. Use calculus techniques to verify that your time corresponds to a maximum rate.
[6 marks]The rate is the slope of the amount
Follow the slope of the amount curve from 10:00 am to 11:40 am. It rises, peaks and falls back to zero.
The amount is greatest at 11:40 am, but the question asks about the rate. The rate peaks at 10:50 am, where the amount curve is steepest.
Decide which function you are maximising
The question gives you the amount $A$ but asks about the rate. Slide through the morning and watch both graphs. Then choose what to maximise and see which time each choice gives.
The blue tangent on the top graph has the same slope as the height of the blue dot below.
Choose a function. Each choice runs the usual routine of setting a derivative to zero and testing the next one.
Setting $\frac{dA}{dt}=0$ gives $t=0$ or $t=\frac{5}{3}$, and $\frac{d^2A}{dt^2}=-20$ at $t=\frac{5}{3}$, so the amount is greatest at 11:40 am. Look at the blue graph at that time. The rate has fallen to 0. This answers a different question.
The rate is $\frac{dA}{dt}=20t-12t^2$. Its derivative $\frac{d^2A}{dt^2}=20-24t$ is 0 at $t=\frac{5}{6}$, and $\frac{d^3A}{dt^3}=-24$ is negative, so the rate is a maximum 50 minutes after the start, at 10:50 am.
Work through the solution one mark at a time
Try each step yourself before you reveal it. Every step is the usual optimisation routine, applied to the rate instead of the amount.
The question asks about the rate of absorption. $A$ is the amount, so the rate is its derivative with respect to $t$.
$$\frac{dA}{dt}=20t-12t^2$$
This is the function you maximise from here on.
Marker: correctly determines the first derivative
To find where the rate is greatest, differentiate the rate. That makes it the second derivative of $A$.
$$\frac{d^2A}{dt^2}=20-24t$$
If your first derivative had a slip, you can still earn this mark by follow-through.
Marker: determines the second derivative
A turning point of the rate is where the derivative of the rate is zero.
$$20-24t=0$$
Marker: equates the second derivative to zero
Solve for $t$ and keep it as an exact fraction.
$$\begin{aligned}24t&=20\\t&=\frac{20}{24}=\frac{5}{6}\ \text{hours}\end{aligned}$$
Equivalent values are accepted, such as $\frac{20}{24}$. This time sits inside the domain $0\le t\le 1\tfrac{2}{3}$.
Marker: determines time when second derivative is zero
Verify with the second derivative test, applied to the rate. The derivative of $\frac{d^2A}{dt^2}$ is the third derivative of $A$.
$$\frac{d^3A}{dt^3}=-24$$
This is negative, so the rate of absorption is a maximum at $t=\frac{5}{6}$. The marker line asks for a calculus test, so showing that $\frac{d^2A}{dt^2}$ changes from positive to negative either side of $t=\frac{5}{6}$ is another way to verify it. For example, $\frac{d^2A}{dt^2}=8$ at $t=\frac{1}{2}$ and $-4$ at $t=1$.
Marker: performs a calculus test to confirm the time corresponds to a maximum for $\frac{dA}{dt}$
The question wants a time of day. Turn the hours into minutes and add them to the 10:00 am start.
$$\begin{aligned}\frac{5}{6}\ \text{hours}&=\frac{5}{6}\times60\\&=50\ \text{minutes}\end{aligned}$$
Stopping at $t=\frac{5}{6}$ or 0.83 hours leaves this mark behind.
Marker: determines the time for maximum rate of absorption in minutes
Each phrase of the question gave you something
You have to spot that the rate of absorption is $\frac{dA}{dt}$, then run the whole optimisation routine one level higher than usual. The function you maximise is already a derivative, so its turning point comes from the second derivative and the verification needs the third. Then the answer in hours has to become a time of day. It is still familiar because the wording points at the method. “Rate”, “maximum” and “use calculus techniques to verify” name every tool you need.
If I had to guess where you might lose marks, it would be maximising the wrong function. The subject report says many students ran the second derivative test on $A$ itself. That leads to $t=1\tfrac{2}{3}$, or 11:40 am, the moment the most chemical has been absorbed and the rate has dropped to zero. Read “rate of absorption” as $\frac{dA}{dt}$ before you differentiate anything. The last mark also needs the time in minutes, so turn $\frac{5}{6}$ of an hour into 50 minutes and write 10:50 am.
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 2023 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2023, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.
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