Question 16
A species of fish is being raised in a fish pond. The number of fish, $M$, in the pond can be modelled by a function $M=100b^{t}$, where $t$ is the time (days) since the fish were initially introduced into the pond and $b$ is a constant to be determined.
After seven days, there are 150 fish in the pond.
Find the rate of population growth when the pond contains five times the initial number of fish.
[4 marks]One point fixes the curve, and the tangent gives the rate
The day 7 count pins down $b$. Then find where the pond holds 500 fish and read the gradient of the curve there.
The question gives you a number of fish, so you first find where on the curve that is. The rate of growth is the gradient of the tangent at that point.
The growth rate depends on how many fish there are
Move along the curve and watch the orange tangent. Its gradient is always $M\ln b$, so once you know the number of fish you know the rate. Then test what “five times the initial number of fish” should mean.
The initial number is $M(0)=100$, so five times it is 500 fish. The pond gets there at $t\approx27.7855$ days, and the tangent there has gradient $500\ln b\approx28.96$ fish per day.
750 is five times the 150 fish at day 7. That is not the initial number. The initial number is the count at $t=0$, which is 100. This point gives a rate of about 43.44 fish per day.
Five times seven days is $t=35$, but the question multiplies the number of fish, not the time. At 35 days there are $100\times1.5^5\approx759$ fish, far more than 500.
This is the data point you were given. It fixes $b=1.5^{\frac{1}{7}}$. The rate here is only $150\ln b\approx8.69$ fish per day, so the rate is not the same along the curve.
This is the start. There are 100 fish, the initial number, and the rate is $100\ln b\approx5.79$ fish per day. Five times this many fish will grow five times as fast.
The rate here is the number of fish multiplied by $\ln b\approx0.0579$. Double the fish and the tangent is twice as steep. Choose what “five times the initial number” should mean.
Work through the solution one mark at a time
Try each step yourself before you reveal it. These steps follow the QCAA’s first method, which finds the time. The second method, which skips the time, appears in steps 3 and 4.
Find the unknown first. After seven days, $t=7$ and $M=150$. Substitute and solve for $b$.
$$\begin{aligned}150&=100b^7\\b^7&=1.5\\b&=(1.5)^{\frac{1}{7}}\approx1.0596\end{aligned}$$
$$\therefore M=100\times(1.5)^{\frac{t}{7}}$$
Writing $b=\sqrt[7]{1.5}$ is the same value. Keep it exact, or store the full decimal in your calculator, so the later answers don’t drift.
Marker: correctly determines the constant $b$
The initial number of fish is the number at $t=0$, and $b^0=1$. Five times that is your target.
$$\begin{aligned}M&=5\times M(0)\\&=5\times100\\&=500\end{aligned}$$
The 150 is the number after seven days, so five times 150 answers a different question.
Marker: correctly determines five times the initial number of fish
Find when the pond holds 500 fish. Set the model equal to 500 and use your calculator’s equation solver.
$$\begin{aligned}500&=100(1.5)^{\frac{t}{7}}\\t&\approx27.7855\text{ days}\end{aligned}$$
By hand this is $t=\dfrac{7\ln5}{\ln1.5}$, the same value.
The QCAA’s second method does not find $t$. It earns this mark by writing the model with base $e$:
$$\begin{aligned}M&=100b^t\\&=100e^{\ln b^t}\\&=100e^{t\ln b}\end{aligned}$$
Marker: determines when the pond contains five times the initial number of fish
Marker (method 2): correctly expresses the model in terms of the exponential function
The rate of population growth is $\dfrac{dM}{dt}$. Use your calculator to evaluate the first derivative at $t=27.7855$.
$$\left.\frac{dM}{dt}\right|_{t\,=\,27.7855}\approx28.9618$$
The second method gets the same rate straight from the number of fish:
$$\begin{aligned}M'&=100\ln b\,e^{t\ln b}\\&=M\ln b\\&=500\ln b\\&\approx28.9618\text{ fish per day}\end{aligned}$$
Marker: determines the rate of growth
Marker (method 2): determines the rate of growth when the pond contains five times the initial number
Each phrase of the question gave you something
Four small jobs are chained together, and each one feeds the next. The data point gives you $b$. The phrase “five times the initial number” has to become a number of fish, and that number has to become a time before your calculator can give the derivative. The question asks for the rate at a population, so you have to work out where that is yourself. It is still complex familiar because the model is given and “rate of population growth” points straight at the derivative.
If I had to guess where you might lose marks, it would be the 500. The 150 is the last number you read, so $5\times150=750$ is tempting, and it gives a rate of about 43.44 fish per day. The other risk is differentiating $b^t$ as if it were a power of $t$. The power rule does not apply when the variable is in the index. Rewrite it with base $e$ first, or let your calculator do the derivative. The 2025 subject report says students responded well to modelling with exponential functions in complex familiar contexts, so these are marks to bank.
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 2025 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2025, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.
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