Mathematical Methods — Question 16

QCAA 2025, Paper 2 · 4 marks

Q16 · 2025 · Technology-activeComplex familiar4 marks

QUESTION 16 (4 marks)

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.

Question linkOriginal paper

WORKED SOLUTION

QCAA guide · typeset solution

4 marks

ANSWER

28.9618

fish per day.

Worked solution

Method 1: Time is calculated

t=7

M=150

150=100b^7,

b=(1.5)^{1/7}.

Therefore

M=100(1.5)^{t/7}.

Five times the initial population is

M=5M(0)=500.

Solving

500=100(1.5)^{t/7}

gives

t=27.7855\text{ days}.

Using a GDC to evaluate the first derivative at this time gives

28.9618\text{ fish per day}.

Method 2: Time is not calculated

Again

b=(1.5)^{1/7}

and

M=500

. Write

M=100b^t=100e^{t\ln b}.

Then

M'=100\ln(b)e^{t\ln b}=M\ln b.

M=500

M'=500\ln b=28.9618\text{ fish per day}.

correctly determines the constant

b

[1 mark]

correctly determines five times the initial number of fish

[1 mark]

determines when the pond contains five times the initial number of fish, or correctly expresses the model in exponential form

[1 mark]

determines the rate of growth when the pond contains five times the initial number

[1 mark]

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Mathematical Methods — Question 16

QUESTION 16 (4 marks)

QCAA guide · typeset solution

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