Question 2(e) Merit to Excellence

An exponential integral, a negative-power differential equation, an upper limit that appears on both sides, a trig quotient whose top is a multiple of the derivative of its bottom, and a proportional-to-the-square model built from scratch.

A cake factory pumps liquid chocolate out of a container so that the rate of change of the volume remaining is proportional to the square of the volume remaining. After one hour of use, p litres remain, where p is a positive constant. After a further one hour, only (4/5) p litres remain.

Write a differential equation modelling this, and solve it to calculate how much liquid chocolate was in the container at the start of the day, giving your answer in terms of p.

The 2023 paper, page 8 ↗ Marking schedule, page 4 ↗

Work it out

Try it on paper first. Stuck? Take one hint at a time.

1. 1 Write the sentence as a differential equation. The rate of change of the volume V is proportional to what?

   To the square of the volume: dV/dt = kV², with t in hours from the start of the day.

2. 2 Separate and integrate.

   int V⁻² dV = int k dt, so −1/V = kt + c.

3. 3 Use the two readings: V = p at t = 1, and V = (4/5) p at t = 2.

   −1/p = k + c
   −5/4p = 2k + c

4. 4 Subtract the first from the second to find k, then find c.

   k = −5/4p + 1/p = −1/4p
   c = −1/p − k = −1/p + 1/4p = −3/4p

5. 5 The start of the day is t = 0.

   −1/V = c = −3/4p, so V = 4p/3.

What the marker wanted

• Achieved Reaching −1/V = kt + c.
• Merit Either k or the constant found in terms of p.
• Excellence The starting volume, (4/3) p litres, with correct integrations, is E8. Reaching 1/V = 3/4p at t = 0 without turning it into V, or a solution with one minor error, is E7; finding k or the constant is the Merit cell. Writing dV/dt = −kV², as the schedule does, gives k = 1/4p instead, and the same answer.

In our words, from NZQA’s 2023 marking schedule.

Common mistake

How did it go?

Got it moves it further out. Review brings it back tomorrow.