AS 91578 · 2025 paper · logs, exponentials, tangents, radius of curvature

Question 2(b) Achieved

Practises Tangents, normals and curve features

A reciprocal minus a log, a product that has to be differentiated to decide increasing or decreasing, two tangents of the same gradient to a rational curve, a parametric tangent in sec and tan, and a given curvature formula to apply.

The depth of water, in metres, in a particular harbour is P = (30t − 5t²)e^−t, where t is measured in hours and 0 < t ≤ 4.

Show whether the depth of the water in the harbour is increasing or decreasing after 2 hours.

P	(30t − 5t²)e^−t
0 < t ≤ 4
t	2 hours

The 2025 paper, page 5 ↗ Marking schedule, page 3 ↗

Work it out

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

1 Whether something is rising or falling is the sign of which quantity?
Of the derivative dP/dt: positive means the depth is rising, negative means it is falling.
2 P is a product. What are u and v, and their derivatives?
u = 30t − 5t², so u' = 30 − 10t
v = e^−t, so v' = −e^−t
3 Use the product rule.
dP/dt = (30 − 10t)e^−t − (30t − 5t²)e^−t = (5t² − 40t + 30)e^−t
4 Put t = 2 in, and read the sign.
dP/dt = (20 − 80 + 30)e⁻² = −30e⁻² ≈ −4.06
It is negative, so the depth is going down.

Answer Decreasing: at t = 2, dP/dt = −30e⁻² ≈ −4.06 metres per hour, which is negative.

What the marker wanted

Achieved Says the depth is decreasing, backed by a correct derivative (−4.06 at t = 2).

In our words, from NZQA’s 2025 marking schedule.

How did it go?

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