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

Question 2(e) Merit to Excellence

Practises The rules & standard derivatives

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 radius of curvature measures how much a curve is bending at a point, and is given by p = ((1 + (dy/dx)²)^3/2)/((d²y)/dx²).

Find the radius of curvature for the curve y = cos²(2x) when x = π/3.

p	((1 + (dy/dx)²)^3/2)/((d²y)/dx²)

The 2025 paper, page 8 ↗ Marking schedule, page 4 ↗

Work it out

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

1 The formula needs dy/dx and d²y/dx². How do you differentiate cos²(2x)?
It is (cos 2x)², so the chain rule, twice:
dy/dx = 2 cos 2x × (−2 sin 2x) = −4 sin 2x cos 2x
2 Before differentiating again, can a double-angle identity make it shorter?
2 sin 2x cos 2x = sin 4x, so dy/dx = −2 sin 4x and (d²y)/dx² = −8 cos 4x.
(The product rule on −4 sin 2x cos 2x gives 8 sin² 2x − 8 cos² 2x, which is the same thing.)
3 Find both at x = π/3.
4x = 4π/3, where sin = −√3/2 and cos = −1/2.
dy/dx = −2 × (−√3/2) = √3 and (d²y)/dx² = −8 × (−1/2) = 4
4 Put them into the formula.
p = ((1 + (√3)²)^3/2)/4 = 4^3/2/4 = 8/4 = 2

Answer p = 2.

What the marker wanted

Achieved A correct dy/dx.
Merit Correct first and second derivatives.
Excellence p = 2, with the calculus shown (E8). A correct solution with one minor error is E7.

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

Common mistake

Getting the values from a graphics calculator instead of showing the derivatives. NZQA’s 2025 report names this part among those where too many answers leaned on the calculator; Excellence needs the calculus on the page.

How did it go?

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