← Differentiation methods: past papers

AS 91578 · 2025 paper · derivatives, normal, tangents

Question 1(e) Merit to Excellence

Practises Parametric differentiation

A chain-rule power, an oven cooling written in half-powers of t, a trig normal, a tangent to an exponential-plus-quadratic, and a parametric curve where inflection and stationary have to coincide.

A curve is defined by the parametric equations x = 3t2 + 6t + 1 and y = 2t3. Find the coordinates of any points of inflection on this curve that are also stationary points. You may assume that any inflection points you find are inflection points.

The 2025 paper, page 4 ↗ Marking schedule, page 2 ↗

Work it out

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

  1. 1 Differentiate x and y with respect to t, then find dy/dx.

    dx/dt = 6t + 6 and dy/dt = 6t2
    dy/dx = 6t2/(6t + 6) = t2/(t + 1)

  2. 2 For the second derivative, differentiate dy/dx with respect to t. Then what must you do with the result?

    Quotient rule: (d/dt)(t2/(t + 1)) = (2t(t + 1) − t2)/(t + 1)2 = (t2 + 2t)/(t + 1)2
    Then divide by dx/dt again:
    (d2y)/dx2 = ((t2 + 2t)/(t + 1)2) ÷ (6t + 6) = (t2 + 2t)/(6(t + 1)3)

  3. 3 Where could the points of inflection be?

    (d2y)/dx2 = 0 when t2 + 2t = t(t + 2) = 0: at t = 0 or t = −2.

  4. 4 Which of those is also a stationary point?

    Stationary means dy/dx = 0.
    t = 0: dy/dx = 0/1 = 0. Stationary.
    t = −2: dy/dx = 4/(−1) = −4. An inflection, but not a stationary one.

  5. 5 Turn the t you kept into coordinates.

    t = 0 gives x = 0 + 0 + 1 = 1 and y = 0.
Answer One point: (1, 0), at t = 0. The other inflection, at t = −2, has gradient −4, so it is not stationary.

What the marker wanted

  • Achieved A correct dy/dx.
  • Merit Correct first and second derivatives, and the second derivative set to zero and solved for both values of t.
  • Excellence The single point (1, 0), with the reason the other value of t is dropped. That full answer is E8; a correct solution with one minor error is E7. Finding the stationary points first and then testing which are inflections is accepted too.

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

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Common mistake

Stopping at (d/dt)(dy/dx) and calling it the second derivative, without dividing by dx/dt again. NZQA’s 2025 report says many students did, leaving an answer resting on a method that does not work. Here the wrong method even gives the same two values of t, which is exactly why it is easy to miss.

How did it go?

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