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

Question 2(d) Merit

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.

Find the equation of the tangent to the graph defined by the pair of equations x = 2 sec t and y = 5 tan t, at the point where t = π/6.

The 2025 paper, page 7 ↗ Marking schedule, page 3 ↗

Work it out

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

1 Differentiate x and y with respect to t. What is the derivative of sec t?
dx/dt = 2 sec t tan t and dy/dt = 5 sec² t
2 Divide to get dy/dx. Can it be simplified?
dy/dx = (5 sec² t)/(2 sec t tan t) = (5 sec t)/(2 tan t) = 5/(2 sin t)
because (sec t)/(tan t) = 1/(cos t) × (cos t)/(sin t) = 1/(sin t).
3 At t = π/6, find the gradient and the point.
sinπ/6 = 1/2, so dy/dx = 5/(2 × 1/2) = 5.
secπ/6 = 2/√3, so x = 4/√3 ≈ 2.3094.
tanπ/6 = 1/√3, so y = 5/√3 ≈ 2.8868.
4 Write the tangent, then rearrange it into y = mx + c.
y − 5/√3 = 5(x − 4/√3)
y = 5x − 15/√3 = 5x − 5√3 ≈ 5x − 8.66

Answer y = 5x − 5√3, or about y = 5x − 8.66.

What the marker wanted

Achieved dy/dx found. It need not be simplified.
Merit A correct equation for the tangent, in any equivalent form.

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

Common mistake

Leaving the tangent as y − 2.8868 = 5(x − 2.3094). NZQA’s 2025 report says a tangent’s equation should be finished off in gradient–intercept or general form, fully evaluated, to get full credit.

How did it go?

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