Question 3(d) Merit

Reading differentiability, stationary points and a limit straight off a drawn graph, then a log quotient, a related-rates sphere, a parametric gradient, and a rectangle inscribed in a curve.

The equation of a curve is given by the parametric equations x = 5/e^2t and y = 5e^2t. Find the coordinates of the point(s) on the curve where the gradient is −2.

The 2025 paper, page 11 ↗ Marking schedule, page 5 ↗

Work it out

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

1. 1 Write x as a power of e. Then differentiate x and y with respect to t.

   x = 5e^−2t, so dx/dt = −10e^−2t
   y = 5e^2t, so dy/dt = 10e^2t

2. 2 Find dy/dx.

   dy/dx = 10e^2t/(−10e^−2t) = −e^4t

3. 3 Set the gradient to −2 and solve for t.

   −e^4t = −2, so e^4t = 2, 4t = ln 2 and t = (ln 2)/4 ≈ 0.173.

4. 4 The question wants coordinates. Find x and y.

   e^2t = √e^4t = √2, so x = 5/√2 ≈ 3.54 and y = 5√2 ≈ 7.07.
   e^4t = 2 has only one solution, so there is only one point.

What the marker wanted

• Achieved A correct dy/dx.
• Merit The point (3.54, 7.071), with the derivative shown.

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

How did it go?

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