AS 91578 · 2023 paper · derivatives, tangents, a proof

Question 1(b) Achieved

Practises The rules & standard derivatives

A chain-rule root, a product rate of change, a second tangent parallel to a given one, a parametric circle tangent to be proved in general form, and an optimisation whose answer is a fraction of a given area.

Find the rate of change of the function f(t) = t²e^2t when t = 1.5.

The 2023 paper, page 2 ↗ Marking schedule, page 1 ↗

Work it out

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

1 Which rule does t²e^2t need?
The product rule, with the chain rule on e^2t: u = t², u' = 2t; v = e^2t, v' = 2e^2t.
2 Write the derivative, and take out the common factor.
f'(t) = 2te^2t + 2t²e^2t = 2te^2t(1 + t)
3 Put t = 1.5 in.
f'(1.5) = 2(1.5)e³(2.5) = 7.5e³ ≈ 150.64

Answer f'(1.5) = 7.5e³ ≈ 150.64.

What the marker wanted

Achieved The correct derivative, and the rate of change, 150.64.

In our words, from NZQA’s 2023 marking schedule.

How did it go?

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