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

Question 1(c) Merit

Practises Tangents, normals and curve features

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.

The graph shows the curve y = 2/(x + 1)³ along with the tangent to the curve drawn at x = 1. A second tangent to this curve is drawn which is parallel to the first.

Find the x-coordinate of the point where this second tangent touches the curve.

The 2023 paper, page 3 ↗ Marking schedule, page 1 ↗

Work it out

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

1 Write y as a power, then differentiate.
y = 2(x + 1)⁻³, so dy/dx = −6(x + 1)⁻⁴ = (−6)/(x + 1)⁴.
2 What is the first tangent’s gradient, at x = 1?
(−6)/2⁴ = −6/16 = −3/8
3 Parallel lines have the same gradient. Where else is the gradient −3/8?
(−6)/(x + 1)⁴ = −3/8, so (x + 1)⁴ = 16.
4 How many solutions does that have?
Two: x + 1 = 2 or x + 1 = −2, so x = 1 or x = −3. x = 1 is the tangent already drawn, so the second one touches at x = −3, on the left-hand branch.

Answer x = −3.

What the marker wanted

Achieved A correct dy/dx, and the first tangent’s gradient, −3/8.
Merit x = −3 for the second tangent, worked from the derivative.

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

Common mistake

Taking only the positive root, x + 1 = 2, which gives back the x = 1 you started with. NZQA’s 2023 report reminds students that a square root has two solutions (so does a fourth root), and names this part as one where many could not connect the calculus to the coordinate geometry.

How did it go?

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