AS 91578 · 2024 paper · parametric, velocity, solution & inflection

Question 2(c) Merit

Practises The rules & standard derivatives

A parametric derivative, a velocity from a log displacement, a second-derivative verification, an inflection point on (ln x)/x, and a turning point that survives an unknown constant.

Show that y = sin(x²) − cos(x) is a solution to the equation (d²y)/dx² + 4x²y = 2 cos(x²) + (1 − 4x²) cos x.

The 2024 paper, page 6 ↗ Marking schedule, page 3 ↗

Work it out

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

1 What do you need from y to test the equation?
Its second derivative, so differentiate twice.
2 First derivative. Which rule does sin(x²) need?
The chain rule: (d/dx) sin(x²) = 2x cos(x²). And (d/dx)(−cos x) = sin x.
dy/dx = 2x cos(x²) + sin x
3 Second derivative. 2x cos(x²) is a product, with a chain inside it.
(d/dx)(2x cos(x²)) = 2 cos(x²) + 2x · (−2x sin(x²)) = 2 cos(x²) − 4x² sin(x²)
(d²y)/dx² = 2 cos(x²) − 4x² sin(x²) + cos x
4 Substitute into the left-hand side and simplify until it matches the right.
(d²y)/dx² + 4x²y = 2 cos(x²) − 4x² sin(x²) + cos x + 4x²(sin(x²) − cos x)
= 2 cos(x²) + cos x − 4x² cos x
= 2 cos(x²) + (1 − 4x²) cos x, the right-hand side.

Answer The left-hand side simplifies to 2 cos(x²) + (1 − 4x²) cos x, which is the right-hand side, so y is a solution.

What the marker wanted

Achieved A correct dy/dx.
Merit The left-hand side shown to equal the right, using correct first and second derivatives.

In our words, from NZQA’s 2024 marking schedule.

Common mistake

Dropping the 2x the chain rule brings out of sin(x²), or differentiating 2x cos(x²) without the product rule. NZQA’s 2024 report says many students made one of these slips, and a wrong derivative left nothing to credit. Others had both derivatives right but did not set the substitution out formally enough to count as showing it.

How did it go?

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