Question 3(e) Merit to Excellence

A log chain rule, three readings off a drawn graph, a parametric normal on a Ferris wheel, stationary points of a reciprocal pair, and a catenary to verify by substitution.

A power line hangs between two poles. The curve y = f(x) that models its shape solves the differential equation a(d²y)/dx² = √(1 + (dy/dx)²), where a is a positive constant.

Use differentiation to verify that the function y = (a/2)(e^x/a + e^−x/a) satisfies the above differential equation.

The 2023 paper, page 13 ↗ Marking schedule, page 8 ↗

Work it out

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

1. 1 a is a constant. Differentiate y once.

   y = (a/2)(e^x/a + e^−x/a). The chain rule brings out 1/a and −1/a, which cancel the a in front:
   dy/dx = (1/2)(e^x/a − e^−x/a)

2. 2 Differentiate again.

   (d²y)/dx² = (1/2a)(e^x/a + e^−x/a)

3. 3 Work out the left-hand side.

   a(d²y)/dx² = (1/2)(e^x/a + e^−x/a)

4. 4 Now the right-hand side. Square dy/dx, then add 1.

   (dy/dx)² = (1/4)(e^2x/a − 2 + e^−2x/a), since e^x/ae^−x/a = 1
   1 + (dy/dx)² = (1/4)(e^2x/a + 2 + e^−2x/a) = (1/4)(e^x/a + e^−x/a)²

5. 5 Take the square root, and compare with the left-hand side.

   √(1 + (dy/dx)²) = (1/2)(e^x/a + e^−x/a), which is positive because both exponentials are. It is the left-hand side, so the equation holds.

What the marker wanted

• Achieved A correct dy/dx.
• Merit A correct d²y/dx², and real progress with the substitution: either (dy/dx)² expanded, or a·d²y/dx² worked out.
• Excellence The whole proof, with correct derivatives (E8). Reaching both a·d²y/dx² and the expanded 1 + (dy/dx)², or a correct proof with one minor error, is E7.

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

Common mistake

How did it go?

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