Question 2(e) Merit to Excellence

Expand-then-integrate, a lower limit recovered from a definite integral, a separable equation in y², an area between a curve and a horizontal line, and a physical integral left entirely in symbols.

The mass M of a spherical object of radius p can be approximated by M = int₀^p 4π r² (a/(1 + br³)) dr, where a, b and p are all positive constants.

Using this formula, find an expression for the mass M of a spherical object, giving your answer in terms of a, b, p and π.

The 2024 paper, page 9 ↗ Marking schedule, page 4 ↗

Work it out

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

1. 1 a, b and π are constants. Take them outside, and look at what is left.

   M = 4π aint₀^p (r²/(1 + br³)) dr

2. 2 The top is nearly the derivative of the bottom. What is (d/dr)(1 + br³)?

   3br². So multiply the top by 3b and divide outside by 3b:
   M = ((4π a)/3b)int₀^p (3br²/(1 + br³)) dr

3. 3 The top is now exactly the derivative of the bottom. What does that integrate to?

   ln(1 + br³). No modulus is needed: b and r are positive.
   M = ((4π a)/3b)Bigl[ln(1 + br³)Bigr]₀^p

4. 4 Put in the limits. What is ln 1?

   M = ((4π a)/3b)[ln(1 + bp³) − ln 1] = ((4π a)/3b) ln(1 + bp³)

What the marker wanted

• Merit Reaching ((4π a)/3b)Bigl[ln(1 + br³)Bigr]₀^p.
• Excellence The expression M = ((4π a)/3b) ln(1 + bp³), from correct integration, is E8; a solution with a minor error is E7.

The schedule gives this part no Achieved cell: its first credit is Merit.

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

How did it go?

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