Hint
The original volume is 125 cm³. Expand the volume factor and separate the retained linear term from the discarded terms.
Worked solution
(a)
The cubic Pascal row is 1,3,3,1, so \((1+t)^3=1+3t+3t^2+t^3\), exactly.
Answer: \(1+3t+3t^2+t^3\).
(b)
The first-order volume estimate is \(125(1+3t)\). At t=0.02 it gives 125(1.06)=132.5 cm³. The estimated relative increase is 0.06, or 6%.
Answer: 132.5 cm³; approximately 6% increase.
(c)
The discarded volume is \(125(3t^2+t^3)\). At t=0.02 this is 125(0.0012+0.000008)=0.151 cm³. It is positive, so the estimate is low. The exact volume is 132.651 cm³, agreeing with 5.1³.
Answer: 0.151 cm³; the estimate is low.
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official examination-board mark scheme. Award stated criteria once each, up to the part maximum. Accept valid equivalent methods. Follow-through applies only where explicitly stated; do not award an accuracy criterion for an incorrect final result.
(a) · 2 marks
- 1 mark: Give the correct linear and quadratic coefficients 3.
- 1 mark: Give the complete expansion including the constant and cubic term.
Alternative methods: Direct multiplication of three factors is accepted.
Common error: Using 1+t³ in place of the full expansion.
(b) · 2 marks
- 1 mark: Give the first-order volume estimate 132.5 cm³.
- 1 mark: Give the corresponding estimated increase 6%.
Alternative methods: An estimated increase of 7.5 cm³ added to 125 cm³ is accepted.
Common error: Applying the side-length percentage directly to the volume.
(c) · 3 marks
- 1 mark: Use the discarded terms 125(3t²+t³) or compare with the exact volume.
- 1 mark: Give absolute error 0.151 cm³.
- 1 mark: Identify an underestimate with the sign of the remainder or an exact comparison.
Alternative methods: Compute 5.1³ and subtract 132.5, with the correct units and direction.
Common error: Reporting a relative error as a volume or omitting the original-volume multiplier.
Total: 7 marks.