Unit 5

7 marks · Hard

An environmental scientist measures the length of a rectangular tank as 6.9 m to the nearest 0.1 m and the width as 4.0 m to the nearest 0.1 m. The height is exactly 3 m. Water is pumped into the tank at a constant rate, and the depth of water in the tank follows the quadratic equation where is in hours and is in metres.

(a)

Calculate the lower and upper bounds for the length and width of the tank.

[2 marks]

(b)

Hence, calculate the lower and upper bounds for the cross-sectional area of the base of the tank.

[2 marks]

(c)

After 2 hours, the measured depth of water is 1.8 m. The scientist models the depth by the quadratic equation . Does the model fit the data at ? Show your working.

[3 marks]

Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.

Show answer

(a) Length: lower = 6.85 m, upper = 6.95 m; Width: lower = 3.95 m, upper = 4.05 m

(b) Lower bound area = 6.85 × 3.95 = 27.0575 m²; Upper bound area = 6.95 × 4.05 = 28.1475 m²

(c) Model: h(2) = 0.1×4 + 0.5×2 + 0.2 = 0.4 + 1.0 + 0.2 = 1.6 m. The measured value is 1.8 m. The model underestimates the depth at t=2.

Common mistakes: Confusing upper/lower bounds with ordinary rounding; Multiplying by rounded rather than bound values; Inaccurate evaluation of the quadratic model

Show mark scheme and worked solution

Marking points

M1 (a) · 1 mark · Correctly identifies lower and upper bounds formula for given rounding

A1 (a) · 1 mark · Correct lower and upper bounds for both length and width

M2 (b) · 1 mark · Multiplies correct pairs of bound values to find bound area (follow-through)

A2 (b) · 1 mark · Correctly calculates both lower and upper area bounds (follow-through)

M3 (c) · 1 mark · Substitutes t=2 into quadratic model correctly

A3 (c) · 1 mark · Correctly computes h(2) value

B1 (c) · 1 mark · Makes valid comment comparing model result to measured value

Worked solution

a) For values given to the nearest 0.1 m, bounds are calculated as value ±0.05.

Length: Lower bound = 6.9 - 0.05 = 6.85 m; Upper bound = 6.9 + 0.05 = 6.95 m

Width: Lower bound = 4.0 - 0.05 = 3.95 m; Upper bound = 4.0 + 0.05 = 4.05 m

b) Cross-sectional area: area = length × width

Lower bound area = 6.85 × 3.95 = 27.0575 m²

Upper bound area = 6.95 × 4.05 = 28.1475 m²

c) Substitute t=2 into the model:

Measured depth = 1.8 m, Model predicts 1.6 m, so the model underestimates the actual measured depth at t=2.