Further integration — Question 180

Original QCE Vault practice · 5 marks

Q180 · Practice questionTechnology-freeComplex unfamiliar5 marks

QUESTION 180 (5 marks)

A tank contains

80

L at

t=0

. For

0\le t\le6

minutes, its volume changes at the rate

\dfrac{dV}{dt}=12-\dfrac12t^2

litres per minute. Determine the time at which the volume is greatest, the greatest volume, and the volume at

t=6

WORKED SOLUTION

5 marks

ANSWER

Maximum at

t=2\sqrt6\approx4.899

min; greatest volume

80+16\sqrt6\approx119.19

V(6)=116

Worked solution

V(t)=80+12t-t^3/6

. A maximum occurs when

V'(t)=12-t^2/2=0

, giving

t=\sqrt{24}=2\sqrt6

in the interval. Substitution yields

80+16\sqrt6

. Finally,

V(6)=80+72-36=116

Integrates the rate and uses

V(0)=80

[1 mark]

Solves

V'(t)=0

in the interval.

[1 mark]

Justifies the maximum.

[1 mark]

Determines the greatest volume.

[1 mark]

Determines

V(6)

[1 mark]

Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.

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