Q287 · Practice questionTechnology-freeSimple familiar6 marks
QUESTION 287 (6 marks)
A tank is an inverted square-based pyramid, 6 m deep with top side length 4 m. Water enters at 8 m/min. The water depth h and water-surface side length s satisfy the same similarity ratio as the full tank. A central vertical cross-section is supplied.
a)[2 marks]
Express s and the water volume V in terms of h. Use .
b)[2 marks]
Determine when h=3 m.
c)[2 marks]
On the blank axes, sketch as a function of h for . Label its value at h=3 and describe its behaviour near h=0.
WORKED SOLUTION
6 marksPractice marking scheme
ANSWER
(a) , . (b) m/min. (c) , decreasing from infinity to ; it passes through .
Worked solution
(a) Similar sections give . Substitute into the pyramid-volume formula.
(b) , so . At h=3 the factor is 4, yielding the stated rate.
(c) The function is positive and decreasing, with a vertical asymptote at h=0. This large initial depth rate arises from the very small cross-sectional area near the vertex.
Equivalent justified methods accepted; respect any requested proof method.
Exact values unless specified. Sketches assessed by mathematical features, not artistic quality.
Part a: Use the correct similarity ratio.
Part a: Obtain the volume function.
Part b: Apply the related-rates chain rule.
Part b: Substitute h=3 and determine the rate.
Part c: Draw the correctly shaped positive decreasing curve and point.
Part c: Indicate the excluded zero-depth asymptote and explain the behaviour.
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