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Chain Rule, Log Differentiation & Rate of Change · Question 108 of 130

Question 108

EasyOriginal question

The edge of a cube grows at 2 cm/s. How fast is its volume growing when the edge is 5 cm?

Options for question 108

Show the answer and explanation
CorrectNot quiteAnswer C

Correct answer: C

Explanation

  1. V=a3V = a^{3}, so dVdt=3a2dadt\frac{dV}{dt} = 3a^{2}\frac{da}{dt}.

  2. 3×25×2=1503 \times 25 \times 2 = 150 cm³/s.

Worth noting

75 cm³/s is 3a23a^{2} alone, the rate per cm of edge.

Concept
A rate of change through a related quantity
Typical time
60 s
Your time
—
The rule this testsRate of change of quantities
Practise this chapter →

Ten questions from Chain Rule, Log Differentiation & Rate of Change, at your own pace.

Practise this chapter →

Ten questions from Chain Rule, Log Differentiation & Rate of Change, at your own pace.