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Rate of Change, Marginal Cost & Revenue · Question 17 of 79

How does the rate at which the balloon's radius grows change as the balloon gets bigger?

Question 17

HardOriginal question

Air is pumped into a spherical balloon at a steady 100 cm³ every second. For a sphere of radius rr, V=43πr3V = \frac{4}{3}\pi r^{3} and the surface area is S=4πr2S = 4\pi r^{2}.

How does the rate at which the balloon's radius grows change as the balloon gets bigger?

Options for question 17

Show the answer and explanation
CorrectNot quiteAnswer D

Correct answer: D

Explanation

  1. drdt=1004πr2\frac{dr}{dt} = \frac{100}{4\pi r^{2}}.

  2. The same 100 cm³ spreads over a larger surface each second, so the radius grows more slowly, as 1r2\frac{1}{r^{2}}.

Concept
Reading a related-rates formula
Typical time
150 s
Your time
—
The rule this testsRate of change of a quantity
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Ten questions from Rate of Change, Marginal Cost & Revenue, at your own pace.

Practise this chapter →

Ten questions from Rate of Change, Marginal Cost & Revenue, at your own pace.