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Rate of Change, Marginal Cost & Revenue · Question 15 of 79

How fast is the balloon's surface area growing when the radius is 5 cm?

Question 15

MediumOriginal 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 fast is the balloon's surface area growing when the radius is 5 cm?

Options for question 15

Show the answer and explanation
CorrectNot quiteAnswer B

Correct answer: B

Explanation

  1. dSdt=8πrdrdt\frac{dS}{dt} = 8\pi r\frac{dr}{dt}.

  2. At r=5r = 5, drdt=1π\frac{dr}{dt} = \frac{1}{\pi}: 8π(5)(1π)=408\pi(5)\left(\frac{1}{\pi}\right) = 40 cm²/s.

Worth noting

40π40\pi forgets that drdt\frac{dr}{dt} already carries a 1π\frac{1}{\pi}.

Concept
Related rates: area through the radius
Typical time
120 s
Your time
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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.