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Rate of Change, Marginal Cost & Revenue · Question 14 of 79

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

Question 14

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 radius growing when the radius is 5 cm?

Options for question 14

Show the answer and explanation
CorrectNot quiteAnswer A

Correct answer: A

Explanation

  1. dVdt=4πr2drdt\frac{dV}{dt} = 4\pi r^{2}\frac{dr}{dt}, so 100=4π(25)drdt100 = 4\pi(25)\frac{dr}{dt}.

  2. drdt=100100π=1π\frac{dr}{dt} = \frac{100}{100\pi} = \frac{1}{\pi} cm/s.

Worth noting

5π\frac{5}{\pi} uses rr where the derivative of r3r^{3} needs r2r^{2}.

Concept
Related rates: volume and 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.