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Rate of Change, Marginal Cost & Revenue · Question 16 of 79

How fast is the balloon's radius growing when the radius is 10 cm?

Question 16

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 10 cm?

Options for question 16

Show the answer and explanation
CorrectNot quiteAnswer A

Correct answer: A

Explanation

  1. 100=4π(100)drdt100 = 4\pi(100)\frac{dr}{dt}.

  2. drdt=14π\frac{dr}{dt} = \frac{1}{4\pi} cm/s.

Worth noting

12π\frac{1}{2\pi} halves the rate for twice the radius; the rate goes as 1r2\frac{1}{r^{2}}, so it falls to a quarter.

Concept
Related rates at a new instant
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.