Rates of Change — Question 2

Section 4-1 · Calculus pages 677–678

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Question 2

Problem:

A spherical balloon is being inflated so that its volume increases at a rate of 100 cm3^3/sec. How fast is the radius increasing when the radius is 5 cm?

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Question 2 - Solution

We are given the volume of a sphere:

V=43πr3V = \frac{4}{3}\pi r^3

Differentiate both sides with respect to time tt:

dVdt=4πr2⋅drdt\frac{dV}{dt} = 4\pi r^2 \cdot \frac{dr}{dt}

We are told that:

dVdt=100,r=5\frac{dV}{dt} = 100, \quad r = 5

Substitute into the equation:

100=4π(5)2⋅drdt⇒100=100π⋅drdt⇒drdt=1π100 = 4\pi (5)^2 \cdot \frac{dr}{dt} \Rightarrow 100 = 100\pi \cdot \frac{dr}{dt} \Rightarrow \frac{dr}{dt} = \frac{1}{\pi}

Answer: drdt=1π cm/sec\boxed{\frac{dr}{dt} = \frac{1}{\pi} \text{ cm/sec}}