Rates of Change — Question 4

Section 4-1 · Calculus pages 681–682

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Question 4

Problem:

A spherical balloon is being inflated so that its volume increases at a rate of 100cm3/s100 \, \text{cm}^3/\text{s}. How fast is the radius increasing when the radius is 5cm5 \, \text{cm}?

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Question 4 - Solution

We use the formula for the volume of a sphere: V=43πr3V = \frac{4}{3} \pi r^3

Differentiate both sides with respect to tt: dVdt=4πr2drdt\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}

Substitute: dVdt=100,r=5\frac{dV}{dt} = 100, \quad r = 5 100=4π(5)2drdt⇒100=100πdrdt⇒drdt=100100π=1π100 = 4\pi (5)^2 \frac{dr}{dt} \Rightarrow 100 = 100\pi \frac{dr}{dt} \Rightarrow \frac{dr}{dt} = \frac{100}{100\pi} = \frac{1}{\pi}

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