Answer to Question #242652 in Calculus for Bubu

Question

A spherical balloon is deflated so that its volume is decreasing at a rate of 3 ftଷ/min. How fast is the diameter of the balloon decreasing when the radius is 2 ft? 

Answer 1

$$V=\dfrac{4}{3}\pi R^3=\dfrac{1}{6}\pi D^3$$

Differentiate both sides with respect to $t$ and use the Chain Rule

$$\dfrac{dV}{dt}=\dfrac{1}{6}\pi(3D^2)\dfrac{dD}{dt}$$

$$\dfrac{dD}{dt}=\dfrac{2}{\pi D^2}\dfrac{dV}{dt}$$

Given $\dfrac{dV}{dt}=-3\ {ft}^3/min$

When $R=2\ ft, D=4\ ft$

$$\dfrac{dD}{dt}=\dfrac{2}{\pi (4\ ft)^2}(-3\ ft^3/min)=-\dfrac{3}{8\pi }\ ft/min$$

The diameter of the balloon is decreasing at a rate of $\dfrac{3}{8\pi }\approx0.12$ ft/min when the radius is 2 ft.