Question

Helium is pumped into a spherical balloon at the constant rate of 25 cubic feet/minute. at what rate is the surface area of the balloon increasing at the moment when its radius is 8 feet?

Answer 1

If we let V', A', and r' represent the rates of change of the volume, area, and radius of the sphere, we can differentiate equations for V and A in terms of r, then eliminate r' to get an equation for A' in terms of V'.

$V=\dfrac{4}{3}\pi r^{3}\\V^{\prime} =4\pi r^{2}\cdot r^{\prime}\\\\A=4\pi r^{2}\\A^{\prime} =8\pi r\cdot r^{\prime}\\\\A^{\prime}=8\pi r\cdot \dfrac{V^{\prime}}{4\pi r^{2}}=\dfrac{2}{r}V^{\prime}$

Then, for the given values, the surface area is increasing at the rate ...

... A' = (2/8)·25 ft²/s = 6.25 ft²/s