A balloon, which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the later is 10 cm.
400π
Explanation
This problem is analogous to Example 3 in the provided context, where the rate of change of area with respect to radius is computed by differentiating the area formula. For a spherical balloon, the volume V is a function of radius r. The rate of increase of volume with respect to radius is dV/dr, obtained by differentiating the volume formula V = (4/3)πr³. At r = 10 cm, dV/dr = 4πr² = 4π(10)² = 400π. The units are cm³ per cm, i.e., cm², consistent with the derivative of volume with respect to length.
Solution Steps
Step 1: Write the formula for volume of a sphere: V = (4/3)πr³.
Step 2: Differentiate V with respect to r: dV/dr = d/dr[(4/3)πr³] = (4/3)π * 3r² = 4πr².
Step 3: Substitute the given radius r = 10 cm: dV/dr = 4π(10)² = 4π * 100 = 400π.
Step 4: State the result: The rate at which volume increases with radius is 400π cm².