The radius of an air bubble is increasing at the rate of 1/2 cm/s. At what rate is the volume of the bubble increasing when the radius is 1 cm?
The volume of the bubble is increasing at the rate of 2π cm³/s when the radius is 1 cm.
Explanation
This is a related rates problem from Application of Derivatives. The air bubble is spherical, so we use the volume formula V = (4/3)πr³. We differentiate with respect to time using the Chain Rule, similar to Example 3 in the context which shows dA/dt = dA/dr × dr/dt for a circle's area. Here, dV/dt = dV/dr × dr/dt = 4πr² × dr/dt. Substituting r = 1 cm and dr/dt = 1/2 cm/s gives the final answer.
Solution Steps
Step 1: Let V be the volume of the spherical air bubble with radius r. The volume is given by V = (4/3)πr³.
Step 2: Differentiating V with respect to time t using the Chain Rule: dV/dt = dV/dr × dr/dt = 4πr² × dr/dt.
Step 3: Given: dr/dt = 1/2 cm/s and r = 1 cm.
Step 4: Substituting the values: dV/dt = 4π(1)² × (1/2) = 4π × 1/2 = 2π cm³/s.