Question

Mothballs tend to evaporate at a rate proportional to their surface area. If V is the volume of a mothball, then its surface area is roughly a constant times V²/³ So the mothball's volume decreases at a rate proportional to V²/³. Suppose that initially a mothball has a volume of 27 cubic centimeters and 4 weeks later has a volume of 15.625 cubic centimeters. Construct and solve a differential equation satisfied by the volume at time t. Then, determine if and when the mothball will vanish.

Answer 1

The question seeks to solve a differential equation representing the volume of a mothball over time as it evaporates, which is proportional to its surface area. After integrating the differential equation, we find when the mothball's volume will reach zero.

Step-by-step explanation:

The question involves constructing and solving a differential equation based on the rate of evaporating mothballs, which is proportional to their surface area. Given that the volume V of a mothball evaporates at a rate proportional to V²/³, we can express this as a differential equation, dv/dt = -kV²/³, where k is the proportionality constant. At t=0, V=27 cm³, and at t=4 weeks, V=15.625 cm³. To solve the differential equation, we separate variables and integrate both sides to find V as a function of t. After deriving the function, we can determine when the mothball will vanish, which happens when V(t) reaches zero.