A mothball of radius 1/2 inch evaporates to leave a mothball of radius 1/4 inch at the end of 6 months. Express the radius of the ball as a function of time.
Make the physical assumption that the rate of change of volume with respect to time is proportional to the surface area.
Make the physical assumption that the rate of change of volume with respect to time is proportional to the surface area.
-
Let r(t) be the radius (in inches) of the mothball at time t > 0. Let t be in months.
Volume V of a sphere of radius r: V = (4π/3)r³
Surface area A of a sphere of radius r: A = 4π r²
You are told that dV/dt = kA for some (negative) constant k.
dV/dt = (dV/dr)(dr/dt) = (4π r²)(dr/dt) = A dr/dt
Putting these into the dV/dt equation,
A dr/dt = k A
dr/dt = k
so r decreases at a constant rate.
Integrate both sides with respect to t:
∫ (dr/dt) dt = ∫ k dt
r = kt + c
r(0) = 0.5 so c = ½
r(6) = ¼ so
¼ = 6k + ½
-¼ = 6k
-1/24 = k
Thus, r(t) = (-1/24)t + ½
Obviously, this equation no longer holds after the time when r = 0, which is easily found to be t = 12. So the complete answer is
r(t) = (-1/24)t + ½ for 0 ≤ t ≤ 12
r(t) = 0, t > 12
Volume V of a sphere of radius r: V = (4π/3)r³
Surface area A of a sphere of radius r: A = 4π r²
You are told that dV/dt = kA for some (negative) constant k.
dV/dt = (dV/dr)(dr/dt) = (4π r²)(dr/dt) = A dr/dt
Putting these into the dV/dt equation,
A dr/dt = k A
dr/dt = k
so r decreases at a constant rate.
Integrate both sides with respect to t:
∫ (dr/dt) dt = ∫ k dt
r = kt + c
r(0) = 0.5 so c = ½
r(6) = ¼ so
¼ = 6k + ½
-¼ = 6k
-1/24 = k
Thus, r(t) = (-1/24)t + ½
Obviously, this equation no longer holds after the time when r = 0, which is easily found to be t = 12. So the complete answer is
r(t) = (-1/24)t + ½ for 0 ≤ t ≤ 12
r(t) = 0, t > 12