Answer:
(a) r = ∛(125 -49/2t) . . . . where r is the radius in mm and t is the time in minutes
(b) r = 0 at t = 250/49 ≈ 5.1 min = 5 min 6 sec
(c) the domain should be limited to [0, 250/49]
Step-by-step explanation:
(a) The rate of change of radius is proportional to the inverse of the square of the radius, so you have ...
dr/dt = k/r²
This is a separable differential equation, so can be solved by integrating the separated parts:
∫r²·dr = ∫k·dt
(1/3)r³ = kt + c
Filling in the two given conditions lets you solve for k and c.
For r in mm and t in minutes, you have ...
(1/3)(5)³ = k(0) +c
c = 125/3
and
(1/3)(3)³ = k(4) +125/3
27 = 12k +125
0 = 12k +98
k = -98/12 = -49/6
Putting these values for k and c into the solution to the differential equation, we have something we can solve for r:
(1/3)r³ = -49/6·t +125/3
Multiplying by 3 and taking the cube root, we get ...
r = ∛(125 -49t/2) . . . . equation linking radius and time
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(b) The value of radius will be zero when ...
0 = ∛(125 -49t/2)
0 = 125 -49t/2 . . . . . cube it
49t = 250 . . . . . . . . . multiply by 2, add 49t
t = 250/49 ≈ 5.102 . . . . minutes
The mint will completely dissolve in 5 minutes 6 seconds.
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(c) The expression for r is defined for all values of t, but the domain might reasonably be limited to [0, 250/49] minutes. The expression is not useful predicting the radius of the mint before it starts to dissolve, and negative values of radius make no sense in this context.
