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/r²This is a separable differential equation, so can be solved by integrating the separated parts: ∫r²·dr = ∫k·dt (1/3)r³ = kt + cFilling 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 = 125/3and (1/3)(3)³ = k(4) +125/3 27 = 12k +125 0 = 12k +98 k = -98/12 = -49/6Putting 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 +125/3Multiplying by 3 and taking the cube root, we get ... r = ∛(125 -49t/2) . . . . equation linking radius and time__(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 . . . . minutesThe mint will completely dissolve in 5 minutes 6 seconds.__(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.