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Mathematics

1 answer:

NikAS [45]3 years ago

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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

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

(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.

(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.