University:
University of Michigan
Course:
MATH 115 | Calculus I
Academic year:

2023
Views:

455

Pages:

1
Author:

Koen D

Math 115 / Final (December 14, 2012) 9. [11 points] A cube of ice is removed from the freezer and begins to melt. Let ℓ(t) be its side length, V (t) its volume, and S(t) its surface area, all dependent upon t, the number of minutes since it was removed from the freezer. The ice cube is melting (its volume is changing) at a rate proportional to its surface area. That is, dV dt = kS(t), for some number k. Initially the ice cube has a side length of 2 inches. a. [4 points] Write V and S in terms of ℓ. Calculate the rate of change (with respect to time) of the side length of the ice cube in terms of ℓ and k. Solution: V = ℓ3 and S = 6ℓ2 . We know, by the chain rule, that dV dV dℓ dℓ = = 3ℓ2 . dt dℓ dt dt We are given that dV = kS(t) = k6ℓ2 . dt Setting these equal, we get 3ℓ2 So dℓ dt dℓ = k6ℓ2 . dt = 2k inches per minute. b. [2 points] How fast is the volume of the ice cube changing immediately after it is removed from the freezer? Your answer will involve k. Solution: The side length at time t = 0 is 2 inches, so dV 2 3 dt t=0 = k · 6 · 2 = 24k in per minute c. [2 points] What is the sign of k? Briefly explain. Solution: From part (b) we see that k has the same sign as dV dt t=0 , which is negative, since the volume is decreasing. Thus, k must be negative as well. d. [3 points] How long will it take the ice cube to melt completely? Your answer may involve k. Solution: The side length is changing at a constant rate of 2k and starts at 2 inches, so it will take T minutes, where 2 + 2kT = 0. So T = − k1 minutes.