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Question

t (minutes)--> 0 1 2 4 6 7

W(t) (ounces)--> 0 8 13 22 25 33

water is dripping into a cup. The amount of water in the cup at time t, 0 less than or equal to t which is less than or equal to 6, is given by the differentiable function W, where t is measured in minutes. Selected values of W(t), measured in ounces are given in the table above.

a) use the data in the table to approximate W'(5). Show the computation that led you to your answer and interpret your answer in the context of the problem using proper units. show all calculus work.

b) for 0 less than or equal to t less than or equal to 7, must there be a time when W9t) =18? Explain

c) for t>7, water leaks out of the cup at a constant rate of 0.2 ounces per minute. At what time will there be no water left in the cup?

Answer 1

a. W'(5) ≈ (W(6) - W(4)) / (6 - 4) = (25 - 22) / 2 = 1.5

After 5 minutes, water is dripping into the cup at a rate of approximately 1.5 ounces per minute.

b. The problem states that W(t) is differentiable. Therefore, W(t) is continuous.

Because W(t) is continuous, we can apply the Intermediate Value Theorem.

W(0) = 0

W(7) = 33

0 < 18 < 33

By the Intermediate Value Theorem, there must be a time t between 0 and 7 such that W(t) = 18.

c. W(7) = 33

For t > 7, W(t) = 33 - 0.2(t - 7)

33 - 0.2(t - 7) = 0

-0.2(t - 7) = -33

t - 7 = 165

t = 172

After 172 minutes, there will be no water left in the cup.