AP cal ab unit 5 FRQ

Question

A graphing calculator may be used for the following problem. On a given workday, the rate, in tons per hour, at which unprocessed ore arrives at a refinery is modeled by O(t) = 20 + 10 cos ( t^2/25), where t is measured in hours and 0 ≤ t ≤ 10.

At the beginning of the workday, t = 0, the refinery has 100 tons of unprocessed ore. During the hours of operation, the refinery processes ore at a constant rate of 25 tons per hour.

a) Find O'(4). Using correct units, interpret your answer in the context of the problem.

b) How much total ore has been processed by the refinery after the plant has been open for 3 hours?

c) Is the amount of unprocessed ore at the refinery increasing or decreasing at time t = 4 hours? Show the work that leads to your answer.

d) Determine the time at which the total rate of ore at the refinery is neither increasing nor decreasing? Justify your answer.

Answer 1

This problem does not ask the fun stuff, which would be to figure out how much has arrived.

a)dO/dt = -4/5 t sin(t^2/25)

At t = 4, this is -4/5 * 4 * sin(4^2/25) = -16/5 sin(16/25) tons/hour²

b) 3 * 25 = 75 tons

c) As the consumption is 25 tons/hr, we find when O(t) = 25

20 + 10 cos(t^2/25) = 25

cos(t^2/25) = 1/2

cos(x) = 0 when x = π/3 + 2nπ, n an integer, or x = 5π/3 + 2nπ, n an integer,

t^2/25 = π/3 + 2nπ, 5π/3 + 2nπ

t = 5 √(π/3 + 2nπ), t = 5 √(5π/3 + 2nπ)

t = 5.1166,

The second possible solution would be greater than 10, as √5 > 2, and 2*5 = 10, so there is only 1 solution.

Note that, as our amount of ore increases for the first five plus hours, we start with 100 tons, and the amount it can decrease is at most 15, and (10-5)*15 = 75, we will never have a shortage of ore.