Calculus FRQ AP Practice Question

Question

Water is flowing into and out of a small reservoir so that the total amount of water can be modeled by R, where 𝑅(𝑡) = 20 − 15sin( 𝑡^2/25) cubic feet, t is measured in hours, and 0 ≤ t ≤ 8. Show your work.

a) What is the average rate of change of R over the interval 0 ≤ t ≤ 8? Indicate the units of measure.

b) Find the value of R′(2). Using correct units, interpret the meaning of the value in the context of the problem.

c) Find all times, t, for which the rate of change in the amount of water is equal to the average rate of change of R over the interval 0 ≤ t ≤ 8.

d) The reservoir can hold a maximum of 30 cubic feet. What is the first time after t = 8 hours the reservoir will reach its capacity?

Answer 1

R = 20 - 15sin(t²/25) ft³ from t=0 to 8 hours

a) Avg. R' = (Integral of R'dt from t = 0 to 8 )/ 8 hours

Or just (R(8) - R(0))/8 = (-15sin(64/25))/8 = -1.03 ft³/hr

b) R'(2) = -15cos(t²/25)(2t/25) at t = 2 or (-6/5)(2)cos(4/25) = -(12/5)cos(4/25) ft³/hour

c) The times when R'(t) = R'_avg : (-6/5)tcos(t²/25) = -1.03 from t = 0 to 8

Graph and trace zeroes of R'(t) + 1.03 (t = .859 and 5.972) hours

d) R holds a max of 30 ft³. What is first time after 8 hours that R(t) reaches this? This will be when

-15sin(t²/25) = 10

t = 5sqrt(sin^-1(-2/3)) Inverse sin is only defined from -pi/2 to pi/2. You get an angle value of -.7297 which you translate to the first positive repeat which happens at an angle of π + .7297 (going around c-clockwise from 4th to 3rd quadrant)

t = 5sqrt(π+.7297) = 9.838 hours

I think that these are all right. Involved problem.

Please consider a tutor. Take care.