Question

Cleaning pumps remove oil at the rate modeled by the function R, given by r of t equals 2 plus the cosine of the quantity pi times t divided by 11 with t measure in hours and and R(t) measured in gallons per hour. How much oil will the pumping stations remove during the 6-hour period from t

Answer 1

Answer:

15.47 gallons

Explanation:

The pumps remove the oil at the rate modelled by

R(t) = 2 + cos(πt / 11) in gallon per hour

The oil removes in 6hours  period is given by integral of  R(t)

from t = 0 to t = 6

$Q = \int\limits^6_0 {R(t)} \, dt = \int\limits^6_0 {(2+ cos(\frac{\pi t}{11} })) \, dt$

$Q = [2t + \frac{sin(\frac{\pi t}{11}) }{\frac{\pi }{11} } ]_0^6$

$Q = [2(6) + \frac{11}{\pi } sin(\frac{6\pi }{11}) ] - [0 + \frac{11sin (0)}{\pi } ]$

$Q = [12 + \frac{11}{\pi } (0.9898)]\\\\Q = 12 + 3.465694\\\\Q = 15.465694\\\\Q = 15.47 gallons$