The flow rate of crude oil into a holding tank can be modeled as r(t) = 10(-3.6H

Question

The flow rate of crude oil into a holding tank can be modeled as r(t) = 10(-3.6H + 93.1t + 50.5) cubic feet per minute where t is the number of minutes after the crude oil begins flowing. (a) If the holding tank contains 6000 ft3 of oil when t 0, find a model for the amount of oil in the tank after t minutes. 3.62 Sir_ 50.5t) +60001 93.1t (b) Use your model in part (a) to find how much oil flowed into the tank during the first 14 minutes. (Round your answer to the nearest integer.) (c) If the capacity of the tank is 120,000 ft3, according to the model, how long can the oil flow into the tank before the tank is full? (Round your answer to one decimal place.) min

Explanation / Answer

(a)

R(t)=6000+r(t) dt

R(t)=6000+(10(-3.6t2+93.1t+50.5)) dt

R(t)=6000+(10(-(3.6/3)t3+(93.1/2)t2+50.5t))

(b)

amount of oil flowed into the tank diring first 14 minutes =[0 to 14](10(-3.6t2+93.1t+50.5)) dt

amount of oil flowed into the tank diring first 14 minutes =[0 to 14]10(-(3.6/3)t3+(93.1/2)t2+50.5t)

amount of oil flowed into the tank diring first 14 minutes =10(-(3.6/3)*14+(93.1/2)142+50.5*14) -10(-(3.6/3)03+(93.1/2)02+50.5*0)

amount of oil flowed into the tank diring first 14 minutes =65380 ft3

(c)

R(t)=120000

=>6000+(10(-(3.6/3)t3+(93.1/2)t2+50.5t)) =120000

=>(10(-(3.6/3)t3+(93.1/2)t2+50.5t)) =114000

=>(-(3.6/3)t3+(93.1/2)t2+50.5t) =11400

=>-(3.6/3)t3+(93.1/2)t2+50.5t -11400=0

by calculater , t=23.8348

tank is full after 23.8 minutes

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