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avatar+956 

An oil tank is being drained for cleaning. After t minutes there are V litres of oil left in the tank, where V(t) = 40(20-t)^2, 0 ≤ t ≤ 20. Determine the rate of change of volume at the time t=10 mins.

 

a) -800 L/min 

b) -600 L/min 

c) -400 L/min 

d) -200 L/min 

 

I choose a. I found the derivative and plugged in 10. I was just a bit confused with the restriction. 

Julius  Feb 23, 2018
 #1
avatar+90968 
+2

V(t)  = 40(20 - t)^2 

 

V(t)  =  40 ( t^2 - 40t + 400)

 

V(t)  =  40t^2  - 1600t  + 16000

 

V ' (t)  =   80t - 1600

 

V' (10)   =  80(10) - 1600  =    -800 L / min

 

You were correct  !!!

 

Note, Julius, that the restriction is due to the fact that at 20 min....all the oil is drained from the tank.....

 

 

cool cool cool

CPhill  Feb 23, 2018

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