Question

THis is for AP calc please help!

Answer 1

Given that the volume of water remaining in the tank after t minutes is given by the function


`V(t)=20,000\left(1- (t)/(20) \right)^2`

where V is in gallons, 0 ≤ t ≤ 20 is in minutes, and t = 0 represents the instant the tank starts draining.

The rate at which water is draining four and a half minutes after it begins is given by


`\left. (dV)/(dt) \right|_{t=4 (1)/(2) = (9)/(2) }=\left[40,000\left(1- (t)/(20) \right)\left(- (1)/(20) \right)\right]_{t= (9)/(2) } \\ \\ =\left[-2,000\left(1- (t)/(20) \right)\right]_{t= (9)/(2) }=-2,000\left(1- (4.5)/(20) \right) \\ \\ =-2,000(1-0.225)=-2,000(0.775)=-1,550\, gallons\ per\ minute`

Therefore, the water is draining at a rate of 1,550 gallons per minute four ans a half minutes after it begins.