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Roman55 [17]
4 years ago
9

THis is for AP calc please help!

Mathematics
1 answer:
Alex_Xolod [135]4 years ago
8 0
Given that the volume of water remaining in the tank after t minutes is given by the function

V(t)=20,000\left(1- \frac{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. \frac{dV}{dt} \right|_{t=4 \frac{1}{2} = \frac{9}{2} }=\left[40,000\left(1- \frac{t}{20} \right)\left(- \frac{1}{20} \right)\right]_{t= \frac{9}{2} } \\  \\ =\left[-2,000\left(1- \frac{t}{20} \right)\right]_{t= \frac{9}{2} }=-2,000\left(1- \frac{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.
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