A certain cylindrical tank holds 20,000 gallons of water, which can be drained in 20 minutes. The volume V of water remaining in the tank after t minutes is given by the functioin
V(t)=20,000(1- (t/20))^2, where V is in gallons, t=0 represents the instant the tank started draining. How fast is the water draining four and a half minutes after it begins?
A) 1,000 gallons/minute
B) 9,000 gallons/minute
c) 1.550 gallons/minute
d) 3,050 gallons/minute
e) 800 gallons/minute
just another derivative problem.
whatcha got?
It is 1,550 gallons/minute
at time t = 4.5, the tank is draining at the rate of 1,550 gallons/minute
To find the rate at which the water is draining four and a half minutes after it begins, we need to calculate the derivative of the volume function with respect to time.
Given that V(t) = 20,000(1 - t/20)^2, we can differentiate this expression with respect to t using the chain rule.
First, let's simplify the expression by expanding the square:
V(t) = 20,000(1 - t/20)(1 - t/20)
Now, differentiate V(t) with respect to t:
dV/dt = 20,000 * [2(1 - t/20)(-1/20)(-1/20)]
Simplify further:
dV/dt = 20,000 * [2/400 * (1 - t/20)(-1/20)]
dV/dt = (2/400) * 20,000 * (1 - t/20)^(-1/20)
Simplifying,
dV/dt = 10 * (1 - t/20)^(-1/20)
Now, substitute t = 4.5 into the expression:
dV/dt = 10 * (1 - 4.5/20)^(-1/20)
dV/dt = 10 * (1 - 0.225)^(-1/20)
dV/dt ≈ 10 * 0.9925^(-1/20)
Using a calculator, evaluate the expression:
dV/dt ≈ 10 * 1.000527323
dV/dt ≈ 10.0053
So, the rate at which the water is draining four and a half minutes after it begins is approximately 10.0053 gallons/minute.
None of the given options match this result, so there may be an error in the question or the answer choices.