A certain cylindrical tank holds 20,000 gallons of water, which can be drained from the bottom of the tank in 20 minutes. The volume V of water remaining in the tank after t minutes is given by the function V(t) = 20,000(1- t/20)^2, where V is in gallons, t = 0 represents the instant the tank starts draining. The average rate of change in volume of water in the tank is -1,000 from t=0 to t = 20. At what time t is the instantaneous rate of the water draining from the tank at 1,000 gallons per minute?

a. t = 12 minutes
b. t = 11 minutes
c. t = 10 minutes
d. t = 8 minutes
e. t = 9 minutes

Huh ?

dv /dt = -1000

v = 20,000 (1 - t/20)^2
dv/dt = 20,000 * 2 (1-t/20) (-1/20)
= -2000*(1 - t/20) = -1000 we want
0.5 = 1-t/20
0.5 = t/20
t = 10

Well, well, well! It seems like we have a draining situation here. Let's dive right into it and find the answer!

We know that the instantaneous rate of change of the volume of water is equal to the derivative of the volume function. So, let's find that derivative and set it equal to -1,000 gallons per minute, because that's when the instantaneous rate of water draining is 1,000 gallons per minute.

The derivative of V(t) = 20,000(1- t/20)^2 can be found using the power rule. Are you ready for some math, or should I add a splash of humor before diving in? :D

To find the time at which the instantaneous rate of the water draining from the tank is 1,000 gallons per minute, we need to find the derivative of the volume function.

Given that the volume function is V(t) = 20,000(1- t/20)^2, we differentiate it with respect to t to obtain the derivative dV/dt:

dV/dt = 20,000 * 2 * (1 - t/20) * (-1/20)

Simplifying the expression, we have:

dV/dt = -2,000 * (1 - t/20)

To find the time at which the instantaneous rate of the water draining is 1,000 gallons per minute, we set dV/dt equal to 1,000 and solve for t:

1,000 = -2,000 * (1 - t/20)

Dividing both sides by -2,000:

1/2 = 1 - t/20

Subtracting 1 from both sides:

-1/2 = -t/20

Multiplying both sides by -20:

10 = t

Therefore, the answer is:

c. t = 10 minutes

To find the time (t) at which the instantaneous rate of water draining from the tank is 1,000 gallons per minute, we need to find the value of t that satisfies the equation:

V'(t) = 1,000

Here, V'(t) represents the derivative of the volume function V(t) with respect to time. To find V'(t), we can differentiate the given volume function:

V(t) = 20,000(1 - t/20)^2

Using the chain rule, we can find V'(t):

V'(t) = 20,000 * 2 * (1 - t/20) * (-1/20)

Simplifying further, we get:

V'(t) = -2,000(1 - t/20)

Now, we set V'(t) equal to 1,000 and solve for t:

-2,000(1 - t/20) = 1,000

Dividing both sides by -2,000:

1 - t/20 = -1/2

Subtracting 1 from both sides:

-t/20 = -1/2 - 1

-t/20 = -3/2

Multiplying both sides by -20:

t = (3/2) * 20

t = 30

Therefore, the instantaneous rate of water draining from the tank is 1,000 gallons per minute at t = 30 minutes.

However, t = 30 minutes is not among the options provided. To choose the closest option, we can observe that the volume of water remaining in the tank after 30 minutes is 0, indicating that the tank is empty at that time. Since the tank starts draining at t = 0 and we are looking for a time before the tank is empty, the closest option would be the one immediately before t = 30.

Considering the options given, we find that t = 29 is the closest option to t = 30. Therefore, the answer is:

t = 29 minutes