A water tank is drained. The following function gives the volume, in liters, of the water remaining in the...

Question

A water tank is drained. The following function gives the volume, in liters, of the water remaining in the tank t minutes after the drain is opened:
V(t) = 3000(1 - 0.05t)^2
What is the instantaneous rate of change of the volume after 10 minutes?
Choose 1 answer:
A) -150 liters per minute
B) -150 minutes per liter
C) 750 liters per minute
D) 750 minutes per liter

SnapAndSolve AI · Accepted Answer

To find the instantaneous rate of change of the volume after 10 minutes, we need to find the derivative of V(t) with respect to t and evaluate it at t = 10.

The function is V(t) = 3000(1 - 0.05t)^2.

Using the chain rule, the derivative V'(t) is:

V'(t) = 3000 * 2 * (1 - 0.05t) * (-0.05)

V'(t) = -300 * (1 - 0.05t)

Now, evaluate V'(t) at t = 10:

V'(10) = -300 * (1 - 0.05 * 10)

V'(10) = -300 * (1 - 0.5)

V'(10) = -300 * 0.5

V'(10) = -150

The instantaneous rate of change of the volume after 10 minutes is -150 liters per minute.

A) -150 liters per minute

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A water tank is drained. The following function gives the volume, in liters, of the water remaining in the...

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