If a tank holds 5500 gallons of water, which drains from the bottom of the tank in 50 minutes, then Toricelli's Law gives the volume V of water remaining in the tank after t minutes as

Question

If a tank holds 5500 gallons of water, which drains from the bottom of the tank in 50 minutes, then Toricelli's Law gives the volume V of water remaining in the tank after t minutes as v = 5500(1-(1/50)t)^2  0≤ t ≤ 50

Find the rate at which water is draining from the tank after the following amounts of time. (Remember that the rate must be negative because the amount of water in the tank is decreasing.)

a)  5 min gal/min

b)  10 min gal/min

c) 20 min gal/min

d) 50 min gal/min

e) At what time is the water flowing out the fastest?

t = ? min

f) At what time is the water flowing out the slowest?

t = ? min

Answer 1

To find the rate at which the water is leaving the tank, take the derivative of the volume equation with respect to time. Use the power rule and the chain rule:

The power rule means you bring the exponent (2) out to the front and reduce it by 1. The chain rule means you multiply the result by the derivative of the inside (-1/50).

You can simplify this of course by multiplying -1/50 and 2 and 5500 together to get:

dV/dt = -220(1 - 1/50t) = 4.4t - 220

The rest of the problem you can do on your own. The problem questions you have written do not make sense because you first state that you are going to give time but then you give a rate so it's a bit confusing.

Notice that using the simplified equation (dV/dt = 4.4t - 220) the rate is fastest when t = 0 (at t = 0 the rate is -220 gallons/minute) while it is the slowest at t = 50 (at t = 50, the rate is zero, in other words the water is all gone)