If a tank holds 4700 gallons of water, which drains from the bottoms of the tank in 49 minutes, then Torricelli's Law gives the volume V of water remaining in the tank after t minutes as.

V=4700(1-t/49)^2, 0</= t </= 49. Find rate at which the water is draining from the tank after: @ each 4mi--> rate of change, 14 min, 21min

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`V = 4700(1-t/49)^2`

The rate of change of volume of the tank is given by the first derivative.

`(dV)/dt = 4700xx2(1-t/49)xx(-1/49)` ---(1)

If we need the rate of change at any time what we need to do is to substitute the t value to get the rate of change of volume which is denoted by (dV)/dt.

At t = 4

`((dV)/dt)_(t=4) = 4700xx2(1-4/49)xx(-1/49) = -176.177` gal/min

*So at 4 minutes the rate of draining out water is 176.177 gallons per minute. The negative sign is given because the volume is reducing.*

So at 8min then you have to substitute the value t = 8 for equation (1). Like wise you can get t = 4,8,12,16 up to t = 48 because after 49min tank is fully emptied.

Note

For the other two time what you need is to substitute the t values in to the equation (1). I will left it for you.

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