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
`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.
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.