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An industrial open water tank, as shown in the picture below, has an inlet pipe and an outlet pipe. The depth of the water in the tank changes continually.

The equation $D(t)=4+0.5t^2-0.25t^3$ gives the depth (in metres) of the water, where $t$ represents the time (in hours) that has lapsed since the depth reading was taken at 09:00

Determine:

1. The depth of the water in the tank at 11:00
2. The rate of change of the depth of the water in the tank at 12:00

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1. After 2 hrs the depth = D(2) =  4+ 0,5(2)^2 - 0,25(2)^3 m
= 4 m

2. D = 4 + 0,5t^2 - 0,25t
dD/dt =  t - 0,75t
At 12:00 (3 hours later) we have
(3) - 0,75(3)
= -3,75m/hr
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