#### To determine

**To find:** At which rate the water is draining from the tank after 5 min.

#### Answer

The rate at which the water is draining from the tank after 5 min is, V'(5)=−218.75 kg/m.

#### Explanation

The volume of the remaining water after t minutes is, V(t)=5000(1−t40)2,0≤t≤40.

The rate of change of the volume with respect to time is,

V(t)=5000(1−t40)2V'(t)=2⋅5000(1−t40)(−140)=−250(40−t40)

Substitute *t* = 5 in V'(t),

V'(5)=−250(40−540)=−250(3540)=−250(78)=−218.75

Thus, the rate at which the water is draining from the tank after 5 min is, V'(5)=−218.75 kg/m.

#### To determine

**To find:** The rate at which the water is draining from the tank after 10 min.

#### Answer

The rate at which the water is draining from the tank after 10 min is, V'(10)=−187.5 kg/m.

#### Explanation

From part (a), the rate of change of the volume with respect to time is, V'(t)=−250(40−t40).

Substitute *t* = 10 in V'(t),

V'(10)=−250(40−1040)=−250(3040)=−250(34)=−187.5

Thus, the rate at which the water is draining from the tank after 10 min is, V'(10)=−187.5 kg/m.

#### To determine

**To find:** The rate at which the water is draining from the tank after 20 min.

#### Answer

The rate at which the water is draining from the tank after 20 minutes is, V'(20)=−125 kg/m.

#### Explanation

From part (a), the rate of change of the volume with respect to time is, V'(t)=−250(40−t40).

Substitute *t* = 20 in V'(t),

V'(20)=−250(40−2040)=−250(2040)=−250(12)=−125

Thus, the rate at which the water is draining from the tank after 20 min is, V'(20)=−125 kg/m.

#### To determine

**To find:** The rate at which the water is draining from the tank after 40 min; at what time is the water is draining out fastest and the slowest; summarize the results obtained from the parts (a), (b), (c) and (d).

#### Answer

The rate at which the water is draining from the tank after 40 min is, V'(40)=0 kg/m.

The water is flowing out the fastest at *t *= 5 min. and the slowest at *t* = 40 min.

#### Explanation

From part (a), the rate of change of the volume with respect to time is, V'(t)=−250(40−t40).

Substitute *t* = 40 in V'(t),

V'(40)=−250(40−4040)=−250(040)=0

Thus, the rate at which the water is draining from the tank after 40 min is, V'(40)=0 kg/m.

From the above sub parts, it is identified that the water flow is very fast at *t* = 5 min. and is too slow at *t* = 40 min.

Also, notice that at *t* = 40 min. the velocity is 0 and hence the water flow is very slow.