18. If a tank holds 5000 gallons of water, which drains from the bottom of the tank in 40 minutes, then Torricelli’s Law gives the volume $V$ of water remaining in the tank after $t$ minutes as $$ V=5000\left(1-\frac{1}{40} t\right)^{2} \quad 0 \in t \in 40 $$ Find the rate at which water is draining from the tank after (a) $5 \mathrm{~min}$, (b) $10 \mathrm{~min}$, (c) $20 \mathrm{~min}$, and (d) 40 min. At what time is the water flowing out the fastest? The slowest? Summarize your findings.

Problem 18E

18. If a tank holds 5000 gallons of water, which drains from the bottom of the tank in 40 minutes, then Torricelli’s Law gives the volume $V$ of water remaining in the tank after $t$ minutes as

$$ V=5000\left(1-\frac{1}{40} t\right)^{2} \quad 0 \in t \in 40 $$

Find the rate at which water is draining from the tank after

(a) $5 \mathrm{~min}$, (b) $10 \mathrm{~min}$, (c) $20 \mathrm{~min}$, and

(d) 40 min. At what time is the water flowing out the fastest? The slowest? Summarize your findings.

Step-by-Step Solution

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.