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If a tank holds 4500 gallons of water, which drains from the bottom of the tank in 50 minutes, then Toricelli's Law gives the volume V of water remaining in the tank after t minutes as V = 4500 1 − 1 50 t 2 0≤ t ≤ 50. Find the rate at which water is draining from the tank after the following amounts of time. (Remember that the rate must be negative because the amount of water in the tank is decreasing.) (a) 5 min gal/min (b) 10 min gal/min (c) 20 min gal/min (d) 50 min gal/min At what time is the water flowing out the fastest? t = min At what time is the water flowing out the slowest? t = min

Answer :

jmonterrozar

Answer:

a. - 162 gal / min

b. - 144 gal / min

c. - 108 gal / min

d. 0 gal / min

Step-by-step explanation:

We have that the V is equal to:

V = 4500 * (1 - t / 50) ^ 2

if we derive with respect to time, we are left with:

dV / dt = 9000 * (1 - t / 50) * (1/50)

dV / dt = 180 * (50 - t / 50)

dv / dt = 3.6 * (50 - t)

a. t = 5

dV / dt = 3.6 * (50 - 5)

dV / dt = 162

therefore - 162 gal / min

b. t = 10

dV / dt = 3.6 * (50 - 20)

dV / dt = 144

therefore - 144 gal / min

c. t = 20

dV / dt = 3.6 * (50 - 20)

dV / dt = 108

therefore - 108 gal / min

d. t = 50

dV / dt = 3.6 * (50 - 50)

dV / dt = 0

therefore 0 gal / min

adioabiola

Here, we are required to find the rate at which water is draining from the tank after the varying amounts of time

The answer are as follows;

(a). dV/dt = -162gal/min

(b). dV/dt = -144gal/min

(c). dV/dt = -108gal/min

(d). dV/dt = 0gal/min

(e). at time, t = 0min

(f). at time, t = 50min

  • Toricelli's law gives the volume, V of water remaining in the tank after t minutes as;

V = 4500 (1 - (t/50))²

  • Therefore, the rate at which water is draining is given by;

dV/dt = (4500 × 2) × (1 - (t/

50)) × (-1/50)

  • Therefore, dV/dt = -180 × ((50 - t)/50) i.e

dV/dt = -3.6 × (50 - t).

  • Therefore,

(a) at time, t = 5min, dV/

dt = -3.6 × (50 - 5)

dV/dt = -162gal/min

(b) at time, t = 10min, dV/

dt = -3.6 × (50 - 10)

dV/dt = -144gal/min

(c) at time, t = 20min, dV/

dt = -3.6 × (50 - 20)

dV/dt = -108gal/min

(d) at time, t = 50min, dV/

dt = -3.6 × (50 - 50)

dV/dt = 0gal/min

(e) Then at what time is the

water flowing out the

fastest;

Since the water is draining out from the bottom of the tank, the water is flowing out the fastest just when it first starts draining (point of highest pressure head) i.e at time, t = 0min

where, dV/dt = -180gal/min.

(f) At what point is the

water flowing out the

slowest;

Since, the water is draining out from the bottom of the tank, the water is flowing out the slowest at time, t = 50min when all of the water is completely drained out of the tank.

where, dV/dt = 0gal/min

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