Question

Water is pumped into a tank at a rate of r (t)=30(1−e− 0.16t) gallons per minute, where t is the number of minutes since the pump was turned on. If the tank contained 800 gallons of water when the pump was turned on, how much water, to the nearest gallon, is in the tank after 20 minutes?

Answer 1

Answer:

The total volume of the water in the tank after 20 minutes = 1220 gallons

Step-by-step explanation:

Rate of water pumped into the tank  r (t) = 30 (1 - $e^{-0.16 t}$ )

Initial volume of water in the tank = 800 gallons

The water in the tank after 20 minutes = Initial volume of water in the tank + Volume of water being pumped in the tank

$V_{total} = V_{i} + V_{pump}$

$V_{pump}$ = $\int\limits^a_b {r(t)} \, dt$

Where a = 0 , b = 20

Put the value of r (t) in above equation we get

$V_{pump}$ = $\int\limits^a_b {30 (1 - e^{-0.16t} )} \, dt$

$V_{pump}$ = $30 [ t + \frac{e^{-0.16t} }{0.16} ]$

$V_{pump}$ = $30[ (20- 0) + \frac{1}{0.16}(e^{-0.16 (20)}- e^{0} )$

$V_{pump}$ = 420 gallon

Now, total volume in the tank

$V_{total} = V_{i} + V_{pump}$

$V_{total} = 800 + 420$

$V_{total} = 1220 \ gallon$

Therefore the total volume of the water in the tank after 20 minutes = 1220 gallons