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

Answer 2

The total volume of the water in the tank when the pump is turned on is 1220 gallons.

Given that,

Water is pumped into a tank at a rate of $\rm r (t)=30(1-e^{- 0.16t})$ 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,

We have to determine,

How much water is in the tank after 20 minutes?

According to the question,

Water is pumped into a tank at a rate of $\rm r (t)=30(1-e^{- 0.16t})$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,

Rate; $\rm r (t)=30(1-e^{- 0.16t})$

When the pump is turned on the tank contains 800 gallons of water.

Then,

The tank containing water after 20 minutes is determined by the integration of the rate function with the limits when the time is 0 and after time 20 minutes.

Integration using partial fractions is a chapter that covers some of the most important concepts of integration.

Therefore,

Integrating the function,

$\rm Volume =\int\limits^{20}_0 { 30(1-e^{- 0.16t})} \, dt \\\\Volume = 30\ \left [\int\limits^{20}_0 { 1} \, dt- \int\limits^{20}_0 { e^{- 0.16t}} \, dt\\ \right ]\\\\Volume = 30 \left [ t + \dfrac{e^{-0.16t}}{0.16} \right ]}^{20}_0\\\\Volume = 30 \left [ t + \dfrac{e^{-0.16t}}{0.16} \right ]}^{20}_0\\\\Volume = 30 \left [ (20-0) + \dfrac{e^{-0.16(20)-e^0}}{0.16} \right ]}^$

$\rm Volume = 30 \left [ (20) + \dfrac{e^{-0.16(20)-e^0}}{0.16} \right ]}^$

$\rm Volume = 30 \left [ (20) + \dfrac{e^{-3.2}-e^0}{016} \right ]}^$

$\rm Volume = 30 \left [ (20) + \dfrac{0.40-1}{0.16} \right ]}^$

$\rm Volume = 30 \left [ (20) -3.7\right ]}\\\\ Volume = 420$

Therefore,

The total volume in the tank is,

= 800 gallon + 420 gallon = 1220 gallons

Hence, the total volume in the tank is 1220 gallons.

To know more about Integration click the link given below.

https://brainly.com/question/18651211