Water tank question
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There is an empty tank that has a hole in it. Water can enter the tank at the rate of 1 gallon per second. Water leaves the tank through the hole at the rate of 1 gallon per second for each 100 gallons in the tank. How long, in seconds, will it take to fill the 50 gallons of water.
i think the rate is 1-1/100= 99/100. i dont know how to go from here
calculus
edited Jan 26 at 0:32
Rory Daulton
29.5k63355
asked Jan 25 at 23:33
rohan bagrirohan bagri
132
$endgroup$
add a comment |
$begingroup$
There is an empty tank that has a hole in it. Water can enter the tank at the rate of 1 gallon per second. Water leaves the tank through the hole at the rate of 1 gallon per second for each 100 gallons in the tank. How long, in seconds, will it take to fill the 50 gallons of water.
i think the rate is 1-1/100= 99/100. i dont know how to go from here
calculus
edited Jan 26 at 0:32
Rory Daulton
29.5k63355
asked Jan 25 at 23:33
rohan bagrirohan bagri
132
$endgroup$
$begingroup$
Welcome to StackOverflow. The tags you give a question are important, as it helps people to quickly see if they are able to help you. I changed your tag tocalculussince leaky water tank problems are a standard in calculus, and precalculus is unable to directly solve such problems.
$endgroup$
– Rory Daulton
Jan 26 at 0:34
add a comment |
$begingroup$
There is an empty tank that has a hole in it. Water can enter the tank at the rate of 1 gallon per second. Water leaves the tank through the hole at the rate of 1 gallon per second for each 100 gallons in the tank. How long, in seconds, will it take to fill the 50 gallons of water.
i think the rate is 1-1/100= 99/100. i dont know how to go from here
calculus
edited Jan 26 at 0:32
Rory Daulton
29.5k63355
asked Jan 25 at 23:33
rohan bagrirohan bagri
132
$endgroup$
There is an empty tank that has a hole in it. Water can enter the tank at the rate of 1 gallon per second. Water leaves the tank through the hole at the rate of 1 gallon per second for each 100 gallons in the tank. How long, in seconds, will it take to fill the 50 gallons of water.
i think the rate is 1-1/100= 99/100. i dont know how to go from here
calculus
calculus
edited Jan 26 at 0:32
Rory Daulton
29.5k63355
asked Jan 25 at 23:33
rohan bagrirohan bagri
132
edited Jan 26 at 0:32
Rory Daulton
29.5k63355
asked Jan 25 at 23:33
rohan bagrirohan bagri
132
edited Jan 26 at 0:32
Rory Daulton
29.5k63355
edited Jan 26 at 0:32
Rory Daulton
29.5k63355
edited Jan 26 at 0:32
Rory Daulton
29.5k63355
29.5k63355
asked Jan 25 at 23:33
rohan bagrirohan bagri
132
asked Jan 25 at 23:33
rohan bagrirohan bagri
132
asked Jan 25 at 23:33
rohan bagrirohan bagri
132
132
$begingroup$
Welcome to StackOverflow. The tags you give a question are important, as it helps people to quickly see if they are able to help you. I changed your tag tocalculussince leaky water tank problems are a standard in calculus, and precalculus is unable to directly solve such problems.
$endgroup$
– Rory Daulton
Jan 26 at 0:34
add a comment |
$begingroup$
Welcome to StackOverflow. The tags you give a question are important, as it helps people to quickly see if they are able to help you. I changed your tag tocalculussince leaky water tank problems are a standard in calculus, and precalculus is unable to directly solve such problems.
$endgroup$
– Rory Daulton
Jan 26 at 0:34
$begingroup$
Welcome to StackOverflow. The tags you give a question are important, as it helps people to quickly see if they are able to help you. I changed your tag to
calculus since leaky water tank problems are a standard in calculus, and precalculus is unable to directly solve such problems.$endgroup$
– Rory Daulton
Jan 26 at 0:34
$begingroup$
Welcome to StackOverflow. The tags you give a question are important, as it helps people to quickly see if they are able to help you. I changed your tag to
calculus since leaky water tank problems are a standard in calculus, and precalculus is unable to directly solve such problems.$endgroup$
– Rory Daulton
Jan 26 at 0:34
add a comment |
1 Answer
1
active
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$begingroup$
The water doesn't go out at a constant rate, as it depends on how much water is in the tank. This gives the differential equation $y' = 1-frac{y}{100}$ which is $frac{dy}{dt} +frac{y}{100} = 1$. Multiplying by the integrating factor $e^frac{t}{100}$ gives $frac{dy}{dt}e^frac{t}{100} +frac{y}{100}e^frac{t}{100} = e^frac{t}{100}$. By the product rule this is $frac{d}{dt}[ye^frac{t}{100}]=e^frac{t}{100}$. Integrating both sides with respect to t gives $ye^frac{t}{100}=100e^frac{t}{100} + C$ so $y=100 + Ce^{-frac{t}{100}}$. Plugging in the initial condition of $y(0)=0$ we get that $C=-100$, so $y(t)=100-100e^{-frac{t}{100}}$.
We want to find when $y(t)=50$, so $50=100-100e^{-frac{t}{100}}$ which is $frac12=e^{-frac{t}{100}}$, so $t=100ln(2)$.
answered Jan 26 at 0:20
Erik ParkinsonErik Parkinson
1,17519
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add a comment |
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1 Answer
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$begingroup$
The water doesn't go out at a constant rate, as it depends on how much water is in the tank. This gives the differential equation $y' = 1-frac{y}{100}$ which is $frac{dy}{dt} +frac{y}{100} = 1$. Multiplying by the integrating factor $e^frac{t}{100}$ gives $frac{dy}{dt}e^frac{t}{100} +frac{y}{100}e^frac{t}{100} = e^frac{t}{100}$. By the product rule this is $frac{d}{dt}[ye^frac{t}{100}]=e^frac{t}{100}$. Integrating both sides with respect to t gives $ye^frac{t}{100}=100e^frac{t}{100} + C$ so $y=100 + Ce^{-frac{t}{100}}$. Plugging in the initial condition of $y(0)=0$ we get that $C=-100$, so $y(t)=100-100e^{-frac{t}{100}}$.
We want to find when $y(t)=50$, so $50=100-100e^{-frac{t}{100}}$ which is $frac12=e^{-frac{t}{100}}$, so $t=100ln(2)$.
answered Jan 26 at 0:20
Erik ParkinsonErik Parkinson
1,17519
$endgroup$
add a comment |
$begingroup$
The water doesn't go out at a constant rate, as it depends on how much water is in the tank. This gives the differential equation $y' = 1-frac{y}{100}$ which is $frac{dy}{dt} +frac{y}{100} = 1$. Multiplying by the integrating factor $e^frac{t}{100}$ gives $frac{dy}{dt}e^frac{t}{100} +frac{y}{100}e^frac{t}{100} = e^frac{t}{100}$. By the product rule this is $frac{d}{dt}[ye^frac{t}{100}]=e^frac{t}{100}$. Integrating both sides with respect to t gives $ye^frac{t}{100}=100e^frac{t}{100} + C$ so $y=100 + Ce^{-frac{t}{100}}$. Plugging in the initial condition of $y(0)=0$ we get that $C=-100$, so $y(t)=100-100e^{-frac{t}{100}}$.
We want to find when $y(t)=50$, so $50=100-100e^{-frac{t}{100}}$ which is $frac12=e^{-frac{t}{100}}$, so $t=100ln(2)$.
answered Jan 26 at 0:20
Erik ParkinsonErik Parkinson
1,17519
$endgroup$
add a comment |
$begingroup$
The water doesn't go out at a constant rate, as it depends on how much water is in the tank. This gives the differential equation $y' = 1-frac{y}{100}$ which is $frac{dy}{dt} +frac{y}{100} = 1$. Multiplying by the integrating factor $e^frac{t}{100}$ gives $frac{dy}{dt}e^frac{t}{100} +frac{y}{100}e^frac{t}{100} = e^frac{t}{100}$. By the product rule this is $frac{d}{dt}[ye^frac{t}{100}]=e^frac{t}{100}$. Integrating both sides with respect to t gives $ye^frac{t}{100}=100e^frac{t}{100} + C$ so $y=100 + Ce^{-frac{t}{100}}$. Plugging in the initial condition of $y(0)=0$ we get that $C=-100$, so $y(t)=100-100e^{-frac{t}{100}}$.
We want to find when $y(t)=50$, so $50=100-100e^{-frac{t}{100}}$ which is $frac12=e^{-frac{t}{100}}$, so $t=100ln(2)$.
answered Jan 26 at 0:20
Erik ParkinsonErik Parkinson
1,17519
$endgroup$
The water doesn't go out at a constant rate, as it depends on how much water is in the tank. This gives the differential equation $y' = 1-frac{y}{100}$ which is $frac{dy}{dt} +frac{y}{100} = 1$. Multiplying by the integrating factor $e^frac{t}{100}$ gives $frac{dy}{dt}e^frac{t}{100} +frac{y}{100}e^frac{t}{100} = e^frac{t}{100}$. By the product rule this is $frac{d}{dt}[ye^frac{t}{100}]=e^frac{t}{100}$. Integrating both sides with respect to t gives $ye^frac{t}{100}=100e^frac{t}{100} + C$ so $y=100 + Ce^{-frac{t}{100}}$. Plugging in the initial condition of $y(0)=0$ we get that $C=-100$, so $y(t)=100-100e^{-frac{t}{100}}$.
We want to find when $y(t)=50$, so $50=100-100e^{-frac{t}{100}}$ which is $frac12=e^{-frac{t}{100}}$, so $t=100ln(2)$.
answered Jan 26 at 0:20
Erik ParkinsonErik Parkinson
1,17519
edited Jan 26 at 4:25
answered Jan 26 at 0:20
Erik ParkinsonErik Parkinson
1,17519
answered Jan 26 at 0:20
Erik ParkinsonErik Parkinson
1,17519
answered Jan 26 at 0:20
Erik ParkinsonErik Parkinson
1,17519
1,17519
add a comment |
add a comment |
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$begingroup$
Welcome to StackOverflow. The tags you give a question are important, as it helps people to quickly see if they are able to help you. I changed your tag to
calculussince leaky water tank problems are a standard in calculus, and precalculus is unable to directly solve such problems.$endgroup$
– Rory Daulton
Jan 26 at 0:34