Time taken to fill up the bucket
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An empty bucket has a pipe in it. Fluid enters at x l/second. It exits xl/second for each y l in bucket. How long will it take to fill bucket with z l of fluid?
differential-equations
asked Nov 24 at 16:57
Arjun C
11
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up vote
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down vote
favorite
An empty bucket has a pipe in it. Fluid enters at x l/second. It exits xl/second for each y l in bucket. How long will it take to fill bucket with z l of fluid?
differential-equations
asked Nov 24 at 16:57
Arjun C
11
1
This is not precalculus. This is a differential equations problem.
– Ted Shifrin
Nov 24 at 17:07
@TedShifrin Thanks. My bad. I altered the tags.
– xbh
Nov 24 at 17:21
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
An empty bucket has a pipe in it. Fluid enters at x l/second. It exits xl/second for each y l in bucket. How long will it take to fill bucket with z l of fluid?
differential-equations
asked Nov 24 at 16:57
Arjun C
11
An empty bucket has a pipe in it. Fluid enters at x l/second. It exits xl/second for each y l in bucket. How long will it take to fill bucket with z l of fluid?
differential-equations
differential-equations
asked Nov 24 at 16:57
Arjun C
11
asked Nov 24 at 16:57
Arjun C
11
edited Nov 24 at 22:29
asked Nov 24 at 16:57
Arjun C
11
asked Nov 24 at 16:57
Arjun C
11
asked Nov 24 at 16:57
Arjun C
11
11
1
This is not precalculus. This is a differential equations problem.
– Ted Shifrin
Nov 24 at 17:07
@TedShifrin Thanks. My bad. I altered the tags.
– xbh
Nov 24 at 17:21
add a comment |
1
This is not precalculus. This is a differential equations problem.
– Ted Shifrin
Nov 24 at 17:07
@TedShifrin Thanks. My bad. I altered the tags.
– xbh
Nov 24 at 17:21
1
1
This is not precalculus. This is a differential equations problem.
– Ted Shifrin
Nov 24 at 17:07
This is not precalculus. This is a differential equations problem.
– Ted Shifrin
Nov 24 at 17:07
@TedShifrin Thanks. My bad. I altered the tags.
– xbh
Nov 24 at 17:21
@TedShifrin Thanks. My bad. I altered the tags.
– xbh
Nov 24 at 17:21
add a comment |
1 Answer
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I think this is a first order differential equation. If the amount of water in time $t$ is $g(t)$, then the rate of filling the tank is $(1-frac{g(t)}{100})times t$ gallons per second. so you can write the rate of change as $frac{dg}{dt}=(1-frac{g(t)}{100}) t$. now we need the initial condition to solve this problem which is $g(0)=0$. to solve this you can use this simple trick :
$$frac{dg}{1-frac{g(t)}{100}} = tdt Rightarrow -100 ln (1-frac{g}{100}) = t^2 +Const.$$
the $Const.$ term is calculated from initial condition $g(0)=0$ which gives $Const. = 0$.
so $g(t) = 100times(1-e^{frac{-t^2}{100}}) quad gallons.$
you still need to solve for $g(t) = 50$ to get your desired time.
answered Nov 24 at 17:15
K.K.McDonald
944518
Hey, I don't think this is correct. By putting g(t) = 50 , I am getting t as 8.3 , which is clearly not true.
– Arjun C
Nov 24 at 18:45
add a comment |
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1 Answer
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I think this is a first order differential equation. If the amount of water in time $t$ is $g(t)$, then the rate of filling the tank is $(1-frac{g(t)}{100})times t$ gallons per second. so you can write the rate of change as $frac{dg}{dt}=(1-frac{g(t)}{100}) t$. now we need the initial condition to solve this problem which is $g(0)=0$. to solve this you can use this simple trick :
$$frac{dg}{1-frac{g(t)}{100}} = tdt Rightarrow -100 ln (1-frac{g}{100}) = t^2 +Const.$$
the $Const.$ term is calculated from initial condition $g(0)=0$ which gives $Const. = 0$.
so $g(t) = 100times(1-e^{frac{-t^2}{100}}) quad gallons.$
you still need to solve for $g(t) = 50$ to get your desired time.
answered Nov 24 at 17:15
K.K.McDonald
944518
Hey, I don't think this is correct. By putting g(t) = 50 , I am getting t as 8.3 , which is clearly not true.
– Arjun C
Nov 24 at 18:45
add a comment |
up vote
1
down vote
I think this is a first order differential equation. If the amount of water in time $t$ is $g(t)$, then the rate of filling the tank is $(1-frac{g(t)}{100})times t$ gallons per second. so you can write the rate of change as $frac{dg}{dt}=(1-frac{g(t)}{100}) t$. now we need the initial condition to solve this problem which is $g(0)=0$. to solve this you can use this simple trick :
$$frac{dg}{1-frac{g(t)}{100}} = tdt Rightarrow -100 ln (1-frac{g}{100}) = t^2 +Const.$$
the $Const.$ term is calculated from initial condition $g(0)=0$ which gives $Const. = 0$.
so $g(t) = 100times(1-e^{frac{-t^2}{100}}) quad gallons.$
you still need to solve for $g(t) = 50$ to get your desired time.
answered Nov 24 at 17:15
K.K.McDonald
944518
Hey, I don't think this is correct. By putting g(t) = 50 , I am getting t as 8.3 , which is clearly not true.
– Arjun C
Nov 24 at 18:45
add a comment |
up vote
1
down vote
up vote
1
down vote
I think this is a first order differential equation. If the amount of water in time $t$ is $g(t)$, then the rate of filling the tank is $(1-frac{g(t)}{100})times t$ gallons per second. so you can write the rate of change as $frac{dg}{dt}=(1-frac{g(t)}{100}) t$. now we need the initial condition to solve this problem which is $g(0)=0$. to solve this you can use this simple trick :
$$frac{dg}{1-frac{g(t)}{100}} = tdt Rightarrow -100 ln (1-frac{g}{100}) = t^2 +Const.$$
the $Const.$ term is calculated from initial condition $g(0)=0$ which gives $Const. = 0$.
so $g(t) = 100times(1-e^{frac{-t^2}{100}}) quad gallons.$
you still need to solve for $g(t) = 50$ to get your desired time.
answered Nov 24 at 17:15
K.K.McDonald
944518
I think this is a first order differential equation. If the amount of water in time $t$ is $g(t)$, then the rate of filling the tank is $(1-frac{g(t)}{100})times t$ gallons per second. so you can write the rate of change as $frac{dg}{dt}=(1-frac{g(t)}{100}) t$. now we need the initial condition to solve this problem which is $g(0)=0$. to solve this you can use this simple trick :
$$frac{dg}{1-frac{g(t)}{100}} = tdt Rightarrow -100 ln (1-frac{g}{100}) = t^2 +Const.$$
the $Const.$ term is calculated from initial condition $g(0)=0$ which gives $Const. = 0$.
so $g(t) = 100times(1-e^{frac{-t^2}{100}}) quad gallons.$
you still need to solve for $g(t) = 50$ to get your desired time.
answered Nov 24 at 17:15
K.K.McDonald
944518
answered Nov 24 at 17:15
K.K.McDonald
944518
answered Nov 24 at 17:15
K.K.McDonald
944518
answered Nov 24 at 17:15
K.K.McDonald
944518
944518
Hey, I don't think this is correct. By putting g(t) = 50 , I am getting t as 8.3 , which is clearly not true.
– Arjun C
Nov 24 at 18:45
add a comment |
Hey, I don't think this is correct. By putting g(t) = 50 , I am getting t as 8.3 , which is clearly not true.
– Arjun C
Nov 24 at 18:45
Hey, I don't think this is correct. By putting g(t) = 50 , I am getting t as 8.3 , which is clearly not true.
– Arjun C
Nov 24 at 18:45
Hey, I don't think this is correct. By putting g(t) = 50 , I am getting t as 8.3 , which is clearly not true.
– Arjun C
Nov 24 at 18:45
add a comment |
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1
This is not precalculus. This is a differential equations problem.
– Ted Shifrin
Nov 24 at 17:07
@TedShifrin Thanks. My bad. I altered the tags.
– xbh
Nov 24 at 17:21