mixing problem

**superdude** · August 9th 2009, 12:49 PM

tank contains 1000L of water with 15kg of salt dissolved in it. Pure water enters tank at 10L/min and the mixture leaves at the same rate. How much salt is in the tank after t minutes?

Work:
let y(t)=mass of salt in water
y(0)=15kg
dy/dt = rate in - rate out
rate in = 0 because no salt is entering
rate out = $\frac{y(t)}{1000L}$
$\longrightarrow \frac{dy}{dt}=0-{y(t)}{1000L}\times\frac{10L}{min}=\frac{-y(t)}{100min}$
This becomes a seperatable equation so I get $\int\frac{-1}{y(t)}dy=\int\frac{1}{100t} dt = -ln|y| = \frac{ln|t|}{100}$ solving for y I get $\frac{1}{t^{\frac{1}{100}}}$

The answer given is $15e^{\frac{-t}{100}}kg$

**skeeter** · August 9th 2009, 01:21 PM

Originally Posted by superdude

tank contains 1000L of water with 15kg of salt dissolved in it. Pure water enters tank at 10L/min and the mixture leaves at the same rate. How much salt is in the tank after t minutes?

Work:
let y(t)=mass of salt in water
y(0)=15kg
dy/dt = rate in - rate out
rate in = 0 because no salt is entering
rate out = $\frac{y(t)}{1000L}$
$\longrightarrow \frac{dy}{dt}=0-{y(t)}{1000L}\times\frac{10L}{min}=\frac{-y(t)}{100min}$
This becomes a seperatable equation so I get $\int\frac{-1}{y(t)}dy=\int\frac{1}{100t} dt = -ln|y| = \frac{ln|t|}{100}$ solving for y I get $\frac{1}{t^{\frac{1}{100}}}$

The answer given is $15e^{\frac{-t}{100}}kg$

$\frac{dy}{dt} = -\frac{1}{100} y<br />$

separate variables ...

$\frac{dy}{y} = -\frac{1}{100} dt$

integrate ...

$\ln{y} = -\frac{t}{100} + C$

change to an exponential equation ...

$y = Ae^{-\frac{t}{100}}$

utilize the given initial condition, at t = 0 ... y = 15

$15 = Ae^0$

$y = 15e^{-\frac{t}{100}}$

**superdude** · August 9th 2009, 09:42 PM

I get lost after "change to an exponential equation". Are you doing $e^{\ln y} = e^{\frac{-t}{100}} + e^C$ ? If yes, where does the $A$ come from?

**mr fantastic** · August 10th 2009, 01:55 AM

Originally Posted by superdude

I get lost after "change to an exponential equation". Are you doing $e^{\ln y} = e^{\frac{-t}{100}} + e^C$ ? If yes, where does the $A$ come from?

$\ln{y} = -\frac{t}{100} + C$

$\Rightarrow e^{\ln{y}} = e^{-\frac{t}{100} + C} = e^{-\frac{t}{100}} e^C$

using the usual index law

$\Rightarrow y = e^{-\frac{t}{100}} e^C$ .

Since C is arbitrary $e^C$ is also arbitrary and so can be given a new symbol, like A. Therefore:

$y = A e^{-\frac{t}{100}}$ .

Thread: mixing problem

LinkBack

Thread Tools

Display

mixing problem

Similar Math Help Forum Discussions

[SOLVED] Mixing problem

A Mixing Problem....

mixing problem

D.e mixing problem

Mixing Problem

Search Tags