Math Help - Dillusion Problem

1. Dillusion Problem

An automobile cooling system has a capacity of 17qt. Initially, the system contains a mixture of 2 gal of antifreeze and 9 qt of water. Water runs into the system at the rate of 1 gal/min and the homogeneous mixture runs out a petcock at the same rate. How much anti freeze remains in the system after 5 min?

First I converted the units but after that, I don't know how to solve it.

2. Originally Posted by circuscircus
An automobile cooling system has a capacity of 17qt. Initially, the system contains a mixture of 2 gal of antifreeze and 9 qt of water. Water runs into the system at the rate of 1 gal/min and the homogeneous mixture runs out a petcock at the same rate. How much anti freeze remains in the system after 5 min?

First I converted the units but after that, I don't know how to solve it.
i assume this is for a differential equations class? note that 9 quarts = 2.25 gallons.

Let $A$ be the amount of anti-freeze in the system.

then $A' = \mbox {rate in } - \mbox { rate out}$

$= ( \mbox {concentration in} \times \mbox{rate of flow in}) - (\mbox {concentration out} \times \mbox {rate of flow out})$

$\Rightarrow A' = \frac {0 \mbox { gal}}{\mbox {gal}} \cdot \frac {1 \mbox { gal}}{\mbox {min}} - \frac {A \mbox { gal}}{2.25 \mbox { gal}} \cdot \frac {1 \mbox { gal}}{\mbox {min}}$

$\Rightarrow A' = - \frac 49 A$ .......with the initial condition, $A(0) = 2$

3. Originally Posted by Jhevon
i assume this is for a differential equations class? note that 9 quarts = 2.25 gallons.

Let $A$ be the amount of anti-freeze in the system.

then $A' = \mbox {rate in } - \mbox { rate out}$

$= ( \mbox {concentration in} \times \mbox{rate of flow in}) - (\mbox {concentration out} \times \mbox {rate of flow out})$

$\Rightarrow A' = \frac {0 \mbox { gal}}{\mbox {gal}} \cdot \frac {1 \mbox { gal}}{\mbox {min}} - \frac {A \mbox { gal}}{2.25 \mbox { gal}} \cdot \frac {1 \mbox { gal}}{\mbox {min}}$

$\Rightarrow A' = - \frac 49 A$ .......with the initial condition, $A(0) = 2$

Why is the concentration antifreeze/water and not antifreeze/(antifreeze+water)?

4. Originally Posted by circuscircus
Why is the concentration antifreeze/water and not antifreeze/(antifreeze+water)?
it is not the concentration of antifreeze/water, it is the concentration of antifreeze.

Concentration of a solute in a solvent $= \frac {\mbox {Amount of solute}}{\mbox {Amount of solvent}}$

e.g. what is the concentration of salt solution in which we have 2 grams of salt per gallon of water? 2 grams per gallon, of course, not 2 grams per (gallon + 2 grams)

i think what is confusing you here is the fact that the solute and the solvent have the same units, gallons. but it works out if you think about it

5. $\frac{dA}{dt} = -\frac{4}{9}$

$\int \frac{1}{A} dA = -\int \frac{4}{9} dt$

$ln|A| = -\frac{4}{9}t$

$A = e^{-4/9t}$

$A(5)=e^{-4/9(5)}$

is this right?

6. Originally Posted by circuscircus
$\frac{dA}{dt} = -\frac{4}{9}$

$\int \frac{1}{A} dA = -\int \frac{4}{9} dt$

$ln|A| = -\frac{4}{9}t$

$A = e^{-4/9t}$

$A(5)=e^{-4/9(5)}$

is this right?
no, you forgot the constant of integration (which you have to solve for using the initial data)

7. Originally Posted by Jhevon
no, you forgot the constant of integration (which you have to solve for using the initial data)
what do you mean? I don't quite understand what that is...

8. Originally Posted by circuscircus
what do you mean? I don't quite understand what that is...
when you integrate something, you add +C correct, to replace any arbitrary constant that may have been lost in differentiation. you need to do that to the right hand side of your differential equation when you integrate both sides. then solve for it using the initial data

9. Ohhh

$A = e^{-4/9t} + C$

$0 = A(0) = e^{-4/9*(0)} + C$

$0 = 1 + C$

$C=-1$

$A(t) = e^{-4/9t} -1$

So is it like this?

10. Originally Posted by circuscircus
Ohhh

$A = e^{-4/9t} + C$

$0 = A(0) = e^{-4/9*(0)} + C$

$0 = 1 + C$

$C=-1$

$A(t) = e^{-4/9t} -1$

So is it like this?
I said $A(0) = 2$, remember? and that is still wrong.

$\ln A = - \frac 49t + C$

$\Rightarrow A = e^{- \frac 49 t + C}$

$\Rightarrow A = Ke^{- \frac 49t}$ for some $K \in \mathbb {R}$

now continue

11. $A=Ke^{-4/9t}$

$A(0)=2=Ke^{-4/9*0}$

$2=K(1)$

$A(t)=2*e^{-4/9t}$

like this?

12. Originally Posted by circuscircus
$A=Ke^{-4/9t}$

$A(0)=2=Ke^{-4/9*0}$

$2=K(1)$

$A(t)=2*e^{-4/9t}$

like this?
yes. and remember, the power of e is $-\frac 49t$, it looks like you have $-\frac 4{9t}$