TITLE: If this isn't the rate out then what exactly /is/ this equation?
QUESTION [2 upvotes]: So I'm having some trouble with mixing problems and understanding some of the terminology. In this question, I thought I solved for the rate out of the solution from tank A (i didn't move on yet because then I got confused). I'm confused because we're already given the rate out is 5 gal/s? So what exactly did I solve for then? 
Thank you in advance!

REPLY [1 votes]: Let $Q_A(t)$ and $Q_B(t)$ be the amount of salt in tanks $A$ and $B$, respectively, at time $t$. Then 
$$\frac{dQ_A}{dt} = \text{rate in $A$} - \text{rate out $A$} = 0 - 5\left(\frac{Q_A}{V_A}\right),$$
where $V_A(t)$ is the volume in $A$ at time $t$. Since the rate of volume entering $A$ equals the rate of volume leaving $A$, the net volume in $A$ does not change. So $V_A(t) = 100$, and thus $\frac{dQ_A}{dt} = -5\frac{Q_A}{100}$, i.e., $\frac{dQ_A}{dt} = -0.05Q_A$. Hence $Q_A(t) = Q_A(0)e^{-0.05t} = 20e^{-0.05t}$, which matches what you have so far. 
To find $Q_B$, we need another differential equation:
$$\frac{dQ_B}{dt} = \text{rate in B} - \text{rate out B} = 5\left(\frac{Q_A}{100}\right) - 2.5\left(\frac{Q_B}{V_B}\right),$$ 
where $V_B(t)$ is the volumne in $B$ at time $t$. Since the rate of volume entering $B$ is $5$ gal/s and the rate volume leaving $B$ is $2.5$ gal/s, the net volume rate is $5 - 2.5 = 2.5$ gal/s. Therefore $V_B(t) = 200 + 2.5t$. Since $Q_A(t) = 20e^{-0.05t}$, we have  
$$\frac{dQ_B}{dt} = e^{-0.05t} - \frac{2.5 Q_B}{200 + 2.5t}$$
or
$$\frac{dQ_B}{dt} + \frac{2.5 Q_B}{200 + 5t} = e^{-0.05t}.$$
Solve this equation by method of integrating factors, using the initial condition $Q_B(0) = 40$. Once you've found $Q_B(t)$, find the value $t^*$ for which $V_B(t^*) = 250$; your answer will be $Q_B(t^*)$.
{"set_name": "stack_exchange", "score": 2, "question_id": 1892257}