Finding a differential equation to for the amount of chemicals in a body of water I am taking a differential equations course, and most of the problems relate to physical phenomenon. The calculus is not giving me trouble, but the way of approaching the problems is hanging me up. So, here is the question and what I have done so far.
A pond initially contains 1,000,000 gallons of water and an unknown amount of an undesirable chemical. Water containing 0.01 g of this chemical per gallon flows into the pond at a rate of 300 gal/h. The mixture flows out at the same rate, so the amount of water in the pond remains constant. Assume that the chemical is uniformly distributed throughout the pond. Write a differential equation for the amount of chemicals in the pond at any given time.
I try to first conceptualize the problem by writing the following equation:
$$\text{The amount of chemicals in the pound (in gallons)}=1,000,000 \text{ gallons}-\text{The number of gallons of water in the pond}$$
Next, I introduce the notation $Q(t)=`\text{The amount of chemicals in the the pool at time }t\text{'}$ and $W(t)=`\text{the amount of water in the pond at time } t\text{'}$ so that we can represent the amount of water in the pond at any given time as the equation 
$$Q(t)=1,000,000-W(t)$$
But this is not a differential equation, I worked out the $Q'(t)$  and tried to show that the amount of chemicals in the pond is equal to $Q(t-1)+Q'(t)$, but this is far from the answer in the back of the book. The book gives a solution of 
$$\frac{dq}{dt}=300\bigg(\frac 1 {100}-\frac q {1,000,000}\bigg)$$
Where $q$ is in g and $t$ is in h.
 A: Here, you introduce the notation $Q(t)$ which means 'The amount of chemicals in the the pool at time $t$'. It is an important physical quality to get the equation.
In order to get the differential equation, you have to think about what changes with the time goes by, here it is $Q(t)$.
The reason why $Q(t)$ changes is the difference between the income and outgo of chemicals. The income of chemicals per unit time $\Delta t$ is $300\ gal/h \times 0.01\ g/gal \times \Delta t$, and the outgo of chemicals is $300 \ gal/h \times \frac{Q(t)}{1,000,000} \times \Delta t$ due to the amount water of pond is constant $1,000,000\ gal$.
So, you know $$\Delta Q(t) = Q(t+\Delta t) - Q(t) = income - outgo = 300\ gal/h \times 0.01\ g/gal \times \Delta t - 300 \ gal/h \times \frac{Q(t)}{1,000,000} \times \Delta t$$, that is the equation $\frac{\Delta Q(t)}{\Delta t} = 300 (\frac{1}{100} - \frac{Q(t)}{1,000,000})$.
The last step is to set the $\Delta t\rightarrow 0$, and $\frac{\Delta Q(t)}{\Delta t}\rightarrow\frac{dQ(t)}{dt}$.
Now, you get the differential equation.
