Calculus and Geometry Word Problem The Chemist's Dilemma:
Mary, the chemist, is making a solution in her laboratory, pouring ChemA into ChemB. Mary pours 1 mg of ChemA per second into an inverted 60º cone of vertical length .5 meters containing 10 g of ChemB for 1 minute; however, she does not notice a hole at the bottom of the cone that lets 1 mg of the solution in the cone out every second. When Mary's finished, what percentage of the solution is ChemB? [Assume that ChemA and ChemB mix instantaneously and fully on touch]
In my attempts to solve this problem, I've attempted to construct an integral that would include a function determining the current percentages of ChemA and ChemB in the solution so I could calculate how much of each would be left after losing the 1 mg; however, I soon realized that function, which seemed fairly necessary to me, was actually just the original integral I was trying to find. I didn't know what to do at this recursion so I stopped there, but if anyone cares to help me (I came up with this putzing around and would just love to know the answer), it'd be greatly appreciated.
 A: Let the percentage of ChemB in the fluid at time $t$ be $P(t)$. Then since we apparently start with pure ChemB the initial condition is
$$P(0)=100$$
The shape and size of the cone are irrelevant since the amount of fluid is constant and the chemicals mix instantaneously. Since $1$ mg of the $10$ g of fluid containing ChemB is lost each second (and is replaced with ChemA), the amount of ChemB loses one ten-thousandth of its previous amount each second (as a rate) . So our equation is
$$\frac{dP}{dt}=-0.0001P$$
That is an exponential decay equation, which I'm sure you can solve by separation of variables or by the usual technique for first-order linear ODEs. The ending condition in seconds is $t=60$, so evaluate
$$P(60)$$
Let me know if you cannot solve that differential equation.
(Hat tip to @Aretino: his answer reminded me that I got the units wrong in my first answer.)
A: Let $a(t)$ and $b(t)$ the quantities of ChemA and ChemB present in the cone at time $t$. Let then $v$ be the constant rate at which ChemA is poured, and also the rate at which the solution goes out of the cone. Finally, let $b_0$ be the initial quantity of ChemB in the cone.
Notice that $a(t)+b(t)=b_0$ (constant), so that the percentage of ChemB present at time $t$ is $b(t)/b_0$. It follows that we have the following differential equation for $b(t)$:
$$
{db\over dt}=-{b\over b_0}v,
$$
which has the standard solution:
$$
b(t)=b_0e^{-(v/b_0)t}.
$$
With your data: $b_0=10\ \hbox{g}=10000\ \hbox{mg}$, $v=1\ \hbox{mg}/\hbox{s}$, $t=60\ \hbox{s}$, one gets:
$$
b(t=60\ \hbox{s})=10000\cdot e^{-60/10000}\ \hbox{mg}\approx 9940\ \hbox{mg}.
$$
