This problem requires us to solve a first-order linear differential equation. This type of problem, and this method of solving it, requires "only" a basic understanding of integration, and would be demonstrated in the first week or so of an undergraduate class in differential equations.

To put this in a framework that we are used to, suppose we add antifreeze at the rate of 1 quart per unit of time. Then all we have to do is find how long it takes to get to our desired concentration and we automatically get our amount.

Let Q = "the number of quarts of antifreeze in the system at any point in time."

Remember our previous discussion? We start with `Q = 4` quarts of antifreeze in the system. We want to see how much antifreeze we have to add to end up with `Q = 6` quarts of antifreeze in the system.

First, we need to look at the rate of change of `Q`, which we call `Q'` or `dQ/dt`. That rate of change of the amount of antifreeze is determined by how much antifreeze we add minus how much antifreeze we take away. Sound familiar? Anyway, if 1 drop of antifreeze goes in, 1 drop of mixture goes out. That 1 drop of mixture contains `Q/12` drops of antifreeze.

We now have our differential equation in `dQ/dt`. We want an equation in `Q`. We use a carefully timed integration to get rid of the `dQ` and the `dt`.

dQ/dt = 1 - Q/121 drop in minus Q/12 drops outdQ/dt = -(1/12)(Q-12)factor out -(1/12) from RHS[1/(Q-12)] dQ = -(1/12) dtmultiply both sides by dt/(Q-12)ln (Q-12) + C_{1}= -(t/12) + C_{2}integrate both sidesln (Q-12) = -(t/12) + C_{2}- C_{1}move constants to RHSQ-12 = e^{-[t/12]+C2-C1}raise e to both sidesQ-12 = e^{C2-C1}e^{-t/12}adding exponents = multiplyingQ-12 = C e^{-t/12}e^{C2-C1}is just a constantQ = 12 + C e^{-t/12}add 12 to both sides

Now all we have to do is figure out what the constant `C` is. When `t = 0, Q = 4`, so

4 = 12 + C e^{0}C = -8since e^{0}= 1, just subtract 12 from both sides

So, our final solution to the initial value problems is:

Q = 12 - 8e^{-t/12}

We want to know how much time must elapse (and therefore how much antifreeze we will need to add) before `Q = 6`. So we substitute that in and solve for `t`:

6 = 12 - 8e^{-t/12}6 = 8e^{-t/12}subtract 12 from both sides and multiply by -16/8 = e^{-t/12}divide both sides by 88/6 = e^{t/12}invert both sides, since e^{-m}is really 1/(e^{m})ln (8/6) = t/12take the natural log of both sidest = 12 ln (8/6)multiply both sides by 12t = 12 ln (4/3)simplify the fraction (engineers do not seem to like to do this)

You may recall that `t` is the amount of time, but also the number of quarts of antifreeze. Mathematicians will be quite happy with this answer. For all you engineers out there, the answer is `3.4521848694213711292706280719259` quarts, more or less.

We cannot really check our work, at least not in the way Ray proved that his original answer was correct. We can, however, make educated guesses about whether our answer of 3.45 quarts seems right.

First, let's look at our answer. We could expect the answer to be more than 3, since we already know it takes 3 quarts if we let out the mixture all at once. I would not expect it to require a lot more, though. So 3.45 looks good.

Second, we want to make sure that it works when `t = 0`. At that point, we have not added any extra antifreeze, and `Q = 12 - 8 = 4`. So far, so good.

Finally, we can look at what happens if we wait a long time. We would expect that the amount of antifreeze would approach, but never
quite equal, 12. As `t` becomes very large, `e ^{-t/12}` becomes very small, so

So, our three reality checks confirmed what we expected. So from that standpoint, the answer looks right.

Home |

Original Puzzler |

Answer |

How to Solve It |

Challenge Problem |

Differential Equations |

Last modified: |

© 2006 |

Peter C. Scott |