# Apostol Calculus, Volume I, Chapter 8.7, Exercises 9 & 10 (p. 320)

I'm actually only having trouble with #10, but #10 relies on the statement and results of #9, so I'll include both questions for completeness.

#9. In a tank are $100$ gallons of brine containing $50$ pounds of dissolved salt. Water runs into the tank at the rate of $3$ gallons per minute, and the concentration is kept uniform by stirring. How much salt is in the tank at the end of one hour if the mixture runs out at a rate of $2$ gallons per minute?

#10. Refer to Exercise 9. Suppose the bottom of the tank is covered with a mixture of salt and insoluble material. Assume that the salt dissolves at a rate proportional to the difference between the concentration of the solution and that of a saturated solution (3 pounds of salt per gallon), and that if the water were fresh 1 pound of salt would dissolve per minute. How much salt will be in solution at the end of one hour?

First, I define $y = f(t) =$ number of pounds of salt in solution at time $t$. To write an equation for $y'$ we want to determine the amount of salt being added by the dissolving substance and the amount being removed by the solution running off. The amount added by the dissolving substance is given by

$$k \left(3 - \dfrac{f(t)}{100+t}\right)$$

Since we are told that the rate is proportional to the difference between $3$ pounds of salt per gallon and the current concentration of salt. Further, we are told that if the concentration of salt is $0$ then the rate is $1$ pound per minute; hence, $3k = 1 \implies k = \frac{1}{3}$.

Then, from #9 we have the rate at which salt is exiting the solution given by

$$-2 \left(\dfrac{f(t)}{100+t}\right)$$

Putting this together we have

$$y' = -2\left(\dfrac{f(t)}{100+t}\right) + 1 - \dfrac{f(t)}{3(100+t)}$$

Thus, giving us the first-order linear differential equation:

$$y' + \dfrac{7}{3(100+t)}y = 1$$

From this we have the unique solution $y = f(t)$ with $f(0) = 50$ given by

\begin{align*} y &= 50 \dfrac{100^{7/3}}{(100+t)^{7/3}} + \dfrac{1}{(100+t)^{7/3}} \int_0^t (100+x)^{7/3} dx\\ &= 50 \dfrac{100^{7/3}}{(100+t)^{7/3}} + \dfrac{3t}{10}\\ \implies f(60) &= 34.7 \text{ pounds of salt} \end{align*}

Unfortunately, the solution Apostol gives is $54.7$ pounds of salt. I cannot seem to find the error that is causing me to be off. The answers have an obvious similarity, but I don't know if that is just coincidence or if it means there is a small(ish) error in there somewhere.

Thanks for your help.

-

## 1 Answer

Your calculation in the third to last displayed equation is off. You should have \eqalign{ {1\over (100+t)^{7/3}}\int_0^t(100+x)^{7/3}\,dx&= \color{maroon}{1\over (100+t)^{7/3}}\color{darkgreen}{ {3\over10}(100+x)^{10/3}}\Bigl|_0^t\cr &={3\over10 (100+t)^{7/3}}\Bigl[ (100+t)^{10/3} - 100^{10/3}\Bigr]\cr &= {3\over10}(100+t)- {3\cdot 100^{10/3}\over10 (100+t)^{7/3}}. }

I suspect you tried to cancel the maroon and darkgreen terms, which is not valid.

Wolfram returns $54.6796$ as the value of your $f$, with this correction, at $t=60$, then.

I did not check your work up to this point.

-
Thanks. That's the mistake. I did it from scratch at least 3 or 4 times and repeated that mistake each time... argh. – user23784 Apr 23 '12 at 22:54
@rar You're welcome. Glad to help. – David Mitra Apr 23 '12 at 23:11