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Topic:
Re: Applications of a simple differential equation:]
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Re: Applications of a simple differential equation:]
Posted:
Sep 23, 1997 11:41 PM


Richard Sisley wrote:
>In a posting about the dimension of the constant "k" in y'=ky when >applied to radioactive decay Mark Snyder said: > >"Yes. One way to do that is to find the #decays in one second at >various times, and graph the ln of the # of decays in one second at time >t against the time t (in sec). The slope of that graph will be negative >k. Note that the dimensions of that slope (and hence of k) will be >1/sec." > >Here is an alternative use of "k" as a slope. The differntial equation >y'=ky implies that the rates of change vary directly as the amounts of >substance yet to decay. This means that a graph of (amounts of >substance yet to decay at a certain of time, rate of decay at that time) >pairs would be a line through the origin. The slope of that line would >be "k." The dimension of the second coordinates of the points on that >line would be "grams per second (or some other units of mass per some >other unit of time)" These numbers would be the ever increasing (yes, >because they are negative numbers moving toward 0) rates of decay as the >decay process continues. The dimensions of the first coordinates would >be "grams" (or some other unit of mass).
No. Or Yes. With your choice, k must be negative in order to represent radioactive decay. But then y' does not represent the "rate of decay," which is customarily considered as positive (no big deal here: it's like considering a positive loss as a negative gainthen losses can decrease by getting less and less positive); y' represents the rate at which the amount of radioactive element is changing, and that's negative. To be sure, this negative quantity is increasing, but I don't think that's a good way of thnking about it. The rate of decay (which is the negative of y') is positive, and decreasing. > >Should not the slope of the line passing through these (remaining mass, >rates of decay) data points have dimension "grams per second per gram"?
Yes, it does, but that's the same as "1/sec." Note that you have to be careful here: it's "(grams/sec)/(grams)" which equals "1/sec." If it were "grams/(sec/gram)," that would be dimensionally equal to "gram^2/sec." This is an important distinction that is often ignored.
> >Approximations of "k" could then be obtained by measuring the mass of >material yet to decay at two different times to get the average rate of >decay in say grams per second and then divide by the mass at the earlier >of the two times.
In principle, yes, but that's hard to measure in applications: for radioactive decay, C12 and C14 are chemically similar, so you can't easily tell them apart. You might think that since C14 is heavier than C12, you could maybe do it by weight, but there's so little C14 to begin with, it's not really practical to do so. It's much easier to just measure the decay rate at various times, then graph as in my earlier post.
> >I think this exchange about dimension has raised a new question. When >we perform some manipulations to "solve" a differential equation or to >create a series approximation of a solution, must we expect the numbers >used in these solutions or series to retain the dimensionality they had >in their "original" context? Is there some kind of metarule about the >retention of dimensionality regardless of the mathematical gymnastics >performed?
Absolutely. But as this discussion has pointed out (and probably to a greater degree than anyone ever wanted to know), it's important to be clear about what the units of physical quantities are. It's also important to realize that when you write down an algebraic expression for some physical quantity, there is an implicit choice of units inherent in the algebraic expression itself. Since, in calculus books, these units are often (usually?) unstated, it is easy to fall into traps that, had the units been explicitly stated, would not have been set.
mark snyder



