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I am working on the following problem:

In recitation a population model was studied in which the natural growth rate of the population of oryx was a constant $k > 0$, so that for small time intervals $\Delta t$ the population change $x(t + \Delta t) − x(t)$ is well approx­imated by $kx(t)\Delta t$. (You also studied the effect of hunting them, but in this problem we will leave that aside.) Measure time in years and the population in kilo-oryx (ko).

A mysterious virus infects the oryxes of the Tana River area in Kenya, which causes the growth rate to decrease as time goes on according to the formula $k(t) = \dfrac{k_0}{(a + t)^2}$ for $t \ge 0$, where $a$ and $k_0$ are certain positive constants.

(a) What are the units of the constant $a$ in “$a + t$,” and of the constant $k_0$?

The problem set gives the following solution to a):

The growth rate $k(t)$ has units years$^{−1}$ (so that $k(t)x(t)\Delta t$ has the same units as $x(t)$). The variable $t$ has units years, so the $a$ added to it must have the same units, and $k_0$ must have units years in order for the units of the fraction to work out.

The part about $a$ and $t$ having the same units makes sense. However, I have doubts about the correctness of this solution. The solution itself says that $k(t)$ has units years$^{−1}$. If $k_0 = k(0)$ ($k(t)$ where $t = 0$), then is it not inconsistent to then say it has units years in the equation $k(t) = \dfrac{k_0}{(a + t)^2}$? Should it not still have units years$^{−1}$?

I would greatly appreciate it if people could please take the time to clarify this.

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  • $\begingroup$ In fact $k_0$ and $a$ should have units of years in order to get $\frac{k_0}{(a+t)^2}$ to have units of years$^{-1}$. Note the denominator is squared. $\endgroup$ – hardmath Jun 26 '17 at 16:46
  • $\begingroup$ @hardmath I agree with that, but is it not inconsistent for the reasons I described: $k_0 = k(0)$ is just $k(t = 0)$, and we know that $k(t)$ has units years$^-1$? So how can we say that $k(t)$ has units years$^{-1}$ and $k_0$ has units years? Is this not inconsistent, since they are the same thing, just with different values for $t$? $\endgroup$ – The Pointer Jun 26 '17 at 16:53
  • $\begingroup$ Clearly $k(0)$ is $k_0/a^2$, not $k_0$. $\endgroup$ – hardmath Jun 26 '17 at 17:19
  • $\begingroup$ @hardmath So $k(t)$ has units years$^{-1}$ only for $t > 0$? Otherwise, it has units years? $\endgroup$ – The Pointer Jun 26 '17 at 17:22
  • $\begingroup$ I don't follow your argument. $k(t)$ should have by definition units years$^{-1}$ for all time $t$. My point is that if you substitute $t=0$ into the formula $k(t) = k_0/(a+t)^2$, then you get $k(0)$. The claim that $k(0)=k_0$ is not justified by repeating that this is so. $\endgroup$ – hardmath Jun 26 '17 at 17:25
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$k_0\neq k(t)|_{t=0}.$

First, one can check this by plugging in 0 into the expression:

$k(0)=\frac{k_0}{a^2}\neq k_0.$

So we conclude that $k_0$ does not represent $k$ at $t=0.$ $k_0$ can be understood as a constant that does not have any "meaning" by itself but rather as a constant used to create a formula, attached to whatever units make the formula work.

I am not knowledgeable at all about this, but perhaps a similar example would be the Gravitational Constant.

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