I understand that $\frac{dy}{dx} = k*y$ and when $k>0$ this is the law of natural growth and when $k<0%$ this is the law of natural decay, but my textbook gives an example of radioactive decay as follows which confuses me:

Radioactive substances decay by spontaneously emitting radiation. If is the mass remaining from an initial mass of the substance after time t, then the relative decay rate

$\frac{-1}{m}\frac{dm}{dt}$ (1)

has been found experimentally to be constant. (Since $\frac{dm}{dt}$ is negative, the relative decay rate is positive.) It follows that:

$\frac{dm}{dt}=km$ (2)

where k is a negative constant. In other words, radioactive substances decay at a rate proportional to the remaining mass. This means that we can use to show that the mass decays exponentially:

$m(t)=m(0)e^{kt}$ (This equation I understand and accept, the above two confuses me)

Now eventually when the example becomes numerical, then of course k becomes a negative number since it is natural decay. However, the above general notation explanation confuses me because it keeps changing the sign of $k$=relative decay rate(from negative(1) to positive(2), but eventually when numerically worked it turns out to be a negative constant since it's natural decay). I know that $k$ must be less than zero for decay but I'm just trying to fully grasp the notation signs that the textbook uses in the explanation above. Please clarify this. Thank you.

  • 1
    $\begingroup$ The relative decay rate given by (1) is positive. The value of $k$ in equation (2) is negative. This isn't a contradiction because $k$ isn't actually the relative decay rate! You may find it instructive to start from equation (2) and attempt to find the value of (1), that is, $-\frac{1}{m}\frac{dm}{dt}$. $\endgroup$ – Rahul Nov 10 '14 at 7:29
  • $\begingroup$ What is your question, exactly? $\endgroup$ – Najib Idrissi Nov 10 '14 at 9:11
  • $\begingroup$ @Rahul Is the relative decay rate positive in (1) because $\frac{dm}{dt}$ is negative, thereby making it double negative which is positive? Also could you please clarify what you mean when you say k isn't the relative decay rate, I thought it was. $\endgroup$ – Eric Nov 10 '14 at 12:25

On one hand you can say k is defined positive. Then -k is negative. The differential equation in case of decay is then $\frac{dm}{dt}=-km$. The solution then is $m=C\cdot e^{-kt}$

On the other hand you can say that k can be positive or negative. Then the differential eqution is $\frac{dm}{dt}=km$ The solution then is $m=C\cdot e^{kt}$ If you have an exponential decay k gets negative.

In both cases you get a negative expoenent, if you have an exponential decay.


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