Page:Calculus Made Easy.pdf/178

or $\log _{\epsilon }y=\log _{\epsilon }b-ax\log _{\epsilon }\epsilon =\log _{\epsilon }b-ax$,

and, differentiating, ${\frac {1}{y}}\,{\frac {dy}{dx}}=-a$;

hence ${\frac {dy}{dx}}=b\epsilon ^{-ax}\times (-a)=-ay$; or, in words, the slope of the curve is downward, and proportional to $y$ and to the constant $a$.

We should have got the same result if we had taken the equation in the form

${\begin{aligned}y&=bp^{x};\\{\text{for then}}\quad \quad \quad {\frac {dy}{dx}}&=bp^{x}\times \log _{\epsilon }p.\\{\text{But}}\quad \quad \quad \log _{\epsilon }p&=-a;\\{\text{giving us}}\quad \quad \quad {\frac {dy}{dx}}&=y\times (-a)=-ay,\\{\text{as before.}}\quad \quad \quad \quad \;\end{aligned}}$

The Time-constant. In the expression for the “die-away factor” $\epsilon ^{-at}$, the quantity $a$ is the reciprocal of another quantity known as “the time-constant,” which we may denote by the symbol $T$. Then the die-away factor will be written $\epsilon ^{-{\frac {t}{T}}}$; and it will be seen, by making $t=T$ that the meaning of $T$ (or of ${\dfrac {1}{a}}$) is that this is the length of time which it takes for the original quantity (called $\theta _{0}$ or $Q_{0}$ in the preceding instances) to die away ${\dfrac {1}{\epsilon }}$th part—that is to $0.3678$—of its original value.