Page:Calculus Made Easy.pdf/180

of the experiment (i.e. when $t=0$) it is $72^{\circ }$ hotter than the surrounding objects, and if the time-constant of its cooling is $20$ minutes (that is, if it takes $20$ minutes for its excess of temperature to fall to ${\dfrac {1}{\epsilon }}$ part of $72^{\circ }$), then we can calculate to what it will have fallen in any given time $t$. For instance, let $t$ be $60$ minutes. Then ${\dfrac {t}{T}}=60\div 20=3$, and we shall have to find the value of $\epsilon ^{-3}$, and then multiply the original $72^{\circ }$ by this. The table shows that $\epsilon ^{-3}$ is $0.0498$. So that at the end of $60$ minutes the excess of temperature will have fallen to $72^{\circ }\times 0.0498=3.586^{\circ }$.

 

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Further Examples.

(1) The strength of an electric current in a conductor at a time $t$ secs. after the application of the electromotive force producing it is given by the expression $C={\dfrac {E}{R}}\left\{1-\epsilon ^{-{\frac {Rt}{L}}}\right\}$.

The time constant is ${\dfrac {L}{R}}$.

If $E=10$, $R=1$, $L=0.01$; then when $t$ is very large the term $\epsilon ^{-{\frac {Rt}{L}}}$ becomes $1$, and $C={\dfrac {E}{R}}=10$; also

${\frac {L}{R}}=T=0.01$.

Its value at any time may be written:

$C=10-10\epsilon ^{-{\frac {t}{0.01}}}$,