Page:Calculus Made Easy.pdf/180

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

Further Examples.

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

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

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

Its value at any time may be written: