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 $$72^\circ \times 0.0498=3.586^\circ$$.

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

Its value at any time may be written: