# (b)The Die-away Curve

If we were to take $p$ as a proper fraction (less than unity), the curve would obviously tend to sink downwards, as in Figure 42, where each successive ordinate is $\frac{3}{4}$ of the height of the preceding one. The equation is still $y=bp^x;$ but since $p$ is less than one, $\log_\epsilon p$ will be a negative quantity, and may be written $-a$; so that $p=\epsilon^{-a}$, and now our equation for the curve takes the form $y=b\epsilon^{-ax}.$

The importance of this expression is that, in the case where the independent variable is time, the equation represents the course of a great many physical processes in which something is gradually dying away. Thus, the cooling of a hot body is represented (in Newton's celebrated “law of cooling”) by the equation $\theta_t=\theta_0 \epsilon^{-at};$ where $\theta_0$ is the original excess of temperature of a hot body over that of its surroundings, $\theta_t$ the excess of temperature at the end of time $t$, and $a$ is a constant–namely, the constant of decrement, depending on the amount of surface exposed by the body, and on its coefficients of conductivity and emissivity, etc.

A similar formula, $Q_t=Q_0 \epsilon^{-at},$ is used to express the charge of an electrified body, originally having a charge $Q_0$, which is leaking away with a constant of decrement $a$; which constant depends in this case on the capacity of the body and on the resistance of the leakage-path.

Oscillations given to a flexible spring die out after a time; and the dying-out of the amplitude of the motion may be expressed in a similar way.

In fact $\epsilon^{-at}$ serves as a die-away factor for all those phenomena in which the rate of decrease is proportional to the magnitude of that which is decreasing; or where, in our usual symbols, $\dfrac{dy}{dt}$ is proportional at every moment to the value that $y$ has at that moment. For we have only to inspect the curve, Figure 42 above, to see that, at every part of it, the slope $\dfrac{dy}{dx}$ is proportional to the height $y$; the curve becoming flatter as $y$ grows smaller. In symbols, thus $y=b\epsilon^{-ax}$ or $\log_\epsilon y = \log_\epsilon b - ax \log_\epsilon \epsilon = \log_\epsilon b - ax,\\ \text{and, differentiating,}\; \frac{1}{y}\, \frac{dy}{dx} = -a;\\ \text{hence}\; \frac{dy}{dx} = b\epsilon^{-ax} × (-a) = -ay;$ 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{align*} y &= bp^x; \\ \text{for then}\; \frac{dy}{dx} &= bp^x × \log_\epsilon p. \\ \text{But}\; \log_\epsilon p &= -a; \\ \text{giving us}\; \frac{dy}{dx} &= y × (-a) = -ay, \end{align*} as before.

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 $T$ $\left(\text{or of} \dfrac{1}{a}\right)$ 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.

The values of $\epsilon^x$ and $\epsilon^{-x}$ are continually required in different branches of physics, and as they are given in very few sets of mathematical tables, some of the values are tabulated here for convenience.

$x$ $\epsilon^x$ $\epsilon^{-x}$ $1-\epsilon^{-x}$
$0$$1.0000$$1.0000$$0.0000 0.10$$1.1052$$0.9048$$0.0952$
$0.20$$1.2214$$0.8187$$0.1813 0.50$$1.6487$$0.6065$$0.3935$
$0.75$$2.1170$$0.4724$$0.5276 0.90$$2.4596$$0.4066$$0.5934$
$1.00$$2.7183$$0.3679$$0.6321 1.10$$3.0042$$0.3329$$0.6671$
$1.20$$3.3201$$0.3012$$0.6988 1.25$$3.4903$$0.2865$$0.7135$
$1.50$$4.4817$$0.2231$$0.7769 1.75$$5.755$$0.1738$$0.8262$
$2.00$$7.389$$0.1353$$0.8647 2.50$$12.182$$0.0821$$0.9179$
$3.00$$20.086$$0.0498$$0.9502 3.50$$33.115$$0.0302$$0.9698$
$4.00$$54.598$$0.0183$$0.9817 4.50$$90.017$$0.0111$$0.9889$
$5.00$$148.41$$0.0067$$0.9933 5.50$$244.69$$0.0041$$0.9959$
$6.00$$403.43$$0.00248$$0.99752 7.50$$1808.04$$0.00055$$0.99947$
$10.00$$22026.5$$0.000045$$0.999955$

As an example of the use of this table, suppose there is a hot body cooling, and that at the beginning of the experiment (i.e.: when $t = 0$) it is $72°$ 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 $72°$), then we can calculate to what it will have fallen in any given time $t$. For instance, let $t$ be $60$ minutes. Then $\dfrac{t}{T} = 60 ÷ 20 = 3$, and we shall have to find the value of $\epsilon^{-3}$, and then multiply the original $72°$ 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 $72° × 0.0498 = 3.586°$.

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: $C = 10 - 10\epsilon^{-\frac{t}{0.01}},$ the time-constant being $0.01$. This means that it takes $0.01$ sec. for the variable term to fall by $\dfrac{1}{\epsilon} = 0.3678$ of its initial value $10\epsilon^{-\frac{0}{0.01}} = 10$.

To find the value of the current when $t = 0.001 \text{sec.}$, say, $\dfrac{t}{T} = 0.1$, $\epsilon^{-0.1} = 0.9048$ (from table).

It follows that, after $0.001$ sec., the variable term is $0.9048 × 10 = 9.048$, and the actual current is $10 - 9.048 = 0.952$.

Similarly, at the end of $0.1$ sec., $\frac{t}{T} = 10;\quad \epsilon^{-10} = 0.000045;$ the variable term is $10 × 0.000045 = 0.00045$, the current being $9.9995$.

(2) The intensity $I$ of a beam of light which has passed through a thickness $l$ cm. of some transparent medium is $I = I_0\epsilon^{-Kl}$, where $I_0$ is the initial intensity of the beam and $K$ is a “constant of absorption.”

This constant is usually found by experiments. If it be found, for instance, that a beam of light has its intensity diminished by 18% in passing through $10$ cms. of a certain transparent medium, this means that $82 = 100 × \epsilon^{-K×10}$ or $\epsilon^{-10K} = 0.82$, and from the table one sees that $10K = 0.20$ very nearly; hence $K = 0.02$.

To find the thickness that will reduce the intensity to half its value, one must find the value of $l$ which satisfies the equality $50 = 100 × \epsilon^{-0.02l}$, or $0.5 = \epsilon^{-0.02l}$. It is found by putting this equation in its logarithmic form, namely, $\log 0.5 = -0.02 × l × \log \epsilon,$ which gives $l = \frac{-0.3010}{-0.02 × 0.4343} = 34.7 \text{centimetres nearly}.$

(3) The quantity $Q$ of a radio-active substance which has not yet undergone transformation is known to be related to the initial quantity $Q_0$ of the substance by the relation $Q = Q_0 \epsilon^{-\lambda t}$, where $\lambda$ is a constant and $t$ the time in seconds elapsed since the transformation began.

For “Radium $A$,” if time is expressed in seconds, experiment shows that $\lambda = 3.85 × 10^{-3}$. Find the time required for transforming half the substance. (This time is called the “mean life” of the substance.)

We have $0.5 = \epsilon^{-0.00385t}$. \begin{align*} \log 0.5 &= -0.00385t × \log \epsilon; \\ \text{and}\; t &= 3\text{ minutes very nearly}. \end{align*}

### Exercises XIII

(1) Draw the curve $y = b \epsilon^{-\frac{t}{T}}$; where $b = 12$, $T = 8$, and $t$ is given various values from $0$ to $20$.

(2) If a hot body cools so that in $24$ minutes its excess of temperature has fallen to half the initial amount, deduce the time-constant, and find how long it will be in cooling down to $1$ per cent. of the original excess.

(3) Plot the curve $y = 100(1-\epsilon^{-2t})$.

(4) The following equations give very similar curves: \begin{align*} \text{(i)}\ y &= \frac{ax}{x + b}; \\ \text{(ii)}\ y &= a(1 - \epsilon^{-\frac{x}{b}}); \\ \text{(iii)}\ y &= \frac{a}{90°} \arctan \left(\frac{x}{b}\right). \end{align*}

Draw all three curves, taking $a= 100$ millimetres; $b = 30$ millimetres.

(5) Find the differential coefficient of $y$ with respect to $x$, if $(a) y = x^x;\quad (b) y = (\epsilon^x)^x;\quad (c) y = \epsilon^{x^x}.$

(6) For “Thorium $A$,” the value of $\lambda$ is $5$; find the “mean life,” that is, the time taken by the transformation of a quantity $Q$ of “Thorium $A$” equal to half the initial quantity $Q_0$ in the expression $Q = Q_0 \epsilon^{-\lambda t};$ $t$ being in seconds.

(7) A condenser of capacity $K = 4 × 10^{-6}$, charged to a potential $V_0 = 20$, is discharging through a resistance of $10,000$ ohms. Find the potential $V$ after (a ) $0.1$ second; (b ) $0.01$ second; assuming that the fall of potential follows the rule $V = V_0 \epsilon^{-\frac{t}{KR}}$.

(8) The charge $Q$ of an electrified insulated metal sphere is reduced from $20$ to $16$ units in $10$ minutes. Find the coefficient $\mu$ of leakage, if $Q = Q_0 × \epsilon^{-\mu t}$; $Q_0$ being the initial charge and $t$ being in seconds. Hence find the time taken by half the charge to leak away.

(9) The damping on a telephone line can be ascertained from the relation $i = i_0 \epsilon^{-\beta l}$, where $i$ is the strength, after $t$ seconds, of a telephonic current of initial strength $i_0$; $l$ is the length of the line in kilometres, and $\beta$ is a constant. For the Franco-English submarine cable laid in 1910, $\beta = 0.0114$. Find the damping at the end of the cable ($40$ kilometres), and the length along which $i$ is still $8$% of the original current (limiting value of very good audition).

(10) The pressure $p$ of the atmosphere at an altitude $h$ kilometres is given by $p=p_0 \epsilon^{-kh}$; $p_0$ being the pressure at sea-level ($760$ millimetres).

The pressures at $10$, $20$ and $50$ kilometres being $199.2$, $42.2$, $0.32$ respectively, find $k$ in each case. Using the mean value of $k$, find the percentage error in each case.

(11) Find the minimum or maximum of $y = x^x$.

(12) Find the minimum or maximum of $y = x^{\frac{1}{x}}$.

(13) Find the minimum or maximum of $y = xa^{\frac{1}{x}}$.

(1) Let $\dfrac{t}{T} = x$ ($\therefore t = 8x$), and use the Table above.

(2) $T = 34.627$; $159.46$ minutes.

(3) Take $2t = x$; and use the Table above.

(5) (a ) $x^x \left(1 + \log_\epsilon x\right)$;   (b ) $2x(\epsilon^x)^x$;   (c ) $\epsilon^{x^x} × x^x \left(1 + \log_\epsilon x\right)$.

(6) $0.14$ second.

(7) (a ) $1.642$;   (b ) $15.58$.

(8) $\mu = 0.00037$, $31^m \frac{1}{4}$.

(9) $i$ is $63.4$% of $i_0$, $220$ kilometres.

(10) $0.133$, $0.145$, $0.155$, mean $0.144$; $-10.2$%, $-0.9$%, $+77.2$%.

(11) Min. for $x = \dfrac{1}{\epsilon}$.

(12) Max. for $x = \epsilon$.

(13) Min. for $x = \log_\epsilon a$.

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