On true Compound Interest and the Law of Organic Growth

Excerpts

On true Compound Interest and the Law of Organic Growth
It is easy to see that if the value of the yearly interest is $\dfrac{1}{n}$ of the capital, he must go on hoarding for $n$ years in order to double his property.
On true Compound Interest and the Law of Organic Growth
Another reason why $\epsilon$ is important is because it was made by Napier, the inventor of logarithms, the basis of his system.
On true Compound Interest and the Law of Organic Growth
To this mysterious number $2.7182818$ etc., the mathematicians have assigned as a symbol the Greek letter $\epsilon$ (pronounced *epsilon*). All schoolboys know that the Greek letter $\pi$ (called *pi*) stands for $3.141592$ etc.; but how many of them know that *epsilon* means $2.71828$? Yet it is an even more important number than $\pi$!
On true Compound Interest and the Law of Organic Growth
Suppose we were to let $1$ grow at simple interest till it became $2$; then, if at the same nominal rate of interest, and for the same time, we were to let $1$ grow at true compound interest, instead of simple, it would grow to the value *epsilon*.
On true Compound Interest and the Law of Organic Growth
But this mode of reckoning compound interest once a year, is really not quite fair; for even during the first year the £$100$ ought to have been growing. At the end of half a year it ought to have been at least £$105$, and it certainly would have been fairer had the interest for the second half of the year been calculated on £$105$.
On true Compound Interest and the Law of Organic Growth
This process of growing proportionately, at every instant, to the magnitude at that instant, some people call *a logarithmic rate* of growing. Unit logarithmic rate of growth is that rate which in unit time will cause $1$ to grow to $2.718281$.
On true Compound Interest and the Law of Organic Growth
The great reason why $\epsilon$ is regarded of importance is that $\epsilon^x$ possesses a property, not possessed by any other function of $x$, that *when you differentiate it its value remains unchanged*unchanged; or, in other words, its differential coefficient is the same as itself.
On true Compound Interest and the Law of Organic Growth
Note that $x^{-1}$ is a result that we could never have got by the rule for differentiating powers.
On true Compound Interest and the Law of Organic Growth
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.
On true Compound Interest and the Law of Organic Growth
For the benefit of those who have no tutor at hand it may be of use to state that $\epsilon^x$ is read as “*epsilon to the eksth power*;” or some people read it “*exponential eks*.”

Equations

On true Compound Interest and the Law of Organic Growth
y + n\dfrac{y}{n} = 2y
At simple interest, a capital y that earns a yearly interest of y/n for n years has doubled.
On true Compound Interest and the Law of Organic Growth
y_n = y_0\left(1 + \frac{1}{n}\right)^n
Compound interest added n times over the period multiplies the original capital by (1 + 1/n) at each operation, giving the capital after n operations.
On true Compound Interest and the Law of Organic Growth
\frac{d(\log_\epsilon x)}{dx} = x^{-1}
The differential coefficient of the natural logarithm of x with respect to x is 1/x.
On true Compound Interest and the Law of Organic Growth
\epsilon = 1 + 1 + \dfrac{1}{2!} + \dfrac{1}{3!} + \dfrac{1}{4!} + \text{etc}.\ldots
Epsilon, the limit of (1 + 1/n) to the n as n grows, equals the sum of the series 1 + 1 + 1/2! + 1/3! + ...
On true Compound Interest and the Law of Organic Growth
y = \log_\epsilon x
y is the natural (Naperian) logarithm of x, the power to which epsilon must be raised to give x.
On true Compound Interest and the Law of Organic Growth
\log_\epsilon a + \log_\epsilon b = \log_\epsilon ab
The natural logarithm of a product is the sum of the natural logarithms of the factors.
On true Compound Interest and the Law of Organic Growth
y = b\epsilon^{ax}
A quantity growing in geometrical progression with a constant ratio per unit of x follows the exponential curve with initial height b.
On true Compound Interest and the Law of Organic Growth
\log_\epsilon \frac{y}{b}=ax
The natural logarithm of y divided by its initial value b is proportional to x, with constant a.
On true Compound Interest and the Law of Organic Growth
y=b\epsilon^{-ax}
A quantity that dies away exponentially: y starts at b and decays with constant a as x increases.
On true Compound Interest and the Law of Organic Growth
\theta_t=\theta_0 \epsilon^{-at}
The excess of temperature of a hot body over its surroundings falls exponentially with time at a constant rate a.
On true Compound Interest and the Law of Organic Growth
Q_t=Q_0 \epsilon^{-at}
The charge of an electrified body leaking away through a resistance decays exponentially with time.
On true Compound Interest and the Law of Organic Growth
C = \dfrac{E}{R}\left\{1 - \epsilon^{-\efrac{Rt}{L}}\right\}
The strength of an electric current in a conductor rises toward E/R as the exponential term dies away, with time constant L/R.
On true Compound Interest and the Law of Organic Growth
I = I_0\epsilon^{-Kl}
The intensity of a light beam decreases exponentially with the thickness of the transparent medium it passes through.
On true Compound Interest and the Law of Organic Growth
y + n\dfrac{y}{n} = 2y.
At simple interest, if the yearly interest is y/n, then after n years the hoarded property is the original capital y plus n times y/n, which equals 2y, so it has doubled.
On true Compound Interest and the Law of Organic Growth
y_n = y_0\left(1 + \frac{1}{n}\right)^n.
Compound interest added n times, each time by the fraction 1/n of the capital, multiplies the original capital by (1 + 1/n) raised to the power n.
On true Compound Interest and the Law of Organic Growth
y_n = £100 \left( 1 + \tfrac{1}{100} \right)^{100};
Compounding 1 per cent for each of 100 tenth-year periods over ten years grows the £100 capital to about £270 9s 7½d.
On true Compound Interest and the Law of Organic Growth
y_n = £100 \left( 1 + \tfrac{1}{1000} \right)^{1000};
Compounding one-tenth of a per cent for each of 1000 periods over ten years grows the £100 capital to about £271 13s 10d.
On true Compound Interest and the Law of Organic Growth
y_n = £100 \left( 1 + \tfrac{1}{10,000} \right)^{10,000};
Compounding in 10,000 periods of one-thousandth of a year over ten years grows the £100 capital to about £271 16s 3½d.
On true Compound Interest and the Law of Organic Growth
\epsilon^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \text{etc.}\dots
Epsilon to the power x is the infinite series 1 + x + x^2/2! + x^3/3! + ..., called the exponential series.
On true Compound Interest and the Law of Organic Growth
\epsilon^x = 1 + \dfrac{x}{1} + \dfrac{x^2}{1·2} + \dfrac{x^3}{1· 2· 3} + \dfrac{x^4}{1· 2· 3· 4} + \text{etc}.
The series 1 + x + x^2/2! + x^3/3! + ... is shown to equal epsilon to the power x, since it is unchanged by differentiation and equals epsilon when x is 1.
On true Compound Interest and the Law of Organic Growth
y = \log_\epsilon x.
Y is defined as the natural logarithm, to base epsilon, of x, so that x equals epsilon to the power y.
On true Compound Interest and the Law of Organic Growth
\frac{d(\log_\epsilon x)}{dx} = x^{-1}.
The differential coefficient of the natural logarithm of x with respect to x is 1/x.
On true Compound Interest and the Law of Organic Growth
\log_\epsilon a + \log_\epsilon b = \log_\epsilon ab.
The sum of natural logarithms of a and b is the natural logarithm of their product.
On true Compound Interest and the Law of Organic Growth
n × \log_\epsilon a = \log_\epsilon a^n.
n times the natural logarithm of a equals the natural logarithm of a to the power n.
On true Compound Interest and the Law of Organic Growth
p=\epsilon^{-a}
A proper fraction p, less than one, can be written as epsilon to the power minus a, where a is minus the natural logarithm of p.
On true Compound Interest and the Law of Organic Growth
y=bp^x;
A curve whose successive ordinates are in geometrical progression: y equals b times p to the power x.
On true Compound Interest and the Law of Organic Growth
y = b\epsilon^{ax}.
The logarithmic curve, the geometric-progression curve written with epsilon as base, has y equal to b times epsilon to the power a x.
On true Compound Interest and the Law of Organic Growth
\log_\epsilon \frac{y}{b}=ax,
The natural logarithm of y divided by b is a times x, so the logarithm of the ordinate is a straight line in x.
On true Compound Interest and the Law of Organic Growth
y=b\epsilon^{-ax}.
The die-away curve: y equals b times epsilon to the power minus a x, a quantity that decreases exponentially with x.
On true Compound Interest and the Law of Organic Growth
\theta_t=\theta_0 \epsilon^{-at};
The excess of temperature of a cooling hot body above its surroundings falls exponentially with time, as theta_0 times epsilon to the power minus a t.
On true Compound Interest and the Law of Organic Growth
Q_t=Q_0 \epsilon^{-at},
The charge of an electrified body leaking away decays exponentially with time, as Q_0 times epsilon to the power minus a t.
On true Compound Interest and the Law of Organic Growth
Q = Q_0 \epsilon^{-\lambda t}
The quantity of a radio-active substance not yet transformed falls exponentially with time, at a constant rate lambda.

Problems

Exercise XII

Exercise XII, problem 1, p. 153
Differentiate $y=b(\epsilon^{ax} -\epsilon^{-ax})$.
Printed answer:
- $ab(\epsilon^{ax} + \epsilon^{-ax})$.
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes a*b*(exp(a*x) + exp(-a*x))
Exercise XII, problem 10, p. 154
$y=(3x^2-1)(\sqrt{x}+1)$.
Printed answer:
- $\left(\dfrac{6x}{3x^2-1} + \dfrac{1}{2\left(\sqrt x + x\right)}\right) \left(3x^2-1\right)\left(\sqrt x + 1\right)$.
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes (6*x/(3*x**2-1) + 1/(2*(sqrt(x)+x)))*(3*x**2-1)*(sqrt(x)+1)
Exercise XII, problem 11, p. 154
$y=\dfrac{\log_\epsilon(x+3)}{x+3}$.
Printed answer:
- $\dfrac{1 - \log_\epsilon \left(x + 3\right)}{\left(x + 3\right)^2}$.
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes (1 - log(x+3))/(x+3)**2
Exercise XII, problem 12, p. 154
$y=a^x × x^a$.
Printed answer:
- $a^x\left(ax^{a-1} + x^a \log_\epsilon a\right)$.
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes a**x*(a*x**(a-1) + x**a*log(a))
Exercise XII, problem 13, p. 154
It was shown by Lord Kelvin that the speed of signalling through a submarine cable depends on the value of the ratio of the external diameter of the core to the diameter of the enclosed copper wire. If this ratio is called $y$, then the number of signals $s$ that can be sent per minute can be expressed by the formula s=ay^2 _1y; where $a$ is a constant depending on the length and the quality of the materials. Show that if these are given, $s$ will be a maximum if $y=1 ÷ \sqrt{\epsilon}$.
Printed answer:
- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
- other: no printed answer to check
Exercise XII, problem 14, p. 154
Find the maximum or minimum of y=x^3-_x.
Printed answer:
- Min.: $y = 0.7$ for $x = 0.694$.
unverified: no computed check settled this one (yet)
How it was checked
- other: no printed answer to check
- evaluate: passes 0.7
Exercise XII, problem 15, p. 154
Differentiate $y=\log_\epsilon(ax\epsilon^x)$.
Printed answer:
- $\dfrac{1 + x}{x}$.
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes (1 + x)/x
Exercise XII, problem 16, p. 154
Differentiate $y=(\log_\epsilon ax)^3$.
Printed answer:
- $\dfrac{3}{x} (\log_\epsilon ax)^2$.
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes 3/x*log(a*x)**2
Exercise XII, problem 2, p. 153
Find the differential coefficient with respect to $t$ of the expression $u=at^2+2\log_\epsilon t$.
Printed answer:
- $2at + \dfrac{2}{t}$.
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes 2*a*t + 2/t
Exercise XII, problem 3, p. 153
If $y=n^t$, find $\dfrac{d(\log_\epsilon y)}{dt}$.
Printed answer:
- $\log_\epsilon n$.
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes log(n)
Exercise XII, problem 4, p. 153
Show that if $y=\dfrac{1}{b}·\dfrac{a^{bx}}{\log_\epsilon a}$, $\dfrac{dy}{dx}=a^{bx}$.
Printed answer:
- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
- differentiate: no printed answer to check
Exercise XII, problem 5, p. 153
If $w=pn^v$, find $\dfrac{dw}{dv}$.
Printed answer:
- $npv^{n-1}$.
unverified: no computed check settled this one (yet)
How it was checked
- differentiate: the printed answer does not match the problem n*p*v**(n-1)
Exercise XII, problem 6, p. 154
$y=\log_\epsilon x^n$.
Printed answer:
- $\dfrac{n}{x}$.
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes n/x
Exercise XII, problem 7, p. 154
$y=3\epsilon^{-\efrac{x}{x-1}}$.
Printed answer:
- $\dfrac{3\epsilon^{- \frac{x}{x-1}}}{(x - 1)^2}$.
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes 3*exp(-x/(x-1))/(x-1)**2
Exercise XII, problem 8, p. 154
$y=(3x^2+1)\epsilon^{-5x}$.
Printed answer:
- $6x \epsilon^{-5x} - 5(3x^2 + 1)\epsilon^{-5x}$.
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes 6*x*exp(-5*x) - 5*(3*x**2+1)*exp(-5*x)
Exercise XII, problem 9, p. 154
$y=\log_\epsilon(x^a+a)$.
Printed answer:
- $\dfrac{ax^{a-1}}{x^a + a}$.
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes a*x**(a-1)/(x**a + a)

Exercise XIII

Exercise XIII, problem 1, p. 162
Draw the curve $y = b \epsilon^{-\efrac{t}{T}}$; where $b = 12$, $T = 8$, and $t$ is given various values from $0$ to $20$.
Printed answer:
- Let $\dfrac{t}{T} = x$ ($\therefore t = 8x$), and use the Table on [page]littletable.
unverified: no computed check settled this one (yet)
How it was checked
- other: no printed answer to check
Exercise XIII, problem 101, p. 162
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.
Printed answer:
- $0.133$, $0.145$, $0.155$, mean $0.144$; $-10.2$%, $-0.9$%, $+77.2$%.
unverified: no computed check settled this one (yet)
How it was checked
- solve: passes log(760/Rational(1992,10))/10
- evaluate: PASS-LOOSE log(760/199.2)/10
- evaluate: the printed answer does not match the problem 100*(760*exp(-0.144*10) - 199.2)/199.2
Exercise XIII, problem 102, p. 164
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.
Printed answer:
- $0.133$, $0.145$, $0.155$, mean $0.144$; $-10.2$%, $-0.9$%, $+77.2$%.
unverified: no computed check settled this one (yet)
How it was checked
- other: no printed answer to check log(760/42.2)/20
- evaluate: passes log(760/42.2)/20
- evaluate: the printed answer does not match the problem 100*(760*exp(-0.144*20) - 42.2)/42.2
Exercise XIII, problem 103, p. 162
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.
Printed answer:
- $0.133$, $0.145$, $0.155$, mean $0.144$; $-10.2$%, $-0.9$%, $+77.2$%.
unverified: no computed check settled this one (yet)
How it was checked
- solve: passes log(760/Rational(32,100))/50
- evaluate: passes log(760/0.32)/50
- evaluate: the printed answer does not match the problem 100*(760*exp(-0.144*50) - 0.32)/0.32
Exercise XIII, problem 104, p. 164
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.
Printed answer:
- $0.133$, $0.145$, $0.155$, mean $0.144$; $-10.2$%, $-0.9$%, $+77.2$%.
unverified: no computed check settled this one (yet)
How it was checked
- other: no printed answer to check (log(760/199.2)/10 + log(760/42.2)/20 + log(760/0.32)/50)/3
- evaluate: PASS-LOOSE (log(760/199.2)/10 + log(760/42.2)/20 + log(760/0.32)/50)/3
Exercise XIII, problem 11, p. 164
Find the minimum or maximum of $y = x^x$.
Printed answer:
- Min. for $x = \dfrac{1}{\epsilon}$.
verified: the printed answer passed a computed check
How it was checked
- extremum: passes exp(-1)
Exercise XIII, problem 12, p. 164
Find the minimum or maximum of $y = x^{\efrac{1}{x}}$.
Printed answer:
- Max. for $x = \epsilon$.
verified: the printed answer passed a computed check
How it was checked
- extremum: passes E
Exercise XIII, problem 13, p. 164
Find the minimum or maximum of $y = xa^{\efrac{1}{x}}$.
Printed answer:
- Min. for $x = \log_\epsilon a$.
unverified: no computed check settled this one (yet)
How it was checked
- extremum: the printed answer does not match the problem log(a)
Exercise XIII, problem 2, p. 162
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.
Printed answer:
- $T = 34.627$; $159.46$ minutes.
unverified: no computed check settled this one (yet)
How it was checked
- other: no printed answer to check
- evaluate: the printed answer does not match the problem 24/log(2)
- evaluate: PASS-LOOSE 24*log(100)/log(2)
Exercise XIII, problem 3, p. 163
Plot the curve $y = 100(1-\epsilon^{-2t})$.
Printed answer:
- Take $2t = x$; and use the Table on [page]littletable.
unverified: no computed check settled this one (yet)
How it was checked
- other: the record may be misread
Exercise XIII, problem 41
The following equations give very similar curves: align* (i) y &= axx + b; (ii) y &= a(1 - ^-xb); (iii) y &= a90° (xb). align* Draw all three curves, taking $a= 100$ millimetres; $b = 30$ millimetres.
Printed answer:
- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
- other: the record may be misread
Exercise XIII, problem 42
The following equations give very similar curves: align* (i) y &= axx + b; (ii) y &= a(1 - ^-xb); (iii) y &= a90° (xb). align* Draw all three curves, taking $a= 100$ millimetres; $b = 30$ millimetres.
Printed answer:
- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
- other: the record may be misread
Exercise XIII, problem 43
The following equations give very similar curves: align* (i) y &= axx + b; (ii) y &= a(1 - ^-xb); (iii) y &= a90° (xb). align* Draw all three curves, taking $a= 100$ millimetres; $b = 30$ millimetres.
Printed answer:
- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
- other: the record may be misread
Exercise XIII, problem 5a, p. 163
Find the differential coefficient of $y$ with respect to $x$, if (*a*) y = x^x; (*b*) y = (^x)^x; (*c*) y = ^x^x.
Printed answer:
- (*a*) $x^x \left(1 + \log_\epsilon x\right)$; (*b*) $2x(\epsilon^x)^x$; (*c*) $\epsilon^{x^x} × x^x \left(1 + \log_\epsilon x\right)$.
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes x**x*(1 + log(x))
Exercise XIII, problem 5b, p. 163
Find the differential coefficient of $y$ with respect to $x$, if (*a*) y = x^x; (*b*) y = (^x)^x; (*c*) y = ^x^x.
Printed answer:
- (*a*) $x^x \left(1 + \log_\epsilon x\right)$; (*b*) $2x(\epsilon^x)^x$; (*c*) $\epsilon^{x^x} × x^x \left(1 + \log_\epsilon x\right)$.
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes 2*x*exp(x)**x
Exercise XIII, problem 5c, p. 163
Find the differential coefficient of $y$ with respect to $x$, if (*a*) y = x^x; (*b*) y = (^x)^x; (*c*) y = ^x^x.
Printed answer:
- (*a*) $x^x \left(1 + \log_\epsilon x\right)$; (*b*) $2x(\epsilon^x)^x$; (*c*) $\epsilon^{x^x} × x^x \left(1 + \log_\epsilon x\right)$.
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes exp(x**x)*x**x*(1 + log(x))
Exercise XIII, problem 6, p. 162
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 ^-t; $t$ being in seconds.
Printed answer:
- $0.14$ second.
verified: the printed answer passed a computed check
How it was checked
- solve: passes log(2)/5
- evaluate: passes log(2)/5
Exercise XIII, problem 7a, p. 163
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 $V = V_0 \epsilon^{-\efrac{t}{KR}}$.
Printed answer:
- (*a*) $1.642$; (*b*) $15.58$.
unverified: no computed check settled this one (yet)
How it was checked
- other: no printed answer to check
- evaluate: passes V0*exp(-t/(K*R))
Exercise XIII, problem 7b, p. 163
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 $V = V_0 \epsilon^{-\efrac{t}{KR}}$.
Printed answer:
- (*a*) $1.642$; (*b*) $15.58$.
unverified: no computed check settled this one (yet)
How it was checked
- other: no printed answer to check
- evaluate: passes V0*exp(-t/(K*R))
Exercise XIII, problem 81, p. 162
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.
Printed answer:
- $\mu = 0.00037$
- $\mu = 0.00037$, $31^m \frac{1}{4}$. %[** Time units]
verified: the printed answer passed a computed check
How it was checked
- solve: passes log(Rational(5,4))/600
- evaluate: passes log(Rational(20,16))/600
Exercise XIII, problem 82, p. 162
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.
Printed answer:
- $\mu = 0.00037$, $31^m \frac{1}{4}$. %[** Time units]
unverified: no computed check settled this one (yet)
How it was checked
- solve: PASS-ERRATUM log(Rational(5,4))/600
- evaluate: the printed answer does not match the problem 10*log(2)/log(Rational(5,4))
Exercise XIII, problem 91, p. 164
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).
Printed answer:
- $i$ is $63.4$% of $i_0$, $220$ kilometres.
unverified: no computed check settled this one (yet)
How it was checked
- other: no printed answer to check 100*exp(-Rational(114,10000)*40)
- evaluate: passes 100*exp(-Rational(114,10000)*40)
Exercise XIII, problem 92, p. 164
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).
Printed answer:
- $i$ is $63.4$% of $i_0$, $220$ kilometres.
unverified: no computed check settled this one (yet)
How it was checked
- other: no printed answer to check log(Rational(100,8))/Rational(114,10000)
- evaluate: the printed answer does not match the problem log(Rational(100,8))/Rational(114,10000)