Calculus Made Easy · Silvanus P. Thompson
LET there be a quantity growing in such a way that the increment of its growth, during a given time, shall always be proportional to its own magnitude. This resembles the process of reckoning interest on money at some fixed rate; for the bigger the capital, the bigger the amount of interest on it in a given time.
Now we must distinguish clearly between two cases, in our calculation, according as the calculation is made by what the arithmetic books call "simple interest," or by what they call "compound interest." For in the former case the capital remains fixed, while in the latter the interest is added to the capital, which therefore increases by successive additions.
(1) At simple interest . Consider a concrete case. Let the capital at start be £ 100 , and let the rate of interest be 10 per cent. per annum. Then the increment to the owner of the capital will be £ 10 every year. Let him go on drawing his interest every year, and hoard it by putting it by in a stocking, or locking it up in his safe. Then, if he goes on for 10 years, by the end of that time he will have received 10 increments of £ 10 each, or £ 100 , making, with the original £ 100 , a total of £ 200 in all. His property will have doubled itself in 10 years. If the rate of interest had been 5 per cent., he would have had to hoard for 20 years to double his property. If it had been only 2 per cent., he would have had to hoard for 50 years. It is easy to see that if the value of the yearly interest is 1/n of the capital, he must go on hoarding for n years in order to double his property.
Or, if y be the original capital, and the yearly interest is y/n , then, at the end of n years, his property will be y+n y/n=2 y.
(2) At compound interest . As before, let the owner begin with a capital of £ 100 , earning interest at the rate of 10 per cent. per annum; but, instead of hoarding the interest, let it be added to the capital each year, so that the capital grows year by year. Then, at the end of one year, the capital will have grown to £ 110 ; and in the second year (still at 10 \% ) this will earn £ 11 interest. He will start the third year with £ 121 , and the interest on that will be £ 12.2 s . ; so that he starts the fourth year with £ 133.2 s . , and so on. It is easy to work it out, and find that at the end of the ten years the total capital will have grown to £ 259. 7s. 6d. . In fact, we see that at the end of each year, each pound will have earned 1/10 of a pound, and therefore, if this is always added on, each year multiplies the capital by 11/10 ; and if continued for ten years (which will multiply by this factor ten times over) will multiply the original capital by 2.59374. Let us put this into symbols. Put y_0 for the original capital; 1/n for the fraction added on at each of the n operations; and y_n for the value of the capital at the end of the n^(th) operation. Then y_n=y_0(1+1/n)^n.
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 . This would be equivalent to calling it 5 \% per half-year; with 20 operations, therefore, at each of which the capital is multiplied by 21/20 . If reckoned this way, by the end of ten years the capital would have grown to £ 265. 6s. 7d. ; for (1+1/20)^(20)=2.653.
But, even so, the process is still not quite fair; for, by the end of the first month, there will be some interest earned; and a half-yearly reckoning assumes that the capital remains stationary for six months at a time. Suppose we divided the year into 10 parts, and reckon a one-percent. interest for each tenth of the year. We now have 100 operations lasting over the ten years; or y_n=£ 100(1+1/100)^(100) which works out to £ 270. 9s. 71/2 d. .
Even this is not final. Let the ten years be divided into 1000 periods, each of 1/100 of a year; the interest being 1/10 per cent. for each such period; then y_n=£ 100(1+1/1000)^(1000) which works out to £ 271. 13s. 10d. .
Go even more minutely, and divide the ten years into 10,000 parts, each 1/1000 of a year, with interest at 1/100 of 1 per cent. Then y_n=£ 100(1+1/10,000)^(10,000) which amounts to £ 271. 16s. 31/2 d. .
Finally, it will be seen that what we are trying to find is in reality the ultimate value of the expression (1+1/n)^n , which, as we see, is greater than 2; and which, as we take n larger and larger, grows closer and closer to a particular limiting value. However big you make n , the value of this expression grows nearer and nearer to the figure 2.71828 … a number never to be forgotten .
Let us take geometrical illustrations of these things. In Fig. 36 , OP stands for the original value. O T is the whole time during which the value is growing. It is divided into 10 periods, in each of which there is an equal step up. Here (d y)/(d x) is a constant; and if each step up is 1/10 of the original OP , then, by 10 such steps, the height is doubled. If we had taken 20 steps, each of half the height shown, at the end the height would still be just doubled. Or n such steps, each of 1/n of the original height O P , would suffice to double the height. This is the case of simple interest. Here is 1 growing till it becomes 2.
In Fig. 37 , we have the corresponding illustration of the geometrical progression. Each of the successive ordinates is to be 1+1/n , that is, (n+1)/n times as high as its predecessor. The steps up are not equal, because each step up is now 1/n of the ordinate at that part of the curve. If we had literally 10 steps, with (1+1/10) for the multiplying factor, the final total would be (1+1/10)^(10) or 2.594 times the original 1. But if only we take n sufficiently large (and the corresponding 1/n sufficiently small), then the final value (1+1/n)^n to which unity will grow will be 2.71828.
Epsilon . To this mysterious number 2.7182818 etc., the mathematicians have assigned as a symbol the Greek letter ε (pronounced epsilon ). All schoolboys know that the Greek letter π (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 π !
What, then, is epsilon ?
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 .
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. It might also be called the organic rate of growing: because it is characteristic of organic growth (in certain circumstances) that the increment of the organism in a given time is proportional to the magnitude of the organism itself.
If we take 100 per cent. as the unit of rate, and any fixed period as the unit of time, then the result of letting 1 grow arithmetically at unit rate, for unit time, will be 2, while the result of letting 1 grow logarithmically at unit rate, for the same time, will be 2.71828 … .
A little more about Epsilon . We have seen that we require to know what value is reached by the expression (1+1/n)^n , when n becomes indefinitely great. Arithmetically, here are tabulated a lot of values (which anybody can calculate out by the help of an ordinary table of logarithms) got by assuming n=2; n=5; n=10 and so on, up to n=10,000 . (1+1/2)^2 =2.25. ; (1+1/5)^5 =2.489. ; (1+1/10)^(10) =2.594. ; (1+1/20)^(20) =2.653. ; (1+1/100)^(100) =2.704. ; (1+1/1000)^(1000) =2.7171. ; (1+1/10,000)^(10,000) =2.7182.
It is, however, worth while to find another way of calculating this immensely important figure.
Accordingly, we will avail ourselves of the binomial theorem, and expand the expression (1+1/n)^n in that well-known way.
The binomial theorem gives the rule that (a+b)^n=a^n +n (a^(n-1) b)/1!+n(n-1) (a^(n-2) b^2)/2! ; +n(n-1)(n-2) (a^(n-3) b^3)/3!+ etc.
Putting a=1 and b=1/n , we get (1+1/n)^n=1+1 +1/2!((n-1)/n)+1/3! ((n-1)(n-2))/(n^2) ; +1/4! ((n-1)(n-2)(n-3))/(n^3)+ etc.
Now, if we suppose n to become indefinitely great, say a billion, or a billion billions, then n-1, n-2 , and n-3 , etc., will all be sensibly equal to n ; and then the series becomes ε=1+1+1/2!+1/3!+1/4!+ etc. …
By taking this rapidly convergent series to as many terms as we please, we can work out the sum to any desired point of accuracy. Here is the working for ten terms:
1.000000 ; dividing by 1 1.000000 ; dividing by 2 0.500000 ; dividing by 3 0.166667 ; dividing by 4 0.041667 ; dividing by 5 0.008333 ; dividing by 6 0.001389 ; dividing by 7 0.000198 ; dividing by 8 0.000025 ; dividing by 9 0.000002 ; Total 2.718281
ε is incommensurable with 1, and resembles π in being an interminable non-recurrent decimal.
The Exponential Series . We shall have need of yet another series.
Let us, again making use of the binomial theorem, expand the expression (1+1/n)^(n x) , which is the same as ε^x when we make n indefinitely great.
ε^x= 1^(n x)+n x (1^(n x-1)(1/n))/1!+n x(n x-1) (1^(n x-2)(1/n)^2)/2! ; +n x(n x-1)(n x-2) (1^(n x-3)(1/n)^3)/3!+ etc. ; = 1+x+1/2! · (n^2 x^2-n x)/(n^2)+1/3! · (n^3 x^3-3 n^2 x^2+2 n x)/(n^3)+ etc. ; = 1+x+(x^2-x/n)/2!+(x^3-(3 x^2)/n+(2 x)/(n^2))/3!+ etc.
But, when n is made indefinitely great, this simplifies down to the following: ε^x=1+x+(x^2)/2!+(x^3)/3!+(x^4)/4!+ etc....
This series is called the exponential series .
The great reason why ε is regarded of importance is that ε^x possesses a property, not possessed by any other function of x , that when you differentiate it its value remains unchanged ; or, in other words, its differential coefficient is the same as itself. This can be instantly seen by differentiating it with respect to x , thus: (d(ε^x))/(d x)=0+1+(2 x)/(1 · 2)+(3 x^2)/(1 · 2 · 3) +(4 x^3)/(1 · 2 · 3 · 4) ; +(5 x^4)/(1 · 2 · 3 · 4 · 5)+ etc. ; or =1+x+(x^2)/(1 · 2)+(x^3)/(1 · 2 · 3) +(x^4)/(1 · 2 · 3 · 4)+ etc. which is exactly the same as the original series.
Now we might have gone to work the other way, and said: Go to; let us find a function of x , such that its differential coefficient is the same as itself. Or, is there any expression, involving only powers of x , which is unchanged by differentiation? Accordingly; let us assume as a general expression that y=A+B x+C x^2+D x^3+E x^4+ etc., (in which the coefficients A, B, C , etc. will have to be determined), and differentiate it. (d y)/(d x)=B+2 C x+3 D x^2+4 E x^3+ etc .
Now, if this new expression is really to be the same as that from which it was derived, it is clear that A must =B ; that C=B/2=A/(1 · 2) ; that D=C/3=A/(1· 2 · 3) ; that E=D/4=A/(1 · 2 · 3· 4) , etc.
The law of change is therefore that y=A(1+x/1+(x^2)/(1 · 2)+(x^3)/(1 · 2 · 3)+(x^4)/(1 · 2 · 3 · 4)+ etc.).
If, now, we take A=1 for the sake of further simplicity, we have y=1+x/1+(x^2)/(1 · 2)+(x^3)/(1 · 2 · 3)+(x^4)/(1 · 2 · 3 · 4)+ etc.
Differentiating it any number of times will give always the same series over again.
If, now, we take the particular case of A=1 , and evaluate the series, we shall get simply when x=1, y=2.718281 etc. ; that is, y=ε; ; when x=2, y=(2.718281 etc.)^2; that is, y=ε^2; ; when x=3, y=(2.718281 etc.)^3; that is, y=ε^3 and therefore when x=x, y=(2.718281 etc.)^x; that is, y=ε^x, thus finally demonstrating that ε^x=1+x/1+(x^2)/(1 · 2)+(x^3)/(1 · 2 · 3)+(x^4)/(1 · 2 · 3 · 4)+ etc.
[ Note .-- How to read exponentials . For the benefit of those who have no tutor at hand it may be of use to state that ε^x is read as " epsilon to the eksth power ;" or some people read it " exponential eks ." So ε^(p t) is read " epsilon to the pee-teeth-power " or " exponential pee tee ." Take some similar expressions:--Thus, ε^(-2) is read " epsilon to the minus two power " or " exponential minus two ." ε^(-a x) is read " epsilon to the minus ay-eksth " or " exponential minus ay-eks ."]
Of course it follows that ε^y remains unchanged if differentiated with respect to y . Also ε^(a x) , which is equal to (ε^a)^x , will, when differentiated with respect to x , be a ε^(a x) , because a is a constant.
Natural or Naperian Logarithms.
Another reason why ε is important is because it was made by Napier, the inventor of logarithms, the basis of his system. If y is the value of ε^x , then x is the logarithm , to the base ε , of y . Or, if then y=ε^x, ; x= log _(ε) y.
The two curves plotted in Figs. 38 and 39 represent these equations.
The points calculated are:
{For Fig. 38
x 0 0.5 1 1.5 2
y 1 1.65 2.71 4.50 7.69
{For Fig. 39
x 1 2 3 4 8
y 0 0.69 1.10 1.39 2.08
It will be seen that, though the calculations yield different points for plotting, yet the result is identical. The two equations really mean the same thing.
As many persons who use ordinary logarithms, which are calculated to base 10 instead of base ε , are unfamiliar with the "natural" logarithms, it may be worth while to say a word about them. The ordinary rule that adding logarithms gives the logarithm of the product still holds good; or log _(ε) a+ log _(ε) b= log _(ε) a b.
Also the rule of powers holds good; n × log _(ε) a= log _(ε) a^n.
But as 10 is no longer the basis, one cannot multiply by 100 or 1000 by merely adding 2 or 3 to the index. One can change the natural logarithm to the ordinary logarithm simply by multiplying it by 0.4343; or log _(10) x =0.4343 × log _(ε) x ; and conversely , log _(ε) x =2.3026 × log _(10) x
Exponential and Logarithmic Equations.
Now let us try our hands at differentiating certain expressions that contain logarithms or exponentials.
Take the equation: y= log _(ε) x First transform this into ε^y=x whence, since the differential of ε^y with regard to y is the original function unchanged (see p. 139 ), (d x)/(d y)=ε^y,
A USEFUL TABLE OF "NAPERIAN LOGARITHMS" (Also called Natural Logarithms or Hyperbolic Logarithms)
Number log _(ε) Number log _(ε)
1 0.0000 6 1.7918
1.1 0.0953 7 1.9459
1.2 0.1823 8 2.0794
1.5 0.4055 9 2.1972
1.7 0.5306 10 2.3026
2.0 0.6931 20 2.9957
2.2 0.7885 50 3.9120
2.5 0.9163 100 4.6052
2.7 0.9933 200 5.2983
2.8 1.0296 500 6.2146
3.0 1.0986 1,000 6.9078
3.5 1.2528 2,000 7.6009
4.0 1.3863 5,000 8.5172
4.5 1.5041 10,000 9.2103
5.0 1.6094 20,000 9.9035
and, reverting from the inverse to the original function, (d y)/(d x)=1/((d x)/(d y))=1/(ε^y)=1/x.
Now this is a very curious result. It may be written (d(log _(ε) x))/(d x)=x^(-1).
Note that x^(-1) is a result that we could never have got by the rule for differentiating powers. That rule ( page 24 ) is to multiply by the power, and reduce the power by 1. Thus, differentiating x^3 gave us 3 x^2 ; and differentiating x^2 gave 2 x^1 . But differentiating x^0 does not give us x^(-1) or 0 × x^(-1) , because x^0 is itself =1 , and is a constant. We shall have to come back to this curious fact that differentiating log _(ε) x gives us 1/x when we reach the chapter on integrating.
Now, try to differentiate y= log _(ε)(x+a), that is ε^y=x+a we have (d(x+a))/(d y)=ε^y , since the differential of ε^y remains ε^y . This gives (d x)/(d y)=ε^y=x+a hence, reverting to the original function (see p. 128 ), we get (d y)/(d x)=1/((d x)/(d y))=1/(x+a).
Next try y= log _(10) x.
First change to natural logarithms by multiplying by the modulus 0.4343. This gives us y =0.4343 log _(ε) x; ; whence (d y)/(d x) =0.4343/x.
The next thing is not quite so simple. Try this: y=a^x.
Taking the logarithm of both sides, we get log _(ε) y =x log _(ε) a, ; or x=(log _(ε) y)/(log _(ε) a) =1/(log _(ε) a) × log _(ε) y.
Since 1/(log _(ε) a) is a constant, we get (d x)/(d y)=1/(log _(ε) a) × 1/y=1/(a^x × log _(ε) a) hence, reverting to the original function. (d y)/(d x)=1/((d x)/(d y))=a^x × log _(ε) a.
We see that, since (d x)/(d y) × (d y)/(d x)=1 and (d x)/(d y)=1/y × 1/(log _(ε) a), 1/y × (d y)/(d x)= log _(ε) a.
We shall find that whenever we have an expression such as log _(ε) y= a function of x , we always have 1/y (d y)/(d x)= the differential coefficient of the function of x , so that we could have written at once, from log _(ε) y= x log _(ε) a , 1/y (d y)/(d x)= log _(ε) a and (d y)/(d x)=a^x log _(ε) a.
Let us now attempt further examples.
Examples.
(1) y=ε^(-a x) . Let -a x=z ; then y=ε^z . (d y)/(d x)=ε^z; (d z)/(d x)=-a; hence (d y)/(d x)=-a ε^(-a x).
Or thus: log _(ε) y=-a x; 1/y (d y)/(d x)=-a; (d y)/(d x)=-a y=-a ε^(-a x).
(2) y=ε^fracx^23 . Let (x^2)/3=z ; then y=ε^z . (d y)/(d z)=ε^z; (d z)/(d x)=(2 x)/3; (d y)/(d x)=(2 x)/3 ε^fracx^23.
Or thus: log _(ε) y=(x^2)/3; 1/y (d y)/(d x)=(2 x)/3; (d y)/(d x)=(2 x)/3 ε^fracx^23.
(3) y=ε^((2 x)/(x+1)) . log _(ε) y =(2 x)/(x+1), 1/y (d y)/(d x)=(2(x+1)-2 x)/((x+1)^2); ; hence (d y)/(d x) =2/((x+1)^2) ε^((2 x)/(x+1)).
Check by writing (2 x)/(x+1)=z .
(4) y=ε^sqrtx^2+a. log _(ε) y=(x^2+a)^(1/2) . 1/y (d y)/(d x)=x/((x^2+a)^(1/2)) and (d y)/(d x)=fracx × ε^sqrtx^2+a(x^2+a)^(1/2).
For if (x^2+a)^(1/2)=u and x^2+a=v, u=v^(1/2) , (d u)/(d v)=1/(2 v^(1/2)); (d v)/(d x)=2 x; (d u)/(d x)=x/((x^2+a)^(1/2)).
Check by writing √(x^2+a)=z .
(5) y= log (a+x^3) . Let (a+x^3)=z ; then y= log _(ε) z . (d y)/(d z)=1/z; (d z)/(d x)=3 x^2; hence (d y)/(d x)=(3 x^2)/(a+x^3).
(6) y= log _(ε)\3 x^2+√(a+x^2)\ . Let 3 x^2+√(a+x^2)=z ; then y= log _(ε) z . (d y)/(d z)=1/z; (d z)/(d x)=6 x+x/(√(x^2+a)); ; (d y)/(d x)=(6 x+x/(√(x^2+a)))/(3 x^2+√(a+x^2))=(x(1+6 √(x^2+a)))/((3 x^2+√(x^2+a)). √(x^2+a))
(7) y=(x+3)^2 √(x-2) . log _(ε) y =2 log _(ε)(x+3)+1/2 log _(ε)(x-2). ; 1/y (d y)/(d x) =2/((x+3))+1/(2(x-2)); ; (d y)/(d x) =(x+3)^2 √(x-2)\2/(x+3)+1/(2(x-2))\.
(8) y=(x^2+3)^3(x^3-2)^(2/3) . log _(ε) y =3 log _(ε)(x^2+3)+2/3 log _(ε)(x^3-2); ; 1/y (d y)/(d x) =3 (2 x)/((x^2+3))+2/3 (3 x^2)/(x^3-2)=(6 x)/(x^2+3)+(2 x^2)/(x^3-2).
For if y= log _(ε)(x^2+3) , let x^2+3=z and u= log _(ε) z . (d u)/(d z)=1/z; (d z)/(d x)=2 x; (d u)/(d x)=(2 x)/(x^2+3).
Similarly, if v= log _(ε)(x^3-2), (d v)/(d x)=(3 x^2)/(x^3-2) ) and (d y)/(d x)=(x^2+3)^3(x^3-2)^(2/3)\(6 x)/(x^2+3)+(2 x^2)/(x^3-2)\.
(9) y=dfracsqrt[2]x^2+asqrt[3]x^3-a . log _(ε) y =1/2 log _(ε)(x^2+a)-1/3 log _(ε)(x^3-a). ; 1/y (d y)/(d x) =1/2 (2 x)/(x^2+a)-1/3 (3 x^2)/(x^3-a)=x/(x^2+a)-(x^2)/(x^3-a) and (d y)/(d x)=fracsqrt[2]x^2+asqrt[3]x^3-a\x/(x^2+a)-(x^2)/(x^3-a)\.
(10) y=1/(log _(ε) x) (d y)/(d x)=(log _(ε) x × 0-1 × 1/x)/(log _(ε)^2 x)=-1/(x log _(ε)^2 x).
(11) y=sqrt[3] log _(ε) x=(log _(ε) x)^(1/3) . Let z= log _(ε) x; y=z^(1/3) . (d y)/(d z)=1/3 z^(-2/3); (d z)/(d x)=1/x; (d y)/(d x)=frac13 x sqrt[3] log _(ε)^2 x.
(12) y=(1/(a^x))^(a x) . log _(ε) y =a x(log _(ε) 1- log _(ε) a^x)=-a x log _(ε) a^x. ; 1/y (d y)/(d x) =-a x × a^x log _(ε) a-a log _(ε) a^x. ; and (d y)/(d x) =-(1/(a^x))^(a x)(x × a^(x+1) log _(ε) a+a log _(ε) a^x).
Try now the following exercises.
Exercises XII. (See page 260 for Answers.)
(1) Differentiate y=b(ε^(a x)-ε^(-a x)) .
(2) Find the differential coefficient with respect to t of the expression u=a t^2+2 log _(ε) t .
(3) If y=n^t , find (d(log _(ε) y))/dt .
(4) Show that if y=1/b · (a^(b x))/ log _(ε)a, (d y)/(d x)=a^(b x) .
(5) If w=p v^n , find (d w)/(d v) .
Differentiate
(6) y= log _(ε) x^n .
(7) y=3 ε^(-x/(x-1)) .
(8) y=(3 x^2+1) ε^(-5 x) .
(9) y= log _(ε)(x^a+a) .
(10) y=(3 x^2-1)(√(x)+1) .
(11) y=(log _(ε)(x+3))/(x+3) .
(12) y=a^x × x^a .
(13) 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=a y^2 log _(ε) 1/y 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 ÷ √(ε) .
(14) Find the maximum or minimum of y=x^3- log _(ε) x.
(15) Differentiate y= log _(ε)(a x ε^x) .
(16) Differentiate y=(log _(ε) a x)^3 .
The Logarithmic Curve.
Let us return to the curve which has its successive ordinates in geometrical progression, such as that represented by the equation y=b p^x .
We can see, by putting x=0 , that b is the initial height of y . Then when x=1, y=b p; x=2, y=b p^2; x=3, y=b p^3, etc.
Also, we see that p is the numerical value of the ratio between the height of any ordinate and that of the next preceding it. In Fig. 40 , we have taken p as 6/5 ; each ordinate being 6/5 as high as the preceding one.
If two successive ordinates are related together thus in a constant ratio, their logarithms will have a constant difference; so that, if we should plot out a new curve, Fig. 41 , with values of log _(ε)y as ordinates, it would be a straight line sloping up by equal steps. In fact, it follows from the equation, that log _(ε) y= log _(ε) b+x · log _(ε) p, ; whence log _(ε) y- log _(ε) b=x · log _(ε) p.
Now, since log _(ε) p is a mere number, and may be written as log _(ε) p=a , it follows that log _(ε) y/b=a x, and the equation takes the new form y=b ε^(a x).
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 Fig. 42 , where each successive ordinate is 3/4 of the height of the preceding one.
The equation is still y=b p^x but since p is less than one, log _(ε) p will be a negative quantity, and may be written -a ; so that p=ε^(-a) , and now our equation for the curve takes the form y=b ε^(-a x).
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 θ_t=θ_0 ε^(-a t) where θ_0 is the original excess of temperature of a hot body over that of its surroundings, θ_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 ε^(-a t), 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 ε^(-a t) 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, (d y)/(d t) is proportional at every moment to the value that y has at that moment. For we have only to inspect the curve, Fig. 42 above, to see that, at every part of it, the slope (d y)/dx is proportional to the height y ; the curve becoming flatter as y grows smaller. In symbols, thus y=b ε^(-a x) or log _(ε) y= log _(ε) b-a x log _(ε) ε= log _(ε) b-a x, and, differentiating, 1/y (d y)/(d x)=-a hence (d y)/(d x)=b ε^(-a x) ×(-a)=-a y 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 for then y =b p^x; ; (d y)/(d x) =b p^x × log _(ε) p.
But log _(ε) p=-a giving us (d y)/(d x)=y ×(-a)=-a y, as before.
The Time-constant . In the expression for the "die-away factor" ε^(-a t) , 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 ε^(-t/T) ; and it will be seen, by making t=T that the meaning of T (or of 1/a ) is that this is the length of time which it takes for the original quantity (called θ_0 or Q_0 in the preceding instances) to die away 1/ε th part--that is to 0.3678--of its original value.
The values of ε^x and ε^(-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 on p. 157 for convenience.
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 1/ε 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 t/T=60 ÷ 20=3 , and we shall have to find the value of ε^(-3) , and then multiply the original 72^(°) by this. The table shows that ε^(-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^(°) .
THE LAW OF ORGANIC GROWTH
x ε^x ε^(-x) 1-ε^(-x)
0.00 1.0000 1.0000 0.0000
0.10 1.1052 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
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=E/R\1-ε^-Rt/L\ .
The time constant is L/R .
If E=10, R=1, L=0.01 ; then when t is very large the term ε^(-(R t)/L) becomes 1, and C=E/R=10 ; also L/R=T=0.01.
Its value at any time may be written: C=10-10 ε^(-t/0.01), the time-constant being 0.01. This means that it takes 0.01 sec. for the variable term to fall by 1/ε=0.3678 of its initial value 10 ε^(-0/0.01)=10 .
To find the value of the current when t=0.001 sec. , say, t/T=0.1 , ε^(-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., t/T=10; ε^(-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 ε^(-K l) , 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 × ε^(-K × 10) or ε^-10 K=0.82 , and from the table one sees that 10 K=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 × ε^(-0.02 l) , or 0.5=ε^(-0.02 l) . It is found by putting this equation in its logarithmic form, namely, log 0.5=-0.02 × l × log ε, which gives l=(-0.3010)/(-0.02 × 0.4343)=34.7 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 ε^-λ t , where λ 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 λ=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=ε^(-0.00385 t) . log 0.5=-0.00385 t × log ε and t=3 minutes very nearly.
Exercises XIII. (See page 260 for Answers.)
(1) Draw the curve y=b ε^(-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-ε^(-2 t)) .
(4) The following equations give very similar curves: (i) y =(a x)/(x+b); ; (ii) y =a(1-ε^(-x/b)); ; (iii) y =a/(90^(°)) arctan (x/b).
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; (b) y=(ε^x)^x; (c) y=ε^x^x .
(6) For "Thorium A , " the value of λ 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 ε^(-λ 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_0ε^(-t/(K R)) .
(8) The charge Q of an electrified insulated metal sphere is reduced from 20 to 16 units in 10 minutes. Find the coefficient μ of leakage, if Q=Q_0 × ε^(-μ 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 ε^(-β 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 β is a constant. For the Franco-English submarine cable laid in 1910, β=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 ε^(-k h); 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^(1/x) .
(13) Find the minimum or maximum of y=x a^(1/x) .
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End of part 17 of 27
Continue readingChapter XV: How to Deal with Sines and Cosines
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