Page:The New International Encyclopædia 1st ed. v. 04.djvu/32

CALCULUS. 18 y) required to reach the point is therefore (•l + x2)^ o — x „, .. . , . ,, _ -^ — -- — !■ — _i_ — - — The question is, what must be the uunierical value of x in order that this expression for time may have the smallest possible value? As in Example I., the question may be answered by ascertaining the form of the differential coefficient -j- and making it equal zero, so that it may correspond to the minimum magnitude of y. The result is as follows: <^ _ J 1 dx 4i/ir+^ 5 When v is a minimum,-^ = 0. and therefore, die g _l = n 4-/ + j;' a Ui&nce X = 4. To reach the point of destination in iiiiniiiiuui time, the person must therefore land 1 mile (Ti — 4 = 1) from that point. Differentiation and Integration. — It may^)e seen from the examples thus far discussed that the problem of the differential calculus is to obtain for every function considered the differ- dit ential coefficient -^; i.e. the rate of change of the dx function y with respect to its variable x. The process may also be described as follows: To dif- ferentiate a given magnitude, _(/. expressed in terms of its variable, x. is to obtain the value of the in- tinitesimal element dy in terms of the infinitesi- mal element dx of the variable J-. The converse process, called integration, may be described as follows: To integrate means to obtain the func- tion itself, when its rate of change with respect to the variable is given : i.e. to find y in terms of X when— = is given. The problem of the integral calculus may also be stated in the following terms: To integrate is to find the magnitude y in terms of the variable x, when dy (an infinitesi- mal element of j/ 1 is given in terms of d.c (an infinitesimal clement of x) . In the latter defini- tion integration apjjcars as a process of summa- tion, the addends being an infinite number of infinitely small elements. The symbol of inte- gration is r (the media;val S, standing for sitmma). Thus, the symbol dx indicates that it is required to integrate the differential of x. The Constant of Inteyration. — Since a fixed (constant) quantity neither increases nor de- creases, its differential is nothing at all. If, therefore, c denote any constant whatever, we may write rfc = 0. For this reason the difieren- tiai of X + c is simply </x, the same as the differ- ential of jr. Comparing x -- c and x, we see that X is one of an infinite numl)er of possible values of x + c; the latter, namely, equals x in the particular ease where c = 0. When it is required to find in general the integral of dx, we therefore write not x, but x + c. So that I dx = x + c. The constant c is then called the 'constant of integration.' and may either have a finite fixed value, or else may equal zero. 111. l<olulion of the Third Problem. — The last of the cited problems may now be attacked, viz. to determine the work performed when a gas is com- CALCULUS. |>ressed at constant temperature. The difficulty of this problem is in the fact that during com- pression the force is variable, i.e. it must be con- A tinually increased. If we were to suppose the force constant, we would, in calculating the work, commit the greater an error, the greater the amount of compression. Hut suppose the piston to be moved inward only an infinitely small distame. If we then calculate the work required, on the hypothesis that withir that dis- tance the pressure remains constant, we commit only an infinitely small error. In other words, our result is infinitely near the truth. Let dl therefore stand for an infinitely small distance traversed by the piston, let the area of the piston be (1, and let the variable pressure be denoted by /). The work is then padl. But as adl is the rolnnic of the infinitely small cylinder traversed by the piston, it may lie denoted by dr and re- garded as an infinilcsiinal clement of our cylin- drical vessel. The work is thus jidv. To determine now the finite amount of work required to compress the gas from some initial volume, r„ to some final volume, i;, we will first answer the question: How much work would be required in compressing a given :miount of gas from any volume r to unit volume'.' This is ac- complished by 'integrating" pdr, i.e. by ])erform- ing the operation denoted by the symbol | /ido; and our result will be infinitely near the truth, because the error involved in assuming that the pressure p remains constant through the infini- tesimal compression dr is infinitely small. To integrate jidr we must remember that at constant temperature the product of pressure and olumc of a gas is constant: //r = A', whence /; P=-- kdv Substituting this in i>dv, we have / kdp rlcdv Now might be shown, by the meth- od of limit, repeatedly employed in this article, to be the exact differential of either Alogr, or Iclogo + c. c being any constant. We may therefore write, convcr^elv. kdv = /.'logr 4' <■■ the constant of integration c being retained so as to give the solution its more general form. The integral calculus has thus performed for us a wonderful task. Figuratively speaking, it employed a cylinder of volume r filled with gas and compressed it. an infinitesimal amount at a time, until unit volume was attained; it then summed uji the infinitesimal amounts of work jierformed, and it has told us that the total amount of work done is tlogt' + c. The result is algebraic, since r may have any finite value whatever. lUU to obtain a desired particular result, all we have to do is to substitute for u some numerical value. Further, the work per- formed in compressing the gas from some par- ticular volume r, to some jiarticular volume r, evidently equals the difference between the work required to reduce the gas from volume i:, to unit volume, and that reipiired to reduce the gas from volume i'; to unit volume. The rc<iuired