4 hours. The distance of the landing point the point of destination is denoted by 5 — x. To walk this distance at the rate of 5 miles an hour requires 5 —x hours. The total time I call it g) required to reach the point is therefore — ) 2 5 X = 4 + 5 The question is, what must be the numerical 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 ri'ti and making it equal dx zero, so that it may correspond to the minimum magnitude of g. The result is as follows: y 1 dr — 4 When y is a minimum, 0, and therefore, dx 1 = 41/9 whence x = 4 . To reach the point of destination in minimum time, the person must therefore land 1 mile (.1— 4 = 1) from that point.
Diff(renlinlion end Integrntion.—Its seen from the examples thus far discussed that the problem of the differential calculus is to obtain for every function considered the differ ential coefficient i.e. the rate of change of the function y with respect to its variable .r. The process may also be described as follows: To dif ferentiate a given magnitude., (t. expressed in terms of its variable. x. is to obtain the value of the in finitesimal element dy in terms of the infinitesi mal element dr of the variable .r. 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: Le. to find it in terms of djt x when.= is given. The problem of the calculus may also he stated in the following terms: To integrate is to find the magnitude y in terms of the variable x, when dy tan infinitesi mal element of y t is given in terms of infinitesimal element of x). In the latter defini tion integration appears as a process of summa being an infinite nunilwr of infinitely sinall element-. The symbol of inte gration is (the m•dia.val S. standing for SU/HMC) Thus. the symboldx indicates that it is required to integrate the'differential of x.
The constant of hitt !leaf ion.—inee a fixed Icon-tam) quantity neither inereases nor de creases. its differ( OM is nothing at all. lf, therefore, c denote any constant whatever, we may write do = O. For this reason the differen tial of ,e e is simply dr., the sanw as the diner ential I if s. Comparing x c and .r, we sec that (MP of an infinite number of possible value.; (.1 x-r c: the latter. namely. equal, ,r in the
particular ease where e = O. When it is required to find in yi it( riff the integral of dx, we therefore write not x. hut x c. So that r 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.
III. Nolo ion of at Third PriAlt m.—The last of the cited prohlems may now be attacked, viz. to determine the work performed when a gas is coin pressed at •onstara temperature. The difficulty of this problem is in the fact that during pression the force is variable. i.e. it must be eon tinually increased. If we were to suppose the force constant. we would, in calculating the work, commit the greater an error. the greater the aimmeint of compression. But 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 tanve the pressure remains constant, we commit only an infinitely small error. III other words, our result is infinitely near the truth. Let di therefore stand for ;III infinitely small distance traversed by the piston, let the area of the piston be a. and let the variable pressure be denoted by p. The work is then padl. But as ad/ is the rolnine of the infinitely small cylinder traversed by the piston, it. may be denoted by de and re garded as an infinitesimal element of our cylin drical vessel. The work is thus pdr.
'It, determine now the finite :intim)» of work required to compress the gas from some initial volume, r„ to some final volume, We pill first answer the question: How much work would be required in compressing a given amount of gas from any volume r to unit volume? This is ac complished by 'integrating' pdr, i.e. by perform ing the operation denoted by the symbol f pdr; and our result will be infinitely Mal' the truth, because the error involved in assuming that the pressure p remains constant through the infini tesimal dr is infinitely small.
To integrate pile we must remember that at constant temperatn•e the product of pressure and a is constant: pr = I-. whence p=—. Substituting this in pd•., we have o NOW" bir be shown. by the method of limit, repeatedly employed in this article. to he the exact differential of either klogr, ur klcigv e. e being any constant. \\ I. may therefore write. cons ly, kdv • loge the constant of integration c being retained so as to give the solution its more general form.