AC=x2; and let BF represent the increment which x would receive in the next unit of time. Now, let the square be checked in its increasing course as soon as it has arrived at the value x2. The rate of increase of the square (since it is moving with accelerated motion) will not be represented by the increment which it would receive in the next unit of time, but by the increment it would receive if it increased with uniform motion at the rate which it had at the instant at which it was stopped. Therefore, in order that the motion may be uniform, as the sides BC, DC move outwards, they must remain of the same length. Hence, BF or DH representing the increase in the variable, the corresponding increase in the square will be represented by the two rectangles BE and CH. i.e., by 2x BE. But BE-side of square × rate of increase of x, since BF rate of increase of x. ... rate of increase of square i.e., or = =2xx rate of increase of x, rate of increase of x2 rate of increase of x =2x, differential coefficients of x2=2x, VII. A Falling Body. 31. Firstly. Suppose a body to fall from rest for ", it will have fallen through 16 feet and have acquired a velocity of 32 feet per second. Suppose it then to receive a check which brings it to rest, and then let it, without loss of time, fall, as before, for "; it will, as before, fall 16 feet, and again acquire a velocity of 32 feet per second. Let the same process be repeated until, in all, the body has been let fall for 10", that is 100 times; then the body will have passed through 16 feet, and the velocity at the end of the time will be 3.2 feet per second. 32. Secondly. Suppose that, after the body has been arrested at the end of ", we give it an impulse equal to the velocity it had acquired before it was arrested, viz., a velocity of 3.2 feet per second. Then at the end of the second" it will have a velocity of 64 feet per sec., and the space described will be the original space of 16 feet +that which the body would have described moving uniformly with a velocity of 3.2 feet per sec. +the space which it would have described without that impulse = (in feet) ·16+3·2×1+32 × (15)2 10 2 10 If the body had not been arrested, the space fallen through from rest would have been 32×(2)*f ='64 feet. 10 feet Now let the same process be repeated for the third tenth of a second. The starting velocity will be 64 feet per second, and the velocity at the end of the third" will be 96 feet per second; and the space travelled through will be that arrived at at the end of the second ′′ +the space which it would have described 1 32 =(in feet) 64+64× + X 10 2 If the body had not been arrested, the space fallen through would have been 32 × (3)2 X 2 2 feet 1.44 feet. If this process be repeated 100 times, the time of falling will be 10′′, and the velocity acquired will be=320 ft. per sec. and the space described = 16 × 100 ft. =1600 ft. 33. In the following table the first column represents the time in seconds during which the body is falling; the second column gives the corresponding spaces through which the body falls (in feet); the third column is obtained from the second by subtracting each number from the one immediately above it, and gives the spaces fallen through in each "; the fourth column is obtained from the third in the same manner in which the third is obtained from the second, and gives the difference between the spaces fallen through in the consecutive "s seconds, and it will be remarked that these last are all the same. Thus we see that the space fallen through in the interval between any two consecutive tenths of seconds is 32 feet. This space for 100ths secs. = 1000ths secs. = 0032 feet, 1000000ths secs. = '000000000032 feet, etc. = etc., and, when the intervals are made infinitesimally small, the space becomes infinitesimally small, but is always a multiple of 32. We may say, then, that when there is a continuous fall, without any interruption, the motion becomes continuous, losing its jerks and impulses (the jerks becoming inappreciable), the space fallen through is increasing, at any instant, by an infinitesimal multiple of 32. (See also Art. 48, etc.) VIII. Differential Co-efficient of 12, 22, 32, and 42. 34. (1) Here 1 is supposed to receive small increments of 01; therefore 1 will be the variable.* The function considered is the square of the variable. *When variable is mentioned independent variable is implied, (2) Here 2 is supposed to receive small increments of 001; therefore 2 will be the variable. The function under consideration in this case is also the square of the independent variable. (3) Here 3 is supposed to receive small increments of 0001 and the function again is the square. (4) Here 4 is supposed to receive small increments of '00001, and, as before, the function is the square. The first column in each case represents the independent variable, as it increases uniformly by increments of 01, 001, 0001 and 00001 respectively. |