square has an increment of two shaded rectangles (one of whose sides is equal to the side of the original square) and a small square-i.e., small when compared with the original square. The second square, whose side AC receives an increment CD, receives an increment of two rectangles such as DF (one of whose sides is equal to the side of the second square) and the square 6, which is even smaller, when compared with the square on AC, than a is when compared with the square on AB. Now let the side AD receive a further increment DE, then, as before, the square receives an increment of two rectangles such as EG (one of whose sides is equal to the side of the square on AD) and the square c, which is very small when compared with the square on AD. Suppose, now, that the side AE receives a very small increment indeed; then the square receives an increment of the two very narrow rectangles (one of whose sides is equal to the side of the square on AE) and the minute square at K. Finally, when the breadth of the rectangles is indefinitely diminished, or, which is the same thing, when the side receives an infinitesimally small increment, the rectangles become coincident with the sides of the square (see Art. 10), and the small square vanishes, when compared with the square on AE, and the increment in the square corresponding to the infinitesmal increase in the side is made up of two rectangles coincident with the sides-i.e., the ratio of the rate of increase of the square to the rate of increase of the side, when the increment to the side is infinitesimal, is 2x side: 1, as before. Now, let us look at this from a different point of view. Let AC be a square on AB-AB being a variable; and suppose the square to be growing continuously as AB increased, having originally been Ac, and let AB have arrived at the value x; in consequence of which 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 = 2x × rate of increase of x, = 31. Firstly. Suppose a body to fall from rest for ", it will have fallen through 16 feet and have acquired a velocity of 3.2 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 If the body had not been arrested, the space fallen through from rest would have been 32×(2) ='64 feet. feet Now let the same process be repeated for the third tenth of a second. The starting velocity will be 6'4 feet per second, and the velocity at the end of the third" will be 9.6 feet per second; and the space travelled through will be 1 that arrived at at the end of the second " +the space which it would have described =(in feet) ·64+64 × 1 32 2 =64+64+*16 =1.44 feet. If the body had not been arrested, the space fallen through would have been 32×(3) feet-1-44 feet. 10 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 f"; 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. |