THE TEN PILLARS 



15 



The question is: What is the velocity of the falling body at the instant 

 it passes the point, S ? 



At S, S = 1/2 gt 2 -d-1) 



At 5 + AS, S + AS = 1/2 g(t + A/) 2 . 

 Multiplying out the square, 



S + AS = 1/2 gt 2 + gtAt + 1/2 gAt 2 - -(1-2) 



Between the two points, then, the value for AS is given by Eq. (1-2)- 

 Eq. (1-1): 



AS = gtAt + 1/2 gAt 2 

 The average rate, over a small increment of time is: 



AS/At = gt + 1/2 gAt 

 Hence, the instantaneous rate is: 



dS/dt = Lim AS/At = gt + 1/2 g x = gt 



A/— 



That is, the instantaneous rate of change of distance with time (or velocity) 

 at the point, S, is: 



dS/dt = gt (1-3) 



For example, 5 sec after free fall starts, Eq. (1-1) says that the distance fallen 

 is 400 ft; and Eq. (1-3) says that the velocity as it passes the 400- ft mark is 

 160 ft per sec. 



Maximum and minimum values of functions with changing slope and 

 curvature must be given by the values of the function for which the instan- 

 taneous rate of change, or slope, is zero. This can be visualized in the 

 periodic function of Figure 1-2, for example. 



6. The Differential and Integral Calculus 



It has been seen that, given the explicit form of the "mother" function, it 

 is possible by the method of increments to determine the explicit form of the 



s 



■ t 



^ 



t + ^t 



Figure 1-5. The Falling Body. 



