Higher Concepts of Mathematics CALCULUSThe value of the derivative ds/dt for the caseFigure 4 Graph of Distance vs. Timeplotted in Figure 4 can be understood byconsidering small changes in the two variabless and t.DsDt(sDs)s(tDt)tThe values of (s + Ds) and s in terms of (t +Dt) and t, using Equation 5-4 can now besubstituted into this expression. At time t, s= 40t; at time t + Dt, s + Ds = 40(t + Dt).DsDt40(tDt)40t(tDt)tDsDt40t40(Dt)40ttDttDsDt40(Dt)DtDsDt40The value of the derivative ds/dt in the case plotted in Figure 4 is a constant. It equals 40 ft/s.In the discussion of graphing, the slope of a straight line on a graph was defined as the changein y, Dy, divided by the change in x, Dx. The slope of the line in Figure 4 is Ds/Dt which, in thiscase, is the value of the derivative ds/dt. Thus, derivatives of functions can be interpreted interms of the slope of the graphical plot of the function. Since the velocity equals the derivativeof the distance s with respect to time t, ds/dt, and since this derivative equals the slope of the plotof distance versus time, the velocity can be visualized as the slope of the graphical plot ofdistance versus time.For the case shown in Figure 4, the velocity is constant. Figure 5 is another graph of thedistance traveled by an object as a function of the elapsed time. In this case the velocity is notconstant. The functional relationship shown is given by the following equation:s = 10t2(5-5)Rev. 0 Page 35 MA-05
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