Slope of a Curve:

The value of the derivative ds/dt in the case plotted in Figure is a constant.  It equals 40 ft/s. In the discussion of graphing, the slope of a straight line on a graph was describe as the change in y, Δy, divided by the change in x, Δx.  The slope of the line in Figure is Δs/ Δt which, in that case, is the value of the derivative ds/dt.   Therefore, derivatives of functions can be interpreted in terms of the slope of the graphical plot of the function.  Because the velocity equals the derivative of the distance s along with respect to time t, ds/dt, and since this derivative equals the slope of the plot of distance versus time, the velocity can be visualized as the slope of the graphical plot of distance versus time.

For the case display in Figure, the velocity is constant. Figure is another graph of the distance traveled through an object as a function of the elapsed time.  In that case the velocity is not constant.  The functional relationship display is given through the subsequent equation:

s = 10t²

The instantaneous velocity again equals the value of the derivative ds/dt. This value is changing along with time. Therefore, the instantaneous velocity at any specified time can be determined.  First, make little changes in s and t are considered.

Δs/Δt = (s + Δs)-s/(t + Δt)-t

The values of (s + Δs) and s in terms of (t +   Δt) and t using Equation 5, can then be substituted into this expression.  At time t, s = 10t²; at time t +Δt, s + Δs = 10(t + t)². The  value  of  (t  + Δt)²   equivalent  t²   + 2t(  Δt)  +  ( Δ t)²;  however,  for  incremental values of t, the term (Δt)²   is so small, it can be neglected. Therefore, (t + Δt)²   = t²   + 2t( Δt).

Δs/Δt = 10[t² + 2t(Δt)] - 10t²/(t + Δt) -t

Δs/Δt = 10t² + 20t(Δt)] -10t²/t +Δt-t

Δs/Δt =20t

The value of the derivative ds/dt in the case plotted in Figure equals 20t.  Therefore, at time t = 1 s, the instantaneous velocity equals 20 ft/s; at time t = 2 s, the velocity equals 40 ft/s, and many more.

Figure: Slope of a Curve