If velocity is the slope of a position graph, how can we find the velocity at a [i]single [/i]point in time, when it takes [i]two [/i]points to make a line?

You may have learned in Physics class that the [i]slope [/i]of a position versus time graph gives us the [i]velocity[/i] of the object. So, if the object moves [math]\Delta d=10[/math] meters in [math]\Delta t=2[/math] seconds, we calculate its velocity as [math]v=\frac{10 m}{2 sec}=5m/s[/math]. But suppose this object is not moving at a constant speed - perhaps it is falling and therefore constantly speeding up. In this case, it is not going 5 m/s for the entire 2 seconds. We could try to get closer to its velocity at a single point in time by measuring [math]\Delta d[/math] for a shorter [math]\Delta t[/math], since we expect less variation in the velocity in a shorter time interval. As we make [math]\Delta t[/math] smaller, our estimate of the instantaneous velocity improves. But as long as [math]\Delta t \neq 0[/math], any such calculation is still an approximation over a time [i]interval[/i], and not the instantaneous velocity at a [i]single point[/i] in time![br][br]At first the answer seems simple - just set [math]\Delta t=0[/math]! But then [math]\Delta d=0[/math] too - the object moves zero meters in zero time. We cannot calculate [math]\frac{\Delta d}{\Delta t}=\frac{0}{0}[/math], because division by zero is undefined.[br][br]In this demonstration, the object's [i]position [/i](its distance from the origin) varies with time according to [math]d(t)=t^3[/math], which is graphed above. We want to know what its velocity is at time [math]t=2[/math] seconds by using the slope at that point. To do this, we'll construct a line (in green) between the point at our target (in blue) and a nearby point along the curve (in red). We'll use the line's slope as our estimate of the velocity. We can adjust the movable point's location with the slider. As we do, we see that [math]\Delta d[/math] and [math]\Delta t[/math] both change, and that the slope [math]\frac{\Delta d}{\Delta t}[/math] of the green line - our estimate of the velocity - changes with it. The closer we bring the movable point to the target point, the closer we get to the velocity at [math]t=2[/math]. But if we set the movable point to exactly the same position as the target, we have a problem.[br][br]But notice that as you position the movable point closer to the target, the slope (velocity) seems to be approaching a certain number. If we approach from the left, the velocity estimate is increasing, and if we approach from the right, it is decreasing. Is it possible that the "real" instantaneous velocity is some number between these values? The answer is "Yes"![br][br]We can write a formula for the velocity estimate [math]\frac{\Delta d}{\Delta t}[/math] and use our calculator to get very close estimates. From this exercise, perhaps we could deduce the "true" instantaneous velocity. We have two points for our line: [math](2,2^3)[/math] for the target point, and [math](x,x^3)[/math] for the movable point, for which time [math]t=x[/math]. Thus the slope (velocity) is [math]\frac{(x)^3-\left(2\right)^3}{(x)-(2)}[/math], using the slope formula. Now, plug in values of [math]x[/math] slightly larger than and slightly smaller than 2 into [math]\frac{x^3-8}{x-2}[/math] and see that "12" seems to be a reasonable result.[br][br]We call this number the [i]limit[/i] of [math]\frac{x^3-8}{x-2}[/math] as [math]x[/math] approaches [math]2[/math], and we write it as [math]\lim_{x \to 2}\frac{x^3-8}{x-2}=12[/math].