Suppose we have a linear function such as y = 2x+3. The graph of this function is as follows:
Assume that to be a graph of the distance traveled by a car from home base, such that at time x = 0, the car is 3 miles from home. Now consider two points on the graph P(1,5) and Q(2,7). If we wish to find the rate of change of distance between these two points, we use the formula:
Rate of change = (y2-y1)/(x2-x1) = (7-5)/(2-1) = 2/1 = 2
That was fairly simple. That is because our graph was a straight line. Now suppose if the graph is not a straight line. And it is a curve instead.
The above graph as we see is not a straight line, but it is a curve. This time the co-ordinates of the points P and Q are (1,7) and (2,13) respectively. The average rate of change from P to Q can be found using the same formula above:
Average rate of change = (13-7)/(2-1) = 6/1 = 6. But we call this average rate of change since, it cannot be exact because the line between P and Q is not a straight line.
With this back ground let us now try to understand what is instantaneous rate of change. Now suppose the point Q moves closer and closer to point P, such that the distance between P and Q is infinitesimally small.
Therefore if co-ordinates of P are (x0, f(x0)) then those of Q would be (x0+h, f(x0+h)). Now, as Q moved closer and closer to P, the value of h goes on decreasing till it finally becomes 0 when Q coincides with P. That does not actually happen, h goes on decreasing to an infinitesimally small value. So we say that h tends to 0. Symbolically, h -> 0. Then the rate of change of the function f would be given as:
Rate of change = lim (h->0) [f(x+h) – f(x)]/[x+h – x] = lim(h->0) [f(x+h) - f(x)]/[h]
Stated above is the instantaneous rate of change equation. The term ‘rate of change’ now becomes ‘instantaneous rate of change’. We call it instantaneous because, at the instant when x = x0, the rate of change of the function is given by the limit:
instantaneous rate of change = lim(h->0) [f(x0+h) – f(x0)]/h
Instantaneous rate of change examples:
Find instantaneous rate of change of f(x) = x^2 at x = 4.
Ir = lim(h->0) [f(4+h) – f(4)]/h = lim(h->0)[16+8h+h^2-16]/h = lim(h->0)[8h+h^2]/h = lim(h->0)[h(8+h)]/h = lim(h->0)[8+h] = 8+0 = 8
