Rates of Change with Examples

Rates of Change Suppose it took you 45 minutes to drive to class today, and you know that you drove 30 miles to class. With this information it is not difficult to calculate your average speed was 30 miles = 40 miles . In general, we calculate the average rate of change of a function f (x) over .75 hours hour an interval [x1 , x2 ] by dividing the change in f (∆f = f (x2 ) − f (x1 )) by the length of the interval over which it occured (∆x = x2 − x1 ). Thus, f (x2 ) − f (x1 ) f (x1 + ∆x) − f (x1 ) ∆f = = ∆x x2 − x1 ∆x Example 1 As we move into studying calculus, we will concern ourselves with continuoustime dynamical systems, as calculus is just the tool we need to study them. Let b(t) = 2t denote the population of a colony of bacteria (in millions), as a continuous function of time (in hours), defined for all t ≥ 0. Find the average rate of change of the population between t = 1 and t = 2. Solution In order to calculate the average rate of change of the population, we need to find b(1) and b(2). b(1) = 21 = 2 and b(2) = 22 = 4 Now we can find the average rate of change b(t2 ) − b(t1 ) 4−2 ∆b = = =2 ∆t t2 − t1 2−1 Now suppose we drew a line connecting the points (t1 , b(t1 )) and (t2 , b(t2 )) as follows. 5 4.5 4 b(t) (millions) 3.5 3 2.5 2 1.5 1 0.5 1 1.5 t (hours) 2 2.5 The slope of such a line is exactly the same as the average rate of change of the function between those two points. This leads us to the following definition. Secant Line A secant line of the function f (x) is a line connecting two points of the function, say (x0 , f (x0 )) and (x1 , f (x1 )). The equation of such a line is fs (x) = f (x0 ) + m(x − x0 ) where m = f (x1 ) − f (x0 ) x1 − x0 Example 2 Find the equation of the secant line between the points (1, 2) and (2, 4) of the function b(t) = 2t . Solution From example 1, we already know that m = 2. Let us use (1, 2) as our point to find the equation of the secant line. fs (x) = b(t0 ) + m(t − t0 ) = 2 + 2(t − 1) = 2t Now suppose that rather than the average rate of change between two points, we are interested in the instantaneous rate of change at a point. It’s unlikely that average rate of change will be a very accurate estimate of the instantaneous rate of change, because it is averaged over such a large interval. However, if we were to decrease the size of the interval over which we calculate the average rate of change, we should be able to get a better estimate. The closer the length of the interval comes to 0, the closer our estimate will come to exact; we are interested in the average rate of change over an interval as the length of that interval approaches 0 (it cannot equal 0 as dividing by 0 is undefined). Thus, the instantaneous rate of change f 0 (x0 ) of a function f (x) at a point x0 is f (x0 + ∆x) − f (x0 ) ∆x→0 ∆x f 0 (x0 ) = lim provided that the limit exists. When we take the limit of a function at a point, we look at the behavior of the function at points arbitrarily close to that point, but not at that point. That is, limx→x0 f (x) does not depend on the value of f (x0 ). Soon we will see how to calculate the limit of a function at a point. The instantaneous rate of change of a function is a very important mathematical concept, and is called the derivative of a function. We have seen how to calculate the instantaneous rate of change at a point, which is called the derivative of the function at that point. The df derivative of a function (f 0 (x) or dx ) in general is a function that returns the instananeous rate of change (or derivative) of a function at every point of that function where the derivative is defined. For now it suffices to know that it is possible to calculuate the derivative of a function, even if we currently do not possess the tools to do so. Recall that the average rate of change between two points is equal to the slope of the secant line that intersects those two points. As we move the two points of intersection closer together, and take the limit when the distance between them is 0, we find the line that is tangent to the curve at that point. The tangent line is the best possible linear approximation to a curve at a point. The slope of a tangent line is equal to the instantaneous rate of change, or derivative of a function at that point. The equation of the line tangent to a function at a point (x0 , f (x0 )) is y(x) = f 0 (x)(x − x0 ) + f (x0 ) Example 3 Suppose b(t) = 2t and b0 (1) ≈ 1.386. Find the equation of the line tangent to b(t) at t = 1. Solution We want to find the equation of the line tangent at the point (1, b(1)) = (1, 2). It follows y(t) = b0 (1)(t − 1) + b(1) = 1.386(t − 1) + 2 = 1.386t + .614 The function plotted with the tangent line looks as follows 4 3.5 3 b(t) 2.5 2 1.5 1 0.5 0 0.5 1 t 1.5 2 In some situations, we may know a relationship between the rate at which a quantity grows, and the quantity itself. For example, let’s suppose we have a population, and we know the instantaneous rate of change of the population is twice the current population size. We could express such a relationship as db = 2b(t) dt The above equation is a differential equation, because it expresses a relationship between the value of a function, and the value of one or more of its derivatives. If we can solve the above equation, then we can convert the knowledge of the relationship between the population and its derivative into an explicit equation for the population. b0 (t) = 2b(t) or