Math 1241, Spring 2014 Section 3.3. Rates of Change Average vs. Instantaneous Rates

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Math 1241, Spring 2014 Section 3.3 Rates of Change Average vs. Instantaneous Rates

Average Speed The concept of speed (distance traveled divided by time traveled) is a familiar instance of a rate of change. Example: To drive the 15.5 miles from Clayton State to Turner field, it takes 18 minutes. 15.5 miles 18 min. = 0.88 mi/ min = 52.7 mph Question: If you drove to Turner Field, would your speedometer always read 52.7 mph?

Average Speed The 52.7 mph is an average speed. Your speedometer measures something else. In pre-calculus courses, you solve problems assuming that speed (or some other rate of change) is constant: it does not change. One of the main features of Calculus: we can solve problems where speed (or some other rate or change) is not necessarily constant.

Average Rate of Change For a function y = f(x), we define the average rate of change from x = a to x = b as: f b f(a) b a In the case of speed: We often use t instead of x (for obvious reasons). f(t) is the total distance we ve travelled at time t. Numerator = Change/difference in distance Denominator = Change/difference in time

Average Rate of Change = Slope The blue curve is the graph of y = f(x). The two blue dots show the function s values at x = a and x = b. The red line is called a secant line. Its slope equals the average rate of change of f(x) from a to b.

Instantaneous Speed Your speedometer measures your speed at a given time. What does this mean? Average speed: Change in distance divided by change in time. We can t do this at a given time, because the change in time is zero (in the denominator). Solution: Take the limit as change in time approaches zero!

Instantaneous Rates of Change Take the limit of (average rate of change), as the change in the independent (x) variable approaches zero. There are two ways to do so: lim b a f b f(a) b a lim h 0 f a + h f(a) h Although these appear to be different formulas, note that h = b a (thus b = a + h).

Graphical Demonstration It s somewhat difficult to do dynamic graphs in Graph, so we ll use the following link: https://www.desmos.com/calculator/irip8pnpdf Left-click on one of the dots and hold down the button. Drag your mouse to see how the secant line (in red) changes. As the dots get closer together, the slope of the secant line approaches the instantaneous rate of change.

Tangent Lines Consider what happens to the secant line as b a (or as h 0). Any secant line contains the point (a, f a ). The slope of the secant line approaches the instantaneous rate of change (at x = a). The line through (a, f a ) with slope equal to the instantaneous rate of change (at x = a) is called the tangent line (at x = a). The tangent line of a circle is a special case.

Graphical Demonstration Using the link from earlier: https://www.desmos.com/calculator/irip8pnpdf Change the function definition to sqrt(4-x^2). This is the upper half of the circle centered at (0,0) with radius 2. Drag the two points close together. The secant line is very close to a tangent line of the circle.

Tangent Lines in Graph Fortunately, Graph will draw a tangent line. Start by graphing a function. I ll use x + 1 x. Select Function -> Insert Tangent/Normal from the menu (or press F2, or use the toolbar button). In the x = field, type the x-value where you want the tangent line (I ll use x = 1). Use a different color than the original function. Zoom in on the point where the function touches the tangent line. What do you see?

An important note The instantaneous rate of change of the function f(x) at x = a is a limit. To actually compute it, we need to know the function value for x values closer and closer to x = a. This would mean infinitely many values! We can avoid this if we have a formula for f(x) that is valid near x = a. We ll usually take this approach.

Algebraic Example Find the instantaneous rate of change of the function f x = x 2 at the point a = 1. Before computing the limit, use Graph to draw the tangent line. What is the slope? We need to evaluate one of the following: x 2 1 2 1 + h 2 (1) 2 lim x 1 lim h 0 x 1 h Try both forms; which one is easier?