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!n$t@nt@n30u$ @nd @v3r@g3 R@t3$ 0f Ch@ng3. @l3x J0n3$ @nd m@ry b33ch @k@: A + @13X & M@1!c!0u$ M@rY. http://bedesblog.files.wordpress.com/2009/04/pennies1.jpg. @v3r@g3 R@t3 0f Ch@ng3. Definition: rate of change of the dependent variable on an interval of the independent variable.
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!n$t@nt@n30u$ @nd @v3r@g3 R@t3$ 0f Ch@ng3 @l3x J0n3$ @nd m@ry b33ch @k@: A + @13X & M@1!c!0u$ M@rY http://bedesblog.files.wordpress.com/2009/04/pennies1.jpg
@v3r@g3 R@t3 0f Ch@ng3 • Definition: rate of change of the dependent variable on an interval of the independent variable For example, in the graph to the left, the average rate of change of f(x) from x = 0 to x = 1 is the change in f(x) divided by the change in x.
@v3r@g3 R@t3 0f Ch@ng3 • For a car moving forward, the position of the car as a function of time can be graphed. The average rate of change is the change in distance divided by the change in time. • This represents the average speed of the car. http://s.chakpak.com/se_images/136705_-1_564_none/cool-car-wallpaper.jpg
Gr@ph!c@1 R3pp!n’ The average rate of change can be represented graphically by drawing a secant line (a line between two different points-shown in green) and calculating its slope. Calculate the slope using: where a and b are the endpoints of the interval Substitute in values found on graph to get: The average change from x = 0 to x = 1 for this graph is 4
@1g3br@!c R3pp!n’ • Use the function to evaluate the slope of the secant line from x = 0 to x = 1 by calculating the slope between the two points • f(x) = x3+2x2+x+1 • f(0) = 03+2(02)+0+1 = 1 • f(1) = 13+2(12)+1+1 = 5 • Calculate the slope using these values: Picture from http://images.buycostumes.com/mgen/merchandiser/27253.jpg
Num3r!c@1 R3pp!n’ • For the function: f(x) = x3+2x2+x+1 a table of values is given below: Use the equation for slope to evaluate the slope from x = 0 to x = 1 using the values given in the table:
!n$t@nt@n30u$ R@t3 0f Ch@ng3 • Definition: the rate of change of the dependent variable at a specific value of the independent variable For example, for the graph to the left, the instantaneous rate of change at x = 0 is the exact slope of the graph at x = 0. This value cannot be determined using only the function, but there are numerous ways to approximate this value (and later you will learn how to calculate this exactly).
!n$t@nt@n30u$ R@t3 0f Ch@ng3 • For a car moving forward, the instantaneous rate of change of its position over time is the exact speed of the car at a specific time. This is impossible to calculate exactly using the position-time graph, but can be approximated. However, for a velocity-time graph for the car, the speed would be the y-value at a point, but the instantaneous rate of change of velocity would be the acceleration. http://bigpicture.typepad.com/writing/images/peugeot_908_rc.jpg
Gr@ph!c@1 R3pp!n’ The instantaneous rate of change at x = 0 can be represented graphically by drawing a tangent line (a straight line that touches the curve only at one point) at a given point and calculating its slope. Calculate the slope for two estimated points on the tangent line using the equation: Substitute in values found on graph for the points (-1, 0) and (1, 2) to get: The approximated instantaneous rate of change at x = 0 for this graph is 1
Gr@ph!c@1 R3pp!n’ • The instantaneous rate of change can be approximated from the average rate of change by using the slope of the secant line as the change in x goes to 0 (giving you the tangent line mentioned before).
@1g3br@!c R3pp!n’ • Use the method for algebraically calculating average rate of change. Do this multiple times with smaller and smaller intervals around the given point to estimate the instantaneous rate of change.
@1g3br@!c R3pp!n’ • f(x) = x3+2x2+x+1 • x = 0 to x = 0.01: • x = 0 to x = 0.001: • As the x-interval around x = 0 gets smaller, the values of the average rate of change go from 1.0201 to 1.002001, so they approach 1. • This is the limit of the average rate of change as the interval of x-values decreases to 0.
Num3r!c@1 R3pp!n’ • For a mass oscillating on a spring, the distance (in ft) of the mass from the ground is recorded each second. A sample of the data is shown below: Approximate the instantaneous rate of change of the mass’s position at t = 4 s. Use the equation for slope to evaluate the slope around t = 4 using the values given in the table (this will give you a rough estimate of the instantaneous rate of change): The instantaneous rate of change at t = 4 s is approximately -2 ms-1 (the negative signifies that the distance between the mass and the ground is decreasing). *Note that this is also the average rate of change from t = 2 s to t = 4 s