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Approximate rate of change from graphs and tables

Page history last edited by PBworks 13 years, 10 months ago
When a function is not explicitly defined, it is impossible to find the exact rate of change of that function by taking its derivative.  To approximate the rate of change of an undefined function one may use either a table of values or the graph of the function.  By examining either of these representations of a function, one can easily approximate the rate of change at or between any points.
 

Approximating rates of change from a table:

To approximate a rate of change from a table, use the difference quotient:  (change in value of the function / the change in the variable)

ex.

 Find the approximate rate of change of the function f(x) at x=2 using the table below.

x

f(x)

1

2

2

5

3

10

4

17

 

There are several methods methods to determine the approximate rate of change at x=2.  We will describe all of these, but only solve one.

Choose x=1 and x=3 as the change in the x variable.  By choosing these points, you will utilize the difference quotient = (10 - 2) / 2 = 4.

The approximate rate of change at x=2 is 4. 

 

One may also find the approximate the rate of change at x=2 by selecting x=1 and x=2 or x=2 and x=3 as the points at which you evaluate the function.  All of these choices yield correct results, but keep in mind that they are all approximations and will give slightly different answers.

 

This applet illustrates the method to approximate rates of change from a table.

 

 

Approximating rates of change from a graph:

 

When given a graph of an unknown function, one must use a tangent line to approximate the instantaneous rate of change at a point. 

 The slope of the tangent line represents the instantaneous rate of change of the function at the point of tangency. 

 

To find the average rate of change of a function, draw a secant line between the two point you are evaluating.

The slope of the secant line on this graph represents the average rate of change of the function on the interval x=1 to x=2.

 

This applet allows users to draw secant lines on a variety of graphs and evaluate their slopes.

 

N.B.- One cannot determine the rate of change of a function where the graph shows a cusp, such as point (0,0) in the absolute value graph below.  At the location of a cusp, the derivative of a function is said to be undefined.

 

 

 

 

 

Example Problems:

 

 

 

 Below is a graph of the movement of a particle (position vs. time):

 

During which time interval did the particle have the highest average velocity?

a. From 0 to 5 sec.

b. From 5 to 10 sec.

c. From 10 to 15 sec.

d. From 15 to 20 sec.

 

Problem from a Ms. Groppe test

 

 

Answer: a.

Because we know that the rate of change of a position function represents the velocity of an object, we merely need to find the interval of the function with the highest average rate of change.  To do this, draw secant lines for each interval given and find the value of the slope of each.  The secant line with the greatest slope will show the interval in which the particle has the highest average velocity.

 

 

 

The number of races that Peter Minnig has won is given in the form of a table below where x represents the number of years since he started racing and f(x) represents the cummulative number of races he has won.

 

 

 

x

f(x)

1

0

2

7

3

15

4

24

5

29

6

35

 

Based on the information above, calculate Peter's approximate yearly winning rate on the interval x=2 to x=5 and explain this number's significance.

 

 

 vrooom!!

 

 

Answer: Between his second and fifth years racing, Peter had an average of ~7.33 wins per year.

To find this answer, use the difference quotient .  By this formula, you get the expression  that is equivalent to (29-7)/3=~7.33By calculating the average rate of change between years 2 and 5, you are finding the average winning rate in wins per year over that interval.

 

 

Curling Caleb decides to model the velocity of one of his "draw" shots on the graph below where x represents the time elapsed since he bagan his delivery (sec.) and y represents the speed of the stone (m/s).  At approximately what time x does Caleb release the stone? Approximate the acceleration of the rock at time x=1.

 

 

 

 

 Impressive, Eh?

 

 

Answer: Caleb releases the stone at approximately time x=0.8 seconds.  At time x=1, the stone undergoes an acceleration of -3m/s/s.

To find the time that Caleb releases the stone, look for when the derivative of the speed graph is 0 by finding a spot on the graph where the tangent line's slope is equal to 0.  Caleb releases the stone at this point because it is only his push the propels the rock forward and gives it positive acceleration.  Once he lets go, the stone experiences the friction of the ice and begins to slow.  To find the acceleration at time x=1, draw a tangent line to the graph at that point.  Then calculate the slope of that line.  In this problem, calculate the slope using slope =  = -3 / 1.  This gives you an acceleration of -3m/s/s.

 

 

To find the time that Caleb releases the stone, look for when the derivative of the speed graph is 0 by finding a spot on the graph where the tangent line's slope is equal to 0. Caleb releases the stone at this point because it is only his push the propels the rock forward and gives it positive acceleration. Once he lets go, the stone experiences the friction of the ice and begins to slow. To find the acceleration at time = -3 / 1. This gives you an acceleration of -3m/s/s.

 

 

 

Below is a graph of the approval rating of former President Bill Clinton.  During what interval did Mr. Clinton's approval rating fall the fastest?

 

a. 1996-1997

b. 1998-1999

c. 1993.75-1994.25

d. 2000-2001

 

 

Answer: c.

The secant line between 1993.75 - 1994.25 has the most negative slope.

 

 

 

The table below represents the number of pieces of candy in the college counselling office over the course of a day.  If the school day starts at 8:00 am, and x represents the number of hours since the begining of the school day, approximate the rate a which candy disapearing at time x=3.

 

x

f(x)

1

10

2

8

3

0

4

6

5

2

 

 

Answer: -10/3 pieces per hour.

Use the difference quotient to find the rate of change at x=3.

 

 

Find the average rate of change of the function f(x) along the interval x=-2, x=2.

 

 

 

a. -1

b. 1

c. 0

d. -3

e. 2

 

Answer: c.

Draw a secant line from x=-2 to x=2.  Find this line's slope to determine that the average rate of change is 0.

 

 

Images from:

http://www.coolschool.ca/lor/CALC12/unit2/U02L01/graph_01.png

http://nsmmathstat1.math.csus.edu:8080/webMathematica/Math030/Images/parabline.gif

http://www.math.rutgers.edu/~greenfie/mill_courses/math135/gifstuff/abs.gif

Jen Groppe

Stephen Minnig

http://www.analyzemath.com/calculus/Problems/tangent_2.gif

http://www.belvedere-wengen.ch/fileadmin/user_upload/Curling.JPG

http://online.wsj.com/media/info-presapp0605-clinton.gif

http://hotmath.com/images/gt/lessons/genericalg1/abs_value_graph.gif

 

 

 

 

Comments (6)

Anonymous said

at 1:54 am on Jan 31, 2008

thanks! :)
~Marissa

ps i thought the marshmallow problem was integral to my understanding of the concept

Anonymous said

at 7:05 pm on Feb 7, 2008

Your visual representations (IE images) were very helpful. The distinction between the tangent line and the secant line was especially useful, and it was good to see a real world application of this technique with your graph of Clinton's approval ratings.

Anonymous said

at 9:55 am on Feb 8, 2008

I really like your page, Petsey. Your explanations are very easy to understand, and the pictures are helpful (especially your racecar with the caption "vroom" and the curler...even though it is not you). KUDOS!!!!

Anonymous said

at 9:55 am on Feb 8, 2008

That comment was by Rachel

Anonymous said

at 8:38 am on Feb 21, 2008

I love the variety of problems on the page -- many drawn from "real life." I also like the pictures, although they do make loading the page rather slow. Excellent work.

Alex Schneider said

at 1:14 am on Feb 3, 2009

Since this page is attributed to "Anonymous," I'm not sure who to address my comments to. Whoever you are, though, excellent job. The explanation is concise and clear, and the clever problems and fun graphics add to the whole experience. My one critique regards the first word problem: unless I'm very mistaken, the correct answer should be B, not A, as the graph seems to be changing at a much faster rate during that second interval. Other than that, excellent job - a great reference.

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