Embed Size (px)
Citation preview
The AcroTEX Web Site, 2000The AcroTEX Web Site, 2000
A Slide ShowA Slide ShowDemonstrating theDemonstrating the
Tangent Line ProblemTangent Line Problem
D. P. StoryD. P. StoryThe Department of MathematicsThe Department of Mathematics
and Computer Scienceand Computer ScienceThe University of Akron, Akron, OHThe University of Akron, Akron, OH
Tangent Line ProblemProblem: Given a point P ( a, f(a) ), we want to define andcalculate . . .
|
a
P ( a, f(a) )
� �
Example: f(x) = 5− (x − 1)2 and a = 1.5.
x
msec
Tangent Line ProblemProblem: Given a point P ( a, f(a) ), we want to define andcalculate the slope of the line tangent the graph at P .
|
a
P ( a, f(a) )
� �
Example: f(x) = 5− (x − 1)2 and a = 1.5.
x
msec
Tangent Line ProblemProblem: Given a point P ( a, f(a) ), we want to define andcalculate the slope of the line tangent the graph at P .
Q( x, f(x) )
|
x|
a
P ( a, f(a) )
� �
• Choose a point x near a andplot Q(x, f(x) ).
Example: f(x) = 5− (x − 1)2 and a = 1.5.
x 3msec
Tangent Line ProblemProblem: Given a point P ( a, f(a) ), we want to define andcalculate the slope of the line tangent the graph at P .
Q( x, f(x) )
|
x|
a
P ( a, f(a) )
� �
• Choose a point x near a andplot Q(x, f(x) ).
• Draw the secant line through Pand Q. The slope of this secantline is
msec =f(x)− f(a)
x − a
Example: f(x) = 5− (x − 1)2 and a = 1.5.
x 3msec −2.5
Tangent Line ProblemProblem: Given a point P ( a, f(a) ), we want to define andcalculate the slope of the line tangent the graph at P .
Q( x, f(x) )
|
x|
a
P ( a, f(a) )
� �
• Choose a point x near a andplot Q(x, f(x) ).
• Draw the secant line through Pand Q. The slope of this secantline is
msec =f(x)− f(a)
x − a
• Repeat.
Example: f(x) = 5− (x − 1)2 and a = 1.5.
x 3msec −2.5
Tangent Line ProblemProblem: Given a point P ( a, f(a) ), we want to define andcalculate the slope of the line tangent the graph at P .
b
|
x
Q( x, f(x) )
|
a
P ( a, f(a) )
� �
• Choose a point x near a andplot Q(x, f(x) ).
Example: f(x) = 5− (x − 1)2 and a = 1.5.
x 3 2.5msec −2.5
Tangent Line ProblemProblem: Given a point P ( a, f(a) ), we want to define andcalculate the slope of the line tangent the graph at P .
b
|
x
Q( x, f(x) )
|
a
P ( a, f(a) )
� �
• Choose a point x near a andplot Q(x, f(x) ).
• Draw the secant line through Pand Q. The slope of this secantline is
msec =f(x)− f(a)
x − a
Example: f(x) = 5− (x − 1)2 and a = 1.5.
x 3 2.5msec −2.5 −2
Tangent Line ProblemProblem: Given a point P ( a, f(a) ), we want to define andcalculate the slope of the line tangent the graph at P .
b
|
x
Q( x, f(x) )
|
a
P ( a, f(a) )
� �
• Choose a point x near a andplot Q(x, f(x) ).
• Draw the secant line through Pand Q. The slope of this secantline is
msec =f(x)− f(a)
x − a
• Repeat.
Example: f(x) = 5− (x − 1)2 and a = 1.5.
x 3 2.5msec −2.5 −2
Tangent Line ProblemProblem: Given a point P ( a, f(a) ), we want to define andcalculate the slope of the line tangent the graph at P .
b
b
|
x
Q( x, f(x) )
|
a
P ( a, f(a) )
� �
• Choose a point x near a andplot Q(x, f(x) ).
Example: f(x) = 5− (x − 1)2 and a = 1.5.
x 3 2.5 2.25msec −2.5 −2
Tangent Line ProblemProblem: Given a point P ( a, f(a) ), we want to define andcalculate the slope of the line tangent the graph at P .
b
b
|
x
Q( x, f(x) )
|
a
P ( a, f(a) )
� �
• Choose a point x near a andplot Q(x, f(x) ).
• Draw the secant line through Pand Q. The slope of this secantline is
msec =f(x)− f(a)
x − a
Example: f(x) = 5− (x − 1)2 and a = 1.5.
x 3 2.5 2.25msec −2.5 −2 −1.75
Tangent Line ProblemProblem: Given a point P ( a, f(a) ), we want to define andcalculate the slope of the line tangent the graph at P .
b
b
|
x
Q( x, f(x) )
|
a
P ( a, f(a) )
� �
• Choose a point x near a andplot Q(x, f(x) ).
• Draw the secant line through Pand Q. The slope of this secantline is
msec =f(x)− f(a)
x − a
• Repeat.
Example: f(x) = 5− (x − 1)2 and a = 1.5.
x 3 2.5 2.25msec −2.5 −2 −1.75
Tangent Line ProblemProblem: Given a point P ( a, f(a) ), we want to define andcalculate the slope of the line tangent the graph at P .
b
b
b
|
x
Q( x, f(x) )
|
a
P ( a, f(a) )
� �
• Choose a point x near a andplot Q(x, f(x) ).
Example: f(x) = 5− (x − 1)2 and a = 1.5.
x 3 2.5 2.25 2msec −2.5 −2 −1.75
Tangent Line ProblemProblem: Given a point P ( a, f(a) ), we want to define andcalculate the slope of the line tangent the graph at P .
b
b
b
|
x
Q( x, f(x) )
|
a
P ( a, f(a) )
� �
• Choose a point x near a andplot Q(x, f(x) ).
• Draw the secant line through Pand Q. The slope of this secantline is
msec =f(x)− f(a)
x − a
Example: f(x) = 5− (x − 1)2 and a = 1.5.
x 3 2.5 2.25 2msec −2.5 −2 −1.75 −1.5
Tangent Line ProblemProblem: Given a point P ( a, f(a) ), we want to define andcalculate the slope of the line tangent the graph at P .
b
b
b
|
x
Q( x, f(x) )
|
a
P ( a, f(a) )
� �
• Choose a point x near a andplot Q(x, f(x) ).
• Draw the secant line through Pand Q. The slope of this secantline is
msec =f(x)− f(a)
x − a
• Repeat.
Example: f(x) = 5− (x − 1)2 and a = 1.5.
x 3 2.5 2.25 2msec −2.5 −2 −1.75 −1.5
Tangent Line ProblemProblem: Given a point P ( a, f(a) ), we want to define andcalculate the slope of the line tangent the graph at P .
b
b
b
b
|
x
Q( x, f(x) )
|
a
P ( a, f(a) )
� �
• Choose a point x near a andplot Q(x, f(x) ).
Example: f(x) = 5− (x − 1)2 and a = 1.5.
x 3 2.5 2.25 2 1.75msec −2.5 −2 −1.75 −1.5
Tangent Line ProblemProblem: Given a point P ( a, f(a) ), we want to define andcalculate the slope of the line tangent the graph at P .
b
b
b
b
|
x
Q( x, f(x) )
|
a
P ( a, f(a) )
� �
• Choose a point x near a andplot Q(x, f(x) ).
• Draw the secant line through Pand Q. The slope of this secantline is
msec =f(x)− f(a)
x − a
Example: f(x) = 5− (x − 1)2 and a = 1.5.
x 3 2.5 2.25 2 1.75msec −2.5 −2 −1.75 −1.5 −1.25
Tangent Line ProblemProblem: Given a point P ( a, f(a) ), we want to define andcalculate the slope of the line tangent the graph at P .
b
b
b
b
|
x
Q( x, f(x) )
|
a
P ( a, f(a) )
� �
• Choose a point x near a andplot Q(x, f(x) ).
• Draw the secant line through Pand Q. The slope of this secantline is
msec =f(x)− f(a)
x − a
• Repeat.
Example: f(x) = 5− (x − 1)2 and a = 1.5.
x 3 2.5 2.25 2 1.75msec −2.5 −2 −1.75 −1.5 −1.25
Tangent Line ProblemProblem: Given a point P ( a, f(a) ), we want to define andcalculate the slope of the line tangent the graph at P .
b
b
b
b
b
|
x
b Q( x, f(x) )
|
a
P ( a, f(a) )
� �
• Choose a point x near a andplot Q(x, f(x) ).
Example: f(x) = 5− (x − 1)2 and a = 1.5.
x 3 2.5 2.25 2 1.75 1.6msec −2.5 −2 −1.75 −1.5 −1.25
Tangent Line ProblemProblem: Given a point P ( a, f(a) ), we want to define andcalculate the slope of the line tangent the graph at P .
b
b
b
b
b
|
x
b Q( x, f(x) )
|
a
P ( a, f(a) )
� �
• Choose a point x near a andplot Q(x, f(x) ).
• Draw the secant line through Pand Q. The slope of this secantline is
msec =f(x)− f(a)
x − a
Example: f(x) = 5− (x − 1)2 and a = 1.5.
x 3 2.5 2.25 2 1.75 1.6msec −2.5 −2 −1.75 −1.5 −1.25 −1.1
Tangent Line ProblemProblem: Given a point P ( a, f(a) ), we want to define andcalculate the slope of the line tangent the graph at P .
b
b
b
b
b
|
x
b Q( x, f(x) )
|
a
P ( a, f(a) )
� �
• Choose a point x near a andplot Q(x, f(x) ).
• Draw the secant line through Pand Q. The slope of this secantline is
msec =f(x)− f(a)
x − a
• Repeat.
Example: f(x) = 5− (x − 1)2 and a = 1.5.
x 3 2.5 2.25 2 1.75 1.6msec −2.5 −2 −1.75 −1.5 −1.25 −1.1
Tangent Line ProblemProblem: Given a point P ( a, f(a) ), we want to define andcalculate the slope of the line tangent the graph at P .
b
b
b
b
b
b
|
x
b
Q( x, f(x) )
|
a
P ( a, f(a) )
� �
• Choose a point x near a andplot Q(x, f(x) ).
Example: f(x) = 5− (x − 1)2 and a = 1.5.
x 3 2.5 2.25 2 1.75 1.6 1.55msec −2.5 −2 −1.75 −1.5 −1.25 −1.1
Tangent Line ProblemProblem: Given a point P ( a, f(a) ), we want to define andcalculate the slope of the line tangent the graph at P .
b
b
b
b
b
b
|
x
b
Q( x, f(x) )
|
a
P ( a, f(a) )
� �
• Choose a point x near a andplot Q(x, f(x) ).
• Draw the secant line through Pand Q. The slope of this secantline is
msec =f(x)− f(a)
x − a
Example: f(x) = 5− (x − 1)2 and a = 1.5.
x 3 2.5 2.25 2 1.75 1.6 1.55msec −2.5 −2 −1.75 −1.5 −1.25 −1.1 −1.05
Tangent Line ProblemProblem: Given a point P ( a, f(a) ), we want to define andcalculate the slope of the line tangent the graph at P .
b
b
b
b
b
b
|
x
b
Q( x, f(x) )
|
a
P ( a, f(a) )
� �
• Choose a point x near a andplot Q(x, f(x) ).
• Draw the secant line through Pand Q. The slope of this secantline is
msec =f(x)− f(a)
x − a
• Repeat.
Example: f(x) = 5− (x − 1)2 and a = 1.5.
x 3 2.5 2.25 2 1.75 1.6 1.55msec −2.5 −2 −1.75 −1.5 −1.25 −1.1 −1.05
Tangent Line ProblemProblem: Given a point P ( a, f(a) ), we want to define andcalculate the slope of the line tangent the graph at P .
b
b
b
b
b
bb
|
x
b
Q( x, f(x) )
|
a
P ( a, f(a) )
� �
• Choose a point x near a andplot Q(x, f(x) ).
Example: f(x) = 5− (x − 1)2 and a = 1.5.
x 3 2.5 2.25 2 1.75 1.6 1.55 1.501msec −2.5 −2 −1.75 −1.5 −1.25 −1.1 −1.05
Tangent Line ProblemProblem: Given a point P ( a, f(a) ), we want to define andcalculate the slope of the line tangent the graph at P .
b
b
b
b
b
bb
|
x
b
Q( x, f(x) )
|
a
P ( a, f(a) )
� �
• Choose a point x near a andplot Q(x, f(x) ).
• Draw the secant line through Pand Q. The slope of this secantline is
msec =f(x)− f(a)
x − a
Example: f(x) = 5− (x − 1)2 and a = 1.5.
x 3 2.5 2.25 2 1.75 1.6 1.55 1.501msec −2.5 −2 −1.75 −1.5 −1.25 −1.1 −1.05 −1.001
Tangent Line ProblemProblem: Given a point P ( a, f(a) ), we want to define andcalculate the slope of the line tangent the graph at P .
b
b
b
b
b
bb
|
x
b
Q( x, f(x) )
|
a
P ( a, f(a) )
� �
• Choose a point x near a andplot Q(x, f(x) ).
• Draw the secant line through Pand Q. The slope of this secantline is
msec =f(x)− f(a)
x − a
• Repeat.• Continue . . .
Example: f(x) = 5− (x − 1)2 and a = 1.5.
x 3 2.5 2.25 2 1.75 1.6 1.55 1.501msec −2.5 −2 −1.75 −1.5 −1.25 −1.1 −1.05 −1.001
DiscussionExample: f(x) = 5 − (x − 1)2 and a = 1.5. As we choose values ofx getting closer and closer to a = 1.5, the corresponding secant linesrotate around the point P and become more and more “tangent-like”.Therefore, it is not too surprising that the slopes of these secant linesare approaching a value we would want to call “the slope of the linetangent to the graph at P”.
There are more calculations for those who want to see more.
x < 1.5x msec
1 −0.51.4 −0.91.45 −0.951.49 −0.991.499 −0.9991.4999 −0.99991.49999 −0.99999
x > 1.5x msec
2 −1.51.6 −1.11.55 −1.051.51 −1.011.501 −1.0011.5001 −1.00011.50001 −1.00001
The values of msecappear to be gettingclose and closer to−1. In this case, wewrite:
limx→1.5
msec = −1
