Lesson 2.1 The Tangent Line Problem

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```					Lesson 2.1
The Tangent Line
Problem
By
Darren Drake
05/16/06
History

   Calculus grew from 4 major problems
   Velocity/acceleration problem
   Max and min value problem
   Area problem
   THE TANGENT LNE PROBLEM
History

 Pierre de Format, Rene’ Descartes,
Christian Huygens, and Isaac Barrow are
given credit for finding partial solutions
 Isaac Newton(1642-1727) is credited for
finding the general solution to the tangent
line problem
What is a Tangent Line?

 The Tangent line touches the curve at
one point
 But this doesn’t work for all curves
How do you find the
tangent line?
 To find the tangent line at point c, you must
first find the slope of the tangent line at point c
 You can approximate this slope by of the
secant line containing the points (c, f (c)) and
(c  x, f (c  x)

y2  y1
 slope of the secant =
Recall m                f (c  x)  f (c) f (c  x)  f (c)
x2  x1                         
(c  x)  c            x
 The closer x is to 0, the closer c  x is to c
 The more accurate the tangent line
approximation will be

 Hmmmm….
this sounds like a limit!!!
Soo…

 If f is defined on an openinterval containing
c, and if the limit
y         f (c  x)  f (c)
lim       lim                      m
x 0 x   x          x
 Exists, then the line passing through the
point (c, f (c)) with the slope m is the line
tangent to the graph f(x) at the point (c, f (c))
Example 1
The slope of the graph of a
linear function
Given f(x) = 2x-3, find the slope at (2,1)

f (2  x)         [2(2  x)  3]  [2(2)  3]
lim               lim
x 0             x 0             x
4  2x  3  4  3
 lim
x 0         x
2x
 lim
x 0 x

 lim 2
x 0

2
Example 2
Tangent Lines to the graph
of a nonlinear function
 Find the slope at (0,1) of the tangent line to
the graph of f ( x)  x  1 and write an equation
2

for the tangent line at this point
( x  x) 2  1  ( x 2  1)
lim
x 0              x
x 2  2 x(x)  (x) 2  1  x 2  1
y  y1  m( x  x1 )
 lim
x  0                  x                    y  1  2(0)( x  0)
2 x(x)  (x) 2
 lim
x  0         x
y 1  0
 lim (2 x  x)
x  0                                        y 1
 2x
Example 3
Finding the derivative by the
limit process
The limit used to define the slope of the tangent line is also used to
define differentiation. The derivative of f at x is given by
f ( x  x)  f ( x)                  f ( x )  f (c )
f ( x)  lim                      or f (c)  lim
x 0           x                                       x c       xc
Find the derivative of
f ( x)  x3  2 x                                    x 3  3 x 2 x  3 x(x) 2  (x)3  2 x  x 3  2 x
 lim
x  0                          x
( x  x)3  2( x  x)  ( x 3  2 x)
3 x 2 x  3 x(x) 2  (x) 2  2
f ( x)  lim                                             lim
x 0                  x
x  0                 x
x [3 x 2  3 xx  (x) 2  2
 lim
x  0                x
 lim[3 x 2  3 xx  (x) 2  2]
x  0

 3x 2  2
Example 4
Differentiablilty
 Derivatives have
certain rules on
when they exist
Continuity
Difeferentiability
Vertical tangent
Example 5                                   f ( x)  lim
x 0
x  x  x
x

Applications                                 lim 

x 0 

x  x)  x   x  x  x 
x

  x  x  x 
 


( x  x)  x
                                           lim
                
There is a hill whose cross
section forms the equation
x 0
x       x  x  x
f ( x )  x . Your car just died so
x
you have to push your car up the        lim
hill. But the hill is too steep at
first. So You make a ramp to
x 0
x      x  x  x   
make it up the hill but you don’t       lim
1
know how steep it is. You won’t            x 0    x  x  x
be able to push your car up the
ramp if it is steeper than 1/3               1

When the ramp is at your feet, it          2 x
touches the hill at one point 4                                     1
feet from the start of the hill. How                        m
steep is the ramp? Will you be                                    2 (4)
able to push you car up it? Use
the definition of the tangent line                             1
4
Example 6                                             f ( x)   x  3
Application                                                 f ( x)  f (0)
f (0)  lim
A see saw sits on the                                 x 0      x0
pivot formed by the
 lim
  x  3  3  1         Deriv.
equation f ( x)   x  3
x  0     x0                  from left
What is the slope of the
of the see saw at x=0?                               and
Use the definition of the
lim
  x  3  3  1      Deriv. from
Derivative to find                                                     right
x  0    x0
The derivative from the left and right do
f ( x )  f (c )      not equal eachother; thereofre, f is not
Recall f (c)  lim
x c        xc              differentiable at x = 0 so we don’t know
what the slope of the see saw is at x = 0
Trick Question!!!!!!
The slope of the see saw
is indeterminate, we
don’t what it is!!
The End

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