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					  Tangents


Calc AB- Section2.7
If we want to find the tangent line at a point P of a curve, we
consider a nearby point and find the SLOPE of the secant line
                                       _______
through P and Q
       Qx0  h, f x0  h
   y                                   y

   Px0 , f x0                                    QQ
                     Q
                                                 Q
                                             P
           P
                 h
                                                                  x
                                x
            x0                      Tangent line at P
                                    as h goes to zero
                     x0  h
Definitions: TANGENT LINE
                                                  Px0 , f x0 
 The slope of the curve y  f  x  at the point ____________ is
the number


               f x0  h   f x0 
    m    lim
          h 0
                                       (Provided the limit exists)
                        h
                                           Line Through
The tangent line to the curve at P is the ____________ P with
the slope m defined above
Find the line tangent to the curve y  f  x  at the point  x0 , y0 
                                                           3
Example 1: Find the line tangent to the hyperbola f  x 
at the point  3,1
                                                           x

                   f x0     f x0  h 
Step 1: Calculate ______ and _________

   f 3   1           f 3  h  
          3                            3
          3                           3 h
Step 2: Find the _________ of the tangent line
                  SLOPE
       3
           1        3  3  h           h               1
      3 h     lim                lim             lim
                                    h 0 h3  h    h 0 3  h 
lim
 h 0    h       h 0 h3  h 

                                                         1        1
                   EQUATION                                   
                                                       3  0 3
Step 3: Find the ____________
        of the tangent line.
                   1
            y 1     x  3
                   3
Rates of Change

AVERAGE RATE OF CHANGE
The average rate of change is the SLOPE of the SECANT line.
                                  ______       ________
It’s an _________________!
        APPROXIMATION
                                  Through the points
        y f x0  h   f x0          x0 , f x0  x0  h, f x0  h
           
        x          h


INSTANTANEOUS RATE OF CHANGE
The instantaneous rate of change is the ________ of the
                                        SLOPE
  TANGENT
____________ line.
It’s __________!
      EXACT
                        f x0  h   f x0 
                   lim
                   h 0          h
Example 2:
The distance traveled, in feet, by a free falling rock from rest
is given by the function f  t   16t 2 , where t  seconds.



a) Estimate the rocks speed in the interval from t  1 to t  1.1

    d f 1  h   f 1 161  h   161 16h 2  2h   16h  2
                                   2       2

                                          
    t         h                    h              h
  Since t goes from 1 to 1.1 the change in time is 0.1 = h

              160.1  2  33.6 ft sec

  The estimated speed of the rock at t =1 is 33.6 ft/sec
b) Find the actual speed of the rock at t  1

     161  h   16        161  2h  h 2   16        16  32h  16h 2  16
             2
lim                   lim                          lim
h 0        h          h 0          h               h 0          h

            h32  16h 
      lim                lim 32  16h   32  16  0  32 ft sec
       h 0      h         h 0




 The ACTUAL speed of the rock at t =1 is 32 ft/sec
Example 3: Find the slope of the tangent line at x  0
for the function f  x   x 1 3
     f 0  h   f 0        h1 3  0         1
lim                     lim           lim 2 3  
h 0         h            h 0    h       h 0 h
Vertical Tangents
A curve y  f  x  has a vertical tangent at the point x  x0
         f x0  h   f x0 
 if lim
    h 0
                                   or  
                  h
Example 4: Find the slope of the tangent line at x  0
for the function g  x   x
                             23



     f 0  h   f 0          h2 3  0         1
lim                     lim              lim 1 3  
h 0         h            h 0             h 0 h
                                    h

      1
lim 1 3  
h 0 h




No vertical tangent at x = 0 because the limit DNE

				
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