Chapter 3 Introduction to the Derivative Sections 3'4

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Chapter 3Introduction to the DerivativeSections
3.4
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Introduction to the Derivative

• Average Rate of Change
• The Derivative
• Derivatives of Powers, Sums and Constant
  asMultiples
• Marginal Analysis

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Average Rate of Change
The change of f (x) over the interval a,b is
The average rate of change of f (x) over the
interval a , b is
Difference Quotient
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Average Rate of Changefrom a Table
The following table lists the approximate value
of the Standard and Poors (SP) 500 stock market
during the 6-year period 1995-2001 (t5
represents 1995)
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• What was the average rate of change in the SP
  over the 4-year period 1996-2000 (the period 6?t
  ? 10 or 6,10),
• 2-year period 1999-2001 (the period 9?t ?11 or
  9,11),
• and over the period 8,11?
• Graph the values shown in the table. How are the
  rates of change reflected on the graph?

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• Solution During the 4-year period 6,10, the
  SP changed as follows
• At the start of the period (t 6) S(6)
  650
• At the end of the period (t 10) S(10)
  1450
• Change during the period 6,10 S(10) -
  S(6) 800

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Thus, the SP increased by 800 points in 4 years,
giving an average of 800/4 200 points per year.
We write this calculation in the following way
Interpretation During the period 1996-2000, the
SP increased at an average rate of 200 points
per year.
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Similarly, during the period 9,11, the average
rate of change was
Interpretation During the period 1999-2001, the
SP decreased at an average rate of 75 points
per year.
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Similarly, during the period 8,11, the average
rate of change was
Interpretation During the period 1998-2001, on
average, the SP did not change.
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• Graph of the values shown in the table.

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Average Rate of Change
Is equal to the slope of the secant line through
the points (a , f (a)) and (b , f (b)) on the
graph of f (x)
secant line has slope mS
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Units

• The units of ?f , change in f , are the units of
  f (x).
• The units of the average rate of change of f are
  nunits of f (x) per units of x.

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Notation
The average rate of change of f over the interval
a , b can be written in two different
ways The first one is
The second one comes from letting b? a ? h,
which means b ? a h. In this case we get,
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Notation
The second one comes from letting b? a ? h,
which means b ? a h. In this case we write,
Average Rate of Change of f over a , a h
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Notation
Both versions represent the slope of the secant
line through the points (a , f (a)) and (b , f
(b)).
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Average Rate of Change
Example Compute the average rate of change of
over 0,2.
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This means that the slope of the secant line to
the graph of parabola through the points (0 , 0)
and (2 , 8) is mS ? 4
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Geometric Interpretation
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Example The figure shows the numbers of SUVs
sold in the US each year from 1990 to 2003.
Notice t ? 0 represents the year 1990, and N(t)
represents sales in year t in thousands of
vehicles.
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• Use the graph to estimate the average rate of
  change of N(t) with respect to t over 6 , 11
  and interpret the result.
• Over which one-year period(s) was the average
  rate of N(t) the greatest?

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• The average rate of change of N over 6 , 11 is
  given by the slope of the line through the points
  P and Q.

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• Sales of SUVs were increasing at an average rate
  of 200,000 SUVs per year from 1996 to 2001.

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• The rates of change of annual sales over
  successive one-year periods are given by the
  slopes of the individual segments that make up
  the graph of N(t). Thus, the greatest average
  rate of change over a single year corresponds to
  the segment(s) with the largest slope.

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Average Velocity
For an object that is moving in a straight line
with position s(t) (in feet, miles, etc.) at time
t (in sec, hours, etc.), the average velocity
from time t to time t h is the average rate of
change of position with respect to time
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Average Velocity
Initial position
Final position
Distance traveled
Time elapsed is
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Average Velocity
Initial velocity vi
Final velocity vf
During the motion, the velocity of the object
over the time interval t , th may not be
uniform, that is, the velocity may have different
values at different points on the trajectory.
However, it is intuitive, that if
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Average Velocity
Initial velocity vi
Final velocity vf
the time elapsed between the initial and final
position is small, then the average velocity over
the time interval t , th should be
approximately equal to vi, initial velocity, and
vf , the final velocity. That is,
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Average Velocity
Initial velocity vi
Final velocity vf
therefore, as the time elapsed, ?t ? h , between
the initial and final position gets smaller and
smaller, the value of the average velocity gets
closer and closer to the value of velocity v (t)
of the object at time t.
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Average Velocity
Initial velocity vi
Final velocity vf
In symbols, as (delta t
approaches zero), the average velocity over the
interval t , th approaches the velocity v (t)
of the object at time t.
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Example Velocity Problem

• Investigate the example of a falling ball.
• Suppose that a ball is dropped from the upper
  observation deck of the CN Tower in Toronto,
  450 m above the ground.
• Find the velocity of the ball
• after 5 seconds.

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Example Velocity Problem

• Through experiments carried out four centuries
  ago,
• Galileo discovered that the distance fallen by
  any freely
• falling body is proportional to the square of the
  time it
• has been falling. If the distance fallen after t
  seconds is
• denoted by s(t) and measured in meters, then
  Galileos
• law is expressed by the following equation.
• s(t) 4.9t2
• Remember, this model neglects air resistance.

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Example Velocity Problem

• The difficulty in finding the velocity after 5
  seconds is
• that you are dealing with a single instant of
  time (t 5).
• No time interval is involved.
• However, we can approximate the desired quantity
  by
• computing the average velocity over the brief
  time
• interval of one second (from t 5 to t 6).

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Example Velocity Problem

• The table shows the results of similar
  calculations of the
• average velocity over successively smaller time
  periods.
• It appears that, as we shorten the time period,
  the average velocity is becoming closer to 49
  m/s.

h ?1 h ?0.1 h ?0.05 h ?0.01 h ?0.001
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Example Velocity Problem

• The instantaneous velocity when t 5 is defined
  to be
• the limiting value of these average velocities
  over
• shorter and shorter time periods that start at t
  5.
• Thus, the instantaneous velocity after 5 s is v
  49 m/s

h ?1 h ?0.1 h ?0.05 h ?0.01 h ?0.001