Transcript Slide 1

CHAPTER 2
LIMITS AND DERIVATIVES
LIMITS AND DERIVATIVES
 The idea of a limit underlies the various
branches of calculus.
 It is therefore appropriate to begin our study of
calculus by investigating limits and their properties.
 The special type of limit used to find tangents and
velocities gives rise to the central idea in differential
calculus—the derivative.
LIMITS AND DERIVATIVES
2.1
The Tangent and
Velocity Problems
In this section, we will learn:
How limits arise when we attempt to find the
tangent to a curve or the velocity of an object.
THE TANGENT PROBLEM
 The word tangent is derived from the Latin
word tangens, which means ‘touching.’ Thus,
a tangent to a curve is a line that touches the
curve.
 In other words, a tangent line should have the same
direction as the curve at the point of contact.
THE TANGENT PROBLEM
 For a circle, we could simply follow Euclid and
say that a tangent is a line that intersects the
circle once and only once.
THE TANGENT PROBLEM
 For more complicated curves, that definition is
inadequate.
 The figure displays two lines l and t passing through a
point P on a curve.
 The line l intersects only once, but it certainly does not
look like what is thought of as a tangent.
THE TANGENT PROBLEM
 In contrast, the line t looks like a tangent,
but it intersects the curve twice.
THE TANGENT PROBLEM

Example 1
Find an equation of the tangent line to the
parabola y = x2 at the point P(1,1).
 We will be able to find an equation of the tangent
line as soon as we know its slope m.
 The difficulty is that we know only one point, P,
on t, whereas we need two points to compute the
slope.
THE TANGENT PROBLEM
Example 1
 However, we can compute an approximation to
m by choosing a nearby point Q(x, x2) on the
parabola and computing the slope mPQ of the
secant line PQ.
THE TANGENT PROBLEM
Example 1
We choose x  1 so that Q  P.
 Then,
x 2  1 f ( x)  f (1)
mPQ 

x 1
x 1
 For instance, for the point Q(1.5, 2.25),
we have:
mPQ
2.25  1 1.25


 2.5
1.5  1
0.5
THE TANGENT PROBLEM
Example 1
 The tables below the values of mPQ for several
values of x close to 1. The closer Q is to P, the
closer x is to 1 and, it appears from the tables,
the closer mPQ is to 2.
 This suggests that the slope
of the tangent line t should
be m = 2.
THE TANGENT PROBLEM
Example 1
 The slope m of the tangent line is said to be the
limit of the slopes of the secant lines. This is
expressed symbolically as follows.
m  lim mPQ
QP
x 1
m  lim
2
x 1 x  1
2
f ( x)  f (1)
x 1
m  lim
 lim
2
x 1
x 1 x  1
x 1
2
THE TANGENT PROBLEM
Example 1
 Assuming that the slope of the tangent line is
indeed m  2, we can use the point-slope form
of the equation of a line to write the equation
of the tangent line through (1, 1) as:
y  1  2( x  1)
or
y  2x 1
THE TANGENT PROBLEM
Example 1
 The figure illustrates the limiting process that
occurs in this example.
THE TANGENT PROBLEM
Example 1
 As Q approaches P along the parabola, the
corresponding secant lines rotate about P and
approach the tangent line t.
DEFINITIONS AND SOME
NOTATION FOR THE
GENERIC CASE
Average Rate of Change
The change of f (x) over the interval [a,x] is
f  f ( x)  f (a)
The average rate of change of f (x) over the
interval [a, x] is
change in f f
f ( x)  f (a)


xa
change in x x
Difference
Quotient
Average Rate of Change
Is equal to the slope of the secant line through the
points (a, f (a)) and (x, f (x)) on the graph of f (x)
Q   x, f ( x) 
P   a, f (a) 
mS  mPQ
f  f ( x)  f (a)
x  x  a
secant line
has slope mS
f
f ( x)  f (a )


x
xa
Units for f /x
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.
f
x
is given in
units of f ( x)
units of x
Alternative Notation for f /x
The average rate of change of f over the interval
[a, x] can be written in two different ways:
The first one is
f
f ( x)  f (a )

x
xa
Alternative Notation for f /x
The second one comes from letting x  a  h,
which means x  a + h. In this case we write,
f
f ( x )  f ( a ) f ( a  h)  f ( a )


x
xa
h
Average Rate of Change of f over [a, a + h]
f
f ( a  h)  f (a )

x
h
Alternative Notation for f /x
Both versions represent the slope of the secant
line through the points (a, f (a)) and (x, f (x)).
Q   x, f ( x)    a  h, f (a  h) 
P   a, f (a) 
mS  mPQ
f  f (b)  f (a)  f (a  h)  f (a)
x  x  a  h
f
f ( x )  f ( a ) f ( a  h)  f (a )



x
xa
h
Average Rate of Change as h0
We now look at the behavior of the average rate
of change of f (x) as x a  h gets smaller and
smaller, that is, we will let h tend to 0 (h0) and
look for a geometric interpretation of the result.
For this, we consider an example of values of
mS  mPQ
f
f ( a  h)  f ( a )


x
h
and their corresponding secant lines.
Average Rate of Change as h0
Secant line tends to become the Tangent line
h  2.57
Average Rate of Change as h0
Secant line tends to become the Tangent line
h2
Average Rate of Change as h0
Secant line tends to become the Tangent line
h  1.5
Average Rate of Change as h0
Secant line tends to become the Tangent line
h 1
Average Rate of Change as h0
Secant line tends to become the Tangent line
h  0.5
Average Rate of Change as h0
Secant line tends to become the Tangent line
h  0.2
Average Rate of Change as h0
Secant line tends to become the Tangent line
h  0.1
Zoom in
Average Rate of Change as h0
Secant line tends to become the Tangent line
h  0.1
Average Rate of Change as h0
Secant line tends to become the Tangent line
h  0.05
Average Rate of Change as h0
We observe that as h approaches zero,
1. The secant line through P and Q approaches the
tangent line to the point P on the graph of f.
2. and consequently, the slope mS of the secant line
approaches the slope mT of the tangent line to the
point P on the graph of f.
f ( a  h)  f ( a )
mT  mS 
h
h small
Instantaneous Rate of Change of f at
x=a
That is, the limiting value, as h gets increasingly
smaller, of the difference quotient
f ( a  h)  f ( a )
mS 
h
(slope of secant line)
is the slope mT of the tangent line to the graph of the
function f (x) at x = a.
Instantaneous Rate of Change of f at
x=a
This is usually written as
f ( a  h)  f ( a )
mS 
 mT
h
as h  0
tends to, or
approaches
That is, mS approaches mT as h tends to 0.
Instantaneous Rate of Change of f at
x=a
More briefly using symbols by writing
f ( a  h)  f ( a )
mT  lim
h 0
h
and read lim as “the limit as h approaches 0 of ”
h 0
Instantaneous Rate of Change of f at
x=a
that is, the limit symbol indicates that the value of
f ( a  h)  f ( a )
mS 
h
can be made arbitrarily close to mT by taking h to be
a sufficiently small number.
Instantaneous Rate of Change of f at
x=a
Definition: The instantaneous rate of change of f (x)
at x = a is defined to be the slope of the tangent line
to the graph of the function f (x) at x = a. That is,
f ( a  h)  f ( a )
mT  lim
 "instantaneous rate of change at a "
h 0
h
Remark: The slope mT gives a precise indication of
how fast the graph of f (x) is increasing or decreasing
at x = a.
Equation of Tangent Line at x = a
Using the point-slope form of the line
y  f (a)  mT ( x  a)
 a, f (a)
tangent line at a
THE TANGENT PROBLEM
 Many functions that occur in science are not
described by explicit equations, but by
experimental data.
 The next example shows how to estimate the slope
of the tangent line to the graph of such a function.
THE TANGENT PROBLEM
Example 2
 The flash unit on a camera operates by
storing charge on a capacitor and releasing it
suddenly when the flash is set off.
 The data in the table describe
the charge Q remaining on
the capacitor (measured in
microcoulombs) at time t
(measured in seconds after
the flash goes off).
THE TANGENT PROBLEM
Example 2
 Using the data, you can draw the graph of this
function and estimate the slope of the tangent
line at the point where t = 0.04.
 Remember, the slope of the
tangent line represents the
electric current flowing from
the capacitor to the flash bulb
measured in microamperes.
THE TANGENT PROBLEM
Example 2
THE TANGENT PROBLEM
Example 2
 In the figure, the given data are plotted and
used to sketch a curve that approximates the
graph of the function.
THE TANGENT PROBLEM
Example 2
 Given the points P(0.04, 67.03) and R(0, 100), we
find that the slope of the secant line PR is:
mPR
100.00  67.03

 824.25
0.00  0.04
THE TANGENT PROBLEM
Example 2
 The table shows the results of similar
calculations for the slopes of other secant lines.
 From this, we would expect the slope of the tangent line at t
= 0.04 to lie somewhere between –742 and –607.5.
THE TANGENT PROBLEM
Example 2
 In fact, the average of the slopes of the two
closest secant lines is:
1
 742  607.5  674.5
2
 So, by this method, we estimate the slope of
the tangent line to be –675.
THE TANGENT PROBLEM
Example 2
 Another method is to draw an approximation to
the tangent line at P and measure the sides of
the triangle ABC.
THE TANGENT PROBLEM
Example 2
 This gives an estimate of the slope of the
tangent line as:
AB
80.4  53.6


 670
BC
0.06  0.02
THE VELOCITY PROBLEM
 If you watch the speedometer of a car as you
travel in city traffic, you see that the needle
does not stay still for very long. That is, the
velocity of the car is not constant.
 We assume from watching the speedometer that the
car has a definite velocity at each moment.
 How is the ‘instantaneous’ velocity defined?
 To answer this question, we first look at the notion
of average velocity.
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:
vaverage
s (t  h)  s (t ) s


h
t
Average Velocity
Initial
position
s (t )
Final
position
s(t  h)
Distance traveled
s(t  h)  s(t )  s
Time elapsed is
(t  h)  t  h  t
vaverage
s (t  h)  s (t ) s


h
t
Average Velocity
Initial
velocity vi
s (t )
Final
velocity vf
s(t  h)
During the motion, the velocity of the object over the
time interval [t,t+h] may not be uniform, that is, the
velocity may have different values at different points
on the trajectory. However, it is intuitive, that if
Average Velocity
Initial
velocity vi
s (t )
Final
velocity vf
s(t  h)
the time elapsed between the initial and final position is
small, then the average velocity over the time interval
[t,t+h] should be approximately equal to vi, initial
velocity, and vf , the final velocity. That is,
vaverage
s

 vi  v f
t
Average Velocity
Initial
velocity vi
s (t )
Final
velocity vf
s(t  h)
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.
Average Velocity
Initial
velocity vi
s (t )
Final
velocity vf
s(t  h)
In symbols, as t  h  0 (delta t approaches zero),
the average velocity over the interval [t,t+h] approaches
the velocity v(t) of the object at time t.
vaverage
s(t  h)  s(t )

 v(t )
h
as
h0
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.
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 Galileo’s
law is expressed by the following equation.
s(t) = 4.9t2
 Remember, this model neglects air resistance.
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).
vaverage
s s(5  1)  s(5) 4.9(62 )  4.9(52 )



 53.9 m/s
t
1
1
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
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
Example: Velocity Problem
 You may have the feeling that the calculations
used in solving the problem are very similar to
those used earlier to find tangents.
 There is a close connection between the tangent
problem and the problem of finding velocities.
 If we draw the graph of the distance function of
the ball and consider the points P(a, 4.9a2) and
Q(a + h, 4.9(a + h)2), then the slope of the
secant line PQ is:
4.9( a  h)  4.9a

( a  h)  a
2
mPQ
2
 That is the same as the average velocity over the
time interval [a, a + h].
 Therefore, the velocity at time t = a (the limit of these
average velocities as h approaches 0) must be equal to the
slope of the tangent line at P (the limit of the slopes of the
secant lines).
Another Velocity Problem
My friend Eric, an enthusiastic baseball player, claims he
can “probably” throw a ball upward (vertical direction) at
a speed of 100 feet per second (ft/sec). Our physicist
friends tell us that its height in feet, seconds later, would
be s(t )  100t  16t 2.
Find its average velocity over the following intervals:
[2 , 3], [2 , 2.5], [2 , 2.1], [2 , 2.01] and [2 , 2.001].
Interpret your results. Notice that h is1, 0.5, 0.1, 0.01,
and 0.001 seconds, respectively.
Another Velocity Problem
Numerical approach
h [sec]
s(2) [ft]
s(2  h) [ft]
s (2  h)  s (2)  ft 
 sec 
h
1
136
156
20
0.5
136
150
28
0.1
136
139.44
34.4
0.01
136
136.3584
35.84
0.001
136
136.035984
35.984
0.0001
136
136.0359984
35.9984
Algebraic approach
s(t )  100t  16t 2
s(2  h)  s(2)
 36  16h
h
Another Velocity Problem
s(2  h)  s(2)
h
h [sec]
 ft 
36  16h  
 sec 
1
20
0.5
28
0.1
34.4
0.01
35.84
0.001
35.984
0.0001
35.9984
Another Velocity Problem
s(2  h)  s(2)
h
h [sec]
 ft 
36  16h  
 sec 
1
20
0.5
28
0.1
34.4
0.01
35.84
0.001
35.984
0.0001
35.9984
Another Velocity Problem
s(2  h)  s(2)
h
h [sec]
 ft 
36  16h  
 sec 
1
20
0.5
28
0.1
34.4
0.01
35.84
0.001
35.984
0.0001
35.9984
Another Velocity Problem
s(2  h)  s(2)
h
h [sec]
 ft 
36  16h  
 sec 
1
20
0.5
28
0.1
34.4
0.01
35.84
0.001
35.984
0.0001
35.9984
Another Velocity Problem
s(2  h)  s(2)
h
h [sec]
 ft 
36  16h  
 sec 
1
20
0.5
28
0.1
34.4
0.01
35.84
0.001
35.984
0.0001
35.9984
Another Velocity Problem
s(2  h)  s(2)
h
h [sec]
 ft 
36  16h  
 sec 
1
20
0.5
28
0.1
34.4
0.01
35.84
0.001
35.984
0.0001
35.9984
THE VELOCITY PROBLEM

Examples 1 and 3 show that to solve tangent
and velocity problems we must be able to find
limits.
 After studying methods for computing limits
for the next five sections we will return to the
problem of finding tangents and velocities in
Section 2.7.