Download presentation

Presentation is loading. Please wait.

Published byDeshawn Yelder Modified about 1 year ago

1
CHAPTER 2 LIMITS AND DERIVATIVES

2
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.

3
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. 2.1 The Tangent and Velocity Problems LIMITS AND DERIVATIVES

4
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

5
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

6
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

7
In contrast, the line t looks like a tangent, but it intersects the curve twice. THE TANGENT PROBLEM

8
Example 1 Find an equation of the tangent line to the parabola y = x 2 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

9
However, we can compute an approximation to m by choosing a nearby point Q(x, x 2 ) on the parabola and computing the slope m PQ of the secant line PQ. Example 1 THE TANGENT PROBLEM

10
We choose so that. Then, For instance, for the point Q(1.5, 2.25), we have: Example 1 THE TANGENT PROBLEM

11
The tables below the values of m PQ 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 m PQ is to 2. This suggests that the slope of the tangent line t should be m = 2. Example 1 THE TANGENT PROBLEM

12
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. Example 1 THE TANGENT PROBLEM

13
Example 1 THE TANGENT PROBLEM 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:

14
The figure illustrates the limiting process that occurs in this example. Example 1 THE TANGENT PROBLEM

15
As Q approaches P along the parabola, the corresponding secant lines rotate about P and approach the tangent line t. Example 1

16
DEFINITIONS AND SOME NOTATION FOR THE GENERIC CASE

17
The average rate of change of f (x) over the interval [a, x] is The change of f (x) over the interval [a,x] is Difference Quotient Average Rate of Change

18
secant line has slope m S 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)

19
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.

20
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

21
The second one comes from letting x a h, which means x a + h. In this case we write, Average Rate of Change of f over [a, a + h] Alternative Notation for f / x

22
Both versions represent the slope of the secant line through the points (a, f (a)) and (x, f (x)). Alternative Notation for f / x

23
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 and their corresponding secant lines. Average Rate of Change as h 0

24
Secant line tends to become the Tangent line

25
Average Rate of Change as h 0 Secant line tends to become the Tangent line

26
Average Rate of Change as h 0 Secant line tends to become the Tangent line

27
Average Rate of Change as h 0 Secant line tends to become the Tangent line

28
Average Rate of Change as h 0 Secant line tends to become the Tangent line

29
Average Rate of Change as h 0 Secant line tends to become the Tangent line

30
Zoom in Average Rate of Change as h 0 Secant line tends to become the Tangent line

31
Average Rate of Change as h 0 Secant line tends to become the Tangent line

32
Average Rate of Change as h 0 Secant line tends to become the Tangent line

33
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 m S of the secant line approaches the slope m T of the tangent line to the point P on the graph of f. Average Rate of Change as h 0

34
That is, the limiting value, as h gets increasingly smaller, of the difference quotient is the slope m T of the tangent line to the graph of the function f (x) at x = a. Instantaneous Rate of Change of f at x = a (slope of secant line)

35
This is usually written as Instantaneous Rate of Change of f at x = a tends to, or approaches That is, m S approaches m T as h tends to 0.

36
More briefly using symbols by writing Instantaneous Rate of Change of f at x = a and readas “the limit as h approaches 0 of ”

37
that is, the limit symbol indicates that the value of can be made arbitrarily close to m T by taking h to be a sufficiently small number. Instantaneous Rate of Change of f at x = a

38
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, Instantaneous Rate of Change of f at x = a Remark: The slope m T gives a precise indication of how fast the graph of f (x) is increasing or decreasing at x = a.

39
Equation of Tangent Line at x = a tangent line at a Using the point-slope form of the line

40
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.

41
THE TANGENT PROBLEM 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). Example 2

42
THE TANGENT PROBLEM 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. Example 2

43
THE TANGENT PROBLEM Example 2

44
THE TANGENT PROBLEM In the figure, the given data are plotted and used to sketch a curve that approximates the graph of the function. Example 2

45
THE TANGENT PROBLEM Given the points P(0.04, 67.03) and R(0, 100), we find that the slope of the secant line PR is: Example 2

46
THE TANGENT PROBLEM 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. Example 2

47
In fact, the average of the slopes of the two closest secant lines is: So, by this method, we estimate the slope of the tangent line to be –675. THE TANGENT PROBLEM Example 2

48
THE TANGENT PROBLEM Another method is to draw an approximation to the tangent line at P and measure the sides of the triangle ABC. Example 2

49
This gives an estimate of the slope of the tangent line as: THE TANGENT PROBLEM Example 2

50
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.

51
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:

52
Average Velocity Initial position Final position Distance traveled Time elapsed is

53
Average Velocity Initial velocity v i Final velocity v f 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

54
Average Velocity Initial velocity v i Final velocity v f 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 v i, initial velocity, and v f, the final velocity. That is,

55
Average Velocity Initial velocity v i Final velocity v f 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.

56
Average Velocity Initial velocity v i Final velocity v f In symbols, as (delta t approaches zero), the average velocity over the interval [t, t+h] approaches the velocity v (t) of the object at time t.

57
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

58
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) = 4.9t 2 Remember, this model neglects air resistance. Example: Velocity Problem

59
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). Example: Velocity Problem

60
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. Example: Velocity Problem h 1 h 0.1 h 0.05 h 0.01 h 0.001

61
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 Example: Velocity Problem h 1 h 0.1 h 0.05 h 0.01 h 0.001

62
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. Example: Velocity Problem

63
If we draw the graph of the distance function of the ball and consider the points P(a, 4.9a 2 ) and Q(a + h, 4.9(a + h) 2 ), then the slope of the secant line PQ is:

64
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).

65
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. 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

66
113615620 0.513615028 0.1136139.4434.4 0.01136136.358435.84 0.001136136.03598435.984 0.0001136136.035998435.9984 Numerical approachAlgebraic approach

67
Another Velocity Problem 120 0.528 0.134.4 0.0135.84 0.00135.984 0.000135.9984

68
Another Velocity Problem 120 0.528 0.134.4 0.0135.84 0.00135.984 0.000135.9984

69
Another Velocity Problem 120 0.528 0.134.4 0.0135.84 0.00135.984 0.000135.9984

70
Another Velocity Problem 120 0.528 0.134.4 0.0135.84 0.00135.984 0.000135.9984

71
Another Velocity Problem 120 0.528 0.134.4 0.0135.84 0.00135.984 0.000135.9984

72
Another Velocity Problem 120 0.528 0.134.4 0.0135.84 0.00135.984 0.000135.9984

73
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.

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google