Section 2'6 Tangents, Velocities, and Other Rates of Change - PowerPoint PPT Presentation

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Section 2'6 Tangents, Velocities, and Other Rates of Change

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... find an equation of the tangent line to the hyperbola y = 3/x at the point (3, 1) ... followed by a graph of the hyperbola and the tangent line. ... – PowerPoint PPT presentation

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Title: Section 2'6 Tangents, Velocities, and Other Rates of Change


1
Section 2.6Tangents, Velocities, and Other Rates
of Change
  • Goals
  • Use limits to find exact values of
  • slopes of tangent lines
  • velocities
  • other rates of change

2
Tangents
  • If a curve C has equation y f(x) and we
    want to find the tangent line to C at the point
    P(a, f(a)) , then we
  • consider a nearby point Q(x, f(x)) , where x ?
    a , and
  • compute the slope of the secant line PQ

3
Tangents (contd)
  • Then we let Q approach P along the curve C
    by letting x approach a .
  • If mPQ approaches a number m , then we define
    the tangent t to be the line
  • through P
  • with slope m .
  • This is the content of the following definition

4
Tangents (contd)
  • On the next two slides we illustrate this
    definition

5
Tangents (contd)
6
Tangents (contd)
7
Example
  • Earlier we used graphs and tables to find an
    equation of the tangent line to the parabola y
    x2 at the point P(1, 1) .
  • Now we can carry out this calculation with
    precision.
  • Here a 1 and f(x) x2 , so the slope is
    given by the limit calculations on the next slide

8
Example (contd)
9
Example (contd)
  • Using the point-slope form of the equation of a
    line, we find that an equation of the tangent
    line at (1, 1) is
  • y 1 2(x 1) or y 2x 1
  • The next three slides show a zoom in on the
    parabola toward the point (1, 1) .
  • Note that the curve becomes almost the same as
    its tangent line as we zoom in

10
Example (contd)
11
Example (contd)
12
Example (contd)
13
Another Expression
  • There is another expression for the slope of a
    tangent line
  • We let h x a
  • Then x a h
  • Thus the slope of secant line PQ is
  • The next slide illustrates the case h gt 0

14
Another Expression (contd)
15
Another Expression (contd)
  • In this way the expression for the slope of the
    tangent line becomes
  • As an example, we find an equation of the tangent
    line to the hyperbola y 3/x at the point (3,
    1)

16
Example (contd)
  • Solution Let f(x) 3/x . Then the slope of the
    tangent at (3, 1) is ? .
  • The details are given in the calculation on the
    next slide
  • followed by a graph of the hyperbola and the
    tangent line.
  • Therefore an equation of the tangent line is y
    1 ?(x 3) , or x 3y 6 0 .

17
Example (contd)
18
Example (contd)
19
Velocity
  • Suppose an object moves
  • along a straight line, and
  • according to an equation of motion s f(t) ,
    where
  • s is the displacement (directed distance) of the
    object from the origin at time t .
  • The function f that describes the motion is
    called the position function of the object.

20
Velocity (contd)
  • In the time interval from t a to t a h
    the change in position is f(a h) f(a) , as
    illustrated on the next slide.
  • The average velocity over this time interval
    equals the slope of the secant line PQ (shown
    on the subsequent slide)

21
Velocity (contd)
22
Velocity (contd)
23
Instantaneous Velocity
  • Now suppose we compute the average velocities
    over shorter and shorter time intervals a, a
    h .
  • That is, we let h approach 0 .
  • As in the earlier falling ball example, we define
    the instantaneous velocity v(a) at time t a
    to be the limit of the average velocities

24
Instantaneous Velocity (contd)
  • This means that the
  • velocity at time t a is equal to the
  • slope of the tangent line at P .

25
Example
  • Again we consider a ball dropped from a height of
    450 m. Find
  • the velocity of the ball after 5 seconds
  • how fast the ball is traveling when it hits the
    ground.
  • Solution We first use the equation of motion s
    f(t) 4.9t2 to find the velocity v(a) after
    a seconds

26
Example (contd)
  • The velocity after 5 s is
  • v(5) (9.8)(5) 49 m/s .

27
Example (contd)
  • The ball will hit the ground at the time t1
    with s(t1) 450 , that is, 4.9t12 450 .
  • Solving for t1 gives t1 9.6 s .
  • The velocity of the ball as it hits the ground is
    therefore
  • v(t1) 9.8t1 94 m/s

28
Other Rates of Change
  • Suppose y is a quantity that depends on another
    quantity x .
  • Thus, y is a function of x and we writey
    f(x) .
  • If x changes from x1 to x2 , then
  • the change in x is ?x x2 x1 , and
  • the corresponding change in y is
  • f(?x) f(x2) f(x1)

29
Other Rates (contd)
  • The difference quotient
  • is called the average rate of change of y with
    respect to x over the interval x1, x2 .
  • It can be interpreted as the slope of the secant
    line PQ in the figure on the next slide

30
Other Rates (contd)
31
Instantaneous Rate of Change
  • By analogy with velocity
  • we consider the average rate over smaller and
    smaller intervals
  • by letting x2 approach x1
  • and therefore letting ?x approach 0 .
  • The limit of these average rates of change is
    called the (instantaneous) rate of change of y
    with respect to x at x x1 .

32
Instantaneous Rate (contd)
  • The instantaneous rate of change can be
    interpreted as the slope of the tangent to the
    curve y f(x) at P(x1 , f(x1))

33
Example
  • Temperature readings T (in C) were recorded
    every hour starting at midnight on a day in
    Whitefish, Montana.
  • The time x is measured in hours from midnight.
  • The data are given in the table on the next slide

34
Example (contd)
35
Example (contd)
  • Find the average rate of change of temperature
    with respect to time
  • from noon to 3 P.M.
  • from noon to 2 P.M.
  • from noon to 1 P.M.
  • Estimate the instantaneous rate of change at
    noon.

36
Solution to (a)
  • From noon to 3 P.M. the temperature changes from
    14.3 C to 18.2 C, so
  • ?T T(15) T(12) 18.2 14.3 3.9 C
  • while the change in time is ?x 3 h .
  • Therefore, the average rate of change of
    temperature with respect to time is

37
Solution to (a) (contd)
  • From noon to 2 P.M. the average rate of change is
  • From noon to 1 P.M. the average rate of change is

38
Solution to (b)
  • We plot the given data on the next slide and
    sketch a smooth curve of the temperature
    function.
  • Then we draw the tangent at the point P where
    x 12 .
  • By measuring the sides of triangle ABC , we
    estimate the slope of the tangent line to be
    10.3/5.5 1.9 .

39
Solution to (b) (contd)
40
Solution to (b) (contd)
  • Therefore the instantaneous rate of change of
    temperature with respect to time at noon is about
    1.9 C .

41
Review
  • Use of limits to define
  • Tangent lines
  • Instantaneous velocities
  • Other rates of change
  • Homework
  • 1, 2, 3, 4, 5, 7, 9, 14, 15, 17, 19, 21, 22, 24,
    27, 28
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