21 Tangent, Slope and Derivative

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  • 8/12/2019 21 Tangent, Slope and Derivative

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    Aim: The Tangent Problem & the Derivative Course: Calculus

    Do Now: What is the equation of the linetangent to the circle at point (7, 8)?

    Aim: What do slope, tangent and the

    derivative have to do with each other?

    5 10

    10

    8

    6

    4

    2

    -2

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    Aim: The Tangent Problem & the Derivative Course: Calculus

    Tangents & Secants

    A tangent to a circle is a line in the plane of

    the circle that intersects the circle in exactly

    one point.

    O

    A

    A secant of a circle is a line that intersects the

    circle in two points.

    B

    C

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    Aim: The Tangent Problem & the Derivative Course: Calculus

    -1

    Tany

    x1-1

    1radius = 1

    center at (0,0)(x,y)

    cos , sin

    cos

    tan lengthof theleg oppositelength of the leg adjacent to

    x

    y

    cos

    sintan

    1

    slope

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    Aim: The Tangent Problem & the Derivative Course: Calculus

    slope is

    steep!

    slope islevel:

    m= 0

    Tangents to a Graph

    (x1, y1)

    (x2, y2)

    (x

    3,y

    3)

    (x4, y4)slope is

    falling:

    mis (-)

    3

    2

    1

    -1

    2A

    Unlike a tangent to a circle,

    tangent lines of curves can

    intersect the graph at more than

    one point.

    3

    2

    1

    -1

    2A

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    Aim: The Tangent Problem & the Derivative Course: Calculus

    Finding the Slope (tangent) of a Graph at a Point

    10

    8

    6

    4

    2

    5

    h x = x2

    1

    2

    21

    2 x

    y

    mslope

    This is an approximation. How can we be

    sure this line is really tangent to f(x) at (1, 1)?

    (1, 1)

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    Aim: The Tangent Problem & the Derivative Course: Calculus

    3

    2.5

    2

    1.5

    1

    0.5

    1

    Slope and the Limit Process

    x, f(x)

    h

    f(x + h)f(x)

    x

    ymslope

    sec

    (x + h, f(x + h))

    h is the change in x

    f(x+ h)f(x)is thechange in y

    As (x + h, f(x + h))

    moves down the curve

    and gets closer to

    (x, f(x)), the slope of

    the secant moreapproximates the

    slope of the tangent at

    (x, f(x).

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    Aim: The Tangent Problem & the Derivative Course: Calculus

    3

    2.5

    2

    1.5

    1

    0.5

    1

    Slope and the Limit Process

    x, f(x)

    h

    f(x + h)f(x)

    (x + h, f(x + h))

    What is happening to

    h, the change in x?

    Its approaching 0,

    or its limit at x as h

    approaches 0.

    x

    ymslope

    sec

    h is the change in x

    f(x+ h)f(x)is thechange in y

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    Aim: The Tangent Problem & the Derivative Course: Calculus

    Slope and the Limit Process

    3

    2.5

    2

    1.5

    1

    0.5

    1

    As h 0, the slope of the

    secant, which

    approximates the slope

    of the tangent at (x, f(x))

    more closely as (x + h, f(x+ h)) moved down the

    curve. At reaching its

    limit, the slope of thesecant equaled the slope

    of the tangent at (x, f(x)).

    sec0

    tan lim mmslopeh

    h

    xfhxfmslope

    )()(sec

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    Aim: The Tangent Problem & the Derivative Course: Calculus

    Definition of slope of a Graph

    3

    2.5

    2

    1.5

    1

    0.5

    1

    The slopem

    of the graphof fat the point (x, f(x)) ,

    is equal to the slope of its

    tangent line at (x, f(x)),

    and is given by

    provided this limit exists.

    sec0

    tan lim mmslopeh

    h

    xfhxf

    m h

    )()(

    lim0

    difference quotient

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    Aim: The Tangent Problem & the Derivative Course: Calculus

    Model Problem

    Find the slope of the graph f(x) = x2at the

    point (-2, 4).

    h

    xfhxfm

    h

    )()(lim

    0

    h

    fhfm

    h

    )2()2(lim

    0

    set up difference

    quotient

    h

    hm

    h

    22

    0

    )2()2(lim

    Use f(x) = x2

    h

    hh

    m h

    444

    lim

    2

    0

    Expand

    h

    hhm

    h

    2

    0

    4lim

    Simplify

    h

    hhm

    h

    )4(lim

    0

    Factor and divide out

    Simplify)4(lim0

    hmh

    4)4(lim0

    hm h Evaluate the limit

    8

    6

    4

    2

    Slope ED = -4.00

    D: (-2 .00, 4.00)

    q x = x2

    E

    D

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    Aim: The Tangent Problem & the Derivative Course: Calculus

    Slope at Specific Point vs. Formula

    h

    xfhxf

    m h

    )()(

    lim)1( 0

    h

    cfhcfm

    h

    )()(lim)2(

    0

    What is the difference between thefollowing two versions of the difference

    quotient?

    (1) Produces a formula for finding the

    slope of any point on the function.

    (2) Finds the slope of the graph for the

    specific coordinate (c, f(c)).

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    Aim: The Tangent Problem & the Derivative Course: Calculus

    Definition of the Derivative

    The derivative of fat xis

    h

    xfhxfxf

    h

    )()(lim)('

    0

    provided this limit exists.

    The derivative f(x) is a formula for the

    slope of the tangent line to the graph offat the point (x,f(x)).

    The function found by evaluating the limit of

    the difference quotient is called thederivative of fat x. It is denoted by f (x),

    which is read f pr ime of x.

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    Aim: The Tangent Problem & the Derivative Course: Calculus

    Finding a Derivative

    Find the derivative of f(x) = 3x22x.

    0

    ( ) ( )'( ) lim

    h

    f x h f x f x

    h

    h

    xxhxhxxf h

    )23()](2)(3[lim)('

    22

    0

    h

    xxhxhxhxxf

    h

    2322363lim)('

    222

    0

    h

    hhxhxf

    h

    236lim)('

    2

    0

    h

    hxh

    xf h

    )236(

    lim)(' 0

    factor out h

    )236(lim)('0

    hxxfh

    26x

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    Aim: The Tangent Problem & the Derivative Course: Calculus

    Do Now:

    Find the equation of the line tangent to

    Aim: What is the connection between

    differentiability and continuity?

    ( ) 2Find the slope of tangent at = 9f x x

    x

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    Aim: The Tangent Problem & the Derivative Course: Calculus

    6

    5

    4

    3

    2

    1

    0.5 1 1.5 2 2.5

    f(x) is a continuous function

    Differentiability and Continuity

    What is the relationship, if any, between

    differentiability and continuity?

    (c, f(c))

    (x, f(x))

    f(x)f(c)

    xc

    xc

    Is there a limit as x

    approachesc? YES

    x c

    f x f c f c

    x c

    ( ) ( )'( ) lim

    alternative

    form of

    derivative

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    Aim: The Tangent Problem & the Derivative Course: Calculus

    Graphs with Sharp TurnsDifferentiable?

    2.5

    2

    1.5

    1

    0.5

    -0.5

    1 2 3 4 5

    f(x) = |x2|

    Is this function continuous at 2?

    m= 1m= -1

    x

    x2

    lim | 2 | ?

    x

    x2

    lim | 2 | ?

    YES

    One-sided limits are not equal, fis therefore notdifferentiable at 2. There is no tangent line at (2, 0)

    2 2

    2 0( ) (2)lim lim 1

    2 2x x

    xf x f

    x x

    x c

    f x f c

    f c x c

    ( ) ( )

    '( ) lim

    alternative

    form of

    derivative

    2 2

    2 0( ) (2)lim lim 1

    2 2x x

    xf x f

    x x

    Is this function differentiable at 2?

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    Aim: The Tangent Problem & the Derivative Course: Calculus

    Graph with a Vertical Tangent Line1.2

    1

    0.8

    0.6

    0.4

    0.2

    -0.2

    -0.4

    -0.6

    -0.8

    -1

    -1 1

    f(x) = x1/3

    Is f continuous at 0?

    YES

    x

    f x f

    x0

    ( ) (0)lim

    0

    x

    x

    x

    1

    3

    0

    0lim

    x

    x

    20 3

    1lim UND

    x

    x

    13

    0

    lim ?

    Does a limit exist at 0?

    NO

    fis not differentiable at 0; slope

    of vertical line is undefined.

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    Aim: The Tangent Problem & the Derivative Course: Calculus

    Differentiability Implies Continuity

    a b c d

    fis not

    continuous at a

    therefore not

    differentiable

    f is

    continuous at

    b& c, but not

    differentiable

    corner

    vertical

    tangent

    fis continuous at

    dand

    differentiable

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    Aim: The Tangent Problem & the Derivative Course: Calculus

    Summary