28. the Integral as Net Change

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    -100

    -50

    0

    50

    100

    2 4 6 8 10

    ft

    min

    minutes

    A honey bee makes several trips from the hive to a flowergarden. The velocity graph is shown below.

    What is the total distance traveled by the bee?

    200ft

    200ft

    200ft

    100ft

    200 200 200 100 700 700 feet

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    -100

    -50

    0

    50

    100

    2 4 6 8 10

    ft

    min

    minutes

    What is the displacementdisplacement of the bee?

    200ft

    -200ft

    200ft

    -100ft

    200 200 200 100 100 100 feet towards the hive

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    To find the displacementdisplacement (position shift) from the velocityfunction, we just integrate the function. The negative

    areas below the x-axis are subtracted from the totaldisplacement.

    Displacement

    b

    a

    V t d t

    Distance Traveledb

    aV t dt

    To find distance traveleddistance traveled we have to use absolute value.

    Find the roots of the velocity equation and integrate in

    pieces, just like when we found the area between a curveand the x-axis. (Take the absolute value of each integral.)

    Or you can use your calculator to integrate the absolutevalue of the velocity function.

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    -2

    -1

    0

    1

    2

    1 2 3 4 5

    velocity graph

    -2

    -1

    0

    1

    2

    1 2 3 4 5

    position graph

    1

    2

    1

    2

    1

    2

    Displacement:

    1 11 2 1

    2 2

    Distance Traveled:

    1 11 2 42 2

    Many exams have at leastone problem requiringstudents to interpretvelocity and positiongraphs.

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    In the linear motion equation:

    dS

    V tdt

    V(t) is a function of time.

    For a very small change in time, V(t) can beconsidered a constant.

    dS V t dt

    S V t t We add up all the small changes in Sto getthe total distance.

    1 2 3S V t V t V t

    1 2 3S V V V t

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    S V t t We add up all the small changes in Sto getthe total distance.

    1 2 3S V t V t V t

    1 2 3S V V V t

    1

    k

    nn

    S V t

    1

    n

    n

    S V t

    S V t dt

    As the number of subintervals becomesinfinitely large (and the width becomesinfinitely small), we have integration.

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    This same technique is used in many different real-life

    problems.

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    Example: National Potato Consumption

    The rate of potato consumptionfor a particular country was:

    2.2 1.1tC t

    where tis the number of years

    since 1970 and Cis in millionsof bushels per year.

    For a small , the rate of consumption is constant.t

    The amount consumed during that short time is . C t t

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    Example 5: National Potato Consumption

    2.2 1.1t

    C t The amount consumed during that short time is . C t t

    We add up all these smallamounts to get the totalconsumption:

    total consumption C t dt

    4

    2 2.2 1.1

    tdt

    4

    2

    1

    2.2 1.1ln1.1

    t

    t

    From the beginning of 1972 tothe end of 1973:

    7.066

    million

    bushels

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    Work:

    work force distance

    Calculating the work is easywhen the force and distance are

    constant.

    When the amount of force

    varies, we get to use calculus!

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    Hookes law for springs: F kx

    x = distance that

    the spring isextended beyondits natural length

    k= spring

    constant

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    Hookes law for springs: F kx

    Example:

    It takes 10 Newtons to stretch a

    spring 2 meters beyond its naturallength.

    F=10 N

    x=2 M

    10 2k

    5 k 5F x

    How much work is done stretchingthe spring to 4 meters beyond itsnatural length?

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    F(x)

    x=4 M

    How much work is done stretchingthe spring to 4 meters beyond its

    natural length?

    For a very small change inx, the

    force is constant.

    dw F x dx

    5dw x dx

    5dw x dx

    4

    05W x dx

    4

    2

    0

    5

    2W x

    40W newton-meters

    40W joules

    5F x x