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Math Help - Derivative and part of word problems...

  1. #1
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    Derivative and part of word problems...

    1. Given that


    Calculate 6
    Calculate -8
    Calculate ?
    Calculate ?
    All the answers i got for (f*g) and (f/g) are wrong??

    2. A space traveler is moving from left to right along the curve
    When she shuts off the engines, she will continue traveling along the tangent line at the point where she is at that time. At what point
    (x,y) should she shut off the engines in order to reach the point ?
    If she was traveling from right to left she would have to shut off the engines at the point (x,y)?

    3. If an arrow is shot straight upward on the moon with a velocity of 78 m/s, its height (in meters) after t seconds is given by .
    What is the velocity of the arrow (in m/s) after 7 seconds? 66.38 m/s
    After how many seconds will the arrow hit the moon? 94 seconds
    With what velocity (in m/s) will the arrow hit the moon? ?? (How do i calculate for this?)
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  2. #2
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    Quote Originally Posted by Kayla_N View Post
    ...

    2. A space traveler is moving from left to right along the curve
    When she shuts off the engines, she will continue traveling along the tangent line at the point where she is at that time. At what point
    (x,y) should she shut off the engines in order to reach the point ?
    If she was traveling from right to left she would have to shut off the engines at the point (x,y)?

    ...
    Let T(t, t^2) denote the point where the engine is shut off. The slope of the tangent in T is m_t = 2t

    Then the tangent at the curve in T is:

    y - t^2 = 2t(x-t) ~\implies~ y = 2tx-t^2

    This tangent passes throught the point P(4, 15), that means it's coordinates must satisfy the equation of the tangent:

    15 = 2t \cdot 4 - t^2~\implies~t^2-8t+15=0~\implies~\boxed{t = 3~\vee~t = 5}

    Therefore the two tangents are:

    t_1:y = 6x-9 or t_2: y = 10x-25

    and the tangent points are:

    T_1(3, 9) or T_2(5, 25)
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  3. #3
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    Quote Originally Posted by Kayla_N View Post
    ...

    3. If an arrow is shot straight upward on the moon with a velocity of 78 m/s, its height (in meters) after t seconds is given by .
    What is the velocity of the arrow (in m/s) after 7 seconds? 66.38 m/s ....... OK
    After how many seconds will the arrow hit the moon? 94 seconds ....... OK
    With what velocity (in m/s) will the arrow hit the moon? ?? (How do i calculate for this?)
    1. The speed at the time t is calculated by:

    v(t)=h'(t)=78-1.66t

    Therefore calculate v(94)

    2. In this case it is actually not necessary to "calculate" any value because you know that

    v(0) = v(93.9759) = 78\ \frac ms
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  4. #4
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    Wink

    Thank you vey much. I got it now.

    Can soneome please help me out with number 1??
    Last edited by Kayla_N; February 21st 2009 at 03:24 PM.
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