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Math Help - Variatons and Parabolas

  1. #1
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    Unhappy Variatons and Parabolas

    Hey, I have this maths assignment which is due in two days. And I can't seem to figure out Question 10 or Question 11.
    Could anyone help me before its too late?
    And if you want, or if you've done it. You could scan me the graph too

    You can find see the image to Question 10 & 11 below. I scanned the questions.



    If not available. Click the link below.
    http://i21.photobucket.com/albums/b2...y5/mathsss.jpg

    Thanks.
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  2. #2
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    Quote Originally Posted by xlovetee View Post
    Hey, I have this maths assignment which is due in two days. And I can't seem to figure out Question 10 or Question 11.
    Could anyone help me before its too late?
    ...
    I couldn't find an equation describing the distances covered by the heli. Therefore I've done the graph "by hand" (=construction).
    The y-axis had to shortened by the factor 10.

    The small distances at the ship's position and at the port are due to picking up or stting down the crew members.
    Attached Thumbnails Attached Thumbnails Variatons and Parabolas-rescue_heli.gif  
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  3. #3
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    Quote Originally Posted by xlovetee View Post
    And I can't seem to figure out Question 10
    ...
    I placed the origin of a coordinate system at the red dot.

    Then you are looking for a parabola with the equation

    y = ax^2+c

    From your drawing you get:

    B(0,-2)

    A(-50,-8)

    Plug in the coordinates of the points into the equation of the parabola:

    -2=c

    -8=a(-50)^2-2~\implies~a=-\frac6{2500}

    Thus the equation of the parabola is:

    y=-\frac6{2500} x^2-2

    Use this equation to answer the other questions of your problem.
    Attached Thumbnails Attached Thumbnails Variatons and Parabolas-arcofbridge.gif  
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  4. #4
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    Quote Originally Posted by earboth View Post
    I placed the origin of a coordinate system at the red dot.

    Then you are looking for a parabola with the equation

    y = ax^2+c

    From your drawing you get:

    B(0,-2)

    A(-50,-8)

    Plug in the coordinates of the points into the equation of the parabola:

    -2=c

    -8=a(-50)^2-2~\implies~a=-\frac6{2500}

    Thus the equation of the parabola is:

    y=-\frac6{2500} x^2-2

    Use this equation to answer the other questions of your problem.


    Thanks for replying to my thread. But I'm not quit sure how you got
    y = ax^2+c
    Because in our class we have been using the quadratic function
    y=ax^2+bx+c

    and also we have learnt to use the rule :
    y=(x-a)^2

    Thank you for your help.
    I appreciate it.
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