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Math Help - Can you help with this problems????

  1. #1
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    Unhappy Can you help with this problems????

    Let T be a triangle witht vertices (0,0), (0,c^2), (c,c^2) and let R be the region between y =cx and y = x^2 where c>0. Fidn the ratio area T/ area R.
    Last edited by Susie38; May 29th 2006 at 01:48 PM. Reason: Wrong wording.
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  2. #2
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    Hello, Susie38!

    It's a straight-forward area problem.
    Exactly where does your difficulty lie?


    Let T be a triangle with vertices (0,0),\;(0,c^2),\;(c,c^2)
    and let R be the region between y = cx and y = x^2,\; c > 0
    Find the ratio \frac{area\:T}{area\:R}
    Code:
    
                |
          (0,cē)* - - - - - * (c,cē)
                |         /:
                |       /::*
                |     /::::
                |   /::::*
                | /:::*
          - - - **- - - - - -
              (0,0)

    The base of the triangle is c, its height is c^2.
    . . Its area is: A_T \:=\:\frac{1}{2}(c)(c^2) \:=\:\frac{1}{2}c^3

    The area of the parabolic segment is:
    . . \displaystyle{A_R \;= \;\int^c_0(cx - x^2)\ dx \;=\;\frac{c}{2}x^2 - \frac{1}{3}x^3\bigg|^c_0 \;= \;\frac{1}{6}c^3

    Therefore: \displaystyle{\frac{A_T}{A_R}\;=\;\frac{\frac{1}{2  }c^3}{\frac{1}{6}c^3} \;= \;3

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