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Math Help - Centroid help

  1. #1
    Junior Member
    Joined
    Feb 2008
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    Centroid help

    y=e^(5x)
    y=0
    x=0
    x=.2

    Find the exact coordinates of the centroid.

    I got the area to be (e-1)/5.

    I got the x coordinate to be 1/(5e-5).

    I'm having trouble with the y coordinate. Any help?
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  2. #2
    Super Member wingless's Avatar
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    These are the formulas for the centroids of the area under the curve f(x) or f(y):
    (a, b are the bounds of the curve for x and c, d are the bounds for y)


    <br />
\displaystyle{\begin{array}{l|l}<br />
\text{For} ~f(x) ~& ~ \text{For} ~f(y) <br />
\\ \hline \\<br />
M_x = \frac{1}{2}\int^{b}_{a} [f(x)]^2 ~dx ~& ~  M_x = \int^{d}_{c} y~f(y) ~dy <br />
\\  \\<br />
M_y = \int^{b}_{a} x~f(x) ~dx ~& ~   M_y = \frac{1}{2}\int^{d}_{c} [f(y)]^2 ~dy<br />
\\  \\<br />
A = \text{Area}= \int^{b}_{a} f(x)~dx ~& ~  A = \text{Area}= \int^{d}_{c} f(y)~dy<br />
\\ \end{array}}


    Then,
    \begin{array}{ll}\bar{x} = \frac{M_y}{A} & \bar{y} = \frac{M_x}{A}\\\end{array}

    The origin is at O\left ( \bar{x} , \bar{y }\right )

    As our function is a function of x, we'll use the formulas on the left side of the table.

    Firstly, this is the graph:




    y=f(x) = e^{5x}


    Find M_x, M_y, A

    M_x = \frac{1}{2}\int^{b}_{a} [f(x)]^2 ~dx
    M_x = \frac{1}{2}\int^{0.2}_{0} e^{10x} ~dx
    M_x = \frac{e^2-1}{20}

    M_y = \int^{b}_{a} x~f(x) ~dx
    M_y = \int^{0.2}_{0} xe^{5x} ~dx
    M_y = 0.04

    A = \int^{b}_{a} f(x) ~dx
    A = \int^{0.2}_{0} e^{5x} ~dx
    A = \frac{e-1}{5}

    Hence,

    \begin{array}{l}<br />
\bar{x} = \frac{M_y}{A} \\ \bar{y} = \frac{M_x}{A}\\\end{array}

    \begin{array}{l}<br />
\bar{x} = \frac{0.04}{\frac{e-1}{5}} \\ \bar{y} = \frac{\frac{e^2-1}{20}}{\frac{e-1}{5}}\\\end{array}

    \begin{array}{l}\bar{x} = \frac{1}{5e-5} \\ \bar{y} = \frac{e+1}{4}\\\end{array}

    So the centroid is at O~\left ( ~\frac{1}{5e-5} ~, ~\frac{e+1}{4}~\right )
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  3. #3
    Junior Member
    Joined
    Feb 2008
    Posts
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    I forgot to factor the e^2-1 to cancel with the e+1 on the bottom. Thanks a bunch! You guys are always so helpful!

    The grader said that it was wrong. Maybe there is an error in the system... I'll have to talk to my teacher since I'm pretty sure it's right.
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