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Math Help - Can't seem to finish this length of graph problem

  1. #1
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    Can't seem to finish this length of graph problem

    I need help to finish finding the length of the graph of y=\frac{1}{3}x^{\frac{3}{2}}-x^{\frac{1}{2}} from (1,-\frac{2}{3}}) to {4,\frac{2}{3}).


    So far I have: \frac{ds}{dx}=\sqrt{1+f'^2} = \sqrt{1+(\frac{1}{2}\sqrt{x} - \frac{1}{2}*\frac{1}{\sqrt{x}})^2dx =\frac{1}{2}\sqrt{\frac{(x+1)^2}{x}}
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  2. #2
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    Quote Originally Posted by FLTR
    I need help to finish finding the length of the graph of y=\frac{1}{3}x^{\frac{3}{2}}-x^{\frac{1}{2}} from (1,-\frac{2}{3}}) to {4,\frac{2}{3})
    You have,
    y=\frac{1}{3}x^{3/2}-x^{1/2}
    y'=\frac{1}{3}\cdot \frac{3}{2}x^{1/2}-\frac{1}{2}x^{-1/2}=\frac{1}{2}x^{1/2}-\frac{1}{2}x^{-1/2}
    Now,
    [y']^2=\left( \frac{1}{2}x^{1/2}-\frac{1}{2}x^{-1/2} \right)^2
    Open,
    \frac{1}{4}x-\frac{2}{1}\cdot \frac{1}{2} \cdot \frac{1}{2}\cdot x^{1/2-1/2}+\frac{1}{4}x^{-1}
    Thus,
    \frac{1}{4}x+\frac{1}{4}x^{-1}-\frac{2}{4}
    The next step is to add one,
    \frac{1}{4}x+\frac{1}{4}x^{-1}+\frac{2}{4}
    Thus,
    \int_1^4 \sqrt{\frac{1}{4}x+\frac{1}{4}x^{-1}+\frac{2}{4}}dx
    \frac{1}{2}\int_1^4 \sqrt{x+\frac{1}{x}+2} dx
    \frac{1}{2}\int_1^4 \sqrt{\frac{x^2+2x+1}{x}} dx
    You need to make the square,
    <br />
\frac{1}{2}\int_1^4 \sqrt{\frac{(x+1)^2}{x}dx
    Since x>0 you make use the rule,
    \sqrt{a/b}=\sqrt{a}/\sqrt{b}
    Thus,
    \frac{1}{2}\int_1^4 \frac{\sqrt{(x+1)^2}}{\sqrt{x}}dx= \frac{1}{2}\int_1^4 \frac{|x+1|}{\sqrt{x}}dx
    But, x>0\to x+1>0 thus,
    \frac{1}{2}\int_1^4 \frac{x+1}{\sqrt{x}}dx
    Note that the fraction:
    \frac{x+1}{\sqrt{x}}=\frac{x}{ \sqrt{x} }+\frac{1}{ \sqrt{x} }=x^{1/2}+x^{-1/2}
    Finally,
    \frac{1}{2}\int_1^4 x^{1/2}+x^{-1/2}dx
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by FLTR
    I need help to finish finding the length of the graph of y=\frac{1}{3}x^{\frac{3}{2}}-x^{\frac{1}{2}} from (1,-\frac{2}{3}}) to {4,\frac{2}{3}).


    So far I have: \frac{ds}{dx}=\sqrt{1+f'^2} = \sqrt{1+(\frac{1}{2}\sqrt{x} - \frac{1}{2}*\frac{1}{\sqrt{x}})^2dx =\frac{1}{2}\sqrt{\frac{(x+1)^2}{x}}
    This is good, all you have left to do is compute:

    <br />
s=\int_{x=1}^2 \frac{ds}{dx} dx=\int_1^2\frac{1}{2}\sqrt{\frac{(x+1)^2}{x}} dx<br />


    RonL
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