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Thread: 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 $\displaystyle y=\frac{1}{3}x^{\frac{3}{2}}-x^{\frac{1}{2}}$ from $\displaystyle (1,-\frac{2}{3}})$ to $\displaystyle {4,\frac{2}{3})$.


    So far I have:$\displaystyle \frac{ds}{dx}=\sqrt{1+f'^2} = \sqrt{1+(\frac{1}{2}\sqrt{x} - \frac{1}{2}*\frac{1}{\sqrt{x}})^2dx$$\displaystyle =\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 $\displaystyle y=\frac{1}{3}x^{\frac{3}{2}}-x^{\frac{1}{2}}$ from $\displaystyle (1,-\frac{2}{3}})$ to $\displaystyle {4,\frac{2}{3})$
    You have,
    $\displaystyle y=\frac{1}{3}x^{3/2}-x^{1/2}$
    $\displaystyle y'=\frac{1}{3}\cdot \frac{3}{2}x^{1/2}-\frac{1}{2}x^{-1/2}=\frac{1}{2}x^{1/2}-\frac{1}{2}x^{-1/2}$
    Now,
    $\displaystyle [y']^2=\left( \frac{1}{2}x^{1/2}-\frac{1}{2}x^{-1/2} \right)^2$
    Open,
    $\displaystyle \frac{1}{4}x-\frac{2}{1}\cdot \frac{1}{2} \cdot \frac{1}{2}\cdot x^{1/2-1/2}+\frac{1}{4}x^{-1}$
    Thus,
    $\displaystyle \frac{1}{4}x+\frac{1}{4}x^{-1}-\frac{2}{4}$
    The next step is to add one,
    $\displaystyle \frac{1}{4}x+\frac{1}{4}x^{-1}+\frac{2}{4}$
    Thus,
    $\displaystyle \int_1^4 \sqrt{\frac{1}{4}x+\frac{1}{4}x^{-1}+\frac{2}{4}}dx$
    $\displaystyle \frac{1}{2}\int_1^4 \sqrt{x+\frac{1}{x}+2} dx$
    $\displaystyle \frac{1}{2}\int_1^4 \sqrt{\frac{x^2+2x+1}{x}} dx$
    You need to make the square,
    $\displaystyle
    \frac{1}{2}\int_1^4 \sqrt{\frac{(x+1)^2}{x}dx$
    Since $\displaystyle x>0$ you make use the rule,
    $\displaystyle \sqrt{a/b}=\sqrt{a}/\sqrt{b}$
    Thus,
    $\displaystyle \frac{1}{2}\int_1^4 \frac{\sqrt{(x+1)^2}}{\sqrt{x}}dx$=$\displaystyle \frac{1}{2}\int_1^4 \frac{|x+1|}{\sqrt{x}}dx$
    But, $\displaystyle x>0\to x+1>0$ thus,
    $\displaystyle \frac{1}{2}\int_1^4 \frac{x+1}{\sqrt{x}}dx$
    Note that the fraction:
    $\displaystyle \frac{x+1}{\sqrt{x}}=\frac{x}{ \sqrt{x} }+\frac{1}{ \sqrt{x} }=x^{1/2}+x^{-1/2}$
    Finally,
    $\displaystyle \frac{1}{2}\int_1^4 x^{1/2}+x^{-1/2}dx$
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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 $\displaystyle y=\frac{1}{3}x^{\frac{3}{2}}-x^{\frac{1}{2}}$ from $\displaystyle (1,-\frac{2}{3}})$ to $\displaystyle {4,\frac{2}{3})$.


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

    $\displaystyle
    s=\int_{x=1}^2 \frac{ds}{dx} dx=\int_1^2\frac{1}{2}\sqrt{\frac{(x+1)^2}{x}} dx
    $


    RonL
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