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Math Help - Arc Length

  1. #1
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    Arc Length

    Given y=f(x) the formula for arc length L over the interval [a,b]:

    L = \int_a^b \sqrt{1+[f'(x)]^2} dx
    or  \int_a^b \sqrt{1+(\frac{dy}{dx})^2}dx

    With the formula above find the arc length of the following functions over the given interval.

    1. y=1+6x^{3/2}, [0,1]
    Answer: \frac{2}{243}(82\sqrt{82}-1)

    2. y=\frac{x^5}{6}+\frac{1}{10x^3}, [1,2]
    Answer: \frac{1261}{240}

    I found the derivatives of the functions, 9\sqrt{x} and \frac{5x^4}{6}-\frac{3}{10x^4}, respectively, and plugged them in, simplified, substituted and converted limits if I had to, but still don't get the correct answer. Could someone tell me what I'm doing wrong?
    Last edited by c_323_h; July 31st 2006 at 02:49 PM.
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  2. #2
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    Quote Originally Posted by c_323_h

    1. y=1+6x^{3/2}, [1,2]
    Answer: \frac{43}{3}
    You have,
    y=1+6x^{3/2}
    Thus,
    y'=6(3/2)x^{1/2}
    Simplfy,
    y'=9x^{1/2}
    Thus,
    \sqrt{1+[f'(x)]^2}=\sqrt{1+[9x^{1/2}]^2}=\sqrt{1+81x}
    Thus,
    \int_1^2 \sqrt{1+81x}dx
    Can you integrate that? Or do you need the quickest integrator in the west to help you?
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  3. #3
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    Quote Originally Posted by ThePerfectHacker
    You have,
    y=1+6x^{3/2}
    Thus,
    y'=6(3/2)x^{1/2}
    Simplfy,
    y'=9x^{1/2}
    Thus,
    \sqrt{1+[f'(x)]^2}=\sqrt{1+[9x^{1/2}]^2}=\sqrt{1+81x}
    Thus,
    \int_1^2 \sqrt{1+81x}dx
    Can you integrate that? Or do you need the quickest integrator in the west to help you?
    What I tried:

    let u=1+81x, then du=81dx.
    Convert limits: when x=0, u=1 and when  x=1, u=82

    so, \frac{1}{81}\int_1^{82}\sqrt{u}du
    \frac{1}{81}(\frac{2u^{3/2}}{3}) |_1^{82}

    substitute u back in. Evaluate at limits using the Fundamental Theorem of Calculus. Is this correct?
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  4. #4
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    Hello, c_323_h!

    What you've done is correct . . .


    Let u=1+81x, then du=81\,dx
    Convert limits: when x=0, u=1 and when  x=1, u=82

    So: \frac{1}{81}\int_1^{82}\!\!\sqrt{u}\:du \;= \;\frac{1}{81}\left(\frac{2u^{3/2}}{3}\right) \bigg|_1^{82}\quad\Rightarrow\quad\frac{2}{243}u^{  \frac{3}{2}}\bigg|^{82}_1

    Evaulate: . \frac{2}{243}\left(82^{\frac{3}{2}} - 1^{\frac{3}{2}}\right) \;= \;\frac{2}{243}\left(82\sqrt{82} - 1\right)


    I believe you switched the two answers in your original post.

    And is there a typo in #2?
    Is there a plus between the two fractions?

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  5. #5
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    Quote Originally Posted by Soroban

    I believe you switched the two answers in your original post.

    And is there a typo in #2?
    Is there a plus between the two fractions?

    yup it's supposed to be a plus and the answers are indeed switched. i'll edit my post..also the answer to the second question is \frac{1261}{240} and not \frac{46}{3}

    Sorry
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