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Thread: Arc Length

  1. #1
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    Arc Length

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

    $\displaystyle L = \int_a^b \sqrt{1+[f'(x)]^2} dx$
    or $\displaystyle \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. $\displaystyle y=1+6x^{3/2}, [0,1]$
    Answer: $\displaystyle \frac{2}{243}(82\sqrt{82}-1)$

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

    I found the derivatives of the functions, $\displaystyle 9\sqrt{x}$ and $\displaystyle \frac{5x^4}{6}-\frac{3}{10x^4}$, 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; Jul 31st 2006 at 02:49 PM.
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  2. #2
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    Quote Originally Posted by c_323_h

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

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

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

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

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

    Evaulate: .$\displaystyle \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 $\displaystyle \frac{1261}{240}$ and not $\displaystyle \frac{46}{3}$

    Sorry
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