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Math Help - Tough (for me) arc length question

  1. #1
    s3a
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    Tough (for me) arc length question

    Question:
    Find the length of the curve

    = ?

    Any help would be greatly appreciated!
    Thanks in advance!
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  2. #2
    Super Member General's Avatar
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    You have x=f(y) and a\leq y \leq b

    L=\int_a^b \sqrt{ 1 + \left[ f'(y) \right]^2} \, dy
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  3. #3
    Super Member Deadstar's Avatar
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    You want to solve...

    \int_{-125}^{512} \sqrt{1 + (f'(y))^2} dx

    = \int_{-125}^{512} \sqrt{1 + \bigg{(}4y^{1/3} - \frac{1}{16} y^{-1/3}\bigg{)}^2} dx

    = \int_{-125}^{512} \sqrt{1 + 16y^{2/3} - \frac{1}{2}  + \frac{1}{256 y^{2/3}} } dx

    = \int_{-125}^{512} \sqrt{\frac{1}{256} \frac{(64y^{2/3} + 1)^2}{y^{2/3}}} dx

    = \int_{-125}^{512} \frac{1}{16} \frac{64y^{2/3} + 1}{y^{1/3}} dx
    Last edited by Deadstar; March 25th 2010 at 01:43 PM.
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  4. #4
    Super Member General's Avatar
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    Quote Originally Posted by Deadstar View Post
    You want to solve...

    \int_{-125}^{512} \sqrt{1 {\color{red} - } (f'(y))^2} dx
    The red one should be "+" not "-" ..
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  5. #5
    Super Member Deadstar's Avatar
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    Quote Originally Posted by General View Post
    The red one should be "+" not "-" ..
    So it should be. It's just a typo, didn't carry it past the line after.
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  6. #6
    s3a
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    But when I do that I get final answer = 10416.65625 and it's wrong.

    More specifically;

    3(512)^(4/3) + 3/32 * (512)^(2/3) - 3(-125)^(4/3) - 3/32 *(-125)^(2/3) = 10416.65625.
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  7. #7
    Super Member Deadstar's Avatar
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    What should the answer be? I've done this a few times now and I always end up with roughly the same (10415.43750). Check your original equation, limits etc and go over the method yourself to see if you get the same final integral as me.
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  8. #8
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    A correction is needed. This is correct:

    \int_{-125}^{512} \sqrt{\frac{1}{256} \frac{(64y^{2/3} + 1)^2}{y^{2/3}}} \, dy

    However, this simplifies like so:

    \frac{1}{16} \int_{-125}^{512} \frac{64y^{2/3} + 1}{|y^{1/3}|} \, dy

    So we should split it into two intervals to evaluate it:

    \frac{1}{16} \int_{-125}^0 \frac{64y^{2/3} + 1}{-y^{1/3}} \, dy  + \frac{1}{16} \int_0^{512} \frac{64y^{2/3} + 1}{y^{1/3}} \, dy
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  9. #9
    s3a
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    Why is there only an absolute value only on the denominator?

    Edit: I see the top will automatically get rid of the negative. I'm guessing I'll need to involve limits for this. I'll do it and say what happened.
    Last edited by s3a; March 27th 2010 at 11:13 AM.
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  10. #10
    s3a
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    Ok so I did it and got it wrong again.
    (My work is attached)

    Edit: I saw a mistake on the L = (before L = 12) where it should be a negative instead of positive for the thing after the first thing that I slashed off but I still get L = 52 which is wrong as well!
    Attached Thumbnails Attached Thumbnails Tough (for me) arc length question-mywork.jpg  
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  11. #11
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    Reducing it first helps avoid the need for a limit. That is:

    \frac{1}{16} \int_{-125}^0 \frac{64y^{2/3} + 1}{-y^{1/3}} \, dy + \frac{1}{16} \int_0^{512} \frac{64y^{2/3} + 1}{y^{1/3}} \, dy

    = -\frac{1}{16} \int_{-125}^0 \frac{64y^{2/3} + 1}{y^{1/3}} \, dy + \frac{1}{16} \int_0^{512} \frac{64y^{2/3} + 1}{y^{1/3}} \, dy

    = -\frac{1}{16} \int_{-125}^0 \left(64y^{1/3} + y^{-1/3}\right) dy + \frac{1}{16} \int_0^{512} \left(64y^{1/3} + y^{-1/3}\right) dy

    = -\frac{1}{16} \left[ 48y^{4/3} + \tfrac{3}{2}y^{2/3}\right]_{-125}^0 + \frac{1}{16} \left[ 48y^{4/3} + \tfrac{3}{2}y^{2/3}\right]_0^{512}

    = -\frac{1}{16} \left[-48(-125)^{4/3}-\tfrac{3}{2}(-125)^{2/3}\right] + \frac{1}{16} \left[ 48(512)^{4/3} + \tfrac{3}{2}(512)^{2/3}\right]

    = -\frac{1}{16} \left[-48(-5)^4-\tfrac{3}{2}(-5)^2\right] + \frac{1}{16} \left[ 48(8)^4 + \tfrac{3}{2}(8)^2\right]

    = -\frac{1}{16} \left[-48(625)-\tfrac{3}{2}(25)\right] + \frac{1}{16} \left[ 48(4096) + \tfrac{3}{2}(64)\right]

    Letting a calculator finish the work, I get that this equals

    =\frac{453483}{32} \approx 14171

    Assuming I didn't make any errors. :<
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  12. #12
    s3a
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    You're right.

    However, I don't get how you get from here:

    <br />
= -\frac{1}{16} \int_{-125}^0 \left(64y^{1/3} + y^{-1/3}\right) dy + \frac{1}{16} \int_0^{512} \left(64y^{1/3} + y^{-1/3}\right) dy<br />

    to here:

    <br />
= -\frac{1}{16} \left[ 48y^{4/3} + \tfrac{3}{2}y^{2/3}\right]_{-125}^0 + \frac{1}{16} \left[ 48y^{4/3} + \tfrac{3}{2}y^{2/3}\right]_0^{512}<br />
    Last edited by s3a; March 27th 2010 at 01:42 PM.
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  13. #13
    Super Member Deadstar's Avatar
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    Quote Originally Posted by s3a View Post
    You're right.

    However, I don't get how you get from here:

    <br />
= -\frac{1}{16} \int_{-125}^0 \left(64y^{1/3} + y^{-1/3}\right) dy + \frac{1}{16} \int_0^{512} \left(64y^{1/3} + y^{-1/3}\right) dy<br />

    to here:

    <br />
= -\frac{1}{16} \left[ 48y^{4/3} + \tfrac{3}{2}y^{2/3}\right]_{-125}^0 + \frac{1}{16} \left[ 48y^{4/3} + \tfrac{3}{2}y^{2/3}\right]_0^{512}<br />
    Since I'm browsing past I'll answer...

    -\frac{1}{16} \int_{-125}^0 \left( 64y^{1/3} + y^{-1/3} \right) dy = -\frac{1}{16} \left[ 64 \frac{y^{4/3}}{4/3} + \frac{y^{2/3}}{2/3} \right]_{-125}^0 = -\frac{1}{16} \left[ 64 y^{4/3} \cdot \tfrac{3}{4} + \tfrac{3}{2} y^{2/3}  \right]_{-125}^0 = -\frac{1}{16} \left[ 48 y^{4/3} + \tfrac{3}{2}y^{2/3}  \right]_{-125}^0

    Now you do the other integral.
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  14. #14
    s3a
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    Oh my gosh! I'm so stupid, my eyes were thinking that the next step was still an integral! (It's kind of like a blind spot if you close one eye and look at a dot on a piece of graph paper; you won't see the dot because your brain fills in the dot's space with what it thinks it will be which is more graph paper.)
    Last edited by mr fantastic; March 27th 2010 at 05:09 PM. Reason: d --> sh
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