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anonymous

  • 4 years ago

Calculus II - Arc Length Problem Hello, I am having some serious problems calculating the Arc Length of a curve. Here is the problem; (Putting in a new post to use equation calculator)

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  1. anonymous
    • 4 years ago
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    \[x = (y^3/33) + 11/(4y) \] on 3<= y <= 5

  2. anonymous
    • 4 years ago
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    So, I take the derivative. No problem; \[dy/dx = (y^2/11) + (11/(4y^2) \] Then Length = \[\sqrt(1 + (dy/dx)^2) \] So; \[Length = \sqrt(1 + (y^2/11 + 11/(4y))^2)\]

  3. anonymous
    • 4 years ago
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    you are off by a minus sign

  4. anonymous
    • 4 years ago
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    \[\int_3^5\sqrt{1+(\frac{y^2}{11}-\frac{11}{4y^2})^2}dy\]

  5. anonymous
    • 4 years ago
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    Holy god. If that's the problem I have been having this whole time I'm not sure whether I'm going to laugh or cry. Probably a bit of both.

  6. anonymous
    • 4 years ago
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    oh wow isn't that annoying!

  7. anonymous
    • 4 years ago
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    happens to me all the damn time, that is why it is good to get another pair of eyes

  8. anonymous
    • 4 years ago
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    \[\int_3^5\sqrt{\frac{y^4}{121}+\frac{121}{16 y^4}+\frac{1}{2}}\]

  9. anonymous
    • 4 years ago
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    Ok, maybe you can tell me this - when I type ((y^2/11) - (11/(4y^2))^2 on my calculator it gives me; ((4y^4 - 121)^2)/1936y^4 Any idea why?

  10. anonymous
    • 4 years ago
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    Same thing on Wolfram - if I type your above equation (everything under the root) and tell it to factor it gives me the same as above.

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spraguer (Moderator)
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is replying to Can someone tell me what button the professor is hitting...

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