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Homeworksucks

  • one year ago

Find the arc length of the graph x=(y^4/4)+1/(8y^2)

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  1. Homeworksucks
    • one year ago
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    \[x=\frac{ y^4 }{ 4 }+\frac{ 1 }{ 8y^2 }\]

  2. Homeworksucks
    • one year ago
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    I've set up the integral as \[\int\limits_{1}^{2}\sqrt{1+(y^3-\frac{ 1 }{ 4y^3 })^{2}}\] but I'm so lost on what to do from here

  3. Homeworksucks
    • one year ago
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    By the way, I forgot to specify an interval, the question states from y=1 to y=2

  4. anonymous
    • one year ago
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    I apologize, I was typing my response and my laptop just decided to turn off :/

  5. Homeworksucks
    • one year ago
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    That's alright, take your time :)

  6. anonymous
    • one year ago
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    As a tip, though, often times when you have a derivative that ends up looking like yours. For example, derivatives like this: \[y - \frac{ 1 }{ 4y }\] \[\frac{ y^{2} }{ 3 } - \frac{ 1 }{ 4y^{2} }\] Basically derivatives with a minus sign in the middle and the negative fraction has a variable on bottom. In those case, the square root often simplifies to the derivative you have but with a plus sign instead of a minus. So I betcha that the root will simolify and youll have the integral of \[y^{3} + \frac{ 1 }{ 4y^{3} }\] Now of course I plan on showing this, but its always a nice idea to have in mind for these arc length problems. In reality, the integrals on these would be ridiculous in a normal context, so textbooks set them up so that they come out perfectly. The easiest set up is to have a function that has that sort of property where that arc length formula simplifies like that. But yeah, on to simplifying and showing what I said should be the result, lol.

  7. anonymous
    • one year ago
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    I won't rewrite the integral over and over, let's just simplify the root \[\sqrt{1 + (y^{3} - \frac{ 1 }{ 4y^{3} })^{2}}\] \[\sqrt{1 + (\frac{ 4y^{6} - 1 }{ 4y^{3} })^{2}}\] \[\sqrt{1 + \frac{ 16y^{12} - 8y^{6} + 1 }{ 16y^{6} }}\] \[\sqrt{\frac{ 16y^{6} + 16y^{12} - 8y^{6} + 1 }{ 16y^{6} }}\] \[\sqrt{\frac{ 16y^{12} + 8y^{6} +1 }{ 16y^{6} }}\] \[\sqrt{\frac{ (4y^{6}+1)^{2} }{ 16y^{6} }} = \frac{ 4y^{6} + 1 }{ 4y^{3} } = y^{3} + \frac{ 1 }{ 4y^{3} }\] Which is what I assumed would happen :) And that assumption comes from knowing that arc length problems are often set up that way and when your derivatives is in a form like yours, this is often the result. I didnt show every little detail, but can you follow what I did to simplify the root?

  8. Homeworksucks
    • one year ago
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    Yeah it's super clear now! Thanks!

  9. Homeworksucks
    • one year ago
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    My textbook just skips over that step and doesn't explain at all so I was totally lost

  10. Homeworksucks
    • one year ago
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    Thank you so much!

  11. anonymous
    • one year ago
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    Yep, no problem! ^_^

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