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ksaimouli

find the arc lenght x=(y^3/3)((x^2)+2)^3/2 from x=0 to x=3

  • one year ago
  • one year ago

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  1. ksaimouli
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    @zepdrix

    • one year ago
  2. ksaimouli
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    \[x=\frac{ y^3 }{ 6 }+\frac{ 1 }{ 2y }\]

    • one year ago
  3. ksaimouli
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    \[\frac{ dx}{ dy }=\frac{ y^2 }{ 2 }-\frac{ 1 }{ y^2}\]

    • one year ago
  4. ksaimouli
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    \[\int\limits_{1}^{2}\sqrt{1+((\frac{ y^2 }{ 2 })-\frac{ 1 }{ y^2 })^2} dy\]

    • one year ago
  5. zepdrix
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    Woops the 2 should still be there on the second term. A -1 came down from the power rule.

    • one year ago
  6. ksaimouli
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    u mean for 1/2y

    • one year ago
  7. zepdrix
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    ya

    • one year ago
  8. zepdrix
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    The problem at the very top seems to be different than the one you formatted below it, I'm confused..

    • one year ago
  9. ksaimouli
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    when i take derivative using quotient rule i got -1/2y^2

    • one year ago
  10. zepdrix
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    Yes that's the correct derivative. Don't use quotient rule, easier to use power rule.\[\large \left(\frac{1}{2y}\right)'\quad = \quad \left(\frac{1}{2}y^{-1}\right)' \quad = \quad -\frac{1}{2}y^{-2}\]

    • one year ago
  11. zepdrix
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    Do you have 2 separate problems listed? This is rather confusing....

    • one year ago
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