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ksaimouli

  • 3 years ago

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

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  1. ksaimouli
    • 3 years ago
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    @zepdrix

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

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

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

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

  6. ksaimouli
    • 3 years ago
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    u mean for 1/2y

  7. zepdrix
    • 3 years ago
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    ya

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

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

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

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

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