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ksaimouli
Group Title
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
ksaimouli Group Title
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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ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
@zepdrix
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
\[x=\frac{ y^3 }{ 6 }+\frac{ 1 }{ 2y }\]
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
\[\frac{ dx}{ dy }=\frac{ y^2 }{ 2 }\frac{ 1 }{ y^2}\]
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
\[\int\limits_{1}^{2}\sqrt{1+((\frac{ y^2 }{ 2 })\frac{ 1 }{ y^2 })^2} dy\]
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Woops the 2 should still be there on the second term. A 1 came down from the power rule.
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
u mean for 1/2y
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
The problem at the very top seems to be different than the one you formatted below it, I'm confused..
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
when i take derivative using quotient rule i got 1/2y^2
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
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

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Do you have 2 separate problems listed? This is rather confusing....
 one year ago
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