## 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

1. ksaimouli Group Title

@zepdrix

2. ksaimouli Group Title

$x=\frac{ y^3 }{ 6 }+\frac{ 1 }{ 2y }$

3. ksaimouli Group Title

$\frac{ dx}{ dy }=\frac{ y^2 }{ 2 }-\frac{ 1 }{ y^2}$

4. ksaimouli Group Title

$\int\limits_{1}^{2}\sqrt{1+((\frac{ y^2 }{ 2 })-\frac{ 1 }{ y^2 })^2} dy$

5. zepdrix Group Title

Woops the 2 should still be there on the second term. A -1 came down from the power rule.

6. ksaimouli Group Title

u mean for 1/2y

7. zepdrix Group Title

ya

8. zepdrix Group Title

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

9. ksaimouli Group Title

when i take derivative using quotient rule i got -1/2y^2

10. zepdrix Group Title

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 Group Title

Do you have 2 separate problems listed? This is rather confusing....