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

1. ksaimouli

@zepdrix

2. ksaimouli

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

3. ksaimouli

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

4. ksaimouli

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

5. zepdrix

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

6. ksaimouli

u mean for 1/2y

7. zepdrix

ya

8. zepdrix

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

9. ksaimouli

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

10. zepdrix

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

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