Question

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

Answer 1

@zepdrix

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

u mean for 1/2y

Answer 7

ya

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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