Find Arc Length of x = 1/3√y(y-3) 1 ≤ y ≤ 9

[vande060]

Homework Statement

find the length of the curve

x = 1/3√y(y-3) 1 ≤ y ≤ 9

Homework Equations

L = ∫ √(1 + (dx/dy)^2)

The Attempt at a Solution

x = 1/3√y(y-3) 1 ≤ y ≤ 9

x = 1/3 (y^3/2 - 3y^1/2)

dx/dy = 1/2(y^1/2) - 1/2(y^-1/2)

dx/dy = 1/2(y^1/2 - y^-1/2)

(dx/dy)^2 = 1/4( y -2 + 1/y)

L = ∫ √[(1 + 1/4(y - 2 + 1/y)] 1 ≤ y ≤ 9

L = ∫ √[(1 + 1/4(y^2 - 2y + 1)/y] 1 ≤ y ≤ 9

L = ∫ √[(1 + 1/4(y-1)^2/y] 1 ≤ y ≤ 9

Im tempted to go ahead with the trig substitution, but this is getting a little complex, so I am wondering if I am even on the right track.

 

[SammyS]

$$\sqrt{1+\frac{1}{4}\left(y-2+\frac{1}{y}\right)}=\frac{1}{2}\,\sqrt{4+y-2+\frac{1}{y}\ }=\frac{1}{2\sqrt{y}}\,\sqrt{y^2+2y+1\,}$$

 

[vande060]

I got the rest of the work done, and it matches the answer in the book, thank you