• Support PF! Buy your school textbooks, materials and every day products via PF Here!

Arc length trouble

  • Thread starter vande060
  • Start date
186
0
1. Homework Statement

find the length of the curve

x = 1/3√y(y-3) 1 ≤ y ≤ 9








2. Homework Equations

L = ∫ √(1 + (dx/dy)^2)



3. 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

Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,176
925
[tex]\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\,}[/tex]
 
186
0
I got the rest of the work done, and it matches the answer in the book, thank you
 

Related Threads for: Arc length trouble

  • Posted
Replies
5
Views
212
  • Posted
Replies
5
Views
1K
  • Posted
Replies
1
Views
1K
  • Posted
Replies
2
Views
3K
  • Posted
Replies
9
Views
3K
Replies
8
Views
9K
  • Posted
Replies
2
Views
3K
  • Posted
Replies
4
Views
4K

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top