# Homework Help: Arc Length

1. Feb 2, 2013

### whatlifeforme

1. The problem statement, all variables and given/known data
find the length of the curve.

2. Relevant equations
x=y^3/15 + 5/4y on 3<=y<=5

3. The attempt at a solution
(dy/dx)^2 = Y^4/25 - 1/2 + 25/16y^4

integral (3,5) y^2/5 + 5/4y^2

2. Feb 2, 2013

### LCKurtz

Is that 5/(4y) or (5/4)y. If you don't use Latex, at least use parentheses.

That isn't (dy/dx)^2 although it may be (dx/dy)^2

That isn't the right formula for arc length. You need$$\sqrt{1 + \left(\frac {dx}{dy}\right) ^2}\, dy$$in the integrand.

3. Feb 2, 2013

### whatlifeforme

ok so thus far this should be correct.

L = integral (3,5) sqrt(1 + y^4/25 - 1/2 + 25/16y^4)

this could be further simplified to: sqrt (y^2/5 + 5/4y^2)^2 --->

y^3/15 - 5/4y ] (3 to 5)

the answer is : 67/10 which i'm not getting.

4. Feb 2, 2013

### whatlifeforme

edit: solved.
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}