# Thread: Arc Length of a curve

1. ## Arc Length of a curve

Find the length of arc of a curve $6xy = y^4 + 3$:
from the point where y = 1 to y = 4;

$
x = \frac{y^4 + 3}{6y}$

$\frac{dx}{dy} = \frac{y^2}{2} - \frac{3}{y^2} = \frac{y^4 - 6}{4y^4}$
the problem is that i can't integrate:

$\int \sqrt{1 + \frac{(y^4 - 6)^2}{4y^4}}dy$

2. Sometimes it pays to persevere, or don't surrender early.

6xy = y^4 +3

Find the length of the curve from y=1 to y=4

So we use dy, because the boundaries are given in values of y.

Length of curve, L.
dL = sqrt[1 +(dx/dy)^2]dy

So we find dx/dy.

x = (y^4 +3)/(6y)
x = (y^3)/6 +1/(2y)
dx/dy = (y^2)/2 -1/(2y^2)
dx/dy = (y^4 -1) /(2y^2) ---------------***

So,
dL = sqrt[1 +[(y^4 -1)/(2y^2)]^2]dy

Now the perseverance. We expand the (dx/dy)^2, then continue from there.

dL = sqrt[1 +[(y^8 -2y^4 +1) /(4y^4)]dy
dL = sqrt[{(4y^4) +(y^8 -2y^4 +1)} / (4y^4)]dy
dL = sqrt[(y^8 +2y^4 +1) /(4y^4)]dy
dL = sqrt[(y^4 +1)^2 /(4y^4)]dy
dL = [(y^4 +1) /(2y^2)]dy
dL = [(y^2)/2 +1/(2y^2)]dy
dL = [(1/2)y^2 +(1/2)y^(-2)]dy
dL = (1/2)[y^2 +y^(-2)]dy

Now that can be integrated.

So,
L = (1/2)INT.(1-->4)[y^2 +y^(-2)]dy
L = (1/2)[(y^3)/3 -1/y]|(1-->4)
L = (1/2)[{(4^3)/3 -1/4} -{(1^3)/3 -1/1}]
L = (1/2)[64/3 -1/4 -1/3 +1]
L = 261/24 = 87/8 = 10.875 units long. --------------answer.

3. Ticbol, it would look so great if your posts were 'd

4. Originally Posted by DivideBy0
Ticbol, it would look so great if your posts were 'd
Sorry. I don't want to learn that. Typing this style alone takes me hours to finish. I'm very slow in typing. Can't be bothered by that LaTex.

5. Originally Posted by ticbol
Sorry. I don't want to learn that. Typing this style alone takes me hours to finish. I'm very slow in typing. Can't be bothered by that LaTex.
That's what I used to think (though I can type rather quickly) before coming to this site. The LaTeX is rather easy to learn and doesn't add significantly to typing time if you are used to typing in ASCII format, as you tend to do.

-Dan

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# 6xy=x^4 3

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