Find the length of arc of a curve :
from the point where y = 1 to y = 4;
the problem is that i can't integrate:
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.