Find the arclength of the function.
x = y^4/8 + 1/4y^2
Solution
x = (1/8)y^4 + 1/(4y^2) = (1/8)y^4 + (1/4)y^(-2) dx/dy = (1/2)y^3 - 1/(2y^3). Then, we have:
1 + (dx/dy)^2 = 1 + [(1/2)y^3 - 1/(2y^3)]^2 = 1 + (1/4)y^6 - 2[(1/2)y^3][1/(2y^3)] + 1/(4y^6) =
(1/4)y^6 + 1/(4y^6) + 1/2 = 1/4 * (y^6 + 1/y^6 + 2) = 1/(4y^6) * (y^12 + 2y^6 + 1) = 1/(4y^6) *
(y^6 + 1)^2 = (y^6 + 1)^2/(2y^3)^2. Thus, the arc length is: ? v[1 + (dx/dy)^2] dy (from a to b)
= ? (y^6 + 1)/(2y^3) dy (from 1 to 2) (pick positive value as terms > 0) = ? [(1/2)y^3 + (1/2)y^(-
3)] dy (from 1 to 2) = [(1/8)y^4 - (1/4)y^(-2)] (evaluated from 1 to 2) = [(1/8)(2)^4 - (1/4)(2)^(-
2)] - [(1/8)(1)^4 - (1/4)(1)^(-2)] = 33/16 units..
![Find the arclength of the function.
x = y^4/8 + 1/4y^2
Solution
x = (1/8)y^4 + 1/(4y^2) = (1/8)y^4 + (1/4)y^(-2) dx/dy = (1/2)y^3 - 1/(2y^3). Then, we have:
1 + (dx/dy)^2 = 1 + [(1/2)y^3 - 1/(2y^3)]^2 = 1 + (1/4)y^6 - 2[(1/2)y^3][1/(2y^3)] + 1/(4y^6) =
(1/4)y^6 + 1/(4y^6) + 1/2 = 1/4 * (y^6 + 1/y^6 + 2) = 1/(4y^6) * (y^12 + 2y^6 + 1) = 1/(4y^6) *
(y^6 + 1)^2 = (y^6 + 1)^2/(2y^3)^2. Thus, the arc length is: ? v[1 + (dx/dy)^2] dy (from a to b)
= ? (y^6 + 1)/(2y^3) dy (from 1 to 2) (pick positive value as terms > 0) = ? [(1/2)y^3 + (1/2)y^(-
3)] dy (from 1 to 2) = [(1/8)y^4 - (1/4)y^(-2)] (evaluated from 1 to 2) = [(1/8)(2)^4 - (1/4)(2)^(-
2)] - [(1/8)(1)^4 - (1/4)(1)^(-2)] = 33/16 units.](https://image.slidesharecdn.com/findthearclengthofthefunction-230402060953-d844f93a/85/Find-the-arclength-of-the-function-x-y-48-14y-2Solu-pdf-1-320.jpg)
