xy4814y2 1

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.