How do you find the lengths of the curve x=(y^4+3)/(6y) for 3<=y<=8?

1 Answer
May 19, 2018

Use the arc length formula.

Explanation:

x=(y^4+3)/(6y)=y^3/6+1/(2y)

x'=1/2(y^2-1/y^2)

Arc length is given by:

L=int_3^8sqrt(1+1/4(y^2-1/y^2)^2)dy

Expand the square:

L=int_3^8sqrt(1+1/4(y^4-2+1/y^4))dy

Combine terms:

L=1/2int_3^8sqrt(y^4+2+1/y^4)dy

Factorize:

L=1/2int_3^8sqrt((y^2+1/y^2)^2)dy

Hence

L=1/2int_3^8(y^2+1/y^2)dy

Integrate term by term:

L=1/2[y^3/3-1/y]_3^8

Insert the limits of integration:

L=1/6(8^3-3^3)-1/2(1/8-1/3)

Simplify:

L=485/6+5/48=1295/6