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