How do you find the lengths of the curve #8x=2y^4+y^-2# for #1<=y<=2#?

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Answer 1 · 2017-04-26T08:11:21

#8x=2y^4+y^(-2)#

#implies x'=y^3- 1/4 y^(-3)#

#s = int_1^2 sqrt( 1 + (x')^2) \ dy#

#= int_1^2 sqrt( 1 + (y^3- 1/4 y^(-3))^2) \ dy#

#= int_1^2 sqrt( 1 + y^6 + 1/16 y^(-6)- 1/2) \ dy#

#= int_1^2 sqrt( y^6 + 1/16 y^(-6) + 1/2) \ dy#

#= int_1^2 sqrt( (y^3 + 1/4 y^(-3))^2) \ dy#

#= int_1^2 y^3 + 1/4 y^(-3) \ dy #

#= [ y^4/4 - 1/(8 y^2) ]_1^2#

#= 123/32#