What is the arclength of #(sqrt(t+2),1/t)# on #t in [1,3]#?

Question

1452 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-09T04:05:23

#"Arc Length " approx .863746#

#"Parmetric Arc Length "=int_a^b (sqrt((dx/dt)^2+((dy)/dt)^2))dt#

Differentiate using the chain rule:
#dx/dt=1/2(t+2)^(-1/2)=frac{1}{2sqrt(t+2)}#

#(dy)/dt=-1/t^2#

#"Arc Length " = int_1^3(sqrt((frac{1}{2sqrt(t+2)})^2+(-1/t^2)^2))dt#

Simplify a bit:
#=int_1^3 (sqrt(frac{1}{4(t+2)}+1/t^4))dt#

#=int_1^3 (sqrt(frac{t^4+4t+8}{4t^4(t+2) }))dt#

Plug this into a calculator to get

#"Arc Length " approx .863746#