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

1 Answer
May 9, 2017

"Arc Length " approx .863746Arc Length .863746

Explanation:

"Parmetric Arc Length "=int_a^b (sqrt((dx/dt)^2+((dy)/dt)^2))dtParmetric Arc Length =ba(dxdt)2+(dydt)2dt

Differentiate using the chain rule:
dx/dt=1/2(t+2)^(-1/2)=frac{1}{2sqrt(t+2)}dxdt=12(t+2)12=12t+2

(dy)/dt=-1/t^2dydt=1t2

"Arc Length " = int_1^3(sqrt((frac{1}{2sqrt(t+2)})^2+(-1/t^2)^2))dtArc Length =31((12t+2)2+(1t2)2)dt

Simplify a bit:
=int_1^3 (sqrt(frac{1}{4(t+2)}+1/t^4))dt=31(14(t+2)+1t4)dt

=int_1^3 (sqrt(frac{t^4+4t+8}{4t^4(t+2) }))dt=31(t4+4t+84t4(t+2))dt

Plug this into a calculator to get

"Arc Length " approx .863746Arc Length .863746