What is the arclength of #(t^2lnt,lnt^2)# on #t in [1,2]#?

Question

1277 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-17T17:54:20

Performed by WolframAlpha
#int_1^2 sqrt(4/t^2+(t+2 t log(t))^2) dt = 3.19515#

Let L be the parametric arclength, then the formula is:

#L = int_a^b sqrt((dx/dt)^2 + (dy/dt)^2)dt#

#x = t^2ln(t)#
#dx/dt = 2tln(t) + t#
#y = ln(t^2)#
#dy/dt = 2/t#
#a = 1#
#b = 2#

Substitute the above into the formula:

#L = int_1^2 sqrt((2tln(t) + t)^2 + (2/t)^2)dt#

#int_1^2 sqrt(4/t^2+(t+2 t log(t))^2) dt = 3.19515#