Line integral - finding the arc length

[nb89]

a curve is given as 3 parameters of t:
x=a(3t - t^3), y=3a(t^2), z=a(3t + t^3)

i have to find the arc length measured from origin and curvature as functions of t.

would i be correct in using the integral at the bottom of page 2 here: http://homepages.ius.edu/wclang/m311/fall2005/notes17.2.pdf

 

[arildno]

Sure enough, with f=1.
Then you get:
\int3|a|\sqrt{(1-t^{2})^{2}+(2t)^{2}+(1+t^{2})^{2}}
If you are clever, you'll find a perfect square buried within the radicand.

 

[nb89]

why is f=1?

would my limits be 0 to t?

 

[arildno]

why is f=1?

Because you are to find \int{ds}=\int{1}*ds

would my limits be 0 to t?

Yes.

 

[nb89]

ive ended up with integral 3a√2 (t^2 +1) dt
so i take the 3a√2 out and integrate t^2 +1.

but I am confused since the limits are 0 to t, how would i replace the t with the limit t once i have integrated it?

 

[HallsofIvy]

Don't worry about the limits of integration: you are finding the arclength as a function of t so find an anti-derivative and take the constant so that the integral is 0 when t= 0.

 

[nb89]

then i get 3a√2 (t^3/3 + t). that's wrong though?