Find the length of the curve correct to four decimal places. (Use a calculator to approximate the integral).

Question

r(t) = (cos π t, 2t, sin 2πt), from (1, 0, 0) to (1, 4, 0)
This what I did.
r'(t)=-πsin(πt),2,2πcos(2πt)
lr'(t)l=√π^2sin^2(πt)+4+4π^2cos^2(2πt)
L=∫√π^2sin^2(πt)+4+4π^2cos^2(2πt)
I don't know integrate from what to what for ∫
Did I did any wrong in my steps above?

Answer 1

So far, so good. The curve from (1,0,0) to (1,4,0) is traced as t goes from 0 to 2. Better get out your calculator again.

Answer 2

I don't get how you get 0,2. can you explain further ?

Answer 3

geez - the j component is just 2t, so as y goes from 0 to 4, t goes from 0 to 2.
cos and sin take care of themselves, too, right?
Your final integrand is a function of t, so you need to set limits of integration for t.

Answer 4

Thank you