Asked by #1
Find the length of the curve correct to four decimal places. (Use a calculator to approximate the integral).
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?
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?
Answers
Answered by
oobleck
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.
Answered by
#1
I don't get how you get 0,2. can you explain further ?
Answered by
oobleck
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.
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.
Answered by
Anonymous
Thank you
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.