Asked by Rick
Find the length of the curve r(t)=sqrt(2)ti + e^tj + e^-tk, 0<=t<=1
Please explain how it happened. Thank you
Please explain how it happened. Thank you
Answers
Answered by
Steve
Just review the derivation of arc length in 3 dimensions, as at
http://tutorial.math.lamar.edu/Classes/CalcIII/VectorArcLength.aspx
s = ∫√(f'^2 + g'^2 + h'^2) dt
So, for your problem,
f' = √2
g' = e^t
h' = -e^-t
s = ∫[0,1]√(2 + e^(2t) + e^(-2t)) dt
If that looks hard to do, just note that
2 + e^(2t) + e^(-2t) = (e^2 + e^-t)^2
http://tutorial.math.lamar.edu/Classes/CalcIII/VectorArcLength.aspx
s = ∫√(f'^2 + g'^2 + h'^2) dt
So, for your problem,
f' = √2
g' = e^t
h' = -e^-t
s = ∫[0,1]√(2 + e^(2t) + e^(-2t)) dt
If that looks hard to do, just note that
2 + e^(2t) + e^(-2t) = (e^2 + e^-t)^2
Answered by
Steve
s = ∫[0,1]√( e^(t) + e^(-t))^2 dt
s = ∫[0,1]( e^(t) + e^(-t)) dt
s = [0,1]( e^(t) - e^(-t))
s = ( e^(0) - e^(-0)) - ( e^(1) - e^(-1))
s = e - e^(-1)
s = ∫[0,1]( e^(t) + e^(-t)) dt
s = [0,1]( e^(t) - e^(-t))
s = ( e^(0) - e^(-0)) - ( e^(1) - e^(-1))
s = e - e^(-1)
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