SOLUTION: Find the length of the curve r(t)=(e^(t/6)cos(t/6), e^(t/6)sin(t6), e^(t/6)) for 0≤t≤2. Arc length = My answer is ((e^(2/3)-1)ln^3(3)+(2(3^(2/3))cos(2/3)+2(3^(2/3))-4)ln^2(

You have the line  r(t) = (x(t), y(t), z(t))  in   .



From Calculus, you should know that the length of such curve is the integral of

    sqrt( (x'(t))^2 + (y'(t))^2 + (z'(t))^2 ) * dt.



So you calculate  x'(t) = (1/6)*e^(t/6)*(cos(t/6) - sin(t/6));

                  y'(t) = (1/6)*e^(t/6)*(sin(t/6) + cos(t/6));

                  z'(t) = (1/6)*e^(t/6).



Then you calculate  (x'(t))^2 + (y'(t))^2 + (z'(t))^2 = 

     = (1/6)^2*e^((2t)/6) * (cos(t/6) - sin(t/6))^2 + (1/6)^2*e^((2t)/6) * (sin(t/6) + cos(t/6))^2 + (1/6)^2*e^(2t/6) = 

     = (1/6)^2*e^((2t)/6) * (1 + 1 + 1) = 3*(1/6)^2*e^((2t)/6).



THEREFORE,  

    sqrt( (x'(t))^2 + (y'(t))^2 + (z'(t))^2 ) * dt = sqrt(3)*(1/6)*e^(t/6)*dt.



From this point, the move forward should be clear and easy to you, if you really are a true Calculus student.



I wish you successfully complete your calculations from this point.



ANSWER.  The length of the curve is  sqrt(3)*(e^(2/6)-1) =  =  = 0.68522  (rounded).    ANSWER