Length of a Helical Path

In this video we work through an example in which we use arc length to find the distance an eagle travels as it flies up in a spiral from the ground.

[lonnythompson]

Hi, Very nice video and explanation of length of helical path.

[lonnythompson]

It appears the problem statement in the book is not clearly stated. The trig functions need to be normalized for correct units. x = a* cos(t), and y = a*sin(t),   z = b*t should be stated as,  x = a* cos(t/t_ref}), y = a* sin(t/t_ref),   z = b*t,   where t_ref = 1 minute. Then when computing the velocity derivative of position,  x’ = -(a/t_ref)*sin(t/t_ref),   y’ = (a/t_ref)*cos(t/t_ref)  , so v = sqrt ( (x’)^2 + (y’)^2 + (z’)^2  )  =   sqrt( (a/t_ref)^2 + b^2 ) , and v = sqrt( (250ft / 1 min)^2 + (100ft/min)^2 ) = 269.2 ft/min, then L = v  t  = 269.2 ft/min ( 10 min) = 2692 ft, Units now check.