arc length of helix $(2\cos(t), 2\sin(t), 3t)$

Question

Hey i have to calculate the arc length of a helix $f:[0,2πn]\to R^3: t\mapsto(2\cos(t), 2\sin(t), 3t)$

I think I have to calculate the arc length as an integral over norm of speed vector. So I started like this:

$$f'(t)=(-2\sin(t), 2\cos(t), 3)$$

$$||f'(t)||= \sqrt{(4\sin^2(t)}+4\cos^2(t)+9= \sqrt{(4+9)}= \sqrt{13}.$$

Where is my error, or is it right and the arc length is already $\sqrt{13}$??

score 1 · Answer 1 · answered Jun 07 '20 at 13:34

1

You do indeed have $||f'(t)||=\sqrt{13}$. You still need to multiply this by the total time $2\pi n$ to get the total arc length, which is then $2\pi n\sqrt{13}$.

If you draw the helix along a cylinder and then the cylinder with the helix onto a plane, the component parallel to the rolling direction becomes a line of length $2t$ and the perpendicular component becomes a line oflength $3t$. The combined motion is then represented by the hypoteneuse obtained from these perpendicular components thus having length $\sqrt{13}t$.

answered Jun 07 '20 at 13:34

Oscar Lanzi

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ok youre sure with $2\pi n\sqrt{13}$ ? i never multiplied that before :D and what is n at all? an unknown ? – Jun 07 '20 at 13:59
Note the $t$ value is from $0$ to $2\pi\color{blue}{n}$ in the problem. – Oscar Lanzi Jun 07 '20 at 14:11
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i got it right now! thank you very much Oscar! :) – Jun 07 '20 at 14:35

arc length of helix $(2\cos(t), 2\sin(t), 3t)$

1 Answers1