Problem on the arc length of a helix
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Find the arc length of the helix $x(t) = \cos t, y(t) = \sin t, z(t) = t$ traced from $t=1$ to $t=2$.
Solution
Recall that
Arc length of a parameterization
The length of the curve $\bfx(t)$ traced out between $t=a$ and $t=b$ is given by the integral of the speed.
Precisely, $s = \int_a^b \left | \frac{d \bfx(t)}{dt} \right| dt.$
We identify $a=1$ and $b=2$ and need to compute the speed as a function of $t$.
Velocity of a parameterization
If $\bfx(t) = \langle \cos t, \sin t, t \rangle$, then the velocity vector is $$\frac{d\bfx}{dt}(t) = \langle -\sin t, \cos t, 1 \rangle.$$
Speed of a parameterization
Length of a vector
The speed at time $t$ is the length of this vector \begin{align}\left |\frac{d\bfx}{dt}(t)\right | &= | \langle -\sin t, \cos t, 1 \rangle | \\ &=\sqrt{\sin^2(t) + \cos^2(t) + 1} \\ &= \sqrt{2} \end{align}
Thus, the arc length swept between $t=1$ and $t=2$ is $$ s = \int_1^2 \sqrt{2} dt = \sqrt{2}.$$
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