Question

Calc III Tutorial: Arc Lengths in Three Dimensions

Answer 1

\({\bf{Arc~Length:}}\) take the derivatives, square all the derivatives, find the sum, then take the square root, then take the integral of the whole thing

\[L = \int\limits_{a}^{b}\sqrt{(dx/dt)^{2}+(dy/dt)^{2}+(dz/dt)^{2}}dt = \int\limits_{a}^{b}|v|dt\]

\({\bf{Arc~Length~Parameter:}}\)
\[s(t) = \int\limits_{a}^{b}|v(τ)dτ\] for some base point P(t0) to P(t). using τ here to avoid confusing with time, other than that works the same way as the arc length equation

Answer 2

\({\bf{Supplementary~Equations:}}\)
s = arc length
r = position
v(t) = velocity

ds/dt = |v(t)|

unit tangent vector = T = v/|v|
more or less, the unit vector of velocity

dt/ds = 1 / (ds/dt) = 1/|v(t)|
reciprocal rule, then using the definition of velocity

therefore

dr/ds = (dr/dt)(dt/ds) = v (1/|v|) = T
split the integral according to the chain rule

Answer 3

Source material is section 13.3 of Thomas' Calculus, Early Transcendentals, 14th edition by Hass, Heil, Weir, et. al.