Sec1Sess20

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Arc Length and Velocity
Distance traveled along a curve
s = arc length = distance traveled along a trajectory (curve)
r = position = net displacement (total difference in position, start vs. end)
v = velocity = tangent vector
lots more notation here (not to repeat the list here):
http://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/1.-vectors-and-matrices/part-c-parametric-equations-for-curves/session-20-velocity-and-arc-length/MIT18_02SC_notes_12.pdf

NOTE: <x,y,z> notation for vectors, (x,y,z) notation for points

can be s vs. t
d(s) a.k.a. d(s)/d(t)
= speed = |<math>\vec{v}</math>| (length of velocity vector) = <math>sqrt(v^2)</math>

arc length <math>= int_{t1,t2}(d(s)/d(t))</math> where t1, t2 are time1, time2
note: for solution on continuous integral subtract value at t2 from t1

Example: length of an arch of cycloid is:
<math>int_{0,2pi}(sqrt(2-2*cot(t)))</math>

want to integrate t = 0 to 2*pi
(this will be solved later not by directly taking integral but by another method)

unit tangent vector T
the vector tangent to the curve at a point, converted to a unit length (length of 1) by dividing by the length of <math>\vec{v}</math>.
<math>T_{hat}\mathrm{\ =\ }\vec{v}\mathrm{\ /\ |}\vec{v}\mathrm{|}</math>

<math>\vec{v}\ =\ d(\vec{r})/d(t)\ =\ d(\vec{r})/d(s)\ *\ d(s)/d(t)</math>

<math>\mathrm{d(}\vec{r}\mathrm{)/d(s)\ =\ T_{hat}}</math>
<math>\mathrm{d(s)/d(t)\ =\ |}\vec{v}\mathrm{|}</math>

<math>\vec{v}\mathrm{\ =\ T_{hat}\ *\ |}\vec{v}\mathrm{|}</math>

velocity has:
1. direction: tangent to trajectory (curve), <math>T_{hat}</math>, d(r)/d(s)
2. length: speed, <math>|\vec{v}|</math>, d(s)/d(t)

<math>r(t+t_{change})</math> is an approximation of the distance along the curve because a straight line is used to measure between point 1,t, and point <math>2,t_{change}.</math>

<math>\mathrm{change(}\vec{r}\mathrm{)\ \approx \ T_{hat}*s_{change}}</math>
<math>\mathrm{=change(}\vec{r}\mathrm{)\ /\ change(d)\ \approx \ T_{hat}*(s_{change\ /\ change(t)})}</math>

Limit as <math>change_{t}</math> -> 0
gives d(r)/d(t) = <math>T_{hat}</math>*d(s)/d(t)
d(s)/d(t) = d(r)/d(t) = <math>sqrt((d(x)/d(t))^2+(d(y)/d(t))^2)</math>

when given the x and y of the position vector, take their derivatives to find the
x and y of the velocity vector.
e.g. x(t) = 20t, y(t) = 5t + 6
v = <20, 5>

Examples of finding v, s, r, etc.: http://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/1.-vectors-and-matrices/part-c-parametric-equations-for-curves/session-20-velocity-and-arc-length/MIT18_02SC_we_11_comb.pdf