4212.sagews

We are exploring arc length. We want to understand what is happening. (We do NOT care whether we can integrate symbollically! Any program will integrate for us.)

YouTube Videos: Parameterization     Arc Length

Related Sage Pages: Arc Length of Explicit Curves in 2D     Arc Length of Parametric Curves in 2D     Line Integral of Function - SOLVER

Related Wiki Pages: 

var ('t')

#Define a 3d curve s (parametrically) and an interval for t
s=vector((2*cos(t)^2,3*sin(t),2*t))
t1=0; t2=4*pi

Remember that a curve in 3d can ONLY be defined parametrically!

#Let's plot C
xmin=-3; xmax=3; ymin=-3; ymax=3; zmin=0; zmax=30
C=parametric_plot(s,(t,t1,t2),color='orange',thickness=6, opacity=.7)

#Let's add a reference
var ('x y z')
px0y= implicit_plot3d(z==0,(x,xmin,xmax),(y,ymin,ymax),(z,zmin,zmax), color='grey', opacity=0.3)
Ax=line3d(([xmin,0,0],[xmax,0,0]), thickness=2, color='red')
Ay=line3d(([0,ymin,0],[0,ymax,0]), thickness=2, color='blue')
Az=line3d(([0,0,zmin],[0,0,zmax]), thickness=2, color='green')

show(C+px0y+Ax+Ay+Az, aspect_ratio=[3,3,1])

Look at the curve above and estimate a minimum and maximum value for its length L.

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Arc Length of a Curve given parametrically $C=s$: $\lt x(t),\, y(t),\, z(t) \gt$ for $t \in [t1,t2]$  is  $$L= \int_C \,ds =\int_{t1}^{t2} \, \sqrt{{\dot{x}}^2 + {\dot{y}}^2+ {\dot{z}}^2 } \, dt$$

ds=diff(s,t)
view(ds)

\newcommand{\Bold}[1]{\mathbf{#1}}\left(-4 \, \sin\left(t\right) \cos\left(t\right),\,3 \, \cos\left(t\right),\,2\right)

Lexact=integral(norm(ds),(t,t1,t2))
n(Lexact)

\newcommand{\Bold}[1]{\mathbf{#1}}40.0345236851

So the arc length of this weirdo curve using the formula (which we are calling our "exact" result even though it is being calculated numerically) is $L=40.03$

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Let us approximate this length by finding tangent line segments at regularly spaced values of t along the curve.

This algorithm is exactly the same as for parametric curves in 2d!

• We decide how many steps.
• The program calculates the stepsize of t.
• We draw points on the curve regularly spaced with repect to stepsize. They are the points: (s(j))

steps=4 
stepsize=(t2-t1)/steps

points=sum([point3d(s(t=j), color='purple', size=10) for j in [t1..t2-stepsize,step=stepsize]])
show(C+points, aspect_ratio=[3,3,1])

We draw pieces of tangent line segments starting at these points.

• Start point is s(j)
• Slope is value of the derivative vector ds(j). 
• Length is $\left\| {ds} \right\| \cdot$ stepsize = $\sqrt{{\dot{x}}^2 + {\dot{y}}^2 + {\dot{z}}^2} \cdot$ stepsize.

So parametrically these line segments are: s(j)+λ·ds(j) for λ=[0, stepsize].

pieces=sum([line3d([(s(t=j)),(s(t=j)+ds(t=j)*stepsize)],thickness=4,color='purple') for j in [t1..t2-stepsize,step=stepsize]])

show(C+pieces+points, aspect_ratio=[7,1,1])

We sum the length of these pieces.

Lapprox=sum([norm(ds(t=j))*stepsize for j in [t1..t2-stepsize,step=stepsize]])
n(Lapprox)

\newcommand{\Bold}[1]{\mathbf{#1}}45.3086935965559

So the approximate arc length of this weirdo curve using 4 tangent pieces is: $L_4 =45.31$.

We calculate our error.

error=abs((Lexact-Lapprox)/Lexact)
n(error)

\newcommand{\Bold}[1]{\mathbf{#1}}0.131740543560301

Our error is $\approx 13$%.

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Let us try more or less step sizes - change the value of steps2 and revaluate.

steps2=16 
stepsize2=(t2-t1)/steps2

points2=sum([point3d(s(t=j), color='purple', size=10) for j in [t1..t2-stepsize2,step=stepsize2]])
pieces2=sum([line3d([(s(t=j)),(s(t=j)+ds(t=j)*stepsize2)],thickness=4,color='purple') for j in [t1..t2-stepsize2,step=stepsize2]])

show(C+pieces2+points2, aspect_ratio=[3,3,1])

We sum the length of these pieces.

Lapprox2=sum([norm(ds(t=j))*stepsize2 for j in [t1..t2-stepsize2,step=stepsize2]])
n(Lapprox2)

\newcommand{\Bold}[1]{\mathbf{#1}}39.8247733971104

So the approximate arc length of this weirdo curve using 16 tangent pieces is: $L_{16} =39.82$.

We calculate our new error.

error2=abs((Lexact-Lapprox2)/Lexact)
n(error2)

\newcommand{\Bold}[1]{\mathbf{#1}}0.00523923525710368

Our error is now $\approx 0.5$%.