Lab 4A - Arcs, Parametrizations, and Arclength
Math 2374 - University of Minnesota
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In Lab 2A we worked with parametric equations and curves, which are also called "arcs."  In this lab we're going to examine these objects a little deeper.  We'll talk about arc length, as well as line integrals of scalar functions.  For most of these purposes we're going to use the unit circle, because you're already familiar with it, and yet it's complicated enough to illustrate many different concepts.

FYI:  In your textbook, "curve" and "arc" are defined to be the same thing.  In practice most of us will generally refer to these things as curves.  However when we deal with the length of a curve, we suddenly change terms and talk about arc length as opposed to curve length.  Apparently the term "arc length" is so deeply ingrained in mathematicians' brains that we'll never be able to switch!  Please don't get confused by this change in terminology.

One other reminder: a "path" is actually a function which is a parametrization for some curve.  Sometimes in an abuse of language we'll say "path" when we really mean "curve" and vice versa, but you should remember that there is a (subtle) difference!  (Ask your TA if you don't understand what the difference is.)

Of course, we basically just have to throw our hands up in the air about the terminology when we get to line integrals, which are really integrals along a curve, not a straight line.  Why on earth would they be called line integrals instead of curve integrals?  Let's just agree not to go there...

The rest of this lab assumes that you remember what the derivative of a parametrization f(t) is.  Remember, if f(t) represents the position of the particle, then the tangent vector represents the velocity of the particle.  You can review this at the end of Lab 2A if you'd like.

Estimating Arc Length

In the first part of this lab, we will explore how to estimate the length of a curve by approximating the curve with line segments.  It turns out that the accuracy of these estimates can be tied in with the tangent vector, which is why we asked you to review that if necessary.

To begin, we're going to use larger and larger numbers of line segments to estimate the length of the upper half of the unit circle.  We're using this particular curve because -- as you should definitely know by now -- the arc length of the upper half of the unit circle is π.

π is a very famous number which shows up everywhere.  You probably know that it is irrational, i.e. it cannot be represented as a fraction.  In fact, the decimal expansion of π goes on forever and ever, with no apparent pattern that we've been able to discern.  To 40 decimal places,

π = 3.141592653589793238462643383279502884197.

Historically, people have used many different approximations for π.  The ancient Egyptians and Babylonians both realized that it is slightly bigger than 3.  The Babylonians used an approximation of 3 1/8 = 3.125, a bit low.  The Egyptians' estimate of  256/81≈ 3.16049 is too high.  In the fifth century, a Chinese mathematician named Tsu Chung Chi estimated π as 355/113≈ 3.14159292, which agrees with the true value of π for six decimal places!

In high school you may have used approximations such as 3.14, or 3.14159, etc.  Your calculator probably has the correct value of π stored up to a few dozen digits or so, which is usually more than enough.  In fact, knowing π to just 39 decimal places is sufficient for calculating the circumference of the universe accurate to the radius of a hydrogen atom, but in recent years computer scientists have tried to calculate as many digits of π as possible.  By the late 1990s we knew the value of π to more than 206 billion digits!  In 2002, some of the same researchers involved in that record computed an absolutely ridiculous 1.24 trillion digits of Pi.  (See  If you'd like to know more about π, you also could read a book called A History of π, written by Petr Beckman.)

For over one thousand years the most common method of  estimating π was due to Archimedes, who used it to calculate that 3 10/71< π < 3 1/7.  Essentially, part of his method was to do exactly what we're about to do: estimate the length of a circle using line segments.  Let's get started!

To plot the upper half of the unit circle along with 4 approximating segments, evaluate this command:


Segments[{Cos[t], Sin[t]}, {t, 0, Pi}, 4, AspectRatioAutomatic]




There is another command which will add up the lengths of these segments and give us an estimate of the arclength.


Estimate[{Cos[t], Sin[t]}, {t, 0, Pi}, 4]



You can tell that we're in the right ballpark, but still not very close!  In fact we need to use many more segments before we get anything close to an accurate estimate.  For his lower bound on π, Archimedes would have used about 48 line segments:


Segments[{Cos[t], Sin[t]}, {t, 0, Pi}, 48, AspectRatioAutomatic] Estimate[{Cos[t], Sin[t]}, {t, 0, Pi}, 48]






Exercise 1

Find the minimum number n of segments required such that the estimate of π is accurate to five decimal places, i.e. such that the estimate is 3.14159.  You should hand in what you think the number n is, as well as the estimates with n segments and (n-1) segments.  (In other words, show that you've actually found the smallest such n.)  You do not need to hand in graphs of the line segments; once you use more than about 10 segments, the graph of the circle is nearly indistinguishable from the line segments anyway!

Arc Length and the Derivative

At the end of Lab 2A we looked at two different parametrizations for the unit circle:


f[t_] = {Cos[t], Sin[t]} ; g[t_] = {Cos[t^2], Sin[t^2]} ;

You should go back to that lab and look at the discussion about the differences in the derivatives of f and g.  Let's look at what happens when we try to use these parametrizations to find the circumference (or arc length) of the unit circle.  Initially we'll use 12 line segments for picture.

With f(t), we'll get a picture similar to the one above when we used 4 segments to estimate the length of the upper half circle:


Segments[f[t], {t, 0, 2Pi}, 12, AspectRatioAutomatic] Estimate[f[t], {t, 0, 2Pi}, 12]






(Remember, since we're using the whole circle, the length is really 2π ≈ 6.2831853.)

We get a nice symmetric picture, as we would expect.  But things are different if we use g(t).  Remember that t only goes to (2π)^(1/2) when we use g!


Segments[g[t], {t, 0, Sqrt[2Pi]}, 12, AspectRatioAutomatic] Estimate[g[t], {t, 0, Sqrt[2Pi]}, 12]






As you can see, the picture is all out of whack, and the estimate is much worse than what we obtained using f(t).  The reason for this has to do with the derivatives of f and g.  Remember, if f and g represent the position of a particle at time t, then the derivatives represent the velocity of the particle.

These pictures were obtained by dividing the time interval into 12 equal pieces; each blue segment represents the movement of the particle during one of those subintervals.  The first picture is symmetric because a particle moving according to f(t) has constant speed.  The second picture is uglier because a particle moving according to g(t) picks up speed as time passes, so it moves further during each successive subinterval.

That last paragraph is a little dense, but you should re-read it until you understand it because it holds the key to understanding how the derivative relates to the accuracy of given estimations.  Once you understand what's happening for these pictures, you are ready for problem 2.

Exercise 2

In this problem we'll work with the same two parametrizations for the unit circle,

f(t) = (Cos(t),  Sin(t)),    0≤t≤2π
g(t) = (Cos(t^2), Sin(t^2) ),    0≤t≤(2π)^(1/2).

(a)  Calculate the length of the tangent vector using the first parametrization.  Then do the same for the second tangent vector.  Simplify your answer!

(b)  Look at the pictures above where the circumference is estimated using 12 segments with each parametrization.  Although the estimate using g(t) is clearly the worse of the two, if you look in the first quadrant it's another story.  If you were to add up the lengths of the line segments in the first quadrant of each picture, thereby estimating π/2, it appears that you'd get a better estimate from the second picture, which was made using g(t)!

Confirm this and explain why it is so!  

Hint: to confirm it you can use Segments and Estimate to analyze the quarter circle, but you need the correct number of segments and the correct domain for for each parametrization.  To explain why it is so will take some thought, and we expect you to present a clear explanation of what's going on -- something beyond "the Estimate command shows that it is more accurate."

Your writeup to this problem does not need to be split into parts (a) and (b); rather, it should be one seamless essay which uses the information in (a) and (b) to describe why g(t) gives you a better approximation for π/2.

Dangerous Curve Ahead

In this last part of the lab we'll examine a curve whose length can be tricky to estimate.  The following is a parametrization of the curve:


RowBox[{h[t_], =, RowBox[{{, RowBox[{Cos[t], ,,  , RowBox[{Sin[t], +, RowBox[{0.01,  , Sin[1000t]}]}]}], }}]}]


RowBox[{{, RowBox[{Cos[t], ,, RowBox[{Sin[t], +, RowBox[{0.01,  , Sin[1000 t]}]}]}], }}]

(For our purposes we'll assume 0≤t≤π.)

First use the command Segments with 100 segments to sketch the curve and the approximating segments.  Then use Estimate with 100 segments to calculate the approximate length.  Do you think this is accurate?  (You can use the following commands.)


Segments[h[t], {t, 0, Pi}, 100] ; Estimate[h[t], {t, 0, Pi}, 100]




You should also try Estimate with 250, 500, and 1000 line segments.  You should be building a large amount of evidence that the arc length of this curve is π.  Again, do you think this is accurate?

Now that you're comfortable with this estimate, try using Estimate with 950 segments.  If you've done everything correctly, you should be thinking, "Huh?" when you see the answer.  Now which estimate do you think is the most accurate?

To find out, let's examine a tiny little piece of the curve.  Instead of letting t range from 0 to π, let's look at the section where 0≤t≤π/50.  1000 segments over the whole curve corresponds to 1000/50=20 segments on this tiny piece of the curve.


Segments[h[t], {t, 0, Pi/50}, 1000/50]




Remember, the curve is red, while the approximating line segments are blue.  Now do you understand what's going on?  Look at this tiny piece with 950/50=19 segments instead of 20 segments.  Which estimate is more accurate?

Exercise 3

Experiment with the number of line segments on the small piece of the curve until you think you have an accurate estimation of the arc length.  Then multiply by 50 to find the corresponding number of line segments on the entire curve, and use this to find an estimation of the total arc length.

You should hand in your estimate, the number of segments used, and a picture of this estimate at the "tiny" level (where 0≤t≤π/50) so we can see how good the fit is.

Exercise 4

(a)  Suppose you have a parametrization f(t) (with some bounds on t) for a given curve.
Now suppose you find the following estimates of the curve's length:

100   segments: length ≈ π.
1000 segments: length ≈ π.
850  segments: length ≈ 400.

Can you tell which number of segments gives the most accurate estimate?  Why?

(b)  Suppose now your estimates look like this:

100   segments: length ≈ π.
1000 segments: length ≈ π.
850  segments: length ≈ 1.

Now can you tell which number of segments gives the most accurate estimate?  Why?  (Has your answer changed?)

Line Integrals of Scalar Functions

So far we've only been estimating arc length, but you know from lecture (and your textbook) that we don't have to be satisfied with estimates.  We can use an integral to find the exact length of a curve.  Suppose our curve C is parametrized by f(t), where a≤t≤b.  Then the length of C is:

Length = ∫_a^b∥f ' (t) ∥dt

This is really just a special case of a "Line Integral of a Scalar Function," which you've also learned about in lecture.  Suppose we want to integrate a real-valued function g(x, y, z)on the same curve C.  Then the line integral of g on C is:

∫_Cg dL = ∫_a^bg (f (t)) ∥f ' (t) ∥dt

So the only difference is that we stick g(f(t)) into the integral; essentially, we're trying to add up the values of g along our curve.  If we use the special function g(x, y, z) = 1, then g(f(t)) = 1, and then we get the integral for arc length.  That's actually the reason for the "dL" -- if you integrate 1, you get the Length of C.

There are a few different physical interpretations of line integrals of scalar functions.  Perhaps the most common ones involves mass.  Suppose our curve represents a wire in three-dimensional space.  The wire is made up of different metals and might have a different density at each point.  If we had a straight wire, with a constant density, then the mass is simply length times density.  When the wire is curvy, and the density can vary, we need a line integral:  

Mass = ∫_Cg dL = ∫_a^bg (f (t)) ∥f ' (t) ∥dt


Consider the following curve, a helix.  It could also represent, say, a wire which has been made into a slinky.  Notice that we're using the bounds 0≤t≤10:


f[t_] = {Cos[2 * Pi * t], Sin[2 * Pi * t], 1 + t/5} ParametricPlot3D[f[t], {t, 0, 10}, ViewPoint {0, -10, 1},     PlotPoints120]


{Cos[2 π t], Sin[2 π t], 1 + t/5}




Suppose we know that the density of the wire (in grams per centimeter) at any given point is equal to its distance from the xy-plane, i.e.


g[x_, y_, z_] = z



Then the total mass of the wire is:

Mass = ∫_Cg dL = ∫_0^10g (f (t)) ∥f ' (t) ∥dt

We need to learn how to evaluate this integral with the computer.  First of all, we can have Mathematica find the length of the derivative of f.  The derivative will be a vector, and the Norm function will find its length:


D[f[t], t] Norm[%]//Simplify


{-2 π Sin[2 π t], 2 π Cos[2 π t], 1/5}


1/5 (1 + 100 π^2)^(1/2)

Next we have to plug f(t) into g(x, y, z):


Apply[g, f[t]]


1 + t/5

Huh?  Why did we use Apply[g,f[t]] when we want g[f[t]]?  We had to do this to avoid an annoying problem.  Remember, we have
            f[t] = {t Cos[2 π t], t Sin[2 π t], 2 t}

For our integral, we want to evaluate   g[t Cos[2 π t], t Sin[2 π t], 2 t],
but Mathematica reads g[f[t]] as g[{t Cos[2 π t], t Sin[2 π t], 2 t}]

See the difference?  The Apply command makes this all work correctly.

Now we can put them all together to find the mass of the wire:


Integrate[Apply[g, f[t]] * Norm[D[f[t], t]], {t, 0, 10}] N[%]


4 (1 + 100 π^2)^(1/2)



Exercise 5

Consider the curve C parametrized by f(t) = (t, t^3, t^2), where 0≤t≤1. The engineers building a new airplane realize that they'll have to have a wire in the shape of C running between two systems; because of the different stresses placed on different parts of the hull, the wire has to be denser in some areas.  If the density of the wire is given by g(x, y, z) = 2x + 4y grams/centimeter, find the mass of the wire.

While we'd like to see the exact answer, a decimal answer might be more useful in the real world.

Exercise 6

Follow the instructions of exercise 5 using the following functions:

f(t) = (t, Cos t, Sin t)where 0≤t≤2π,
g(x, y, z) = -x y z.


Created by Mathematica  (November 6, 2004)