# How Do You Find an Arc Length Parametrization for a Given Curve?

• khemist
In summary, the conversation discusses finding an arc length parametrization of a given curve and the use of the parameter s. The author mentions solving for the arc length, L, which is the definite arc length from 0 to pi/2, but it is not applied anywhere else in the problem. The purpose of finding L is to determine the extent of the given curve and the length of the curve. When changing the parameterization to arc length, the ending value is always the length of the curve, and the new formula is given in terms of the new parameter s. The conversation also includes an example of a parameterization of a quarter circle to further explain the concept.
khemist

## Homework Statement

Find an arc length parametrization of the curve r(t) = <e^t(cos t), -e^t(sin t)>, 0 =< t =< pi/2, which has the same orientation and has r(0) as a reference point.

## Homework Equations

s = int[0,t] (||r'(t)||)

## The Attempt at a Solution

So I found the derivative of r(t), and then normalized it. The problem is that I am left with a value that is negative under a radical, or a function that I simply do not have any idea how to integrate.

P.S: Sorry, I have no idea how to use that type of script which makes viewing functions more convenient...

My r'(t) = <(e^t(cos t - sin t)), (-e^t (cos t + sin t))>

I then normalized the derivative, where I got one of two functions:

e^t * (sqrt(-4sin t cost t) or sqrt(-e^t) * 2sqrt(sin t cos t)

But how would one integrate that? using by parts? My teacher claims that there would be nothing extremely difficult to integrate, but this problem seems to be proving otherwise...

Ps: Sorry, I do not know how to use that script which makes viewing functions more convenient..

Check your work for ||r'(t)||. It should be pretty simple; what did you get?

Yea Thanks. While typing it out, I realized that I forgot to remove the (-) from the second e^t. I guess then it should reduce down to simply e^t * sqrt(2), which is a fairly easy integral.

I then solved the s = int[0,t], which I ended up with t = ln(s/sqrt(2) + 1)

So another question arises, as I probably had messed something up, but where does the limiting of t come into play? Should I instead of integrating from 0 -> t, integrate from 0 -> pi/2?

Last edited:
So now substitute your t in terms of s into your original equation. s will range from 0 to the length of the curve. Gotta hit the sack now. Good luck.

Ok thanks. I am still very unsure why they mention that t goes from 0 -> pi/2

I watched a youtube video on this subject, and he solves for the arc length, L, but it does not show up anywhere else in the problem. My question is what is the point of solving for L, the definite arc length from 0 -> pi/2 if it is not applied anywhere else?

khemist said:
Ok thanks. I am still very unsure why they mention that t goes from 0 -> pi/2

I watched a youtube video on this subject, and he solves for the arc length, L, but it does not show up anywhere else in the problem. My question is what is the point of solving for L, the definite arc length from 0 -> pi/2 if it is not applied anywhere else?

A parameterization will give the formula for the curve and the domain, the values of t for which the curve is given. The author could have given you any values for t, he just chose to give you 0 -> pi/2. That, of course, determines the extent of the given curve and the length of the curve. When you change the parameterization to another domain variable such as arc length s, you give a new formula in terms of the new parameter s. To give the same curve, you need to give the specific domain of the new parameter. When you use arc length as the parameter, the ending value is always the length of the curve. 0 ≤ s ≤ L(C). And of course, you can compute L(C) from the original parameterization.

I guess I am still unclear why he would give me that kind of information when, in terms of the question, it is literally useless. I understand the necessity to know how to find the parametrization in terms of s.

When you say L(C), that means t, not the length of the curve, say if the length from o -> pi/2 was pi, the integral would still go from t(not) -> t, not t(not) to pi?

khemist said:
I guess I am still unclear why he would give me that kind of information when, in terms of the question, it is literally useless. I understand the necessity to know how to find the parametrization in terms of s.

When you say L(C), that means t, not the length of the curve, say if the length from o -> pi/2 was pi, the integral would still go from t(not) -> t, not t(not) to pi?

$$\vec R(t) = \langle \sqrt{1 - t^2},t\rangle,\, 0\le t \le 1$$

$$\vec R'(t) = \langle -\frac{t}{ \sqrt{1 - t^2}},1\rangle$$

$$\frac {ds}{dt}=\sqrt{\frac{t^2}{1-t^2}+1}=\sqrt{\frac 1 {1-t^2}}$$

So to calculate the arc length from 0 to t:

$$s =\int_0^t\sqrt{\frac 1 {1-t^2}}\,dt = \arcsin(t)|_0^t = \arcsin(t)$$

Notice if you integrate for t from 0 to 1, you get

$$s = \arcsin{(1)} = \pi/2$$, which is the length of this arc.

Your t can be expressed in terms of s:

$$t = \sin{(s)} \hbox{ and } \sqrt{1 - t^2} = \cos{(s)}$$

and you can parameterize your curve in terms of s:

$$\vec R(s) = \langle \cos(s), \sin(s)\rangle,\ 0 \le s\le \frac \pi 2$$

You now have two parameterizations of the same curve, in the same direction, one of which is the arc length parameterization. Hopefully the arc length parameterization is already familiar to you because of s = rθ.

It makes a little more sense. Thanks for the great help.

Now I understand the use of the ds/dt, as before I was quite confused on what that meant.

So when parametritizing the curve, you must also "redo" the value of the interval that t is in? And to do that, one would simply plug in the previous t values into the new function of s?

khemist said:
It makes a little more sense. Thanks for the great help.

Now I understand the use of the ds/dt, as before I was quite confused on what that meant.

So when parametritizing the curve, you must also "redo" the value of the interval that t is in? And to do that, one would simply plug in the previous t values into the new function of s?

ds/dt is the speed at which you move along the curve. Each parameterization has its own formula and domain. As the parameter varies you move along the curve. Two people may walk the same curve at different rates, arrive at different times, yet walk the same distance. Same idea.

So when I solve for ds/dt, I still use the interval of 0 -> t, regardless of the initial interval that t is restricted to. But, when showing the answer, it is necessary to show that the interval has changed, with respect to the new parametrization.

khemist said:
So when I solve for ds/dt, I still use the interval of 0 -> t, regardless of the initial interval that t is restricted to. But, when showing the answer, it is necessary to show that the interval has changed, with respect to the new parametrization.

I'm not sure what is bothering you. The answer to your original question would be expressed in the form:

$$\vec R(s) = \langle f(s),g(s)\rangle,\, 0 \le s \le \ell (C)$$

where the f and g formulas you have figured out in terms of s and the length $\ell (C)$ of the curve you can calculate from the usual formula.

## What is arc length parametrization?

Arc length parametrization is a method used in mathematics and physics to represent curves in terms of their arc length instead of the traditional parameterization by time or angle. This allows for more accurate and efficient calculations and analysis of curves.

## Why is arc length parametrization important?

Arc length parametrization is important because it provides a more natural and intuitive way to describe and analyze curves. It also allows for easier comparisons and calculations between different curves, as they are all represented in terms of their arc length.

## How is arc length parametrization calculated?

Arc length parametrization is calculated using the arc length formula, which takes into account the distance traveled along a curve and the curvature of the curve. This formula involves integrals and can be solved using calculus.

## What is the difference between arc length parametrization and regular parametrization?

The main difference between arc length parametrization and regular parametrization is the parameter used to represent the curve. Regular parametrization uses a parameter such as time or angle, while arc length parametrization uses the arc length of the curve. This results in a different and more accurate representation of the curve.

## What are the applications of arc length parametrization?

Arc length parametrization has many applications in mathematics and physics, particularly in the fields of geometry, calculus, and mechanics. It is also used in computer graphics and animation to create smooth and realistic curves. Additionally, it is used in engineering and design to accurately model and analyze curved structures and objects.

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