Arc Length Parametrization

Oct 13, 2025 by miodeya.com 25 views

Introduction

In this article, we will explore the concept of arc length parametrization and how to find it for a given curve. We will use the curve $r(t) = \left\langle e^t \sin(t), e^t \cos(t), 8e^t \right\rangle$ as an example and find its arc length parametrization $r_1(s)$ .

Arc Length Parametrization

The arc length parametrization of a curve $r(t)$ is a new parametrization $r_1(s)$ such that the arc length of the curve is equal to the parameter $s$ . In other words, the arc length of the curve is a function of the parameter $s$ , and the curve is parametrized by this arc length.

Finding the Arc Length

To find the arc length parametrization, we first need to find the arc length of the curve. The arc length of a curve $r(t)$ is given by the formula:

L = \int_{a}^{b} \left\| r^{\prime}(t) \right\| dt

where $r^{\prime}(t)$ is the derivative of the curve with respect to the parameter $t$ , and $\left\| r^{\prime}(t) \right\|$ is the magnitude of the derivative.

Calculating the Derivative

To calculate the derivative of the curve, we need to find the derivative of each component of the curve with respect to the parameter $t$ . The derivative of the curve $r(t) = \left\langle e^t \sin(t), e^t \cos(t), 8e^t \right\rangle$ is given by:

r^{\prime}(t) = \left\langle e^t \sin(t) + e^t \cos(t), e^t \cos(t) - e^t \sin(t), 8e^t \right\rangle

Calculating the Magnitude

To calculate the magnitude of the derivative, we need to find the square root of the sum of the squares of the components of the derivative. The magnitude of the derivative is given by:

\left\| r^{\prime}(t) \right\| = \sqrt{(e^t \sin(t) + e^t \cos(t))^2 + (e^t \cos(t) - e^t \sin(t))^2 + (8e^t)^2}

Simplifying the Magnitude

To simplify the magnitude, we can use the trigonometric identity $\sin^2(t) + \cos^2(t) = 1$ . The magnitude of the derivative can be simplified as:

\left\| r^{\prime}(t) \right\| = \sqrt{e^{2t} + 64e^{2t}} = \sqrt{65e^{2t}} = \sqrt{65}e^t

Finding the Arc Length

Now that we have the magnitude of the derivative, we can find the arc length of the curve by integrating the magnitude with respect to the parameter $t$ . The arc length of the curve is given by:

L = \int_{a}^{b} \sqrt{65}e^t dt

Evaluating the Integral

To evaluate the integral, we can use the formula for the integral of $e^t$ :

\int e^t dt = e^t + C

where $C$ is the constant of integration. The arc length of the curve is given by:

L = \sqrt{65} \int_{a}^{b} e^t dt = \sqrt{65} (e^b - e^a)

Finding the Arc Length Parametrization

Now that we have the arc length of the curve, we can find the arc length parametrization $r_1(s)$ by using the formula:

r_1(s) = r(t(s))

where $t(s)$ is the inverse function of the arc length $L(s)$ .

Finding the Inverse Function

To find the inverse function, we need to solve the equation $L(s) = s$ for $t(s)$ . The arc length $L(s)$ is given by:

L(s) = \sqrt{65} (e^b - e^a)

We can solve for $t(s)$ by using the formula:

t(s) = \ln \left( \frac{s}{\sqrt{65} (e^b - e^a)} + 1 \right)

Finding the Arc Length Parametrization

Now that we have the inverse function, we can find the arc length parametrization $r_1(s)$ by using the formula:

r_1(s) = r(t(s))

The arc length parametrization $r_1(s)$ is given by:

r_1(s) = \left\langle e^{\ln \left( \frac{s}{\sqrt{65} (e^b - e^a)} + 1 \right)} \sin \left( \ln \left( \frac{s}{\sqrt{65} (e^b - e^a)} + 1 \right) \right), e^{\ln \left( \frac{s}{\sqrt{65} (e^b - e^a)} + 1 \right)} \cos \left( \ln \left( \frac{s}{\sqrt{65} (e^b - e^a)} + 1 \right) \right), 8e^{\ln \left( \frac{s}{\sqrt{65} (e^b - e^a)} + 1 \right)} \right\rangle

Conclusion

In this article, we have found the arc length parametrization $r_1(s)$ of the curve $r(t) = \left\langle e^t \sin(t), e^t \cos(t), 8e^t \right\rangle$ . The arc length parametrization is given by:

r_1(s) = \left\langle e^{\ln \left( \frac{s}{\sqrt{65} (e^b - e^a)} + 1 \right)} \sin \left( \ln \left( \frac{s}{\sqrt{65} (e^b - e^a)} + 1 \right) \right), e^{\ln \left( \frac{s}{\sqrt{65} (e^b - e^a)} + 1 \right)} \cos \left( \ln \left( \frac{s}{\sqrt{65} (e^b - e^a)} + 1 \right) \right), 8e^{\ln \left( \frac{s}{\sqrt{65} (e^b - e^a)} + 1 \right)} \right\rangle

The arc length parametrization is a new parametrization of the curve such that the arc length of the curve is equal to the parameter $s$ .

Q&A

Q: What is an arc length parametrization?

A: An arc length parametrization is a new parametrization of a curve such that the arc length of the curve is equal to the parameter $s$ . In other words, the arc length of the curve is a function of the parameter $s$ , and the curve is parametrized by this arc length.

Q: Why is arc length parametrization important?

A: Arc length parametrization is important because it allows us to study the properties of a curve in a more natural way. The arc length of a curve is a fundamental property of the curve, and by parametrizing the curve by its arc length, we can gain a deeper understanding of the curve's behavior.

Q: How do I find the arc length parametrization of a curve?

A: To find the arc length parametrization of a curve, you need to follow these steps:

Find the derivative of the curve with respect to the parameter $t$ .
Calculate the magnitude of the derivative.
Integrate the magnitude with respect to the parameter $t$ to find the arc length of the curve.
Solve the equation $L(s) = s$ for $t(s)$ to find the inverse function.
Use the inverse function to find the arc length parametrization $r_1(s)$ .

Q: What is the formula for the arc length parametrization?

A: The formula for the arc length parametrization is given by:

r_1(s) = r(t(s))

where $t(s)$ is the inverse function of the arc length $L(s)$ .

Q: How do I calculate the inverse function?

A: To calculate the inverse function, you need to solve the equation $L(s) = s$ for $t(s)$ . This can be done using the formula:

t(s) = \ln \left( \frac{s}{\sqrt{65} (e^b - e^a)} + 1 \right)

Q: What is the arc length parametrization of the curve $r(t) = \left\langle e^t \sin(t), e^t \cos(t), 8e^t \right\rangle$ ?

A: The arc length parametrization of the curve $r(t) = \left\langle e^t \sin(t), e^t \cos(t), 8e^t \right\rangle$ is given by:

r_1(s) = \left\langle e^{\ln \left( \frac{s}{\sqrt{65} (e^b - e^a)} + 1 \right)} \sin \left( \ln \left( \frac{s}{\sqrt{65} (e^b - e^a)} + 1 \right) \right), e^{\ln \left( \frac{s}{\sqrt{65} (e^b - e^a)} + 1 \right)} \cos \left( \ln \left( \frac{s}{\sqrt{65} (e^b - e^a)} + 1 \right) \right), 8e^{\ln \left( \frac{s}{\sqrt{65} (e^b - e^a)} + 1 \right)} \right\rangle

Q: Can I use the arc length parametrization to find the arc length of a curve?

A: Yes, you can use the arc length parametrization to find the arc length of a curve. The arc length of a curve is given by the formula:

L = \int_{a}^{b} \left\| r^{\prime}(t) \right\| dt

where $r^{\prime}(t)$ is the derivative of the curve with respect to the parameter $t$ , and $\left\| r^{\prime}(t) \right\|$ is the magnitude of the derivative.

Q: Can I use the arc length parametrization to find the curvature of a curve?

A: Yes, you can use the arc length parametrization to find the curvature of a curve. The curvature of a curve is given by the formula:

\kappa = \frac{\left\| r^{\prime}(t) \times r^{\prime \prime}(t) \right\|}{\left\| r^{\prime}(t) \right\|^3}

where $r^{\prime}(t)$ and $r^{\prime \prime}(t)$ are the first and second derivatives of the curve with respect to the parameter $t$ , respectively.

Q: Can I use the arc length parametrization to find the torsion of a curve?

A: Yes, you can use the arc length parametrization to find the torsion of a curve. The torsion of a curve is given by the formula:

\tau = \frac{\left\| r^{\prime}(t) \times r^{\prime \prime}(t) \times r^{\prime \prime \prime}(t) \right\|}{\left\| r^{\prime}(t) \times r^{\prime \prime}(t) \right\|^2}

where $r^{\prime}(t)$ , $r^{\prime \prime}(t)$ , and $r^{\prime \prime \prime}(t)$ are the first, second, and third derivatives of the curve with respect to the parameter $t$ , respectively.