Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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Question
### Arc Length Parametrization of a Vector Function **Problem Statement:** Find an arc length parametrization of \(\mathbf{r}(t) = \left( e^t \sin(t), e^t \cos(t), 2e^t \right)\). **Solution:** The arc length parametrization \(\mathbf{r}_1(s)\) is given by the following expression: \[ \mathbf{r}_1(s) = \left( \ln \left( 1 + \frac{s}{\sqrt{6}} \right) \sin \left( \ln \left( 1 + \frac{s}{\sqrt{6}} \right) \right), \cos \left( \ln \left( 1 + \frac{s}{\sqrt{6}} \right) \right), 2 \right) \] **Explanation:** Here, \(\mathbf{r}(t)\) describes a vector function with three components. The goal is to find its arc length parametrization, which essentially reparametrizes the curve in terms of the arc length \(s\), making \(s\) the new parameter that measures distance along the curve. **Detailed Breakdown of Steps (Illustrative):** 1. **Parameterize the Curve:** Given the original function \(\mathbf{r}(t)\), calculate the magnitude of the derivative, \(\left| \mathbf{r}'(t) \right|\), which represents the rate of change of the vector function's length. 2. **Integral Calculation:** Integrate the magnitude of the derivative to find the arc length as a function of \(t\), denoted by \(s(t)\). 3. **Invert the Relation:** Solve for \(t\) in terms of the arc length \(s\) to get the inverse function, \(t(s)\). 4. **Reparametrize the Function:** Substitute \(t(s)\) back into the original vector function to obtain the arc length parametrization \(\mathbf{r}_1(s)\). The resulting parametrization ensures that the parameter \(s\) directly corresponds to the arc length measured along the curve from a fixed starting point. This reparametrization simplifies computations involving the length of paths and is particularly useful in applications concerning physical distances and curvatures.
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Transcribed Image Text:### Arc Length Parametrization of a Vector Function **Problem Statement:** Find an arc length parametrization of \(\mathbf{r}(t) = \left( e^t \sin(t), e^t \cos(t), 2e^t \right)\). **Solution:** The arc length parametrization \(\mathbf{r}_1(s)\) is given by the following expression: \[ \mathbf{r}_1(s) = \left( \ln \left( 1 + \frac{s}{\sqrt{6}} \right) \sin \left( \ln \left( 1 + \frac{s}{\sqrt{6}} \right) \right), \cos \left( \ln \left( 1 + \frac{s}{\sqrt{6}} \right) \right), 2 \right) \] **Explanation:** Here, \(\mathbf{r}(t)\) describes a vector function with three components. The goal is to find its arc length parametrization, which essentially reparametrizes the curve in terms of the arc length \(s\), making \(s\) the new parameter that measures distance along the curve. **Detailed Breakdown of Steps (Illustrative):** 1. **Parameterize the Curve:** Given the original function \(\mathbf{r}(t)\), calculate the magnitude of the derivative, \(\left| \mathbf{r}'(t) \right|\), which represents the rate of change of the vector function's length. 2. **Integral Calculation:** Integrate the magnitude of the derivative to find the arc length as a function of \(t\), denoted by \(s(t)\). 3. **Invert the Relation:** Solve for \(t\) in terms of the arc length \(s\) to get the inverse function, \(t(s)\). 4. **Reparametrize the Function:** Substitute \(t(s)\) back into the original vector function to obtain the arc length parametrization \(\mathbf{r}_1(s)\). The resulting parametrization ensures that the parameter \(s\) directly corresponds to the arc length measured along the curve from a fixed starting point. This reparametrization simplifies computations involving the length of paths and is particularly useful in applications concerning physical distances and curvatures.
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Step 1: Calculate the derivatives:
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Step 2: Calculate the arc length s:
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Step 3: Calculate t in terms of s:
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Step 4: Conclude for the required parametrization:
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