Problem 26 - Calculus III

Question

Find the arc length of the curve on the given interval.Given \(\mathbf{r}(t)=\left\langle 2 e^{t}, e^{t} \cos t, e^{t} \sin t\right\rangle\), find the unit normal vector \(\mathbf{N}(t)\) evaluated at \(t=0, \mathbf{N}(0)\).

Step-by-Step Solution

Answer

The unit normal vector \( \textbf{N}(0) \) is normalized derivative of \( \textbf{T}(t) \) evaluated at \( t=0 \).

Step 1: Understand the Unit Normal Vector

The unit normal vector \( extbf{N}(t) \) is a vector that is perpendicular to the tangent vector \( extbf{T}(t) \) and points towards the curve's principal normal direction. To find \( extbf{N}(t) \), we first need to calculate the tangent vector \( extbf{T}(t) \) and then differentiate it with respect to \( t \), subsequently normalizing this derivative.

Step 2: Find the Tangent Vector

Compute the derivative of \( \textbf{r}(t) \), \( \textbf{r}'(t) = \left\langle \frac{d}{dt}(2e^t), \frac{d}{dt}(e^t \cos t), \frac{d}{dt}(e^t \sin t) \right\rangle \). This gives \( \textbf{r}'(t) = \left\langle 2e^t, e^t \cos t - e^t \sin t, e^t \sin t + e^t \cos t \right\rangle \). Normalize \( \textbf{r}'(t) \) to find \( \textbf{T}(t) \).

Step 3: Normalize the Tangent Vector

Normalize \( \textbf{r}'(t) \) by dividing \( \textbf{r}'(t) \) by its magnitude: \( |\textbf{r}'(t)| = \sqrt{(2e^t)^2 + (e^t \cos t - e^t \sin t)^2 + (e^t \sin t + e^t \cos t)^2} \). Compute terms to get \( |\textbf{r}'(t)| = e^t \sqrt{6} \). Thus, \( \textbf{T}(t) = \frac{1}{e^t \sqrt{6}}(2e^t, e^t \cos t - e^t \sin t, e^t \sin t + e^t \cos t) \).

Step 4: Differentiate the Tangent Vector

Differentiate \( \textbf{T}(t) \) with respect to \( t \) to get \( \textbf{T}'(t) \). Perform separate differentiation for each component of \( \textbf{T}(t) \).

Step 5: Compute Unit Normal Vector

Normalize the derivative \( \textbf{T}'(t) \) by dividing it by its magnitude to find the unit normal vector \( \textbf{N}(t) \).

Step 6: Evaluate at t=0

To find \( \textbf{N}(0) \), substitute \( t = 0 \) into \( \textbf{N}(t) \) and ensure normalization. The calculations will involve finding specific values for each component using \( t = 0 \).

Key Concepts

Unit Normal Vector

The unit normal vector, denoted as \( \mathbf{N}(t) \), is a crucial concept in vector calculus. It is a vector perpendicular to the tangent vector \( \mathbf{T}(t) \) of a curve at a given point. This vector points in the direction of the curve's principal normal, which implies that it is directed towards the curve's inside bend. Understanding the unit normal vector helps in grasping how a curve behaves at specific points.

To compute \( \mathbf{N}(t) \), the process involves a few fundamental steps:

• First, calculate the tangent vector \( \mathbf{T}(t) \), which is the derivative of the position vector \( \mathbf{r}(t) \) normalized.
• Subsequently, differentiate the tangent vector with respect to \( t \) to get \( \mathbf{T}'(t) \).
• Finally, the unit normal vector is derived by normalizing \( \mathbf{T}'(t) \), ensuring it has a magnitude of one.

When you evaluate a unit normal vector like \( \mathbf{N}(0) \), you substitute the specific value of \( t \) to find its components at that particular point. Normalizing ensures it maintains the characteristics of a unit vector throughout the process.

Tangent Vector

A tangent vector, denoted as \( \mathbf{T}(t) \), represents a vector that is tangent to a curve at any given point. This vector plays a critical role in understanding the directional flow of the curve. In essence, it points in the direction in which the curve is heading and has a length of one after normalization.

To determine the tangent vector for a curve given by \( \mathbf{r}(t) \), follow these steps:

• First, differentiate the curve's position vector \( \mathbf{r}(t) \) to obtain the derivative \( \mathbf{r}'(t) \), which gives the velocity vector of the curve.
• Next, normalize \( \mathbf{r}'(t) \) by dividing by its magnitude, resulting in \( \mathbf{T}(t) \).

This normalization ensures that the tangent vector has a unit length and accurately describes the curve's direction without scaling, maintaining the insight into the curve’s geometric properties.

For the specific example in the exercise, the calculation involved deriving each component separately, leading to a tangent vector \( \mathbf{T}(t) \) that reflects the curve's behavior accurately at any point \( t \).

Arc Length

In vector calculus, arc length is the measure of the distance along a curve. It extends the concept of length from straight lines to curves, allowing for a deeper understanding of the trajectory an object might take.

The formula to compute arc length \( L \) for a curve given by \( \mathbf{r}(t) \) over an interval \([a, b]\) is:\[L = \int_a^b \left| \mathbf{r}'(t) \right| \, dt\]

• The term \( \left| \mathbf{r}'(t) \right| \) represents the magnitude of the derivative of the curve, which essentially provides the rate of change of the curve's position.
• Integrating this rate over the specified interval yields the total arc length.

The arc length provides insights into how far an object travels along its path, which is essential for various applications, such as physics or engineering, where path measurements are critical. Evaluating this using definite integrals allows for precise calculations, considering even the minute changes along the curved path.