Question

Find \(\vec{N}(t)\). $$ \vec{r}(t)=\langle 4 t, 2 \sin t, 2 \cos t\rangle $$

Short answer

\( \vec{N}(t) = \langle 0, -\sin t, -\cos t \rangle \)

Step-by-step solution

Compute the First Derivative

First, we need to calculate the derivative \( \vec{r}'(t) \) of the vector function \( \vec{r}(t) = \langle 4t, 2 \sin t, 2 \cos t \rangle \). This gives us the tangent vector. The derivative is calculated component-wise: \( \vec{r}'(t) = \langle 4, 2 \cos t, -2 \sin t \rangle \).

Compute the Second Derivative

Next, we compute the second derivative, \( \vec{r}''(t) \), which will be used to find the rate of change of the tangent vector. The derivative of the tangent vector \( \vec{r}'(t) = \langle 4, 2 \cos t, -2 \sin t \rangle \) is: \( \vec{r}''(t) = \langle 0, -2 \sin t, -2 \cos t \rangle \).

Compute the Unit Tangent Vector \( \vec{T}(t) \)

The unit tangent vector \( \vec{T}(t) \) is found by normalizing \( \vec{r}'(t) \). First, find the magnitude of \( \vec{r}'(t) \): \( \| \vec{r}'(t) \| = \sqrt{4^2 + (2 \cos t)^2 + (-2 \sin t)^2} = \sqrt{16 + 4 \cos^2 t + 4 \sin^2 t} = \sqrt{20} \). Then, normalize \( \vec{r}'(t) \): \( \vec{T}(t) = \frac{1}{\sqrt{20}} \langle 4, 2 \cos t, -2 \sin t \rangle \).

Compute \( \vec{N}(t) \)

To find the normal vector \( \vec{N}(t) \), compute the derivative of the unit tangent vector \( \vec{T}(t) \). Let \( \vec{T}'(t) = \langle 0, -2 \sin t, -2 \cos t \rangle / \sqrt{20} \). The magnitude of \( \vec{T}'(t) \) is \( \frac{2}{\sqrt{20}} \). Normalize \( \vec{T}'(t) \) to find \( \vec{N}(t) \): \( \vec{N}(t) = \frac{1}{\|\vec{T}'(t)\|} \frac{1}{\sqrt{20}} \langle 0, -2 \sin t, -2 \cos t \rangle = \langle 0, -\sin t, -\cos t \rangle \).

Key concepts

Vector Derivatives

When working with vectors associated with curves, the first step is often to calculate the derivative. Think of it like finding the velocity of a moving object along a path. This is known as the *vector derivative*.

• **First Derivative**: The first derivative of a vector function gives us the *tangent vector*, which points in the direction of the curve as it moves through space. It is a vector that tells us how the curve is changing at any point. For example, given a vector function \( \vec{r}(t)= \langle 4t, 2 \sin t, 2 \cos t \rangle \), the first derivative is \( \vec{r}'(t) = \langle 4, 2 \cos t, -2 \sin t \rangle \).
• **Second Derivative**: The second derivative, \( \vec{r}''(t) \), provides information about the rate of change of the tangent vector itself. In other words, it gives us a sense of how the curvature of the path is changing, which is crucial for understanding acceleration if we think of the path as a motion trajectory. In our example, \( \vec{r}''(t) = \langle 0, -2 \sin t, -2 \cos t \rangle \).

These derivatives are essential to understanding the geometry of the path traced out by the vector function.

Unit Tangent Vector

Once we have the first derivative, we can find the *unit tangent vector* by normalizing the tangent vector. This converts our tangent vector to a unit vector, which is a vector with a magnitude of 1.

• **Normalization**: To normalize a vector, divide it by its magnitude. This keeps the direction, but scales its length to 1. For the vector \( \vec{r}'(t) = \langle 4, 2 \cos t, -2 \sin t \rangle \), the magnitude, \( \| \vec{r}'(t) \| \), is \( \sqrt{20} \). Thus, the unit tangent vector \( \vec{T}(t) \) is \( \frac{1}{\sqrt{20}} \langle 4, 2 \cos t, -2 \sin t \rangle \).

The unit tangent vector provides a standard, uniform way to discuss the direction of the curve at any point. It is crucial in many fields, including physics and engineering, where understanding direction and orientation is fundamental.

Curve Analysis

To fully analyze a curve, we need more than just direction; we also consider the curve's curvature and how it bends. This leads us to the *normal vector* and the process of curve analysis.

• **Normal Vector**: After computing the unit tangent vector, the next step is to derive the normal vector, \( \vec{N}(t) \). This vector is perpendicular to \( \vec{T}(t) \) and provides insight into how the curve is bending at a given point. From our problem, to find \( \vec{N}(t) \), we can take the derivative of \( \vec{T}(t) \) and normalize it. This results in the vector \( \langle 0, -\sin t, -\cos t \rangle \).
• **Applications**: Understanding \( \vec{N}(t) \) and the curvature of a curve helps in various applications such as designing roads, bridges, and even guiding animated paths in computer graphics.

Overall, curve analysis provides a complete picture of the behavior of a curve in space, equipping us with valuable tools for both conceptual understanding and practical applications.