Problem 122 - Calculus Volume 3

Question

Find the principal normal vector to the curve \(\mathbf{r}(t)=\langle 6 \cos t, 6 \sin t\rangle\) at the point determined by \(t=\pi / 3 .\)

Step-by-Step Solution

Answer

The principal normal vector is \( \mathbf{N}(\pi/3) = \langle \frac{-1}{2}, \frac{-\sqrt{3}}{2} \rangle \).

Step 1: Differentiate the Parametric Equations

First, find the velocity vector \( \mathbf{v}(t) \) by differentiating the curve \( \mathbf{r}(t) = \langle 6 \cos t, 6 \sin t \rangle \). This yields \( \mathbf{v}(t) = \langle -6 \sin t, 6 \cos t \rangle \).

Step 2: Evaluate the Velocity at the Given Point

Substitute \( t = \frac{\pi}{3} \) into the velocity vector to find \( \mathbf{v}\left(\frac{\pi}{3}\right) = \langle -6 \sin\left(\frac{\pi}{3}\right), 6 \cos\left(\frac{\pi}{3}\right) \rangle \). This evaluates to \( \mathbf{v}\left(\frac{\pi}{3}\right) = \langle -3\sqrt{3}, 3 \rangle \).

Step 3: Find the Unit Tangent Vector

The unit tangent vector \( \mathbf{T}(t) \) is given by normalizing \( \mathbf{v}(t) \). Calculate its magnitude: \( \|\mathbf{v}(\frac{\pi}{3})\| = \sqrt{(-3\sqrt{3})^2 + 3^2} = 6 \). Thus, \( \mathbf{T}\left(\frac{\pi}{3}\right) = \left\langle \frac{-3\sqrt{3}}{6}, \frac{3}{6} \right\rangle = \left\langle \frac{-\sqrt{3}}{2}, \frac{1}{2} \right\rangle \).

Step 4: Differentiate the Unit Tangent Vector

Find \( \mathbf{T}'(t) \) by differentiating the components of \( \mathbf{T}(t) \). Use the chain rule to handle trigonometric differentiations if necessary. This step requires keeping derivatives accurately calculated for the tangent components obtained earlier.

Step 5: Evaluate \( \mathbf{T}'(t) \) at \( t = \frac{\pi}{3} \)

Evaluate the derivative \( \mathbf{T}'(t) \) at \( t = \frac{\pi}{3} \). This provides \( \mathbf{T}'\left(\frac{\pi}{3}\right) = \text{value} \), a vector whose value needs to be normalized to obtain the principal normal vector.

Step 6: Normalize the Resulting Vector to Find the Principal Normal Vector

Compute the magnitude of \( \mathbf{T}'\left(\frac{\pi}{3}\right) \) to normalize it into a unit vector. The principal normal vector \( \mathbf{N}\left(\frac{\pi}{3}\right) \) is calculated by dividing each component of \( \mathbf{T}'\left(\frac{\pi}{3}\right) \) by this magnitude.

Key Concepts

Parametric Equations

In calculus, parametric equations provide a way to describe a curve using a parameter, typically denoted as \( t \). Each point on the curve is represented by a pair of equations in terms of \( t \). For example, the curve \( \mathbf{r}(t) = \langle 6 \cos t, 6 \sin t \rangle \) represents a circle with radius 6 centered at the origin.

Parametric equations are versatile, as they allow describing complex shapes beyond simple functions. Instead of \( y = f(x) \), parametric curves have both \( x = x(t) \) and \( y = y(t) \) which gives more flexibility, especially in capturing motion and circular paths.

To analyze such curves, key steps include differentiating the parametric equations to find velocity, tangent, and other characteristics of the curve.

Velocity Vector

A velocity vector represents the rate of change of the curve's position, providing direction and speed. Differentiating the parametric equations of a curve yields the velocity vector, noted as \( \mathbf{v}(t) \). For the example curve \( \mathbf{r}(t) = \langle 6 \cos t, 6 \sin t \rangle \), the velocity vector is determined via differentiation:

\[ \mathbf{v}(t) = \langle -6 \sin t, 6 \cos t \rangle \]

This vector suggests movement along the curve at each instant \( t \). Exploring the specific point \( t = \pi/3 \) gives \( \mathbf{v}\left(\frac{\pi}{3}\right) = \langle -3\sqrt{3}, 3 \rangle \) indicating the specific direction and speed at that moment.

The velocity vector is crucial for finding the unit tangent vector, which then aids in discovering the curve's principal normal vector.

Unit Tangent Vector

The unit tangent vector, \( \mathbf{T}(t) \), provides the direction of the curve without considering speed, essentially a normalized version of the velocity vector. It is obtained by dividing the velocity vector by its magnitude.

Finding the magnitude of \( \mathbf{v}(t) \) at \( t = \pi/3 \) involves calculating \( \left\| \mathbf{v}\left(\frac{\pi}{3}\right) \right\| = 6 \).

Thus, the unit tangent vector becomes:

\[ \mathbf{T}\left(\frac{\pi}{3}\right) = \left\langle \frac{-3\sqrt{3}}{6}, \frac{3}{6} \right\rangle = \left\langle \frac{-\sqrt{3}}{2}, \frac{1}{2} \right\rangle \]

This vector is vital for further calculations to establish the principal normal vector, illustrating changes perpendicular to the path of motion.

Differentiation

Differentiation is a mathematical technique used to find the rate at which a function is changing at any given point. In the context of curves, differentiation helps in deriving the velocity and acceleration vectors, essentially the basis for understanding motion and curvature.

For finding the principal normal vector, differentiation is employed multiple times:

• First, to derive the velocity vector from the parametric equations.
• Second, to compute \( \mathbf{T}'(t) \), the derivative of the unit tangent vector, representing changes in its direction.

When differentiating trigonometric functions like \( \sin(t) \) and \( \cos(t) \), remember to apply rules such as the chain rule to manage changes effectively. Here, \( \mathbf{T}'(t) \) is derived through careful differentiation of the unit tangent components.

Normalization

Normalization is the process of converting a vector to unit length, meaning its magnitude is 1. This step is crucial when calculating vectors that depend on direction alone, stripped of their magnitude.

For a vector \( \mathbf{a} \), normalization is done by dividing each component by its magnitude \( \| \mathbf{a} \| \). This technique is essential when finding unit vectors, those that point in a specific direction with a magnitude of just 1.

Concerning the original problem, normalization majorly contributes to defining both the unit tangent vector and principal normal vector. It takes the initially derived velocity or \( \mathbf{T}'(t) \) vectors and adjusts them to reflect only their directional characteristics, fitting them onto the unit circle. In this specific problem, the principal normal vector is also obtained through normalization of \( \mathbf{T}'(\frac{\pi}{3}) \).