Problem 22 - Calculus III

Question

Find the arc length of the curve on the given interval.The position function for a particle is \(\mathbf{r}(t)=a \cos (\omega t) \mathbf{i}+b \sin (\omega t) \mathbf{j}\). Find the unit tangent vector and the unit normal vector at \(t=0 .\)

Step-by-Step Solution

Answer

The unit tangent vector at \(t = 0\) is \(\mathbf{T}(0) = \mathbf{j}\), and the unit normal vector is \(\mathbf{N}(0) = \mathbf{i}\).

Step 1: Find the Derivative of the Position Function

To find the unit tangent vector, we first need the derivative of the position vector \( \mathbf{r}(t) = a \cos(\omega t) \mathbf{i} + b \sin(\omega t) \mathbf{j} \). Using differentiation, we find:\[ \mathbf{r}'(t) = -a \omega \sin(\omega t) \mathbf{i} + b \omega \cos(\omega t) \mathbf{j} \]

Step 2: Calculate the Magnitude of the Derivative

Next, calculate the magnitude of \( \mathbf{r}'(t) \):\[ \left| \mathbf{r}'(t) \right| = \sqrt{(-a \omega \sin(\omega t))^2 + (b \omega \cos(\omega t))^2} = \omega \sqrt{a^2 \sin^2(\omega t) + b^2 \cos^2(\omega t)} \]

Step 3: Evaluate the Magnitude at t = 0

Plug \( t = 0 \) into the expression for \( \left| \mathbf{r}'(t) \right| \):\[ \left| \mathbf{r}'(0) \right| = \omega \sqrt{a^2 \cdot 0 + b^2 \cdot 1} = \omega |b| \]

Step 4: Find the Unit Tangent Vector

The unit tangent vector \( \mathbf{T}(t) \) is given by dividing the derivative vector by its magnitude:\[ \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\left| \mathbf{r}'(t) \right|} \]For \( t = 0 \):\[ \mathbf{T}(0) = \frac{-a \omega \sin(0) \mathbf{i} + b \omega \cos(0) \mathbf{j}}{\omega |b|} = \frac{b \omega \mathbf{j}}{\omega |b|} = \frac{b}{|b|} \mathbf{j} \]

Step 5: Calculate the Unit Normal Vector

The unit normal vector \(\mathbf{N}(t)\) is orthogonal to the unit tangent vector, and since \(\mathbf{T}(t)\) points along \(\mathbf{j}\), the normal vector rotates \(\mathbf{T}(t)\) 90 degrees in the plane:\[ \mathbf{N}(0) = \mathbf{i} \]

Key Concepts

Unit Tangent Vector

In the context of calculus and vector analysis, the unit tangent vector plays a crucial role in understanding the direction of curves. It is a vector of unit length that points in the direction of the curve's motion at any given point.
To find the unit tangent vector, you begin by differentiating the position function. This differentiation gives you the velocity vector, which indicates direction but is not of unit length. By dividing the velocity vector by its magnitude, or length, you get the unit tangent vector.

• First, calculate: \[\mathbf{r}'(t) = -a \omega \sin(\omega t) \mathbf{i} + b \omega \cos(\omega t) \mathbf{j}\]
• Next, find the magnitude of this derivative: \[\left| \mathbf{r}'(t) \right| = \omega \sqrt{a^2 \sin^2(\omega t) + b^2 \cos^2(\omega t)}\]
• Then, for the unit tangent vector: \[ \mathbf{T}(0) = \frac{b}{|b|} \mathbf{j}\]

This process helps visualize how the curve behaves and directs your understanding of the motion happening along it.

Unit Normal Vector

Once you have the unit tangent vector, the next step is to find the unit normal vector. The unit normal vector is perpendicular to the unit tangent vector, helping you understand the curve's instantaneous direction change.
When working with parametrized curves, the unit normal vector is essentially a 90-degree rotation of the tangent vector within the given plane. Consequently, it's particularly helpful in analyzing the curve's shape and its rotation at any given position.

• The unit normal vector for this particular function at \(t = 0\) is calculated as: \[\mathbf{N}(0) = \mathbf{i}\]

Understanding these vector relations allows you to deeply appreciate the geometry and physics involved when analyzing the motion described by equations.

Position Function

A position function in the realm of physics and mathematics describes the location of a particle in space over time. In this case, it is expressed as a vector function:
\[\mathbf{r}(t)=a \cos (\omega t) \mathbf{i}+b \sin (\omega t) \mathbf{j}\]
This function allows you to pinpoint exactly where a particle is situated in a 2D plane at any given time \(t\). Here’s the breakdown of its components:

• \(a \cos(\omega t) \mathbf{i}\): Represents the horizontal component influenced by amplitude \(a\) and oscillation frequency \(\omega\).
• \(b \sin(\omega t) \mathbf{j}\): Represents the vertical component, with amplitude \(b\) and frequency \(\omega\).

This description effectively maps out the trajectory of a particle following a circular or elliptical path, crucial in various physics and engineering applications.

Vector Differentiation

Vector differentiation is a fundamental operation within calculus, especially when dealing with vector functions. It involves computing the derivative of a vector function with respect to time or another variable.
Differentiation of vector functions, like \( \mathbf{r}(t)= a \cos(\omega t)\mathbf{i} + b \sin(\omega t)\mathbf{j} \), requires you to differentiate each component separately:

• For \(a \cos(\omega t)\), differentiation yields \(-a \omega \sin(\omega t) \mathbf{i}\).
• For \(b \sin(\omega t)\), differentiation gives \(b \omega \cos(\omega t) \mathbf{j}\).

Through vector differentiation, you gain insights ranging from the derivative or rate of change of position, known typically as the velocity vector, to higher-order derivatives that can reveal further intricate details about the motion dynamics.