Problem 38 - Calculus Early Transcendentals

Question

\(33-38\) Find the tangential and normal components of the acceleration vector. $$\mathbf{r}(t)=t \mathbf{i}+\cos ^{2} t \mathbf{j}+\sin ^{2} t \mathbf{k}$$

Step-by-Step Solution

Answer

Tangential: 0; Normal: \(2\sqrt{2}|\cos 2t|\).

Step 1: Find the Velocity Vector

The velocity vector \( \mathbf{v}(t) \) is the derivative of \( \mathbf{r}(t) \) with respect to \( t \). Calculate this derivative as follows:\[ \mathbf{v}(t) = \frac{d}{dt}(t \mathbf{i} + \cos^2 t \mathbf{j} + \sin^2 t \mathbf{k}) \]\[ \mathbf{v}(t) = \mathbf{i} - 2\cos t \sin t \mathbf{j} + 2\cos t \sin t \mathbf{k} \]Simplify using the identity \( \sin 2t = 2 \sin t \cos t \):\[ \mathbf{v}(t) = \mathbf{i} - \sin 2t \mathbf{j} + \sin 2t \mathbf{k} \]

Step 2: Compute the Magnitude of the Velocity Vector

The magnitude of \( \mathbf{v}(t) \) is the square root of the sum of the squares of its components:\[ \|\mathbf{v}(t)\| = \sqrt{1 + (-\sin 2t)^2 + (\sin 2t)^2} \ \|\mathbf{v}(t)\| = \sqrt{1 + \sin^2 2t + \sin^2 2t} \ \|\mathbf{v}(t)\| = \sqrt{1 + 2\sin^2 2t} \]

Step 3: Find the Unit Tangent Vector

The unit tangent vector \( \mathbf{T}(t) \) is \( \mathbf{v}(t) \) divided by its magnitude:\[ \mathbf{T}(t) = \frac{\mathbf{v}(t)}{\|\mathbf{v}(t)\|} = \frac{1}{\sqrt{1 + 2\sin^2 2t}}(\mathbf{i} - \sin 2t \mathbf{j} + \sin 2t \mathbf{k}) \]

Step 4: Calculate the Acceleration Vector

The acceleration vector \( \mathbf{a}(t) \) is the derivative of \( \mathbf{v}(t) \) with respect to \( t \):\[ \mathbf{a}(t) = \frac{d}{dt}(\mathbf{i} - \sin 2t \mathbf{j} + \sin 2t \mathbf{k}) \ \mathbf{a}(t) = -2\cos 2t \mathbf{j} + 2\cos 2t \mathbf{k} \]

Step 5: Find the Tangential Component of Acceleration

The tangential component of the acceleration vector \( a_T \) is the dot product of \( \mathbf{a}(t) \) and \( \mathbf{T}(t) \):\[ a_T = \mathbf{a}(t) \cdot \mathbf{T}(t) \ a_T = \left(-2\cos 2t \mathbf{j} + 2\cos 2t \mathbf{k}\right) \cdot \left(\frac{\mathbf{i} - \sin 2t \mathbf{j} + \sin 2t \mathbf{k}}{\sqrt{1 + 2\sin^2 2t}}\right) \ a_T = \frac{-2\cos 2t (-\sin 2t) + 2\cos 2t (\sin 2t)}{\sqrt{1 + 2\sin^2 2t}} \ a_T = 0 \]

Step 6: Find the Normal Component of Acceleration

The normal component \( a_N \) is derived from the formula:\[ a_N = \sqrt{\|\mathbf{a}(t)\|^2 - a_T^2} \] Calculate \( \|\mathbf{a}(t)\| \):\[ \|\mathbf{a}(t)\| = \sqrt{(-2\cos 2t)^2 + (2\cos 2t)^2} = \sqrt{8\cos^2 2t} = 2\sqrt{2}|\cos 2t| \]Since \( a_T = 0 \),\[ a_N = \|\mathbf{a}(t)\| = 2\sqrt{2}|\cos 2t| \]

Key Concepts

Velocity Vector

The velocity vector is a crucial concept in understanding motion. It represents the rate of change of position with respect to time and gives us insight into the direction and speed of an object. In mathematical terms, it is the derivative of the position vector.
To find the velocity vector from the given position vector \( \mathbf{r}(t) \), we differentiate \( \mathbf{r}(t) = t \mathbf{i} + \cos^2 t \mathbf{j} + \sin^2 t \mathbf{k} \) with respect to time \( t \).

• Calculating the derivative gives us \( \mathbf{v}(t) = \mathbf{i} - 2\cos t \sin t \mathbf{j} + 2\cos t \sin t \mathbf{k} \).
• We simplify it using the trigonometric identity \( \sin 2t = 2 \sin t \cos t \), resulting in \( \mathbf{v}(t) = \mathbf{i} - \sin 2t \mathbf{j} + \sin 2t \mathbf{k} \).

This vector not only encodes the direction of motion but also changes as the parameters of the curve change.

Unit Tangent Vector

The unit tangent vector is a normalized form of the velocity vector that indicates direction but not the magnitude. It tells us the precise direction in which the path is moving at any point in time.
To calculate the unit tangent vector \( \mathbf{T}(t) \), we divide the velocity vector \( \mathbf{v}(t) \) by its magnitude \( \|\mathbf{v}(t)\| \).

• The magnitude is computed as \( \|\mathbf{v}(t)\| = \sqrt{1 + 2\sin^2 2t} \).
• Subsequently, the unit tangent vector is \( \mathbf{T}(t) = \frac{1}{\sqrt{1 + 2\sin^2 2t}}(\mathbf{i} - \sin 2t \mathbf{j} + \sin 2t \mathbf{k}) \).

This vector is of great importance when breaking down different components of motion, as it simplifies computations involving tangential and normal directions.

Acceleration Vector

Acceleration is the rate of change of velocity with respect to time, providing insights into how the velocity of an object changes.
The acceleration vector \( \mathbf{a}(t) \) is calculated by differentiating the velocity vector \( \mathbf{v}(t) \) with respect to time \( t \).

• We have \( \mathbf{a}(t) = \frac{d}{dt}(\mathbf{i} - \sin 2t \mathbf{j} + \sin 2t \mathbf{k}) \).
• This further simplifies to \( \mathbf{a}(t) = -2\cos 2t \mathbf{j} + 2\cos 2t \mathbf{k} \).

This vector showcases how the speed and direction of the object change, and forms the basis for calculating the tangential and normal components of acceleration.

Tangential Component

The tangential component of acceleration indicates how much of the acceleration is along the direction of the velocity vector. It marks changes in the speed of the object.
It is computed as the dot product of the acceleration vector \( \mathbf{a}(t) \) and the unit tangent vector \( \mathbf{T}(t) \).

• The formula used is \( a_T = \mathbf{a}(t) \cdot \mathbf{T}(t) \).
• In this case, we find that \( a_T = 0 \), indicating no acceleration along the direction of motion.

Understanding this component is crucial as it portrays how swiftly an object is gaining or losing speed without changing its direction.

Normal Component

The normal component of acceleration represents the part of the acceleration that is perpendicular to the velocity vector, indicating a change in direction.
To calculate this, we start from the magnitude of the acceleration vector \( \|\mathbf{a}(t)\| \) and subtract the square of the tangential component.

• First, find the magnitude \( \|\mathbf{a}(t)\| = \sqrt{8\cos^2 2t} = 2\sqrt{2}|\cos 2t| \).
• Given that \( a_T = 0 \), the normal component is simply \( a_N = \|\mathbf{a}(t)\| = 2\sqrt{2}|\cos 2t| \).

This component helps in understanding how much the direction of an object's movement is changing, essential for complete motion analysis.