Problem 191 - Calculus Volume 3

Question

Find the tangential and normal components of acceleration. \(\mathbf{r}(t)=\left\langle 6 t, 3 t^{2}, 2 t^{3}\right\rangle\)

Step-by-Step Solution

Answer

Tangential component: \( \frac{6t}{\sqrt{1 + t^2 + t^4}} \), Normal component: \( \sqrt{36(1 + 4t^2) - \frac{36t^2}{1 + t^2 + t^4}} \).\)

Step 1: Compute the velocity vector

The velocity vector is the derivative of the position vector, \( \mathbf{v}(t) = \frac{d\mathbf{r}}{dt} \). Differentiate \( \mathbf{r}(t) = \langle 6t, 3t^2, 2t^3 \rangle \) to find \( \mathbf{v}(t) = \langle 6, 6t, 6t^2 \rangle \).

Step 2: Compute the acceleration vector

The acceleration vector is the derivative of the velocity vector. Differentiate \( \mathbf{v}(t) = \langle 6, 6t, 6t^2 \rangle \) to find \( \mathbf{a}(t) = \langle 0, 6, 12t \rangle \).

Step 3: Calculate the magnitude of the velocity vector

The magnitude of the velocity vector \( \mathbf{v}(t) = \langle 6, 6t, 6t^2 \rangle \) is calculated by using the formula \( ||\mathbf{v}(t)|| = \sqrt{6^2 + (6t)^2 + (6t^2)^2} = \sqrt{36 + 36t^2 + 36t^4} \).

Step 4: Simplify the expression for the magnitude of velocity

Factor out the common factor in each term: \( ||\mathbf{v}(t)|| = \sqrt{36(1 + t^2 + t^4)} = 6\sqrt{1 + t^2 + t^4} \).

Step 5: Find the tangential component of acceleration

The tangential component of acceleration \( a_T \) is given by the formula \( a_T = \frac{\mathbf{a}(t) \cdot \mathbf{v}(t)}{||\mathbf{v}(t)||} \). First calculate the dot product \( \mathbf{a}(t) \cdot \mathbf{v}(t) = \langle 0, 6, 12t \rangle \cdot \langle 6, 6t, 6t^2 \rangle = 0 + 36t + 72t^3 = 36t + 72t^3 \). Then substitute: \( a_T = \frac{36t + 72t^3}{6\sqrt{1 + t^2 + t^4}} = \frac{6t(6 + 12t^2)}{6\sqrt{1 + t^2 + t^4}} = \frac{6t(6 + 12t^2)}{6\sqrt{1 + t^2 + t^4}} = \frac{6t(6 + 12t^2)}{6\sqrt{1 + t^2 + t^4}} = \frac{6t}{\sqrt{1 + t^2 + t^4}} \).

Step 6: Find the normal component of acceleration

The normal component of acceleration \( a_N \) is found using the formula \( a_N = \sqrt{||\mathbf{a}(t)||^2 - a_T^2} \). Begin by finding \( ||\mathbf{a}(t)|| = \sqrt{0^2 + 6^2 + (12t)^2} = \sqrt{36 + 144t^2} = 6\sqrt{1 + 4t^2} \). Then, substitute the previously found \( a_T \) into the formula: \( a_N = \sqrt{(6\sqrt{1 + 4t^2})^2 - \left(\frac{6t}{\sqrt{1 + t^2 + t^4}}\right)^2} = \sqrt{36(1 + 4t^2) - \frac{36t^2}{1 + t^2 + t^4}} \).

Key Concepts

Position Vector

The position vector, often denoted as \( \mathbf{r}(t) \), represents the location of a point in space as a function of time. In this exercise, the position vector is given by the equation \( \mathbf{r}(t) = \left\langle 6t, 3t^2, 2t^3 \right\rangle \).
This form gives us the coordinates in a three-dimensional space over time:

• The first component, \( 6t \), describes movement along the x-axis.
• The second component, \( 3t^2 \), describes movement along the y-axis.
• The third component, \( 2t^3 \), describes movement along the z-axis.

This vector helps us understand the path of an object in motion by mapping its precise location at any given time \( t \). Understanding the position vector is crucial, as it sets the foundation for determining velocity and acceleration vectors.

Velocity Vector

The velocity vector is derived from the position vector by taking its derivative with respect to time. This gives us the rate of change of position at any given moment, essentially showing us how fast and in what direction the object is moving.
In our exercise, the velocity vector \( \mathbf{v}(t) \) is calculated as:

• First, differentiate each component of the position vector \( \mathbf{r}(t) = \langle 6t, 3t^2, 2t^3 \rangle \).
• This leads to \( \mathbf{v}(t) = \langle 6, 6t, 6t^2 \rangle \).

The resulting velocity vector indicates:

• A constant speed of 6 units along the x-axis.
• An increasing speed over time along the y-axis, shown by the term \( 6t \).
• Accelerated movement along the z-axis, indicated by \( 6t^2 \).

By studying the velocity vector, you gain insight into the dynamics of motion.

Acceleration Vector

The acceleration vector measures how the velocity changes over time and is found by differentiating the velocity vector. This is key when studying the forces influencing the object's motion.
In this exercise, the acceleration vector \( \mathbf{a}(t) \) is obtained by:

• Differentiating the velocity vector \( \mathbf{v}(t) = \langle 6, 6t, 6t^2 \rangle \).
• This results in \( \mathbf{a}(t) = \langle 0, 6, 12t \rangle \).

Here's what the acceleration vector tells us:

• No acceleration along the x-axis, indicating a consistent linear motion.
• A consistent acceleration along the y-axis, represented by the component \( 6 \).
• An acceleration that grows with time along the z-axis, \( 12t \), suggesting the influence of increasing forces.

Understanding the acceleration vector is crucial for predicting how the object's motion will evolve over time.

Magnitude of Velocity

The magnitude of the velocity vector represents the speed of the object, or how fast it's moving regardless of direction. It's computed by taking the Euclidean norm of the velocity vector.
For the given velocity vector \( \mathbf{v}(t) = \langle 6, 6t, 6t^2 \rangle \), the magnitude is calculated as:

• Use the formula \( ||\mathbf{v}(t)|| = \sqrt{6^2 + (6t)^2 + (6t^2)^2} \).
• This simplifies to \( 6\sqrt{1 + t^2 + t^4} \).

The magnitude of the velocity helps quantify:

• The total speed the object is moving at any time \( t \).
• How speed increases as functions of \( t^2 \) and \( t^4 \), demonstrating accelerating trends.

Having the magnitude gives a scalar measure of speed, simplifying complex motion analysis.

Dot Product

The dot product is a fundamental operation in vector calculus, serving as a tool for finding the angle between vectors and measuring vector projections. In this exercise, it specifically helps find the tangential component of acceleration.
The dot product of the acceleration vector \( \mathbf{a}(t) = \langle 0, 6, 12t \rangle \) and the velocity vector \( \mathbf{v}(t) = \langle 6, 6t, 6t^2 \rangle \) is calculated as:

• \( \mathbf{a}(t) \cdot \mathbf{v}(t) = 0 \cdot 6 + 6 \cdot 6t + 12t \cdot 6t^2 \).
• This simplifies to \( 36t + 72t^3 \).

The dot product is essential for:

• Determining the component of acceleration in the direction of the velocity vector.
• Understanding how much of the acceleration contributes to speed changes in the direction of travel.

The result from the dot product is used in further calculations for both tangential and normal accelerations, helping in-depth analysis of motion dynamics.