Problem 189 - Calculus Volume 3

Question

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

Step-by-Step Solution

Answer

Find velocity and acceleration vectors, then compute their tangential and normal projections.

Step 1: Find the First Derivative of the Position Vector

The position vector given is \( \mathbf{r}(t) = \langle 2t, t^2, \frac{t^3}{3} \rangle \). To find the velocity vector \( \mathbf{v}(t) \), take the derivative of \( \mathbf{r}(t) \) with respect to \( t \). The derivative is \( \mathbf{v}(t) = \langle 2, 2t, t^2 \rangle \).

Step 2: Compute the Magnitude of the Velocity Vector

The magnitude \( \| \mathbf{v}(t) \| \) of the velocity vector \( \langle 2, 2t, t^2 \rangle \) is computed as: \[ \| \mathbf{v}(t) \| = \sqrt{2^2 + (2t)^2 + (t^2)^2} = \sqrt{4 + 4t^2 + t^4} = \sqrt{(t^2 + 2)^2} = t^2 + 2. \]

Step 3: Find the Unit Tangent Vector

The unit tangent vector \( \mathbf{T}(t) \) is calculated by dividing the velocity vector by its magnitude: \( \mathbf{T}(t) = \frac{\mathbf{v}(t)}{\|\mathbf{v}(t)\|} = \langle \frac{2}{t^2 + 2}, \frac{2t}{t^2 + 2}, \frac{t^2}{t^2 + 2} \rangle \).

Step 4: Compute the Second Derivative of the Position Vector

The acceleration vector \( \mathbf{a}(t) \) is the derivative of \( \mathbf{v}(t) \). Thus, \( \mathbf{a}(t) = \langle 0, 2, 2t \rangle \).

Step 5: Project Acceleration onto the Tangent Unit Vector

The tangential component of acceleration \( a_T \) is found by projecting \( \mathbf{a}(t) \) onto \( \mathbf{T}(t) \). This is computed as: \[ a_T = \mathbf{a}(t) \cdot \mathbf{T}(t) = \langle 0, 2, 2t \rangle \cdot \langle \frac{2}{t^2 + 2}, \frac{2t}{t^2 + 2}, \frac{t^2}{t^2 + 2} \rangle = \frac{4t}{t^2 + 2} + \frac{2t^3}{t^2 + 2}. \]

Step 6: Calculate the Normal Component of Acceleration

The normal component \( a_N \) is given by \( a_N = \sqrt{\| \mathbf{a}(t) \|^2 - a_T^2} \). First compute \( \| \mathbf{a}(t) \| = \sqrt{0^2 + 2^2 + (2t)^2} = \sqrt{4 + 4t^2} = 2\sqrt{1 + t^2} \). Find \( a_N = \sqrt{4(1+t^2) - \left(\frac{4t + 2t^3}{t^2 + 2}\right)^2} \) after determining \( a_T^2 \).

Key Concepts

Unit Tangent Vector

In calculus, the unit tangent vector is a crucial concept when analyzing the motion along a path. It provides a direction tangent to the curve at any given point and is normalized to have a unit length. For the position vector \( \mathbf{r}(t) = \langle 2t, t^2, \frac{t^3}{3} \rangle \), the velocity vector \( \mathbf{v}(t) = \langle 2, 2t, t^2 \rangle \) gives the direction of motion. By dividing \( \mathbf{v}(t) \) by its magnitude, we find the unit tangent vector \( \mathbf{T}(t) \). This process ensures that the resulting vector has a magnitude of 1, while maintaining the direction of \( \mathbf{v}(t) \). This concept helps in determining how fast the position vector changes direction.

The formula for the unit tangent vector is given by:

• \( \mathbf{T}(t) = \frac{\mathbf{v}(t)}{\|\mathbf{v}(t)\|} \)
• Where \( \|\mathbf{v}(t) \| \) is the magnitude of the velocity vector.
• \( \mathbf{T}(t) = \langle \frac{2}{t^2 + 2}, \frac{2t}{t^2 + 2}, \frac{t^2}{t^2 + 2} \rangle \)

Understanding this helps in analyzing curves more dynamically, predicting movement directions, and determining how variables change concerning time.

Velocity Vector

The velocity vector represents the rate of change of position with respect to time. It essentially provides both the direction and the speed at which an object moves along a path. Calculus allows us to compute the velocity vector by differentiating the position vector \( \mathbf{r}(t) \). For this exercise, the position vector was \( \mathbf{r}(t) = \langle 2t, t^2, \frac{t^3}{3} \rangle \). Its derivative, the velocity vector, is \( \mathbf{v}(t) = \langle 2, 2t, t^2 \rangle \). It tells us how each component \( x, y, z \) of the position vector changes over time.

The velocity vector explains:

• The speed of the object per unit time.
• The specific direction in a given instant.

By examining the velocity vector, one can predict both how fast and in which direction the object is traveling. This is pivotal for understanding the dynamics of a moving particle or body.

Acceleration Vector

The acceleration vector expresses how the velocity of an object changes with respect to time. This change can be due to either a change in speed or a change in direction, or both. Obtaining the acceleration vector requires differentiating the velocity vector \( \mathbf{v}(t) \). For this exercise, with \( \mathbf{v}(t) = \langle 2, 2t, t^2 \rangle \), the acceleration vector becomes \( \mathbf{a}(t) = \langle 0, 2, 2t \rangle \).

The acceleration vector provides insight into:

• How quickly speed or direction changes over time.
• Whether the motion is speeding up or slowing down.
• Curve behavior, influenced by external forces.

Understanding acceleration helps predict and control motion in physics, engineering, and various real-world applications.

Tangential Component of Acceleration

The tangential component of acceleration indicates the rate of change of speed of an object moving along a path. It is the component of the acceleration that aligns with the unit tangent vector. Calculating this component entails projecting the acceleration vector \( \mathbf{a}(t) \) onto the unit tangent vector \( \mathbf{T}(t) \).

The steps include:

• Finding the dot product between \( \mathbf{a}(t) \) and \( \mathbf{T}(t) \).
• The formula \( a_T = \mathbf{a}(t) \cdot \mathbf{T}(t) \).
• \( a_T = \frac{4t}{t^2 + 2} + \frac{2t^3}{t^2 + 2} \)

This component helps understand whether an object speeds up or slows down along its path. It's a critical concept in mechanics where forces often align or oppose motion direction.

Normal Component of Acceleration

The normal component of acceleration represents the change in direction of the velocity vector, perpendicular to the path. This component doesn't affect the speed but is crucial for understanding how the path curves. It can be found after determining the tangential component.

Steps to calculate include:

• Calculate the magnitude of the acceleration vector, \( \|\mathbf{a}(t)\| \).
• Use the formula: \( a_N = \sqrt{\|\mathbf{a}(t)\|^2 - a_T^2} \).
• In the exercise, \( a_N = \sqrt{4(1+t^2) - \left(\frac{4t + 2t^3}{t^2 + 2}\right)^2} \).

This vector component is essential for designing paths in situations where the direction is critical, such as road curves or amusement park rides.