Problem 39 - Calculus III

Question

For each of the following problems, find the tangential and normal components of acceleration. $$ \mathbf{r}(t)=\left\langle\frac{2}{3}(1+t)^{3 / 2}, \frac{2}{3}(1-t)^{3 / 2}, \sqrt{2} t\right\rangle $$

Step-by-Step Solution

Answer

The tangential component of acceleration is 0, and the normal component depends on \( t \) and requires further simplification.

Step 1: Calculate the Velocity Vector

The velocity vector, \( \mathbf{v}(t) \), is the derivative of the position vector \( \mathbf{r}(t) \) with respect to time \( t \). Differentiate each component of \( \mathbf{r}(t) \): - For the first component, differentiate \( \frac{2}{3}(1+t)^{3/2} \) to get \( (1+t)^{1/2} \). - For the second component, differentiate \( \frac{2}{3}(1-t)^{3/2} \) to get \( -(1-t)^{1/2} \). - For the third component, differentiate \( \sqrt{2}t \) to get \( \sqrt{2} \). Thus, \( \mathbf{v}(t) = \langle (1+t)^{1/2}, -(1-t)^{1/2}, \sqrt{2} \rangle \).

Step 2: Calculate the Acceleration Vector

The acceleration vector, \( \mathbf{a}(t) \), is the derivative of the velocity vector \( \mathbf{v}(t) \) with respect to time \( t \). Differentiate each component of \( \mathbf{v}(t) \): - For the first component, differentiate \( (1+t)^{1/2} \) to get \( \frac{1}{2(1+t)^{1/2}} \). - For the second component, differentiate \( -(1-t)^{1/2} \) to get \( \frac{1}{2(1-t)^{1/2}} \). - The third component is constant \( \sqrt{2} \), so its derivative is \( 0 \). Thus, \( \mathbf{a}(t) = \langle \frac{1}{2(1+t)^{1/2}}, \frac{1}{2(1-t)^{1/2}}, 0 \rangle \).

Step 3: Compute the Speed

The speed is the magnitude of the velocity vector \( \mathbf{v}(t) \). Calculate the magnitude: \[ \| \mathbf{v}(t) \| = \sqrt{((1+t)^{1/2})^2 + (-(1-t)^{1/2})^2 + (\sqrt{2})^2} = \sqrt{(1+t) + (1-t) + 2} = \sqrt{4} = 2. \]

Step 4: Find the Tangential Component of Acceleration

The tangential component of acceleration, \( a_T \), is given by \( a_T = \frac{d}{dt}(\| \mathbf{v}(t) \|) \). Since the speed \( \| \mathbf{v}(t) \| = 2 \) is constant, its derivative with respect to time is zero. Therefore, \( a_T = 0 \).

Step 5: Find the Normal Component of Acceleration

The normal component of acceleration, \( a_N \), is found using the formula: \[ a_N = \sqrt{\| \mathbf{a}(t) \|^2 - a_T^2}. \] Since \( a_T = 0 \), this simplifies to \[ a_N = \| \mathbf{a}(t) \|. \] Calculate \( \| \mathbf{a}(t) \| \) as follows: \[ \| \mathbf{a}(t) \| = \sqrt{\left(\frac{1}{2(1+t)^{1/2}}\right)^2 + \left(\frac{1}{2(1-t)^{1/2}}\right)^2 + 0} = \sqrt{\frac{1}{4(1+t)} + \frac{1}{4(1-t)}}. \]

Key Concepts

Tangential Component of Acceleration

The tangential component of acceleration, often denoted as \( a_T \), refers to the portion of an object's acceleration that is aligned with its velocity vector. This component measures how much an object's speed changes over time. When assessing motion, it's vital to identify whether speed increases, decreases, or remains constant.

The tangential acceleration can be calculated from the derivative of the speed with respect to time. Mathematically, it is expressed as:

• \( a_T = \frac{d}{dt}(\| \mathbf{v}(t) \|) \)

A key observation in our exercise is that the speed of the object is a constant value of '2', as determined in Step 3 from the original solution. Since the speed does not vary, its rate of change is zero, leading to the conclusion that the tangential component of acceleration is zero:
\[ a_T = 0 \]
This outcome indicates that, while the object is in motion, its speed remains unchanged over time, ensuring no tangential acceleration is present. It's essential to note that a zero tangential component can imply constant velocity but not necessarily a lack of movement.

Normal Component of Acceleration

The normal component of acceleration, denoted \( a_N \), is responsible for the change in direction of an object's velocity. Unlike the tangential component, it doesn't affect the speed, but instead, it influences how the path of the object bends or turns.

To determine the normal component, first calculate the magnitude of the acceleration vector \( \mathbf{a}(t) \), and then differentiate between tangential and normal components using the formula:

• \( a_N = \sqrt{\| \mathbf{a}(t) \|^2 - a_T^2} \)

Given that \( a_T = 0 \) as found earlier, this simplifies to:

• \( a_N = \| \mathbf{a}(t) \| \)

In the specific problem at hand, the calculation for \( \| \mathbf{a}(t) \| \) is:\[\| \mathbf{a}(t) \| = \sqrt{\left(\frac{1}{2(1+t)^{1/2}}\right)^2 + \left(\frac{1}{2(1-t)^{1/2}}\right)^2}\]This expression captures the presence of forces acting orthogonally to the velocity, modifying the trajectory but not the speed of the object. Such components are crucial for understanding curvilinear motion or any motion involving path curvature.

Velocity Vector

The velocity vector \( \mathbf{v}(t) \) represents both the direction and speed of an object at any given point in time. It is derived by calculating the time derivative of the position vector \( \mathbf{r}(t) \).

• A comprehensive understanding of \( \mathbf{v}(t) \) is fundamental when attempting to grasp overall motion.

The calculations in the original steps provide the velocity vector as:

• \( \mathbf{v}(t) = \langle (1+t)^{1/2}, -(1-t)^{1/2}, \sqrt{2} \rangle \)

Each component of this vector signifies motion in the respective dimension:

• The first component, \((1+t)^{1/2}\), reflects motion along the x-axis.
• The second component, \(-(1-t)^{1/2}\), describes motion along the y-axis.
• The third constant \(\sqrt{2}\) aligns with the z-axis, emphasizing steady progression in that direction.

Understanding velocity vectors helps disentangle not only how fast an object is moving, but also where it's heading. This vectorization shows how speed varies with direction, guiding us in calculating both tangential and normal components of acceleration accurately.