Problem 38 - Calculus III

Question

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

Step-by-Step Solution

Answer

The tangential component is \(\frac{8t}{3\sqrt{4 + \frac{40t^2}{9}}}\), and the normal component is calculated from the acceleration magnitude less tangential squared.

Step 1: Find the Velocity Vector

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

Step 2: Find the Acceleration Vector

To find the acceleration vector \(\mathbf{a}(t)\), take the derivative of the velocity vector \(\mathbf{v}(t)\). The derivative of \(\left\langle 2, 2t, \frac{2t}{3} \right\rangle\) gives \(\mathbf{a}(t) = \left\langle 0, 2, \frac{2}{3} \right\rangle\).

Step 3: Determine Speed from Velocity

The speed \(\|\mathbf{v}(t)\|\) is the magnitude of the velocity vector. Calculate the magnitude: \(\|\mathbf{v}(t)\| = \sqrt{2^2 + (2t)^2 + \left(\frac{2t}{3}\right)^2}\). Simplify to get \(\|\mathbf{v}(t)\| = \sqrt{4 + 4t^2 + \frac{4t^2}{9}} = \sqrt{4 + \frac{36t^2}{9} + \frac{4t^2}{9}} = \sqrt{4 + \frac{40t^2}{9}} = \sqrt{4 + \frac{40t^2}{9}} \).

Step 4: Calculate Tangential Component of Acceleration

The tangential component \(a_T\) of acceleration is given by \(a_T = \frac{d}{dt}\|\mathbf{v}(t)\|\). Evaluating this derivative, \(a_T = \frac{d}{dt}\sqrt{4 + \frac{40t^2}{9}} = \frac{8t}{3\sqrt{4 + \frac{40t^2}{9}}}\).

Step 5: Calculate Normal Component of Acceleration

The normal component \(a_N\) of acceleration is given by \(a_N = \sqrt{\|\mathbf{a}(t)\|^2 - a_T^2}\). First, calculate \(\|\mathbf{a}(t)\| = \sqrt{0^2 + 2^2 + \left(\frac{2}{3}\right)^2} = \sqrt{4 + \frac{4}{9}} = \sqrt{\frac{36}{9} + \frac{4}{9}} = \sqrt{\frac{40}{9}} = \frac{2\sqrt{10}}{3}\). Use this to find \(a_N = \sqrt{\left(\frac{2\sqrt{10}}{3}\right)^2 - \left(\frac{8t}{3\sqrt{4 + \frac{40t^2}{9}}}\right)^2}\). Simplify to find \(a_N\).

Key Concepts

Velocity Vector

A velocity vector represents the rate of change of position with respect to time, and it's a crucial concept in understanding motion. For a position vector \( \mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle \), the velocity vector \( \mathbf{v}(t) \) is found by differentiating \(|\mathbf{r}(t)|\) with respect to time. Essentially, it tells us how fast and in what direction an object is moving at any given moment.

In the original exercise, the position vector is given as \( \mathbf{r}(t) = \langle 2t, t^2, \frac{t^2}{3} \rangle \). By differentiating each component of this vector, we identify the velocity vector as \( \mathbf{v}(t) = \langle 2, 2t, \frac{2t}{3} \rangle \). This result gives us a snapshot of the object's velocity in the x, y, and z directions.

Key takeaways:

• Velocity vector helps in describing the direction and speed of an object.
• Differentiate the position vector to find the velocity vector.
• Velocity is vectorial in nature, having both a magnitude and direction.

Acceleration Vector

The acceleration vector reveals how the velocity of an object changes over time. By understanding acceleration, we can predict changes in the velocity of a moving object and, by extension, its trajectory. Much like the velocity vector, the acceleration vector \( \mathbf{a}(t) \) is derived by taking the derivative of the velocity vector.

In this exercise, after calculating the velocity vector \( \mathbf{v}(t) = \langle 2, 2t, \frac{2t}{3} \rangle \, differentiating it gives us the acceleration vector \ ( \mathbf{a}(t) = \langle 0, 2, \frac{2}{3} \rangle \).

Helpful tips:

• Acceleration indicates change in velocity, providing insights about the object's motion dynamics.
• Take derivative of the velocity vector to find the acceleration vector.
• Components of acceleration vector show how each aspect of motion (x, y, and z) evolves over time.

Magnitude of a Vector

Understanding the magnitude of a vector is essential, as it represents the length or size of the vector. For any vector \( \mathbf{v}(t) = \langle v_x(t), v_y(t), v_z(t) \rangle \), its magnitude \(|\mathbf{v}(t)|\) is determined by the formula:

\[ |\mathbf{v}(t)| = \sqrt{v_x(t)^2 + v_y(t)^2 + v_z(t)^2} \]

In the context of this exercise, the magnitude of the velocity vector \( \mathbf{v}(t) = \langle 2, 2t, \frac{2t}{3} \rangle \) is found as:

\[ |\mathbf{v}(t)| = \sqrt{2^2 + (2t)^2 + \left(\frac{2t}{3}\right)^2} \]

This simplifies to:

\[ |\mathbf{v}(t)| = \sqrt{4 + 4t^2 + \frac{4t^2}{9}} \]

Understanding this concept helps in comprehending how speed and different vector magnitudes affect motion analysis.

Consider these points:

• The magnitude of a vector describes its length, independent of its direction.
• Magnitude is crucial for understanding how much of something, like speed, is present.
• It's calculated using the square root of the sum of the squares of its components.