Problem 16 - Thomas Calculus

Question

In Exercises \(15-18, \mathbf{r}(t)\) is the position of a particle in space at time \(t .\) Find the angle between the velocity and acceleration vectors at time \(t=0 .\) $$ \mathbf{r}(t)=\left(\frac{\sqrt{2}}{2} t\right) \mathbf{i}+\left(\frac{\sqrt{2}}{2} t-16 t^{2}\right) \mathbf{j} $$

Step-by-Step Solution

Answer

The angle between the velocity and acceleration vectors at \( t = 0 \) is \( 135^{\circ} \).

Step 1: Find the Velocity Vector

The velocity vector \( \mathbf{v}(t) \) is the derivative of the position vector \( \mathbf{r}(t) \) with respect to time \( t \). Compute the derivative:\[ \mathbf{v}(t) = \frac{d}{dt}\left( \frac{\sqrt{2}}{2} t \right) \mathbf{i} + \frac{d}{dt}\left( \frac{\sqrt{2}}{2} t - 16t^2 \right) \mathbf{j} = \frac{\sqrt{2}}{2} \mathbf{i} + \left( \frac{\sqrt{2}}{2} - 32t \right) \mathbf{j} \]

Step 2: Find the Acceleration Vector

The acceleration vector \( \mathbf{a}(t) \) is the derivative of the velocity vector \( \mathbf{v}(t) \) with respect to time \( t \). Compute the derivative:\[ \mathbf{a}(t) = \frac{d}{dt}\left( \frac{\sqrt{2}}{2} \right) \mathbf{i} + \frac{d}{dt}\left( \frac{\sqrt{2}}{2} - 32t \right) \mathbf{j} = 0 \mathbf{i} - 32 \mathbf{j} \]

Step 3: Evaluate Velocity and Acceleration at \( t = 0 \)

Substitute \( t = 0 \) into the expressions for \( \mathbf{v}(t) \) and \( \mathbf{a}(t) \):\[ \mathbf{v}(0) = \frac{\sqrt{2}}{2} \mathbf{i} + \frac{\sqrt{2}}{2} \mathbf{j} \]\[ \mathbf{a}(0) = 0 \mathbf{i} - 32 \mathbf{j} = -32 \mathbf{j} \]

Step 4: Calculate the Dot Product

The angle \( \theta \) between the vectors \( \mathbf{v}(0) \) and \( \mathbf{a}(0) \) is found using the dot product formula:\[ \mathbf{v}(0) \cdot \mathbf{a}(0) = \left( \frac{\sqrt{2}}{2} \right)\cdot 0 + \left( \frac{\sqrt{2}}{2} \right)(-32) = -16\sqrt{2} \]

Step 5: Calculate Magnitudes of Velocity and Acceleration

Find the magnitudes of both \( \mathbf{v}(0) \) and \( \mathbf{a}(0) \):\[ \| \mathbf{v}(0) \| = \sqrt{\left( \frac{\sqrt{2}}{2} \right)^2 + \left( \frac{\sqrt{2}}{2} \right)^2} = \sqrt{1} = 1 \]\[ \| \mathbf{a}(0) \| = \sqrt{0^2 + (-32)^2} = 32 \]

Step 6: Calculate the Cosine of the Angle

Use the dot product and magnitudes to calculate the cosine of the angle:\[ \cos(\theta) = \frac{\mathbf{v}(0) \cdot \mathbf{a}(0)}{\|\mathbf{v}(0)\| \cdot \|\mathbf{a}(0)\|} = \frac{-16\sqrt{2}}{1 \cdot 32} = -\frac{\sqrt{2}}{2} \]

Step 7: Find the Angle Using Cosine Inverse

Solve for the angle \( \theta \) by taking the inverse cosine:\[ \theta = \cos^{-1}\left(-\frac{\sqrt{2}}{2}\right) = 135^{\circ} \]

Key Concepts

Velocity Vector

In particle motion, the velocity vector is crucial as it provides information about both the speed and direction of a moving particle. Mathematically, the velocity vector \( \mathbf{v}(t) \) is found by taking the derivative of the position vector \( \mathbf{r}(t) \) with respect to time \( t \). This derivative represents the rate of change of the position. For our specific problem, the position vector is given by \( \mathbf{r}(t) = \left( \frac{\sqrt{2}}{2} t \right) \mathbf{i} + \left( \frac{\sqrt{2}}{2} t - 16t^2 \right) \mathbf{j} \). By differentiating each component with respect to \( t \), you get the velocity vector:

• The \( \mathbf{i} \)-component: \( \frac{d}{dt} \left( \frac{\sqrt{2}}{2} t \right) = \frac{\sqrt{2}}{2} \)
• The \( \mathbf{j} \)-component: \( \frac{d}{dt} \left( \frac{\sqrt{2}}{2} t - 16t^2 \right) = \frac{\sqrt{2}}{2} - 32t \)

Thus, the velocity vector \( \mathbf{v}(t) = \frac{\sqrt{2}}{2} \mathbf{i} + \left( \frac{\sqrt{2}}{2} - 32t \right) \mathbf{j} \), showing how velocity changes over time.

Acceleration Vector

The acceleration vector is another vital component of particle motion analysis. It represents the rate of change of the velocity vector over time and provides insights into how quickly a particle is speeding up or slowing down. To find the acceleration vector \( \mathbf{a}(t) \), take the derivative of the velocity vector \( \mathbf{v}(t) \) with respect to time \( t \). From the previous step, we know \( \mathbf{v}(t) = \frac{\sqrt{2}}{2} \mathbf{i} + \left( \frac{\sqrt{2}}{2} - 32t \right) \mathbf{j} \).

• The \( \mathbf{i} \)-component derivative is \( 0 \) as \( \frac{\sqrt{2}}{2} \) is a constant.
• The \( \mathbf{j} \)-component derivative is \( -32 \) since the derivative of \( -32t \) is \( -32 \).

Therefore, the acceleration vector \( \mathbf{a}(t) = 0 \mathbf{i} - 32 \mathbf{j} \), indicating constant acceleration in the \( \mathbf{j} \)-direction.

Dot Product

The dot product is a mathematical operation that helps find the angle between two vectors. It involves multiplying corresponding components and adding the results. For our problem, we've calculated the dot product of the velocity vector \( \mathbf{v}(0) % \) and the acceleration vector \( \mathbf{a}(0) \) at \( t = 0 \). At this time, \( \mathbf{v}(0) = \frac{\sqrt{2}}{2} \mathbf{i} + \frac{\sqrt{2}}{2} \mathbf{j} \) and \( \mathbf{a}(0) = 0 \mathbf{i} - 32 \mathbf{j} \). The dot product is:

• The \( \mathbf{i} \)-components: \( \frac{\sqrt{2}}{2} \times 0 = 0 \)
• The \( \mathbf{j} \)-components: \( \frac{\sqrt{2}}{2} \times (-32) = -16\sqrt{2} \)

Adding these gives a total dot product of \( -16\sqrt{2} \). This value is then used in conjunction with the magnitudes of both vectors to find the cosine of the angle between them.

Trigonometry in Physics

Trigonometry is indispensable in physics for understanding angles and relationships between vectors. It allows us to find angles using the cosine function, which is critical when analyzing vector directions. After finding the dot product of the velocity and acceleration vectors, trigonometry helps determine the angle between them.Given the dot product and the magnitudes of both vectors, we can calculate the cosine of the angle \( \theta \). Our calculation gives \( \cos(\theta) = \frac{-16\sqrt{2}}{1 \times 32} = -\frac{\sqrt{2}}{2} \). Using the inverse cosine function, we solve for \( \theta \):

• \( \theta = \cos^{-1}\left(-\frac{\sqrt{2}}{2}\right) = 135^{\circ} \)

This demonstrates how trigonometry helps to find the precise angle between two directionally different vectors, further highlighting its importance in analyzing particle motion.