Problem 15 - 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 \(t=0 .\) \begin{equation} \mathbf{r}(t)=(3 t+1) \mathbf{i}+\sqrt{3} t \mathbf{j}+t^{2} \mathbf{k} \end{equation}

Step-by-Step Solution

Answer

The angle between the velocity and acceleration vectors at \( t=0 \) is \( 90^\circ \).

Step 1: Find the Velocity Vector

The velocity vector \( \mathbf{v}(t) \) of a particle is the derivative of its position vector \( \mathbf{r}(t) \). Differentiate \( \mathbf{r}(t) = (3t+1)\mathbf{i} + \sqrt{3}t\mathbf{j} + t^2\mathbf{k} \) with respect to \( t \):\[ \mathbf{v}(t) = \frac{d}{dt} [(3t+1)\mathbf{i} + \sqrt{3}t\mathbf{j} + t^2\mathbf{k}] = 3\mathbf{i} + \sqrt{3}\mathbf{j} + 2t\mathbf{k}. \]

Step 2: Find the Acceleration Vector

The acceleration vector \( \mathbf{a}(t) \) is the derivative of the velocity vector \( \mathbf{v}(t) \). Differentiate \( \mathbf{v}(t) = 3\mathbf{i} + \sqrt{3}\mathbf{j} + 2t\mathbf{k} \) with respect to \( t \):\[ \mathbf{a}(t) = \frac{d}{dt} [3\mathbf{i} + \sqrt{3}\mathbf{j} + 2t\mathbf{k}] = 0\mathbf{i} + 0\mathbf{j} + 2\mathbf{k}. \]

Step 3: Evaluate Vectors at t=0

Substitute \( t=0 \) into the expressions for \( \mathbf{v}(t) \) and \( \mathbf{a}(t) \) to find the velocity and acceleration at this point:\[ \mathbf{v}(0) = 3\mathbf{i} + \sqrt{3}\mathbf{j} + 0\mathbf{k} = 3\mathbf{i} + \sqrt{3}\mathbf{j}, \]\[ \mathbf{a}(0) = 0\mathbf{i} + 0\mathbf{j} + 2\mathbf{k} = 2\mathbf{k}. \]

Step 4: Compute Dot Product of Velocity and Acceleration

Calculate the dot product \( \mathbf{v}(0) \cdot \mathbf{a}(0) \). Use the result of \( \mathbf{v}(0) = 3\mathbf{i} + \sqrt{3}\mathbf{j} \) and \( \mathbf{a}(0) = 2\mathbf{k} \):\[ \mathbf{v}(0) \cdot \mathbf{a}(0) = (3)(0) + (\sqrt{3})(0) + (0)(2) = 0. \]

Step 5: Compute Magnitudes of Vectors

Find the magnitudes of both \( \mathbf{v}(0) \) and \( \mathbf{a}(0) \):\[ \| \mathbf{v}(0) \| = \sqrt{(3)^2 + (\sqrt{3})^2} = \sqrt{9 + 3} = \sqrt{12} = 2\sqrt{3}, \]\[ \| \mathbf{a}(0) \| = \sqrt{(0)^2 + (0)^2 + (2)^2} = \sqrt{4} = 2. \]

Step 6: Determine Cosine of the Angle

Use the dot product and magnitudes to find the cosine of the angle \( \theta \) between \( \mathbf{v}(0) \) and \( \mathbf{a}(0) \):\[ \cos \theta = \frac{\mathbf{v}(0) \cdot \mathbf{a}(0)}{\| \mathbf{v}(0) \| \| \mathbf{a}(0) \|} = \frac{0}{2\sqrt{3} \times 2} = 0. \]

Step 7: Compute the Angle

Since \( \cos \theta = 0 \), the angle \( \theta \) is \( 90^\circ \) or \( \frac{\pi}{2} \) radians. This indicates that the vectors are perpendicular.

Key Concepts

Dot Product

The dot product is a fundamental operation in vector calculus that allows us to determine the degree to which two vectors align or point in the same direction. It is calculated by multiplying corresponding components of two vectors and summing those products. For two vectors \(\mathbf{u}\) and \(\mathbf{v}\), the dot product is given by:

• \(\mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 + u_3v_3\).

Using dot products, we can efficiently find the cosine of the angle between two vectors. Importantly, if the dot product of two vectors is zero, the vectors are orthogonal (or perpendicular) to each other. This critical property enables us to solve problems involving angles between vectors, such as determining when two paths or forces are at right angles.

Velocity Vector

A velocity vector represents the rate of change of a position vector over time. In calculus terms, it's the derivative of the position vector with respect to time. The velocity vector shows both the speed and direction of a particle moving through space. If a particle’s position is given by a vector \(\mathbf{r}(t)\), then its velocity vector is computed as:

• \(\mathbf{v}(t) = \frac{d}{dt} \mathbf{r}(t)\).

This vector is then evaluated at specific time points to understand the motion of the particle at those instances. In our solution, differentiating the position vector \(\mathbf{r}(t)\) gives the velocity \(\mathbf{v}(t) = 3\mathbf{i} + \sqrt{3}\mathbf{j} + 2t\mathbf{k}\), showing us how the particle's position changes in each spatial dimension over time.

Acceleration Vector

The acceleration vector is a measure of how the velocity of a particle changes over time. Mathematically, it is the derivative of the velocity vector. This vector provides insights into how quickly and in which direction the velocity itself is changing, giving us a deeper understanding of the particle's movement:

• \(\mathbf{a}(t) = \frac{d}{dt} \mathbf{v}(t)\).

In our case, differentiating the velocity vector \(\mathbf{v}(t)\) yields the acceleration vector \(\mathbf{a}(t) = 2\mathbf{k}\), indicating a constant acceleration in the \(\mathbf{k}\)-direction. This computation reveals the force acting on the particle and how that force influences its velocity over time.

Angle Between Vectors

Finding the angle between two vectors can provide crucial information about their relative directions. The relationship is determined using the dot product formula:

• \(\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|}\).

Where \(\theta\) is the angle between vectors \(\mathbf{a}\) and \(\mathbf{b}\), and the denominator is the product of their magnitudes. If the cosine of the angle is zero, we can conclude that the vectors are perpendicular, or \(90^\circ\) to each other. In our step-by-step solution, the dot product of the velocity and acceleration vectors at \(t = 0\) resulted in zero, leading us to determine that the angle between them is \(90^\circ\). This means these vectors are orthogonal, a condition that often simplifies analysis in physics and engineering.