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 \(t=0 .\) $$ \mathbf{r}(t)=(3 t+1) \mathbf{i}+\sqrt{3} t \mathbf{j}+t^{2} \mathbf{k} $$

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 \( \mathbf{v}(t) \) is the first derivative of the position vector \( \mathbf{r}(t) \) with respect to time \( t \). Differentiate each component of \( \mathbf{r}(t) \): \( \mathbf{v}(t) = \frac{d}{dt} (3t+1) \mathbf{i} + \frac{d}{dt} (\sqrt{3}t) \mathbf{j} + \frac{d}{dt} (t^2) \mathbf{k} = 3 \mathbf{i} + \sqrt{3} \mathbf{j} + 2t \mathbf{k} \).

Step 2: Find the acceleration vector

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

Step 3: Evaluate velocity and acceleration at \( t=0 \)

Substitute \( t=0 \) into the velocity and acceleration vectors. \( \mathbf{v}(0) = 3 \mathbf{i} + \sqrt{3} \mathbf{j} + 0 \mathbf{k} = 3 \mathbf{i} + \sqrt{3} \mathbf{j} \). \( \mathbf{a}(0) = 0 \mathbf{i} + 0 \mathbf{j} + 2 \mathbf{k} \).

Step 4: Calculate the dot product

The dot product of \( \mathbf{v}(0) \) and \( \mathbf{a}(0) \) is \( \mathbf{v}(0) \cdot \mathbf{a}(0) = (3 \mathbf{i} + \sqrt{3} \mathbf{j}) \cdot (0 \mathbf{i} + 0 \mathbf{j} + 2 \mathbf{k}) = 0 \).

Step 5: Calculate magnitudes of velocity and acceleration

The magnitude of \( \mathbf{v}(0) \) is \( \|\mathbf{v}(0)\| = \sqrt{(3)^2 + (\sqrt{3})^2 + 0^2} = \sqrt{9 + 3} = \sqrt{12} = 2\sqrt{3} \). The magnitude of \( \mathbf{a}(0) \) is \( \|\mathbf{a}(0)\| = \sqrt{0^2 + 0^2 + 2^2} = \sqrt{4} = 2 \).

Step 6: Find the angle between vectors

Use the dot product formula, \( \mathbf{v}(0) \cdot \mathbf{a}(0) = \|\mathbf{v}(0)\| \cdot \|\mathbf{a}(0)\| \cdot \cos(\theta) \), where \( \theta \) is the angle between them. Since \( \mathbf{v}(0) \cdot \mathbf{a}(0) = 0 \), the vectors are orthogonal: \( \cos(\theta) = \frac{0}{2\sqrt{3} \cdot 2} = 0 \), so \( \theta = 90^\circ \).

Key Concepts

Dot Product

The dot product is a way to multiply two vectors, resulting in a scalar rather than another vector. It is calculated by multiplying the corresponding components of two vectors and summing the results. If you have two vectors \( \mathbf{u} = a_1 \mathbf{i} + b_1 \mathbf{j} + c_1 \mathbf{k} \) and \( \mathbf{v} = a_2 \mathbf{i} + b_2 \mathbf{j} + c_2 \mathbf{k} \), the dot product is:\[\mathbf{u} \cdot \mathbf{v} = a_1a_2 + b_1b_2 + c_1c_2\]One main use of the dot product is to determine angle relationships between vectors. If the dot product is zero, the vectors are orthogonal, meaning they meet at a right angle. In our case, by computing \( \mathbf{v}(0) \cdot \mathbf{a}(0) \), we found the dot product to be zero. This leads us directly to the conclusion that the angle between the velocity and acceleration vectors at \( t=0 \) is 90°, emphasizing the orthogonality of these vectors.

Velocity and Acceleration Vectors

Velocity and acceleration vectors describe the motion of a particle in terms of its speed and direction changes. The velocity vector is the derivative of the position vector with respect to time and represents how fast and in which direction a particle is moving.From the exercise, the position vector is \( \mathbf{r}(t) = (3t+1) \mathbf{i} + \sqrt{3}t \mathbf{j} + t^2 \mathbf{k} \). Differentiating with respect to \( t \), the velocity vector \( \mathbf{v}(t) \) becomes \( 3 \mathbf{i} + \sqrt{3} \mathbf{j} + 2t \mathbf{k} \). This means:- The constant components \( 3 \mathbf{i} \) and \( \sqrt{3} \mathbf{j} \) indicate linear motion in \( x \)- and \( y \)-directions.- The \( 2t \mathbf{k} \) suggests increasing speed in the \( z \)-direction as \( t \) progresses.Acceleration, on the other hand, is the derivative of the velocity vector and shows how the velocity changes with time. In our case, by differentiating the velocity vector, we get the acceleration vector \( \mathbf{a}(t) = 0 \mathbf{i} + 0 \mathbf{j} + 2 \mathbf{k} \). This tells us the acceleration is constant in the \( z \)-direction, confirming uniform speed change along this axis.

Orthogonality of Vectors

Orthogonality between two vectors means they are perpendicular to each other. This can often be expressed through the dot product, as perpendicular vectors yield a dot product of zero.In linear algebra and vector calculus, orthogonal vectors imply no projection of one vector onto the other. This zero projection results in zero work done when considering forces and movements in physics. In the provided exercise, we calculated the dot product of the velocity and acceleration vectors at \( t=0 \) and found it to be zero. This demonstrates orthogonality, indicating that at this specific time, velocity and acceleration vectors meet at a right angle, which is 90°.- **Significance**: This condition is essential for understanding equilibrium states in physical systems and analyzing forces and motions that contribute independently in different spatial directions.Recognizing orthogonality in vector problems helps simplify complex problems and provides insights into the system's dynamics.