Problem 20 - Thomas Calculus

Question

In Exercises \(19-22, \mathbf{r}(t)\) 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 \( 135^\circ \) or \( \frac{3\pi}{4} \) radians.

Step 1: Calculate the Velocity Vector

To find the velocity vector, differentiate the position vector \( \mathbf{r}(t) \) with respect to \( t \). \[ \mathbf{v}(t) = \mathbf{r}'(t) = \left( \frac{d}{dt} \left( \frac{\sqrt{2}}{2} t \right) \right) \mathbf{i} + \left( \frac{d}{dt} \left( \frac{\sqrt{2}}{2} t - 16t^2 \right) \right) \mathbf{j} \] \[ \mathbf{v}(t) = \frac{\sqrt{2}}{2} \mathbf{i} + \left( \frac{\sqrt{2}}{2} - 32t \right) \mathbf{j} \]

Step 2: Calculate the Acceleration Vector

The acceleration vector is the derivative of the velocity vector. Find \( \mathbf{a}(t) = \mathbf{v}'(t) \): \[ \mathbf{a}(t) = \left( 0 \right) \mathbf{i} + \left( -32 \right) \mathbf{j} \] Therefore, the acceleration vector is \( \mathbf{a}(t) = -32 \mathbf{j} \).

Step 3: Evaluate the Velocity and Acceleration at t=0

Substitute \( t = 0 \) into the velocity and acceleration vectors to find their values at that time. \[ \mathbf{v}(0) = \frac{\sqrt{2}}{2} \mathbf{i} + \left( \frac{\sqrt{2}}{2} \right) \mathbf{j} \] \[ \mathbf{a}(0) = -32 \mathbf{j} \]

Step 4: Calculate the Dot Product of Velocity and Acceleration at t=0

The dot product of two vectors \( \mathbf{v} \) and \( \mathbf{a} \) is calculated as follows: \( \mathbf{v}(0) \cdot \mathbf{a}(0) = \left( \frac{\sqrt{2}}{2} \right) \cdot 0 + \left( \frac{\sqrt{2}}{2} \right) \cdot (-32) \). \[ \mathbf{v}(0) \cdot \mathbf{a}(0) = -16\sqrt{2} \]

Step 5: Calculate the Magnitudes of the Velocity and Acceleration at t=0

Calculate the magnitude of each vector: \[ \| \mathbf{v}(0) \| = \sqrt{ \left( \frac{\sqrt{2}}{2} \right)^2 + \left( \frac{\sqrt{2}}{2} \right)^2 } = 1 \] \[ \| \mathbf{a}(0) \| = \sqrt{0^2 + (-32)^2} = 32 \]

Step 6: Calculate the Angle Between Velocity and Acceleration at t=0

Use the dot product formula to find the angle \( \theta \) between \( \mathbf{v}(0) \) and \( \mathbf{a}(0) \): \[ \cos \theta = \frac{\mathbf{v}(0) \cdot \mathbf{a}(0)}{\|\mathbf{v}(0)\| \|\mathbf{a}(0)\|} \] \[ \cos \theta = \frac{-16\sqrt{2}}{32} = -\frac{\sqrt{2}}{2} \] Thus, \( \theta = 135^\circ \) or \( \theta = \frac{3\pi}{4} \) radians.

Key Concepts

Velocity Vector

The velocity vector is crucial in understanding the movement of particles through space. When you have the position of a particle, denoted as \( \mathbf{r}(t) \), you can find its velocity by differentiating the position vector with respect to time \( t \). This gives you the rate of change of the particle's position, or in simpler terms, how fast and in what direction the particle is moving at any given moment. If the position of a particle is given by \( \mathbf{r}(t) = \left( \frac{\sqrt{2}}{2} t \right) \mathbf{i} + \left( \frac{\sqrt{2}}{2} t - 16 t^2 \right) \mathbf{j} \), then the velocity vector is found by taking the derivative:

• Velocity in the \( \mathbf{i} \) direction: \( \frac{d}{dt} \left( \frac{\sqrt{2}}{2} t \right) = \frac{\sqrt{2}}{2} \)
• Velocity in the \( \mathbf{j} \) direction: \( \frac{d}{dt} \left( \frac{\sqrt{2}}{2} t - 16 t^2 \right) = \frac{\sqrt{2}}{2} - 32t \)

Thus, the velocity vector \( \mathbf{v}(t) \) is \( \frac{\sqrt{2}}{2} \mathbf{i} + \left( \frac{\sqrt{2}}{2} - 32t \right) \mathbf{j} \). It's essential to remember that velocity is a dynamic vector that changes as time progresses.

Acceleration Vector

Acceleration is the rate of change of velocity with respect to time. It tells us how a particle's velocity changes, describing the forces acting upon the particle. If the velocity vector is \( \mathbf{v}(t) = \frac{\sqrt{2}}{2} \mathbf{i} + \left( \frac{\sqrt{2}}{2} - 32t \right) \mathbf{j} \), the acceleration vector \( \mathbf{a}(t) \) is obtained by differentiating the velocity vector:

• Acceleration in the \( \mathbf{i} \) direction: \( 0 \) (constant velocity, hence no acceleration)
• Acceleration in the \( \mathbf{j} \) direction: \( -32 \)

So, the acceleration vector is \( \mathbf{a}(t) = -32 \mathbf{j} \). This shows that the particle is accelerating in the \( \mathbf{j} \) direction with a constant rate, implying that some consistent force is affecting its motion.

Dot Product

The dot product is a vector operation that takes two vectors and returns a scalar. It's a measure of how much one vector goes in the direction of another. Mathematically, for two vectors \( \mathbf{u} \) and \( \mathbf{v} \), the dot product is calculated as \( \mathbf{u} \cdot \mathbf{v} = u_1 v_1 + u_2 v_2 + \ldots + u_n v_n \). In the context of the problem, the dot product of the velocity and acceleration vectors at \( t=0 \) was computed. The velocity vector was \( \mathbf{v}(0) = \frac{\sqrt{2}}{2} \mathbf{i} + \left( \frac{\sqrt{2}}{2} \right) \mathbf{j} \) and the acceleration vector was \( \mathbf{a}(0) = -32 \mathbf{j} \). The dot product is:

• \( \mathbf{v}(0) \cdot \mathbf{a}(0) = \left( \frac{\sqrt{2}}{2} \cdot 0 \right) + \left( \frac{\sqrt{2}}{2} \cdot (-32) \right) = -16\sqrt{2} \)

The result, \( -16\sqrt{2} \), indicates that the vectors point in mostly opposite directions.

Angle Between Vectors

The angle between two vectors can tell us about the relationship between them. If two vectors are orthogonal (at a right angle), their dot product is zero. If they're parallel, the angle is 0°. To find the angle \( \theta \) between two vectors, you can use the formula:\[ \cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\| \|\mathbf{v}\|} \]In the given exercise, the angle between the velocity and acceleration vectors at \( t=0 \) was calculated using this formula. The magnitudes were:

• \( \| \mathbf{v}(0) \| = 1 \)
• \( \| \mathbf{a}(0) \| = 32 \)

The dot product was \( -16\sqrt{2} \). Plug these into the formula to find:

• \( \cos \theta = \frac{-16\sqrt{2}}{32} = -\frac{\sqrt{2}}{2} \)

This results in an angle \( \theta \) of \( 135^\circ \) or \( \frac{3\pi}{4} \) radians. This shows the vectors are more opposite than aligned, indicating a larger angle between them.