Problem 11 - Thomas Calculus

Question

In Exercises \(9-14, \mathrm{r}(t)\) is the position of a particle in space at time \(t .\) Find the particle's velocity and acceleration vectors. Then find the particle's speed and direction of motion at the given value of \(t .\) Write the particle's velocity at that time as the product of its speed and direction. $$ \mathbf{r}(t)=(2 \cos t) \mathbf{i}+(3 \sin t) \mathbf{j}+4 t \mathbf{k}, \quad t=\pi / 2 $$

Step-by-Step Solution

Answer

Velocity: \(-2 \mathbf{i} + 4 \mathbf{k}\); Speed: \(2\sqrt{5}\); Direction: \(-\frac{1}{\sqrt{5}} \mathbf{i} + \frac{2}{\sqrt{5}} \mathbf{k}\).

Step 1: Find the Velocity Vector

To find the velocity vector of the particle, we need to differentiate the position vector \( \mathbf{r}(t) \) with respect to \( t \). The position vector is given by \( \mathbf{r}(t) = (2 \cos t) \mathbf{i} + (3 \sin t) \mathbf{j} + 4t \mathbf{k} \). - The derivative of \( 2 \cos t \) with respect to \( t \) is \( -2 \sin t \). - The derivative of \( 3 \sin t \) with respect to \( t \) is \( 3 \cos t \). - The derivative of \( 4t \) with respect to \( t \) is \( 4 \).Thus, the velocity vector \( \mathbf{v}(t) \) is \( -2 \sin t \mathbf{i} + 3 \cos t \mathbf{j} + 4 \mathbf{k} \).

Step 2: Find the Acceleration Vector

To find the acceleration vector, differentiate the velocity vector \( \mathbf{v}(t) = -2 \sin t \mathbf{i} + 3 \cos t \mathbf{j} + 4 \mathbf{k} \) with respect to \( t \). - The derivative of \( -2 \sin t \) is \( -2 \cos t \). - The derivative of \( 3 \cos t \) is \( -3 \sin t \). - The derivative of \( 4 \), which is constant, is \( 0 \).Thus, the acceleration vector \( \mathbf{a}(t) \) is \( -2 \cos t \mathbf{i} - 3 \sin t \mathbf{j} + 0 \mathbf{k} \).

Step 3: Compute Velocity at \( t = \pi/2 \)

To compute the velocity vector at \( t = \pi/2 \), substitute \( \pi/2 \) into the velocity vector \( \mathbf{v}(t) = -2 \sin t \mathbf{i} + 3 \cos t \mathbf{j} + 4 \mathbf{k} \).- \( -2 \sin(\pi/2) = -2(1) = -2 \).- \( 3 \cos(\pi/2) = 3(0) = 0 \).- The \( \mathbf{k} \) component stays \( 4 \).Thus, \( \mathbf{v}(\pi/2) = -2 \mathbf{i} + 0 \mathbf{j} + 4 \mathbf{k} = -2 \mathbf{i} + 4 \mathbf{k} \).

Step 4: Compute Speed at \( t = \pi/2 \)

The speed of the particle is the magnitude of the velocity vector \( \mathbf{v}(t) \). Use the formula \( |\mathbf{v}(t)| = \sqrt{v_x^2 + v_y^2 + v_z^2} \), where \( v_x, v_y, v_z \) are the components of \( \mathbf{v}(t) \).\[ |\mathbf{v}(\pi/2)| = \sqrt{(-2)^2 + 0^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} \].

Step 5: Find Direction of Motion at \( t = \pi/2 \)

The direction of motion is given by the unit vector of the velocity vector \( \mathbf{v}(t) \). It is calculated by dividing \( \mathbf{v}(t) \) by its magnitude:\[ \text{Direction} = \frac{\mathbf{v}(\pi/2)}{|\mathbf{v}(\pi/2)|} = \frac{-2 \mathbf{i} + 4 \mathbf{k}}{2\sqrt{5}} = \left( -\frac{1}{\sqrt{5}} \right) \mathbf{i} + \left( \frac{2}{\sqrt{5}} \right) \mathbf{k} \].

Step 6: Express Velocity as Product of Speed and Direction

Write the velocity at \( t = \pi/2 \) as the product of its speed and the direction:\[ \mathbf{v}(\pi/2) = 2\sqrt{5} \left( -\frac{1}{\sqrt{5}} \mathbf{i} + \frac{2}{\sqrt{5}} \mathbf{k} \right) = -2 \mathbf{i} + 4 \mathbf{k} \].

Key Concepts

Velocity Vector

In particle motion studies, the velocity vector is a crucial concept. The velocity vector provides the rate of change of the particle's position in space. It is derived by differentiating the position vector with respect to time. Given a position vector \( \mathbf{r}(t) = (2 \cos t) \mathbf{i} + (3 \sin t) \mathbf{j} + 4t \mathbf{k} \), the velocity vector \( \mathbf{v}(t) \) can be determined by taking the derivative of each component:

• The derivative of \( 2 \cos t \) is \( -2 \sin t \).
• The derivative of \( 3 \sin t \) is \( 3 \cos t \).
• The derivative of \( 4t \) is \( 4 \).

Thus, the velocity vector is \( \mathbf{v}(t) = -2 \sin t \mathbf{i} + 3 \cos t \mathbf{j} + 4 \mathbf{k} \). This vector indicates both the speed and the direction in which the particle is moving at any given time.

Acceleration Vector

The acceleration vector reveals how the velocity of the particle changes over time. To find this, we differentiate the velocity vector \( \mathbf{v}(t) \) with respect to time \( t \). Starting from the velocity vector \( \mathbf{v}(t) = -2 \sin t \mathbf{i} + 3 \cos t \mathbf{j} + 4 \mathbf{k} \), the acceleration vector \( \mathbf{a}(t) \) is calculated as follows:

• The derivative of \( -2 \sin t \) is \( -2 \cos t \).
• The derivative of \( 3 \cos t \) is \( -3 \sin t \).
• The derivative of the constant \( 4 \) is zero.

Consequently, the acceleration vector is \( \mathbf{a}(t) = -2 \cos t \mathbf{i} - 3 \sin t \mathbf{j} \). This vector helps us understand how quickly and in which direction the velocity is being adjusted as the particle moves over time.

Speed and Direction of Motion

Speed and direction are key elements of a particle's motion. Speed is the magnitude of the velocity vector, which essentially measures how fast the particle is moving. For a particle at time \( t = \pi/2 \), the velocity vector is \( \mathbf{v}(t) = -2 \mathbf{i} + 0 \mathbf{j} + 4 \mathbf{k} \). You can calculate speed by finding the magnitude of this vector:\[ |\mathbf{v}(\pi/2)| = \sqrt{(-2)^2 + 0^2 + 4^2} = \sqrt{20} = 2\sqrt{5}.\]The direction of motion is represented by the unit vector in the direction of the velocity vector. This is found by dividing the velocity vector by its speed:\[\text{Direction} = \frac{-2 \mathbf{i} + 4 \mathbf{k}}{2\sqrt{5}} = \left( -\frac{1}{\sqrt{5}} \right) \mathbf{i} + \left( \frac{2}{\sqrt{5}} \right) \mathbf{k}.\]Ultimately, expressing the velocity vector as the product of its speed and the direction vector provides a clearer depiction of both how fast and in which direction the particle is moving at any moment.