Problem 16 - Essential Calculus Early Transcendentals

Question

What force is required so that a particle of mass \(m\) has the position function \(\mathbf{r}(t)=t^{3} \mathbf{i}+t^{2} \mathbf{j}+t^{3} \mathbf{k} ?\)

Step-by-Step Solution

Answer

The force required is \( 6mt \mathbf{i} + 2m \mathbf{j} + 6mt \mathbf{k} \).

Step 1: Find the Velocity Function

The velocity function \( \mathbf{v}(t) \) is the derivative of the position function \( \mathbf{r}(t) \) with respect to time \( t \). We need to find \( \frac{d}{dt}(t^3 \mathbf{i} + t^2 \mathbf{j} + t^3 \mathbf{k}) \). This gives us \( \mathbf{v}(t) = 3t^2 \mathbf{i} + 2t \mathbf{j} + 3t^2 \mathbf{k} \).

Step 2: Find the Acceleration Function

The acceleration function \( \mathbf{a}(t) \) is the derivative of the velocity function \( \mathbf{v}(t) \). We differentiate \( 3t^2 \mathbf{i} + 2t \mathbf{j} + 3t^2 \mathbf{k} \) with respect to \( t \). This results in \( \mathbf{a}(t) = 6t \mathbf{i} + 2 \mathbf{j} + 6t \mathbf{k} \).

Step 3: Calculate the Required Force

Using Newton's second law \( \mathbf{F} = m \mathbf{a} \), where \( m \) is the mass of the particle, we find the force by substituting \( \mathbf{a}(t) = 6t \mathbf{i} + 2 \mathbf{j} + 6t \mathbf{k} \). Thus, \( \mathbf{F}(t) = m(6t \mathbf{i} + 2 \mathbf{j} + 6t \mathbf{k}) = 6mt \mathbf{i} + 2m \mathbf{j} + 6mt \mathbf{k} \).

Key Concepts

Understanding the Position Function

The position function is a mathematical tool that helps in determining the exact location of a particle at any given time. In mathematical terms, it is expressed as a function of time, and for the given exercise, the position function is \( \mathbf{r}(t) = t^{3} \mathbf{i} + t^{2} \mathbf{j} + t^{3} \mathbf{k} \).
This means that:

• Along the \( \mathbf{i} \) direction (x-axis), the position depends on \( t^3 \).
• Along the \( \mathbf{j} \) direction (y-axis), the position is determined by \( t^2 \).
• Along the \( \mathbf{k} \) direction (z-axis), it again depends on \( t^3 \).

By using this function, you can find where exactly the particle is located in space at any specific time \( t \). Understanding the position function lays the foundation for determining how the particle moves over time and is the first step in solving problems related to motion.

Getting to Know the Velocity Function

The velocity function gives you insight into how fast and in what direction a particle's position changes over time. To find the velocity function, you need to differentiate the position function with respect to time.
In our case, we derived the velocity function \( \mathbf{v}(t) = 3t^2 \mathbf{i} + 2t \mathbf{j} + 3t^2 \mathbf{k} \). What does this tell us?

• In the \( \mathbf{i} \) direction, the velocity is \( 3t^2 \), meaning the speed increases as time squared.
• In the \( \mathbf{j} \) direction, the velocity is linear \( 2t \), growing steadily with time.
• In the \( \mathbf{k} \) direction, it mimics the \( \mathbf{i} \) direction with \( 3t^2 \).

With this function, you can understand the rate of change of the particle's position, indicating how the particle speeds up or slows down.

Decoding the Acceleration Function

Now, down the line of understanding motion, the acceleration function is crucial. It reflects how the velocity of the particle changes with time. To find it, you differentiate the velocity function with respect to time, giving you the acceleration function. In our example, the function is \( \mathbf{a}(t) = 6t \mathbf{i} + 2 \mathbf{j} + 6t \mathbf{k} \).
Here's a breakdown:

• The acceleration in the \( \mathbf{i} \) and \( \mathbf{k} \) directions are both \( 6t \), indicating a steady increase in velocity over time.
• Along the \( \mathbf{j} \) axis, the acceleration is a constant \( 2 \), suggesting a uniform rate of change in velocity.

Understanding acceleration is pivotal because it tells you how quickly a particle's speed is changing, giving you the ability to predict future motion precisely. This knowledge is essential for problem-solving in physics, especially when applying Newton's Second Law, which connects acceleration with force.