Problem 20 - 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 \( \mathbf{F}(t) = 6mt \mathbf{i} + 2m \mathbf{j} + 6mt \mathbf{k} \).

Step 1: Find the Velocity Function

To find the velocity function, take the first derivative of the position function \( \mathbf{r}(t) \) with respect to \( t \). The given position function is \( \mathbf{r}(t) = t^{3} \mathbf{i} + t^{2} \mathbf{j} + t^{3} \mathbf{k} \). Differentiate each component:- \( \frac{d}{dt}(t^{3} \mathbf{i}) = 3t^2 \mathbf{i} \)- \( \frac{d}{dt}(t^{2} \mathbf{j}) = 2t \mathbf{j} \)- \( \frac{d}{dt}(t^{3} \mathbf{k}) = 3t^2 \mathbf{k} \)Thus, the velocity function \( \mathbf{v}(t) \) is \( 3t^2 \mathbf{i} + 2t \mathbf{j} + 3t^2 \mathbf{k} \).

Step 2: Find the Acceleration Function

To find the acceleration function, take the derivative of the velocity function with respect to \( t \). The velocity function is \( \mathbf{v}(t) = 3t^2 \mathbf{i} + 2t \mathbf{j} + 3t^2 \mathbf{k} \). Differentiate each component:- \( \frac{d}{dt}(3t^2 \mathbf{i}) = 6t \mathbf{i} \)- \( \frac{d}{dt}(2t \mathbf{j}) = 2 \mathbf{j} \)- \( \frac{d}{dt}(3t^2 \mathbf{k}) = 6t \mathbf{k} \)Thus, the acceleration function \( \mathbf{a}(t) \) is \( 6t \mathbf{i} + 2 \mathbf{j} + 6t \mathbf{k} \).

Step 3: Apply Newton's Second Law to Find the Force

Newton's second law states that \( \mathbf{F}(t) = m \mathbf{a}(t) \), where \( m \) is the mass of the particle and \( \mathbf{a}(t) \) is the acceleration. Substituting the acceleration from Step 2:\[ \mathbf{F}(t) = m (6t \mathbf{i} + 2 \mathbf{j} + 6t \mathbf{k}) \]Distribute the mass \( m \):\[ \mathbf{F}(t) = 6mt \mathbf{i} + 2m \mathbf{j} + 6mt \mathbf{k} \]

Key Concepts

Understanding Particle Motion

Particle motion refers to the movement of an object, often described in terms of position, velocity, and acceleration. In physics, particularly mechanics, we analyze how a particle moves in space over time. This involves understanding and applying mathematical functions that describe the particle's position at any point in time, helping to predict future positions.

By understanding the position function of a particle, we can also determine its velocity and acceleration, which are critical for solving dynamics problems. For a complete analysis of motion, one must be able to transition seamlessly from one of these descriptions to another, often requiring calculus operations.

• Position function: Describes Where the particle is.
• Velocity function: Shows how fast and where it is going.
• Acceleration function: Indicates how its speed is changing.

Understanding these concepts allows us to predict the future state of a moving particle, which is fundamental in physics.

Breaking Down the Position Function

A position function is a mathematical expression that provides the coordinates of a particle in space as a function of time, denoted by \ \(\mathbf{r}(t)\ \). This function can vary in complexity, depending on the movement's dimensionality and character. Each term of the position function tells us about the particle's path over time along the respective axis.

For example, the position function \ \(\mathbf{r}(t) = t^{3} \mathbf{i} + t^{2} \mathbf{j} + t^{3} \mathbf{k} \ \) describes how a particle's position changes over time in a three-dimensional space. Here, each component represents motion along a specific axis:

• \ \(t^{3} \mathbf{i}\ \) for movement along the x-axis.
• \ \(t^{2} \mathbf{j}\ \) for movement along the y-axis.
• \ \(t^{3} \mathbf{k}\ \) for movement along the z-axis.

By analyzing this function, we can extract valuable insights into the behavior of the particle over time.

Diving Into the Velocity Function

The velocity function signifies the rate and direction of a particle's motion. To derive it, we differentiate the position function with respect to time. Velocity is essentially the speed with a directional component, making it a vector. This makes it an invaluable tool for understanding motion's dynamic nature.

For the particle described by \ \(\mathbf{r}(t) = t^{3} \mathbf{i} + t^{2} \mathbf{j} + t^{3} \mathbf{k}\ \), the velocity function \ \(\mathbf{v}(t) = 3t^2 \mathbf{i} + 2t \mathbf{j} + 3t^2 \mathbf{k}\ \) is obtained by taking derivatives:

• The derivative of \ \(t^{3} \mathbf{i}\ \) yields \ \(3t^2 \mathbf{i}\ \).
• The derivative of \ \(t^{2} \mathbf{j}\ \) yields \ \(2t \mathbf{j}\ \).
• The derivative of \ \(t^{3} \mathbf{k}\ \) yields \ \(3t^2 \mathbf{k}\ \).

This function tells us not just how fast the particle is moving at any given time, but also in what direction.

Exploring the Acceleration Function

The acceleration function provides insight into how the velocity of a particle changes with time. It is the second derivative of the position function or the derivative of the velocity function. Acceleration is crucial as it explains how a particle's speed and direction can vary over time.

For the given example in the problem, the acceleration function \ \(\mathbf{a}(t) = 6t \mathbf{i} + 2 \mathbf{j} + 6t \mathbf{k}\ \) is determined by differentiating the velocity function \ \(\mathbf{v}(t) = 3t^2 \mathbf{i} + 2t \mathbf{j} + 3t^2 \mathbf{k}\ \). Performing this calculus operation gives:

• The derivative of \ \(3t^2 \mathbf{i}\ \) is \ \(6t \mathbf{i}\ \).
• The derivative of \ \(2t \mathbf{j}\ \) is \ \(2 \mathbf{j}\ \).
• The derivative of \ \(3t^2 \mathbf{k}\ \) is \ \(6t \mathbf{k}\ \).

Understanding acceleration helps in predicting how forces applied on the particle will impact its future motion, a foundation of Newton's Second Law.