Problem 2 - Thomas Calculus

Question

\({r}(t)\) is the position of a particle in the \(x y\) -plane at time \(t .\) Find an equation in \(x\) and \(y\) whose graph is the path of the particle. Then find the particle's velocity and acceleration vectors at the given value of \(t .\) \begin{equation} \mathbf{r}(t)=\frac{t}{t+1} \mathbf{i}+\frac{1}{t} \mathbf{j}, \quad t=-\frac{1}{2} \end{equation}

Step-by-Step Solution

Answer

Path: \(y = 1-x\); Velocity at \(t = -\frac{1}{2}\): \(4\mathbf{i} - 4\mathbf{j}\); Acceleration at \(t = -\frac{1}{2}\): \(-16\mathbf{i} - 16\mathbf{j}\).

Step 1: Find the Path Equation

We'll transform the parametric equations of the particle's path into a single equation in terms of \(x\) and \(y\). Given \(x(t) = \frac{t}{t+1}\) and \(y(t) = \frac{1}{t}\), solve the equation for \(t\) in terms of \(x\): \(x = \frac{t}{t+1}\). Rearrange this to get \(t = \frac{x}{1-x}\). Substitute \(t = \frac{x}{1-x}\) into \(y = \frac{1}{t}\) to find the path equation as \(y = 1-x\). Thus, the path is represented by the equation \(y = 1-x\).

Step 2: Calculate the Velocity Vector

The velocity vector \(\mathbf{v}(t)\) is the derivative of the position vector \(\mathbf{r}(t)\). Differentiate both components: \(\frac{d}{dt}\left( \frac{t}{t+1} \right)\) and \( \frac{d}{dt}\left( \frac{1}{t} \right)\). Using the quotient rule, the derivative of \( \frac{t}{t+1} \) is \( \frac{1}{(t+1)^2} \) and the derivative of \( \frac{1}{t} \) is \(-\frac{1}{t^2} \). Therefore, \(\mathbf{v}(t) = \frac{1}{(t+1)^2} \mathbf{i} + -\frac{1}{t^2} \mathbf{j}\).

Step 3: Evaluate the Velocity at Given t

Substitute \(t = -\frac{1}{2}\) into the velocity vector formula: \(\mathbf{v}(-\frac{1}{2}) = \frac{1}{(-\frac{1}{2}+1)^2}\mathbf{i} + -\frac{1}{(-\frac{1}{2})^2}\mathbf{j}\). Simplifying, this gives \(\mathbf{v}(-\frac{1}{2}) = 4 \mathbf{i} - 4 \mathbf{j}\).

Step 4: Calculate the Acceleration Vector

The acceleration vector \(\mathbf{a}(t)\) is the derivative of the velocity vector \(\mathbf{v}(t)\). Differentiate the \(i\) and \(j\) components: \(\frac{d}{dt}\left(\frac{1}{(t+1)^2}\right)\) and \(\frac{d}{dt}\left(-\frac{1}{t^2}\right)\). The derivative of \(\frac{1}{(t+1)^2}\) is \(-\frac{2}{(t+1)^3}\) and the derivative of \(-\frac{1}{t^2}\) is \(\frac{2}{t^3}\). Thus, \(\mathbf{a}(t) = -\frac{2}{(t+1)^3}\mathbf{i} + \frac{2}{t^3}\mathbf{j}\).

Step 5: Evaluate the Acceleration at Given t

Substitute \(t = -\frac{1}{2}\) into the acceleration vector formula: \(\mathbf{a}(-\frac{1}{2}) = -\frac{2}{(-\frac{1}{2}+1)^3}\mathbf{i} + \frac{2}{(-\frac{1}{2})^3}\mathbf{j}\). After simplification, this results in \(-16\mathbf{i} - 16\mathbf{j}\).

Key Concepts

Parametric Equations

Parametric equations provide a way to express a set of related quantities as explicit functions of an independent parameter, often time \( t \). In the context of particle motion, parametric equations allow us to describe the path of a particle in terms of its coordinates in the plane. For example, given the parametric equations \( x(t) = \frac{t}{t+1} \) and \( y(t) = \frac{1}{t} \), the coordinates \( x \) and \( y \) are defined as functions of \( t \). Importance of Parametric Equations:

• They provide a dynamic representation of geometrical figures.
• They are useful in physics and engineering for modeling trajectories.
• They can describe complex behaviors that are not easily captured by standard equations.

In this exercise, we converted these parametric equations into a single equation by eliminating the parameter \( t \). This involves solving one of the parametric equations for \( t \) and then substituting into the other. Thus, we found the path equation \( y = 1 - x \), which suggests the particle follows a straight line.

Velocity Vector

The velocity vector is crucial as it describes the rate and direction of the particle's motion at any point in time. By differentiating the position vector \( \mathbf{r}(t) \) with respect to time \( t \), we obtain the velocity vector \( \mathbf{v}(t) \). Given \( \mathbf{r}(t) = \frac{t}{t+1} \mathbf{i} + \frac{1}{t} \mathbf{j} \), differentiating each component separately gives:Velocity Components:

• For \( x(t) = \frac{t}{t+1} \), the derivative is \( \frac{1}{(t+1)^2} \).
• For \( y(t) = \frac{1}{t} \), the derivative is \( -\frac{1}{t^2} \).

When calculated, this forms the velocity vector \( \mathbf{v}(t) = \frac{1}{(t+1)^2} \mathbf{i} - \frac{1}{t^2} \mathbf{j} \).To find the velocity at a specific time, such as \( t = -\frac{1}{2} \), we substitute \( t \) values. This gives us \( \mathbf{v}(-\frac{1}{2}) = 4 \mathbf{i} - 4 \mathbf{j} \). This vector indicates both the speed and direction of the particle at the given time.

Acceleration Vector

Acceleration is the rate at which velocity changes over time. It's a second derivative, derived from the velocity vector. The acceleration vector, \( \mathbf{a}(t) \), gives insight into how the velocity is changing and is crucial in dynamics.Finding the Acceleration Vector:

• Differentiate the \(i\)-component \(\frac{1}{(t+1)^2}\) to get \(-\frac{2}{(t+1)^3}\).
• Differentiate the \(j\)-component \(-\frac{1}{t^2}\) to get \(\frac{2}{t^3}\).

Therefore, the acceleration vector is \( \mathbf{a}(t) = -\frac{2}{(t+1)^3} \mathbf{i} + \frac{2}{t^3} \mathbf{j} \).To evaluate the acceleration at a specific time such as \( t = -\frac{1}{2} \), we substitute \( t \) into the acceleration equation. The result is \( -16\mathbf{i} - 16\mathbf{j} \), showing how strong and in which direction the particle is accelerating at that moment.