Problem 6 - Thomas Calculus

Question

In Exercises \(5-8, \mathbf{r}(t)\) 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 par- ticle. Then find the particle's velocity and acceleration vectors at the given value of \(t\) . $$ \mathbf{r}(t)=\frac{t}{t+1} \mathbf{i}+\frac{1}{t} \mathbf{j}, \quad t=-\frac{1}{2} $$

Step-by-Step Solution

Answer

Path equation: \(xy = 1-x\). Velocity at \(t=-1/2\): \(4\mathbf{i} - 4\mathbf{j}\). Acceleration at \(t=-1/2\): \(-16\mathbf{i} - 16\mathbf{j}\).

Step 1: Eliminate the parameter

To find an equation in terms of \(x\) and \(y\), we'll first write the expressions for the \(x\) and \(y\) components: \(x = \frac{t}{t+1}\) and \(y = \frac{1}{t}\). To eliminate the parameter \(t\), solve for \(t\) in terms of \(x\) from the first equation: \(t = \frac{x}{1-x}\). Substitute this expression for \(t\) into the equation for \(y\): \(y = \frac{1}{(\frac{x}{1-x})} = \frac{1-x}{x}\). So the equation of the path is \(xy = 1 - x\).

Step 2: Compute the derivative

The velocity vector is the derivative of the position vector \(\mathbf{r}(t)\) with respect to \(t\). Differentiate each component: \(\mathbf{v}(t) = \frac{d}{dt} \left(\frac{t}{t+1}\right) \mathbf{i} + \frac{d}{dt} \left(\frac{1}{t}\right) \mathbf{j}\). Use the quotient rule: \(\frac{d}{dt}(\frac{u}{v}) = \frac{vu' - uv'}{v^2}\). For the \(x\)-component, \(\frac{d}{dt}(\frac{t}{t+1}) = \frac{(t+1)\cdot 1 - t \cdot 1}{(t+1)^2} = \frac{1}{(t+1)^2}\). For the \(y\)-component, \(\frac{d}{dt}(\frac{1}{t}) = -\frac{1}{t^2}\). Thus, \(\mathbf{v}(t) = \frac{1}{(t+1)^2} \mathbf{i} - \frac{1}{t^2} \mathbf{j}\).

Step 3: Calculate the velocity at t = -1/2

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

Step 4: Compute the second derivative

The acceleration vector is the derivative of the velocity vector \(\mathbf{v}(t)\) with respect to \(t\). Differentiate each component: \(\mathbf{a}(t) = \frac{d}{dt} \left(\frac{1}{(t+1)^2}\right) \mathbf{i} + \frac{d}{dt} \left(-\frac{1}{t^2}\right) \mathbf{j}\). For the \(x\)-component, use the chain rule: \(-2\cdot (t+1)^{-3}\cdot 1 = -\frac{2}{(t+1)^3}\). For the \(y\)-component, \(\frac{2}{t^3}\). Thus, \(\mathbf{a}(t) = -\frac{2}{(t+1)^3} \mathbf{i} + \frac{2}{t^3} \mathbf{j}\).

Step 5: Calculate the acceleration at t = -1/2

Substitute \(t = -\frac{1}{2}\) into the acceleration vector \(\mathbf{a}(t) = -\frac{2}{(t+1)^3} \mathbf{i} + \frac{2}{t^3} \mathbf{j}\): \(\mathbf{a}(-\frac{1}{2}) = -\frac{2}{((-\frac{1}{2})+1)^3} \mathbf{i} + \frac{2}{(-\frac{1}{2})^3} \mathbf{j}\). Simplify: \(\mathbf{a}(-\frac{1}{2}) = -16 \mathbf{i} - 16 \mathbf{j} \).

Key Concepts

Position Vector

In the context of parametric equations, a position vector is a mathematical representation that describes the location of a point in space at any given time, often written as \( \mathbf{r}(t) \). Here, "\( t \)" is the parameter, typically representing time. The position vector is expressed in terms of its components in the Cartesian coordinate system: the \( x \)-coordinate as \( \mathbf{i} \) and the \( y \)-coordinate as \( \mathbf{j} \).
In the given problem, the position vector can be written as \( \mathbf{r}(t) = \frac{t}{t+1} \mathbf{i} + \frac{1}{t} \mathbf{j} \). This means that the \( x \)-coordinate is \( \frac{t}{t+1} \) and the \( y \)-coordinate is \( \frac{1}{t} \).
Understanding the path is crucial. By eliminating the parameter \( t \), we deduce the path the particle takes as it moves. Here, solving for \( t \) in terms of \( x \) and replacing in \( y \) reveals the equation \( xy = 1 - x \), indicating the particle's trajectory in the \( xy \)-plane without the need for time as a variable.

Velocity Vector

A velocity vector provides information not only about the speed at which a particle is moving but also the direction of its movement. It is essentially the derivative of the position vector with respect to time, \( \mathbf{v}(t) = \frac{d\mathbf{r}(t)}{dt} \).
In our exercise, the velocity vector is derived as \( \mathbf{v}(t) = \frac{1}{(t+1)^2} \mathbf{i} - \frac{1}{t^2} \mathbf{j} \). This means that each component of the position vector is differentiated separately:

• The \( x \)-component is differentiated using the quotient rule.
• The result is \( \frac{1}{(t+1)^2} \).
• The \( y \)-component results in \( -\frac{1}{t^2} \).

Thus, the velocity at \( t = -\frac{1}{2} \) is calculated to be \( 4 \mathbf{i} - 4 \mathbf{j} \), meaning the particle moves equally in the negative \( y \)- and positive \( x \)-directions at this specific time.

Acceleration Vector

The acceleration vector illustrates how a particle's velocity changes over time, thus offering insights into its dynamics. It is obtained by differentiating the velocity vector, \( \mathbf{a}(t) = \frac{d\mathbf{v}(t)}{dt} \).
For the scenario provided, the corresponding acceleration vector is \( \mathbf{a}(t) = -\frac{2}{(t+1)^3} \mathbf{i} + \frac{2}{t^3} \mathbf{j} \). In differentiation, we apply:

• The chain rule to find the derivative of \( \frac{1}{(t+1)^2} \), resulting in \( -\frac{2}{(t+1)^3} \).
• Similar steps for \( -\frac{1}{t^2} \) leads to \( \frac{2}{t^3} \).

Substituting \( t = -\frac{1}{2} \) gives an acceleration of \( -16 \mathbf{i} - 16 \mathbf{j} \). This indicates that the particle is decelerating in both \( x \) and \( y \) directions equally at this time, suggesting an alteration in the speed and direction of its motion.