Problem 1 - Thomas Calculus

Question

In Exercises \(1-4, \mathbf{r}(t)\) 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 particle. Then find the particle's velocity and acceleration vectors at the given value of \(t .\) \begin{equation} \mathbf{r}(t)=(t+1) \mathbf{i}+\left(t^{2}-1\right) \mathbf{j}, \quad t=1 \end{equation}

Step-by-Step Solution

Answer

The path of the particle is \( y = x^2 - 2x \). Velocity at \( t=1 \) is \( \mathbf{i} + 2\mathbf{j} \). Acceleration at \( t=1 \) is \( 2\mathbf{j} \).

Step 1: Obtain the Parametric Equations

The given position vector \( \mathbf{r}(t) = (t+1) \mathbf{i} + (t^2-1) \mathbf{j} \) can be separated into parametric equations for \( x \) and \( y \). Thus, we have \( x = t + 1 \) and \( y = t^2 - 1 \).

Step 2: Eliminate the Parameter

Our goal is to find an equation in \( x \) and \( y \) only. From the equation \( x = t + 1 \), we isolate \( t \) as follows: \( t = x - 1 \). Substitute \( t \) into \( y = t^2 - 1 \): \( y = (x - 1)^2 - 1 \). Expanding, we get \( y = x^2 - 2x \).

Step 3: Calculate the Velocity Vector

The velocity vector \( \mathbf{v}(t) \) is the derivative of the position vector \( \mathbf{r}(t) \) with respect to time. Differentiate \( \mathbf{r}(t) \) to get \( \mathbf{v}(t) = \frac{d}{dt}((t+1) \mathbf{i} + (t^2-1) \mathbf{j}) = (1) \mathbf{i} + (2t) \mathbf{j} = \mathbf{i} + 2t \mathbf{j} \).

Step 4: Calculate the Velocity Vector at \( t = 1 \)

Substitute \( t = 1 \) into the velocity vector: \( \mathbf{v}(1) = \mathbf{i} + 2 \cdot 1 \mathbf{j} \). This gives \( \mathbf{v}(1) = \mathbf{i} + 2 \mathbf{j} \).

Step 5: Calculate the Acceleration Vector

The acceleration vector \( \mathbf{a}(t) \) is the derivative of the velocity vector \( \mathbf{v}(t) \). Differentiate \( \mathbf{v}(t) = \mathbf{i} + 2t \mathbf{j} \) with respect to time: \( \mathbf{a}(t) = \frac{d}{dt}(\mathbf{i} + 2t \mathbf{j}) = (0) \mathbf{i} + (2) \mathbf{j} = 2 \mathbf{j} \).

Step 6: Calculate the Acceleration Vector at \( t = 1 \)

The acceleration is constant, so \( \mathbf{a}(1) = 2 \mathbf{j} \).

Key Concepts

Velocity Vector

When discussing parametric equations, the velocity vector plays a crucial role in understanding how a particle moves. The velocity vector expresses the rate of change of the particle's position with respect to time. In parametric terms, this means it shows how fast the particle travels along its path in the plane.

To find a velocity vector, simply differentiate each component of the position vector with respect to time. For instance, if the position vector \( \mathbf{r}(t) = (t+1) \mathbf{i} + (t^2-1) \mathbf{j} \), you take the derivative of each part separately:

• The \( x \)-component: Derivative of \( t+1 \) is 1.
• The \( y \)-component: Derivative of \( t^2-1 \) is \( 2t \).

Thus, the velocity vector is \( \mathbf{v}(t) = \mathbf{i} + 2t \mathbf{j} \). This tells you that no matter what \( t \) is, the particle moves at a constant rate in the x-direction.

To find the specific velocity at a given time, say \( t = 1 \), plug it into the velocity vector, resulting in \( \mathbf{v}(1) = \mathbf{i} + 2 \mathbf{j} \). Now you know the particle moves right at 1 unit/second while moving up at 2 units/second at that moment.

Acceleration Vector

The acceleration vector gives us insight into how the velocity of a particle is changing over time. This vector is important for determining the texture of the particle’s motion—whether it is speeding up, or slowing down in any direction.

Calculating the acceleration vector involves taking the derivative of the velocity vector with respect to time. From our velocity vector \( \mathbf{v}(t) = \mathbf{i} + 2t \mathbf{j} \), the derivative is as follows:

• The \( x \)-component from \( \mathbf{i} \) differentiates to 0.
• The \( y \)-component \( 2t \) differentiates to 2.

Therefore, the acceleration vector is \( \mathbf{a}(t) = 2 \mathbf{j} \). This indicates that the acceleration is constant and happens entirely in the y-direction, not affecting the x-motion.

Since the acceleration vector does not depend on \( t \), the particular value \( t = 1 \) does not change it, meaning the acceleration at \( t = 1 \) is still \( 2 \mathbf{j} \). So the velocity is increasing by 2 units/second squared in the upward direction.

Eliminating Parameters

Eliminating parameters allows you to express a curve solely in terms of \( x \) and \( y \), removing the dependence on \( t \). This is often crucial for understanding the geometrical path a particle takes in a plane.

Using the parametric equations derived from the position vector, \( x = t + 1 \) and \( y = t^2 - 1 \), we start by solving for \( t \) in terms of \( x \):

• Solve \( x = t + 1 \) to get \( t = x - 1 \).

Substitute this expression for \( t \) into \( y = t^2 - 1 \), giving

\[ y = (x - 1)^2 - 1 \].

By expanding, the thinking becomes clear as it converts to a more familiar equation, \( y = x^2 - 2x \). This equation describes the path of the particle as a parabola opening upwards. By eliminating the parameter, we have an explicit curve equation in terms of \( x \) and \( y \), making it easier to understand its layout and characteristics on the plane.