Problem 12 - Thomas Calculus

Question

Exercises \(9-12\) give the position vectors of particles moving along various curves in the \(x y\) -plane. In each case, find the particle's velocity and acceleration vectors at the stated times and sketch them as vectors on the curve. Motion on the parabola \(y=x^{2}+1\) $$\mathbf{r}(t)=t \mathbf{i}+\left(t^{2}+1\right) \mathbf{j} ; \quad t=-1,0,\quad and \quad1$$

Step-by-Step Solution

Answer

Velocity: \( \mathbf{v}(t) = \mathbf{i} + 2t \mathbf{j} \). Acceleration: \( \mathbf{a}(t) = 2 \mathbf{j} \).

Step 1: Understand the Given Function

The position vector of a particle moving along a parabola is given by \( \mathbf{r}(t) = t \mathbf{i} + (t^2 + 1) \mathbf{j} \), which describes a particle moving in the xy-plane.

Step 2: Find the Velocity Vector

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

Step 3: Find the Acceleration Vector

The acceleration vector \( \mathbf{a}(t) \) is the derivative of the velocity vector \( \mathbf{v}(t) \). Differentiate \( \mathbf{v}(t) \), \( \frac{d}{dt}(\mathbf{i}) = 0 \) and \( \frac{d}{dt}(2t \mathbf{j}) = 2 \mathbf{j} \). Thus, \( \mathbf{a}(t) = 2 \mathbf{j} \).

Step 4: Evaluate at Given Times

Evaluate the velocity and acceleration vectors at \( t = -1, 0, 1 \).- For \( t = -1 \): \( \mathbf{v}(-1) = \mathbf{i} - 2 \mathbf{j} \), \( \mathbf{a}(-1) = 2 \mathbf{j} \).- For \( t = 0 \): \( \mathbf{v}(0) = \mathbf{i} \), \( \mathbf{a}(0) = 2 \mathbf{j} \).- For \( t = 1 \): \( \mathbf{v}(1) = \mathbf{i} + 2 \mathbf{j} \), \( \mathbf{a}(1) = 2 \mathbf{j} \).

Step 5: Sketch the Vectors

On a sketch of the parabola \( y = x^2 + 1 \), plot the position of the particle at each of the given times. Draw the velocity vectors as arrows from these points, angled consistently with the direction \( \mathbf{v}(t) \) points. Similarly, draw the constant acceleration vector \( \mathbf{a}(t) = 2 \mathbf{j} \), which will be the same at each point, as vertical arrows along the y-direction.

Key Concepts

Position Vector

In vector calculus, a **position vector** is a fundamental concept used to describe the location of a point in space relative to an origin. For the given problem, the position vector is defined as \( \mathbf{r}(t) = t \mathbf{i} + (t^2 + 1) \mathbf{j} \). This expression tells us how the particle moves along the parabola described by \( y = x^2 + 1 \) in the xy-plane.

• The \( t \mathbf{i} \) component represents movement along the x-axis. As time, \( t \), changes, so does the x-coordinate of the particle.
• The \((t^2 + 1) \mathbf{j}\) component represents movement along the y-axis. Here, the position depends on the square of \( t \) plus one, showing how the particle's height changes over time.

Understanding the position vector helps in visualizing the particle's path as it moves over time. This vector ultimately serves as the basis for finding other related vectors, like velocity and acceleration.

Velocity Vector

A **velocity vector** gives the rate of change of the position vector with respect to time. It shows both the speed and direction of the particle's motion. To find this, we take the derivative of the position vector \( \mathbf{r}(t) = t \mathbf{i} + (t^2 + 1) \mathbf{j} \) with respect to time \( t \).

Deriving the Velocity Vector

We compute:

• \( \frac{d}{dt}(t \mathbf{i}) = \mathbf{i} \) indicates a constant unit movement along the x-axis.
• \( \frac{d}{dt}((t^2 + 1) \mathbf{j}) = 2t \mathbf{j} \) shows a change in y-component, which varies with \( t \).

Thus, the velocity vector becomes \( \mathbf{v}(t) = \mathbf{i} + 2t \mathbf{j} \).
This vector provides crucial insights into how quickly and in which direction the particle is moving at any given time.

Acceleration Vector

The **acceleration vector** indicates the rate of change of the velocity vector over time. It captures how the particle's speed and direction are changing. To find the acceleration vector, differentiate the velocity vector \( \mathbf{v}(t) = \mathbf{i} + 2t \mathbf{j} \) with respect to \( t \).

Calculating the Acceleration Vector

Differentiating yields:

• \( \frac{d}{dt}(\mathbf{i}) = 0 \) as the x-component doesn't change with time.
• \( \frac{d}{dt}(2t \mathbf{j}) = 2 \mathbf{j} \) is the constant rate of change in the y-direction.

This results in an acceleration vector \( \mathbf{a}(t) = 2 \mathbf{j} \), which is constant. This tells us that the particle experiences a constant upward acceleration at every point along its path.

Derivative in Calculus

At the heart of vector calculus is the concept of the **derivative**, which is essential for understanding how quantities change. When we discuss derivatives in this context, we're usually referring to the derivative of vector functions, such as position vectors.

Importance of Derivatives

• The derivative helps us find the velocity vector by differentiating the position vector. It tells us how fast the position is changing over time.
• By further differentiating the velocity vector, we derive the acceleration vector, indicating how the velocity is changing at each moment.

The derivative is a critical tool in calculus, allowing us to analyze and predict the motion of particles across various contexts. Understanding how to take and apply derivatives is key to solving real-world problems involving motion.