Problem 1 - Essential Calculus Early Transcendentals

Question

Find the velocity, acceleration, and speed of a particle with the given position function. Sketch the path of the particle and draw the velocity and acceleration vectors for the specified value of \(t .\) $$\mathbf{r}(t)=\left\langle-\frac{1}{2} t^{2}, t\right\rangle, \quad t=2$$

Step-by-Step Solution

Answer

Velocity: \( \langle -2, 1 \rangle \), Acceleration: \( \langle -1, 0 \rangle \), Speed: \( \sqrt{5} \).

Step 1: Find the velocity vector

The velocity vector \( \mathbf{v}(t) \) is the derivative of the position vector \( \mathbf{r}(t) \). Calculate the derivative of \( \mathbf{r}(t) = \left\langle -\frac{1}{2}t^2, t \right\rangle \):\[ \mathbf{v}(t) = \frac{d}{dt}\left\langle -\frac{1}{2}t^2, t \right\rangle = \left\langle -t, 1 \right\rangle \] Now substitute \( t = 2 \):\[ \mathbf{v}(2) = \left\langle -2, 1 \right\rangle \]

Step 2: Find the acceleration vector

The acceleration vector \( \mathbf{a}(t) \) is the derivative of the velocity vector \( \mathbf{v}(t) \).Calculate the derivative of \( \mathbf{v}(t) = \left\langle -t, 1 \right\rangle \):\[ \mathbf{a}(t) = \frac{d}{dt}\left\langle -t, 1 \right\rangle = \left\langle -1, 0 \right\rangle \] Note that the acceleration vector is constant, as it does not depend on \( t \).

Step 3: Calculate the speed

The speed is the magnitude of the velocity vector \( \mathbf{v}(2) \).Calculate the magnitude of \( \mathbf{v}(2) = \left\langle -2, 1 \right\rangle \):\[ \text{Speed} = ||\mathbf{v}(2)|| = \sqrt{(-2)^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5} \]

Step 4: Sketch the path and vectors at t=2

The position of the particle at \( t=2 \) is given by \( \mathbf{r}(2) = \left\langle -2, 2 \right\rangle \).The path of the particle is a parabola opening to the left for \( t \geq 0 \).At \( t=2 \), draw the position vector to the point (-2, 2), the velocity vector \( \mathbf{v}(2) = \left\langle -2, 1 \right\rangle \), which points downwards to the left, and the acceleration vector \( \mathbf{a}(t) = \left\langle -1, 0 \right\rangle \), which is constant and points to the left. These vectors should be drawn starting at the position \( (-2, 2) \).

Key Concepts

Velocity Vector

The velocity vector is a crucial concept in particle motion. It represents the direction and speed of a particle.
To find the velocity vector, we differentiate the position function.

• The position function is given as \( \mathbf{r}(t) = \left\langle -\frac{1}{2}t^2, t \right\rangle \).
• We find the derivative: \( \mathbf{v}(t) = \frac{d}{dt}\left\langle -\frac{1}{2}t^2, t \right\rangle = \left\langle -t, 1 \right\rangle \).
• This derivative tells us the rate of change of the position function.
  We now have the velocity vector for any time \(t\).

For the specified time \( t = 2 \), the velocity vector is \( \mathbf{v}(2) = \langle -2, 1 \rangle \).
This result means the particle is moving to the left and slightly upwards at this point in time.

Acceleration Vector

Acceleration vectors tell us how the velocity of a particle changes over time. The acceleration vector is the derivative of the velocity vector.

• Our velocity vector is found as \( \mathbf{v}(t) = \langle -t, 1 \rangle \).
• To find acceleration, differentiate the velocity vector:
  \( \mathbf{a}(t) = \frac{d}{dt}\langle -t, 1 \rangle = \langle -1, 0 \rangle \)

This tells us that the acceleration is constant.
With the acceleration vector at \( \langle -1, 0 \rangle \), the particle's velocity is decreasing horizontally at a constant rate.

Position Function

The position function gives us the exact location of the particle at any given time. It acts as a roadmap of the path which the particle follows.

• In our exercise, the position function is \( \mathbf{r}(t) = \left\langle -\frac{1}{2}t^2, t \right\rangle \).
• When \( t = 2 \), plug it into the position function: \( \mathbf{r}(2) = \langle -2, 2 \rangle \).

This point \((-2, 2)\) represents the particle's location on the path at time \( t = 2 \).
The graph of this position function is a parabola opening leftwards.
This path is captured by sketching the trajectory, which illustrates the motion across different time points.

Speed Calculation

The speed of a particle is the magnitude of its velocity vector.
It is a scalar quantity that measures how fast the particle is moving.

• Given the velocity vector at \( t=2 \) as \( \mathbf{v}(2) = \langle -2, 1 \rangle \).
• Calculate its magnitude:
  \[ \text{Speed} = ||\mathbf{v}(2)|| = \sqrt{(-2)^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5} \]

The result, \( \sqrt{5} \), gives the particle's speed at that particular time.
Speed differs from velocity as it does not have a direction.