Question

A position function \(\vec{r}(t)\) is given. Find \(\vec{v}(t)\) and \(\vec{a}(t)\). $$ \vec{r}(t)=\left\langle 3 t^{2}-2 t+1,-t^{2}+t+14\right\rangle $$

Short answer

\( \vec{v}(t) = \langle 6t - 2, -2t + 1 \rangle \), \( \vec{a}(t) = \langle 6, -2 \rangle \).

Step-by-step solution

Understanding the Problem

We are given a vector function \( \vec{r}(t) = \langle 3t^2 - 2t + 1, -t^2 + t + 14 \rangle \) that describes the position of a particle over time. We need to find the velocity vector \( \vec{v}(t) \) and the acceleration vector \( \vec{a}(t) \).

Finding the Velocity Function \( \vec{v}(t) \)

The velocity function \( \vec{v}(t) \) is the derivative of the position function \( \vec{r}(t) \). To find \( \vec{v}(t) \), differentiate each component of \( \vec{r}(t) \) with respect to \( t \).For the \( x \)-component: \( \frac{d}{dt}(3t^2 - 2t + 1) = 6t - 2 \).For the \( y \)-component: \( \frac{d}{dt}(-t^2 + t + 14) = -2t + 1 \).Therefore, \( \vec{v}(t) = \langle 6t - 2, -2t + 1 \rangle \).

Finding the Acceleration Function \( \vec{a}(t) \)

The acceleration function \( \vec{a}(t) \) is the derivative of the velocity function \( \vec{v}(t) \). To find \( \vec{a}(t) \), differentiate each component of \( \vec{v}(t) \) with respect to \( t \).For the \( x \)-component: \( \frac{d}{dt}(6t - 2) = 6 \).For the \( y \)-component: \( \frac{d}{dt}(-2t + 1) = -2 \).Therefore, \( \vec{a}(t) = \langle 6, -2 \rangle \).

Summary of Results

We have calculated the velocity vector \( \vec{v}(t) = \langle 6t - 2, -2t + 1 \rangle \) and the acceleration vector \( \vec{a}(t) = \langle 6, -2 \rangle \). These vectors describe the rate of change of the particle's position and the rate of change of velocity, respectively.

Key concepts

Differentiation

In vector calculus, differentiation is used to analyze how something changes over time. When you have a position function, differentiation helps find the velocity and acceleration vectors. Differentiating a vector function involves taking the derivative of each component of the vector separately. It’s like peeling back layers to reveal how each part of your vector changes at every moment.
For example, if we have a function \( \vec{r}(t) = \langle 3t^2 - 2t + 1, -t^2 + t + 14 \rangle \), we differentiate each term with respect to \( t \). By doing so, you can observe how the position changes with time, and then use this to find the velocity by again taking the derivative of the velocity to get the acceleration.

Velocity Vector

The velocity vector \( \vec{v}(t) \) tells us about the speed and direction of a particle's motion at any given time. It's derived from the position function by differentiating each component with respect to time. This provides the rate at which the position changes for each individual dimension.

• The velocity vector indicates how fast and in which direction the particle moves.
• For the given position \( \vec{r}(t) = \langle 3t^2 - 2t + 1, -t^2 + t + 14 \rangle \), the velocity vector \( \vec{v}(t) \) is found as \( \langle 6t - 2, -2t + 1 \rangle \).

Picking apart each element, you find the rate of change for the \( x \)-coordinate and the \( y \)-coordinate.

Acceleration Vector

An acceleration vector \( \vec{a}(t) \) describes how the velocity of a particle changes over time. It is the derivative of the velocity vector and gives insight into how the speed or direction of a particle is changing. In simple terms, it shows how quickly something speeds up, slows down, or changes direction.

• From the velocity vector \( \vec{v}(t) = \langle 6t - 2, -2t + 1 \rangle \), the acceleration vector is derived by differentiating each component.
• This gives us \( \vec{a}(t) = \langle 6, -2 \rangle \).

The components of the acceleration vector provide constant values, indicating constant rates of change in the respective components of velocity.

Position Function

A position function \( \vec{r}(t) \) indicates the location of a particle in a coordinate system at any given time \( t \). It provides vital data about where the particle is and serves as the base to discover velocity and acceleration vectors. The function is often defined in terms of \( t \), which facilitates the mapping of a particle’s path over time.

• The position function \( \vec{r}(t) = \langle 3t^2 - 2t + 1, -t^2 + t + 14 \rangle \) describes the particle's trajectory in a two-dimensional plane.
• This trajectory gives insight into the particle's journey, as both the vector and time evolve.

Understanding the position function is crucial because it sets the groundwork for further explorations of motion.

Particle Motion

Particle motion in physics is the study of how a particle moves through space over time. It involves understanding changes in the position, velocity, and acceleration of a particle along its path. By determining these aspects, you gain a comprehensive view of how particles behave under various forces or influences.

• In the context of a particle’s journey as defined by \( \vec{r}(t) \), the particle’s position is the starting point.
• Velocity and acceleration further elaborate on this motion, each step helping to fully flesh out the path and behavior over time.
• By analyzing the functions derived (\( \vec{v}(t) \) and \( \vec{a}(t) \)), you can predict future states of motion and understand past paths.

Selectively using these concepts illustrates the dynamic language of vector calculus, crucial for unraveling the mysteries of motion.