Question

A position function \(\vec{r}(t)\) is given. Find \(\vec{v}(t)\) and \(\vec{a}(t)\). $$ \vec{r}(t)=\langle 2 t+1,5 t-2,7\rangle $$

Short answer

\( \vec{v}(t) = \langle 2, 5, 0 \rangle \), \( \vec{a}(t) = \langle 0, 0, 0 \rangle \).

Step-by-step solution

Understand the problem

The position function \( \vec{r}(t) \) gives the position of a particle at time \( t \) in a three-dimensional space. Our task is to find the velocity function \( \vec{v}(t) \) and the acceleration function \( \vec{a}(t) \).

Differentiate the position function to find velocity

To find the velocity function \( \vec{v}(t) \), we need to differentiate the position function \( \vec{r}(t) = \langle 2t+1, 5t-2, 7 \rangle \) with respect to \( t \). The derivative of \( 2t+1 \) is \( 2 \), the derivative of \( 5t-2 \) is \( 5 \), and the derivative of the constant \( 7 \) is \( 0 \). Thus, \( \vec{v}(t) = \langle 2, 5, 0 \rangle \).

Differentiate the velocity function to find acceleration

Next, we differentiate the velocity function \( \vec{v}(t) = \langle 2, 5, 0 \rangle \) to get the acceleration function \( \vec{a}(t) \). Since all components of \( \vec{v}(t) \) are constants, their derivatives with respect to \( t \) will be \( 0 \). Therefore, \( \vec{a}(t) = \langle 0, 0, 0 \rangle \).

Interpret the results

The velocity function \( \vec{v}(t) = \langle 2, 5, 0 \rangle \) indicates that the particle is moving at a constant velocity in the direction defined by the vector. The acceleration function \( \vec{a}(t) = \langle 0, 0, 0 \rangle \) indicates that there is no change in velocity, confirming uniform motion.

Key concepts

Position Function

The position function, often denoted as \( \vec{r}(t) \), is a fundamental concept in vector calculus. It describes the location of a point or particle in a space as a function of time. The function maps each time \( t \) to a vector that points from the origin to the position of the particle in a coordinate space. For example, if you have a position function \( \vec{r}(t) = \langle 2t+1, 5t-2, 7 \rangle \), this tells you that at any time \( t \), the position of the particle is described by the coordinates \( (2t+1, 5t-2, 7) \).

• The component \( 2t+1 \) describes the position in the x-direction.
• The component \( 5t-2 \) describes the position in the y-direction.
• The constant component \( 7 \) represents a fixed position in the z-direction, indicating movement occurs only in the xy-plane.

Understanding the position function is crucial because it serves as the starting point for calculating both velocity and acceleration functions by differentiating with respect to time.

Velocity Function

The velocity function, symbolized as \( \vec{v}(t) \), describes how fast and in which direction a particle's position is changing over time. To find this function, you differentiate the position function with respect to \( t \).

In our example, with the position function \( \vec{r}(t) = \langle 2t+1, 5t-2, 7 \rangle \), the derivatives of each component tell us the rate at which they change:

• The derivative of \( 2t+1 \) is \( 2 \), meaning the x-position increases by 2 units per unit time.
• The derivative of \( 5t-2 \) is \( 5 \), indicating the y-position increases by 5 units per unit time.
• The derivative of the constant \( 7 \) is \( 0 \), showing no change in the z-direction.

Thus, the velocity function is \( \vec{v}(t) = \langle 2, 5, 0 \rangle \), signifying a constant velocity with no z-axis movement. This straightforward calculation shows that by understanding and applying derivatives, we can determine the velocity vector directly from the position function.

Acceleration Function

The acceleration function, denoted by \( \vec{a}(t) \), measures the rate of change of the velocity function with respect to time. To find this, we take the derivative of the velocity function. Acceleration indicates whether the velocity of the particle is changing.

For the velocity function \( \vec{v}(t) = \langle 2, 5, 0 \rangle \), each component is a constant, meaning:

• The derivative of the x-component \( 2 \) is \( 0 \), showing no change in velocity.
• The derivative of the y-component \( 5 \) is \( 0 \), also indicating no change in velocity.
• The constant z-component of \( 0 \) remains \( 0 \).

Thus, the acceleration function is \( \vec{a}(t) = \langle 0, 0, 0 \rangle \), evidencing zero acceleration and confirming the particle's movement is uniform. This means the velocity does not change over time, reiterating the concept of constant velocity in vector calculus.