Question

Position functions \(\vec{r}_{1}(t)\) and \(\vec{r}_{2}(s)\) for two objects are given that follow the same path on the respective intervals. (a) Show that the positions are the same at the indicated \(t_{0}\) and \(s_{0}\) values; i.e., show \(\vec{r}_{1}\left(t_{0}\right)=\vec{r}_{2}\left(s_{0}\right) .\) (b) Find the velocity, speed and acceleration of the two objects at \(t_{0}\) and \(s_{0},\) respectively. $$ \begin{array}{l} \vec{r}_{1}(t)=\langle 3 t, 2 t\rangle \text { on }[0,2] ; t_{0}=2 \\ \vec{r}_{2}(s)=\langle 6 s-6,4 s-4\rangle \text { on }[1,2] ; s_{0}=2 \end{array} $$

Short answer

The positions are the same at \( t_0 \) and \( s_0 \), with velocities \( \langle 3, 2 \rangle \) and \( \langle 6, 4 \rangle \). Speeds are \( \sqrt{13} \) and \( 2\sqrt{13} \), accelerations \( \langle 0, 0 \rangle \).

Step-by-step solution

Substitute t and s Values

Begin by calculating the positions \( \vec{r}_{1}(t_0) \) and \( \vec{r}_{2}(s_0) \) by substituting \( t_0 = 2 \) and \( s_0 = 2 \) into their respective position functions.\\( \vec{r}_{1}(t_0) = \langle 3(2), 2(2) \rangle = \langle 6, 4 \rangle \). \( \vec{r}_{2}(s_0) = \langle 6(2) - 6, 4(2) - 4 \rangle = \langle 12 - 6, 8 - 4 \rangle = \langle 6, 4 \rangle \).

Show Positions Are Equal

Compare the two position vectors obtained in the previous step. Since \( \vec{r}_{1}(t_0) = \langle 6, 4 \rangle \) and \( \vec{r}_{2}(s_0) = \langle 6, 4 \rangle \), the positions are indeed the same at \( t_0 \) and \( s_0 \).

Calculate Velocities

Find the velocity vectors by differentiating the position functions with respect to time: \( \vec{v}_{1}(t) = \frac{d}{dt} \langle 3t, 2t \rangle = \langle 3, 2 \rangle \). \( \vec{v}_{2}(s) = \frac{d}{ds} \langle 6s-6, 4s-4 \rangle = \langle 6, 4 \rangle \).

Determine Speeds

Calculate the magnitudes of the velocity vectors to find the speeds:\( \text{Speed of } \vec{r}_1 = \| \vec{v}_{1}(t) \| = \sqrt{3^2+2^2} = \sqrt{13} \).\( \text{Speed of } \vec{r}_2 = \| \vec{v}_{2}(s) \| = \sqrt{6^2+4^2} = \sqrt{52} = 2\sqrt{13} \).

Calculate Accelerations

Determine the acceleration vectors by differentiating the velocity vectors with respect to time: \( \vec{a}_{1}(t) = \frac{d}{dt} \langle 3, 2 \rangle = \langle 0, 0 \rangle \). \( \vec{a}_{2}(s) = \frac{d}{ds} \langle 6, 4 \rangle = \langle 0, 0 \rangle \).

Verify Velocities and Accelerations at Given Points

Since the functions and their derivatives are constant (do not depend on \( t \) or \( s \)), the velocities and accelerations at \( t_0 \) and \( s_0 \) match the previously calculated values. Therefore, the velocity of \( \vec{r}_1 \) at \( t_0 \) is \( \langle 3, 2 \rangle \), speed is \( \sqrt{13} \), and acceleration is \( \langle 0, 0 \rangle \).For \( \vec{r}_2 \) at \( s_0 \), velocity is \( \langle 6, 4 \rangle \), speed is \( 2\sqrt{13} \), and acceleration is \( \langle 0, 0 \rangle \).

Key concepts

Position Vector Functions

Position vector functions are a fundamental concept in multivariable calculus. They are used to describe the path of a moving object in space. Unlike single-variable functions that deal with scalars, position vector functions deal with vectors, providing information on both the magnitude and direction of an object's position.

Each position vector can be represented as a function of one or more variables, often denoted by parameters such as time (t) or another variable (s). In this example, the position vectors are given by \( \vec{r}_{1}(t) = \langle 3t, 2t \rangle \) and \( \vec{r}_{2}(s) = \langle 6s - 6, 4s - 4 \rangle \).

• \( \vec{r}_{1}(t) \) describes the position of the first object as a function of time \( t \).
• \( \vec{r}_{2}(s) \) depicts the position of the second object as a function of a different parameter, \( s \).

By substituting specific values (in this case, \( t_0 = 2 \) and \( s_0 = 2 \)), we can evaluate their exact positions at these points, showing that they both meet at \( \langle 6, 4 \rangle \). This indicates that at these parameters, both objects are at the same point in space.

Velocity and Acceleration

Velocity and acceleration are key components in understanding motion, derivative concepts that stem from position vector functions.

- **Velocity**: This is the rate of change of the position vector with respect to time. It tells how quickly an object is moving along its path, and in which direction.
For \( \vec{r}_{1}(t) \), velocity is calculated by differentiating the position vector function with respect to \( t \):
\( \vec{v}_{1}(t) = \frac{d}{dt} \langle 3t, 2t \rangle = \langle 3, 2 \rangle \).
For \( \vec{r}_{2}(s) \), velocity is:
\( \vec{v}_{2}(s) = \frac{d}{ds} \langle 6s-6, 4s-4 \rangle = \langle 6, 4 \rangle \).

• This shows the first object has a constant velocity of \( \langle 3, 2 \rangle \), and the second a constant velocity of \( \langle 6, 4 \rangle \).

- **Acceleration**: This is the rate of change of velocity over time. Differentiating the velocity vectors gives us the acceleration.
\( \vec{a}_{1}(t) = \frac{d}{dt} \langle 3, 2 \rangle = \langle 0, 0 \rangle \), and similarly,
\( \vec{a}_{2}(s) = \frac{d}{ds} \langle 6, 4 \rangle = \langle 0, 0 \rangle \).
These acceleration vectors imply that both objects are moving at a constant velocity without acceleration, reaffirming the steady movement in their paths.

Differentiation of Vector Functions

Differentiation of vector functions is a technique used to compute velocity and acceleration in multivariable calculus. By differentiating a vector-valued function with respect to a scalar parameter, you find derivatives that inform the dynamics of motion, such as velocities and accelerations.

For a vector function \( \vec{r}(u) = \langle f(u), g(u) \rangle \), the derivative with respect to \( u \) is found by independently differentiating each component function:
\( \frac{d}{du} \vec{r}(u) = \langle f'(u), g'(u) \rangle \).

• In our case, to find velocities:
  - Differentiate \( \vec{r}_{1}(t) = \langle 3t, 2t \rangle \) to get \( \langle 3, 2 \rangle \), showing uniform motion.
• Differentiating \( \vec{r}_{2}(s) = \langle 6s-6, 4s-4 \rangle \) gives \( \langle 6, 4 \rangle \), yielding a different uniform velocity.

Acceleration is obtained by differentiating the velocity vector functions. Here, since the velocities themselves are constant, further differentiation yields zero vectors for acceleration \( \langle 0, 0 \rangle \). This process highlights how differentiation helps quantify the dynamical properties of objects in motion.