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. $$ \begin{array}{l} \vec{r}_{1}(t)=\left\langle t, t^{2}\right\rangle \text { on }[0,1] ; t_{0}=1 \\\ \vec{r}_{2}(s)=\left\langle s^{2}, s^{4}\right\rangle \text { on }[0,1] ; s_{0}=1 \end{array} $$

Short answer

(a) Positions are equal. (b) Velocity: \(\vec{v}_{1}(1) = \langle 1, 2 \rangle\), \(\vec{v}_{2}(1) = \langle 2, 4 \rangle\); Speed: \(\sqrt{5}\), \(2\sqrt{5}\); Acceleration: \(\langle 0, 2 \rangle\), \(\langle 2, 12 \rangle\).

Step-by-step solution

Verify Position Equality

We are given \(\vec{r}_{1}(t) = \langle t, t^2 \rangle \) and \(\vec{r}_{2}(s) = \langle s^2, s^4 \rangle\). For Part (a), substitute \(t_0 = 1\) into \(\vec{r}_{1}\) and \(s_0 = 1\) into \(\vec{r}_{2}\). Calculate \(\vec{r}_{1}(1) = \langle 1, 1^2 \rangle = \langle 1, 1 \rangle\). Similarly, \(\vec{r}_{2}(1) = \langle 1^2, 1^4 \rangle = \langle 1, 1 \rangle\). Since \(\vec{r}_{1}(1) = \vec{r}_{2}(1)\), the positions are the same at \(t_0 = 1\) and \(s_0 = 1\).

Find Velocity

The velocity is the derivative of the position function with respect to its parameter. For \(\vec{r}_{1}(t)\), calculate \(\vec{v}_{1}(t) = \frac{d}{dt}\langle t, t^2 \rangle = \langle 1, 2t \rangle\). Evaluate this at \(t_0 = 1\) to get \(\vec{v}_{1}(1) = \langle 1, 2 \times 1 \rangle = \langle 1, 2 \rangle\). For \(\vec{r}_{2}(s)\), calculate \(\vec{v}_{2}(s) = \frac{d}{ds}\langle s^2, s^4 \rangle = \langle 2s, 4s^3 \rangle\). Evaluate this at \(s_0 = 1\) to get \(\vec{v}_{2}(1) = \langle 2 \times 1, 4 \times 1^3 \rangle = \langle 2, 4 \rangle\).

Compute Speed

The speed is the magnitude of the velocity vector. For \(\vec{v}_{1}(1) = \langle 1, 2 \rangle\), speed is \(\sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5}\). For \(\vec{v}_{2}(1) = \langle 2, 4 \rangle\), speed is \(\sqrt{2^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}\).

Determine Acceleration

Acceleration is the derivative of velocity with respect to time or parameter. For \(\vec{v}_{1}(t) = \langle 1, 2t \rangle\), the acceleration \(\vec{a}_{1}(t) = \frac{d}{dt}\langle 1, 2t \rangle = \langle 0, 2 \rangle\). Thus, \(\vec{a}_{1}(1) = \langle 0, 2 \rangle\). For \(\vec{v}_{2}(s) = \langle 2s, 4s^3 \rangle\), the acceleration \(\vec{a}_{2}(s) = \frac{d}{ds}\langle 2s, 4s^3 \rangle = \langle 2, 12s^2 \rangle\). Thus, \(\vec{a}_{2}(1) = \langle 2, 12 \rangle\).

Key concepts

Position Vector

A position vector represents a specific location in space. It is expressed in terms of its components along the coordinate axes. In our exercise, we have two functions:

• \(\vec{r}_{1}(t) = \langle t, t^2 \rangle\) - a function of the parameter \(t\).
• \(\vec{r}_{2}(s) = \langle s^2, s^4 \rangle\) - a function based on the parameter \(s\).

Each position vector maps a parameter to a point in the 2D plane. The exercise demonstrates how at certain parameter values, both vectors achieve the same spatial point.
At \(t_0 = 1\) and \(s_0 = 1\), both functions output \(\langle 1, 1 \rangle\), confirming their positions coincide. This illustrates that different functions can describe the same paths using different parameters.

Velocity Vector

The velocity vector is derived by differentiating a position vector with respect to its parameter. It represents the direction and speed of an object's movement along its path.

• For \(\vec{r}_{1}(t) = \langle t, t^2 \rangle\), the velocity vector is \(\vec{v}_{1}(t) = \langle 1, 2t \rangle\).
• Substituting \(t_0 = 1\), we get \(\vec{v}_{1}(1) = \langle 1, 2 \rangle\).
• For \(\vec{r}_{2}(s) = \langle s^2, s^4 \rangle\), the velocity becomes \(\vec{v}_{2}(s) = \langle 2s, 4s^3 \rangle\).
• When \(s_0 = 1\), it results in \(\vec{v}_{2}(1) = \langle 2, 4 \rangle\).

These velocity vectors not only show how fast the position changes but also indicate the direction of this change. Although both paths coincide at one point, their velocities differ, showing distinct dynamics at that point.

Acceleration Vector

Acceleration vectors indicate how an object's velocity changes over time or with respect to a parameter. It is a measure of how quickly an object is speeding up, slowing down, or changing direction.

• For \(\vec{v}_{1}(t) = \langle 1, 2t \rangle\), differentiation gives the acceleration \(\vec{a}_{1}(t) = \langle 0, 2 \rangle\).
• Thus, \(\vec{a}_{1}(1) = \langle 0, 2 \rangle\), indicating a constant acceleration in the y-direction.
• For \(\vec{v}_{2}(s) = \langle 2s, 4s^3 \rangle\), the derived acceleration \(\vec{a}_{2}(s) = \langle 2, 12s^2 \rangle\) tells us how it accelerates with changing \(s\).
• Evaluating at \(s_0 = 1\), results in \(\vec{a}_{2}(1) = \langle 2, 12 \rangle\).

The accelerations reveal that although two objects may coincide in position, their rate of change in velocity could be drastically different due to varied underlying functions.

Parametric Equations

Parametric equations express a set of related quantities as functions of independent parameters. They greatly facilitate the study of curves and paths in calculus.
In the given exercise, parametric equations:

• \(\vec{r}_{1}(t) = \langle t, t^2 \rangle\) relies on the parameter \(t\) to express motion.
• \(\vec{r}_{2}(s) = \langle s^2, s^4 \rangle\) uses \(s\) as its varying quantity.

The strength of parametric equations lies in their flexibility to determine complex curves and motion paths, allowing for investigation at specific points like \(t_0\) and \(s_0\) by merely substituting those values. They make it easier to understand multi-dimensional motion, beyond what Cartesian coordinates could sometimes limit us to.