Question

A position function \(\vec{r}(t)\) is given, where \(t=0\) corresponds to the initial position. Find the arc length parameter \(s,\) and rewrite \(\vec{r}(t)\) in terms of \(s ;\) that is, find \(\vec{r}(s)\). \(\vec{r}(t)=\langle 2 t, t,-2 t\rangle\)

Short answer

The arc length parameter is \(s = 3t\), and \(\vec{r}(s) = \left\langle \frac{2s}{3}, \frac{s}{3}, -\frac{2s}{3} \right\rangle\).

Step-by-step solution

Find the Derivative of the Position Function

To determine the arc length parameter, we first need the derivative of the position function with respect to time, which gives us the velocity vector. Differentiate each component of \(\vec{r}(t) = \langle 2t, t, -2t \rangle\). The derivative is \(\vec{r}'(t) = \langle 2, 1, -2 \rangle\).

Calculate the Magnitude of the Velocity Vector

Next, find the magnitude of the velocity vector. The magnitude is given by \( ||\vec{r}'(t)|| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \).

Determine the Arc Length Parameter s(t)

The arc length from \(t = 0\) to \(t = t\) is given by \(s(t) = \int_0^t ||\vec{r}'(u)|| \, du\). Since \(||\vec{r}'(t)|| = 3\) is constant, \(s(t) = \int_0^t 3 \, du = 3t\).

Solve for t in terms of s

The relationship between \(s\) and \(t\) is given by \(s = 3t\). Solving for \(t\), we get \(t = \frac{s}{3}\).

Rewrite \(\vec{r}(t)\) in terms of \(s\)

Substitute \(t = \frac{s}{3}\) into the original position function \(\vec{r}(t) = \langle 2t, t, -2t \rangle\) to find \(\vec{r}(s)\). This gives \(\vec{r}(s) = \left\langle 2\left(\frac{s}{3}\right), \frac{s}{3}, -2\left(\frac{s}{3}\right) \right\rangle = \left\langle \frac{2s}{3}, \frac{s}{3}, -\frac{2s}{3} \right\rangle\).

Key concepts

Position Function

In mathematics and physics, a position function represents the location of an object over time. It's a vector function, often denoted as \( \vec{r}(t) \), where \( t \) represents time. This function provides coordinates in a specific space dimension. For instance, the given position function, \( \vec{r}(t) = \langle 2t, t, -2t \rangle \), is a line in three-dimensional space where:

• "2t" is the component along the x-axis,
• "t" is the component along the y-axis,
• "-2t" is the component along the z-axis.

The position function is crucial for determining the movement path of an object. It provides a direct way to calculate other parameters, such as velocity and arc length.

Velocity Vector

The velocity vector is derived by taking the derivative of the position function concerning time. This vector represents the rate of change of position, essentially describing how fast and in what direction an object is moving. For our position function \( \vec{r}(t) = \langle 2t, t, -2t \rangle \), the derivative is:

• \( \vec{r}'(t) = \langle 2, 1, -2 \rangle \)

Here, constant components indicate uniform motion:

• "2" shows each increase in time adds 2 units in the x-direction,
• "1" adds 1 unit in the y-direction,
• "-2" decreases the z-position by 2 units per time unit.

Understanding the velocity vector is key to analyzing how an object traverses the path outlined by its position function.

Derivative

The concept of a derivative is fundamental in calculus, providing the rate at which a function changes at any given point. When applied to a position function, its derivative yields a velocity vector. This represents instantaneous speed and direction. To differentiate \( \vec{r}(t) = \langle 2t, t, -2t \rangle \), each component is differentiated separately:

• The derivative of "2t" is 2,
• The derivative of "t" is 1,
• The derivative of "-2t" is -2.

By calculating the derivatives, we uncover the constant velocities of motion in each axis. It's the building block for calculating the arc length and parametrization of a curve.

Parameterization

Parameterization involves expressing a curve, like that of a position function, in terms of a parameter – here, the arc length \( s \). This process helps express any point on the curve simply and intuitively. In our case, the arc length from \( t=0 \) to any time \( t \) is calculated using the integral of the velocity's magnitude:

• The magnitude is \( ||\vec{r}'(t)|| = 3 \),
• Thus, the arc length is \( s(t) = \int_0^t 3 \, du = 3t \).

Solving for \( t \) in terms of \( s \) gives \( t = \frac{s}{3} \). Substituting this back into the position function gives \( \vec{r}(s) = \langle \frac{2s}{3}, \frac{s}{3}, -\frac{2s}{3} \rangle \). Parameterization with arc length ensures uniform distance plotting, a critical aspect for motion studies.