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 \(\vec{r}(s)\). \(\vec{r}(t)=\langle 7 \cos t, 7 \sin t\rangle\)

Short answer

The arc length parameter is \( s = 7t \), and \( \vec{r}(s) = \langle 7 \cos \left( \frac{s}{7} \right), 7 \sin \left( \frac{s}{7} \right) \rangle \).

Step-by-step solution

Differentiate the Position Function

To find the arc length parameter, first differentiate the position function \( \vec{r}(t) = \langle 7 \cos t, 7 \sin t \rangle \) with respect to \( t \). The derivative is \( \vec{r}'(t) = \langle -7 \sin t, 7 \cos t \rangle \).

Compute the Magnitude of the Derivative

Find the magnitude of the velocity vector \( \vec{r}'(t) = \langle -7 \sin t, 7 \cos t \rangle \). The magnitude is calculated as \( \left| \vec{r}'(t) \right| = \sqrt{(-7 \sin t)^2 + (7 \cos t)^2} = \sqrt{49 \sin^2 t + 49 \cos^2 t} = 7 \sqrt{ \sin^2 t + \cos^2 t } = 7 \).

Integrate the Magnitude to Find s(t)

Since the magnitude of the velocity is constant, the arc length parameter \( s(t) \) is the integral of \( \left| \vec{r}'(t) \right| \) from 0 to \( t \). Thus, \( s(t) = \int_0^t 7 \ dt = 7t \).

Express t in terms of s

Now solve for \( t \) in terms of \( s \) using the relationship \( s = 7t \). Solving gives \( t = \frac{s}{7} \).

Rewrite the Position Vector in Terms of s

Substitute \( t = \frac{s}{7} \) into \( \vec{r}(t) = \langle 7 \cos t, 7 \sin t \rangle \) to express the position in terms of \( s \): \( \vec{r}(s) = \langle 7 \cos \left( \frac{s}{7} \right), 7 \sin \left( \frac{s}{7} \right) \rangle \).

Key concepts

Position Function

Understanding the position function is crucial in physics and calculus. It helps describe how the coordinates of an object change over time. Here, the position function is represented as a vector function \[ \vec{r}(t) = \langle 7 \cos t, 7 \sin t \rangle \]. This function specifies the path of a particle on a circle of radius 7, starting at an angle related to time \( t \).

• \( 7 \cos t \): Determines the x-coordinate.
• \( 7 \sin t \): Determines the y-coordinate.

These trigonometric functions signify circular motion, linking time \( t \) to the angle in radians through which the particle has traveled since \( t = 0 \). This function can help calculate positions and predict future locations by setting different values of \( t \).

Velocity Vector

To grasp how quickly the position changes with time, we need the velocity vector. It is the derivative of the position function \[ \vec{r}'(t) = \langle -7 \sin t, 7 \cos t \rangle \].The velocity vector provides both the speed and direction of the particle.

• Deriving each part of \( \vec{r}(t) \) shows how the x and y components change over time.
• The negative sign in \( -7 \sin t \) indicates direction, meaning the x-direction changes opposite to the increase of \( t \).

Velocity vector is fundamental to understanding motion characteristics such as when the particle is speeding up, slowing down, or changing its travel direction.

Integration of Velocity

The process of integrating the velocity vector determines the arc length parameter \( s(t) \), an essential step in arc length parameterization. Since the magnitude of the velocity, which is the speed, is constant for circular motion, the arc length equals speed multiplied by time.

• The calculation of \( s(t) = \int_0^t 7 \ dt = 7t \) shows this relationship.
• This integration translates velocity over a time interval into the actual "distance" traveled along the path, or arc length.

Understanding this process helps in converting between the parameters \( t \) and \( s(t) \), crucial when switching expression forms for problems in physics or engineering.

Magnitude of Derivative

Calculating the magnitude of the derivative of the position function \( \vec{r}(t) \) gives the speed of the particle. In this example, the derivative is \( \vec{r}'(t) \), and its magnitude is \[ \left| \vec{r}'(t) \right| = \sqrt{(-7 \sin t)^2 + (7 \cos t)^2} = 7 \].

• This shows us that, regardless of \( t \), the speed or velocity's magnitude is constant at 7 units.
• The use of Pythagorean identity \( \sin^2 t + \cos^2 t = 1 \) simplifies calculations.

A constant speed simplifies finding arc length because it implies linearity in relation to time. Understanding magnitude allows us to interpret the physical understanding of the motion described by the vector function.