Question

A position function \(\vec{r}(t)\) is given. Sketch \(\vec{r}(t)\) on the indicated interval. Find \(\vec{v}(t)\) and \(\vec{a}(t),\) then add \(\vec{v}\left(t_{0}\right)\) and \(\vec{a}\left(t_{0}\right)\) to your sketch, with their initial points at \(\vec{r}\left(t_{0}\right)\) for the given value of \(t_{0}\). $$ \vec{r}(t)=\langle t, \sin t\rangle \text { on }[0, \pi / 2] ; t_{0}=\pi / 4 $$

Short answer

Velocity: \( \langle 1, \sqrt{2}/2 \rangle \), Acceleration: \( \langle 0, -\sqrt{2}/2 \rangle \), both at \( t_0 = \pi/4 \).

Step-by-step solution

Sketch the Position Function

To sketch the position function \( \vec{r}(t) = \langle t, \sin t \rangle \) on the interval \([0, \pi/2]\), plot the values of \( t \) on the x-axis and \( \sin t \) on the y-axis. This interval starts at \( t = 0 \), where \( \vec{r}(t) \) is \( \langle 0, 0 \rangle \), and ends at \( t = \pi/2 \), where \( \vec{r}(t) \) is \( \langle \pi/2, 1 \rangle \). Draw a curve connecting these points to represent the path of \( \vec{r}(t) \).

Compute the Velocity Function

The velocity function \( \vec{v}(t) \) is the derivative of the position function \( \vec{r}(t) \). Differentiate each component: \( \frac{d}{dt} t = 1 \) and \( \frac{d}{dt} \sin t = \cos t \). Therefore, \( \vec{v}(t) = \langle 1, \cos t \rangle \).

Compute the Acceleration Function

The acceleration function \( \vec{a}(t) \) is the derivative of the velocity function \( \vec{v}(t) \). Differentiate each component of \( \vec{v}(t) = \langle 1, \cos t \rangle \): \( \frac{d}{dt} 1 = 0 \) and \( \frac{d}{dt} \cos t = -\sin t \). Therefore, \( \vec{a}(t) = \langle 0, -\sin t \rangle \).

Evaluate Velocity and Acceleration at \(t_0 = \pi/4\)

Substitute \( t_0 = \pi/4 \) into the velocity function: \( \vec{v}(\pi/4) = \langle 1, \cos(\pi/4) \rangle = \langle 1, \sqrt{2}/2 \rangle \). Substitute \( t_0 = \pi/4 \) into the acceleration function: \( \vec{a}(\pi/4) = \langle 0, -\sin(\pi/4) \rangle = \langle 0, -\sqrt{2}/2 \rangle \).

Sketch Velocity and Acceleration Vectors

At \( t_0 = \pi/4 \), the position is \( \vec{r}(\pi/4) = \langle \pi/4, \sin(\pi/4) \rangle = \langle \pi/4, \sqrt{2}/2 \rangle \). Place \( \vec{v}(\pi/4) = \langle 1, \sqrt{2}/2 \rangle \) and \( \vec{a}(\pi/4) = \langle 0, -\sqrt{2}/2 \rangle \) at \( \vec{r}(\pi/4) \). These vectors should originate from the point \( (\pi/4, \sqrt{2}/2) \) on your sketch, indicating the direction and magnitude of velocity and acceleration at \( t_0 = \pi/4 \).

Key concepts

Position Function

The position function, often denoted as \( \vec{r}(t) \), describes the location of a point as it moves over time. In vector calculus, this function is usually expressed in terms of a parameter \( t \), which typically represents time. The components of the function correspond to the coordinates in a given space.

In the original exercise, the position function is given by \( \vec{r}(t) = \langle t, \sin t \rangle \). Here, the first component \( t \) represents the horizontal position, while \( \sin t \) represents the vertical position. When sketching this function over the interval \([0, \pi/2]\), you plot \( t \) on the x-axis and \( \sin t \) on the y-axis, creating a curve that shows how the point moves as time progresses within this interval.

Velocity Function

The velocity function, \( \vec{v}(t) \), describes the rate of change of the position function and tells us how fast and in what direction a point is moving along its path. It is obtained by differentiating the position function with respect to time \( t \).

For the exercise at hand, the velocity function is derived from \( \vec{r}(t) = \langle t, \sin t \rangle \) by differentiating each component with respect to \( t \): \( \frac{d}{dt} t = 1 \) and \( \frac{d}{dt} \sin t = \cos t \). Thus, the velocity function becomes \( \vec{v}(t) = \langle 1, \cos t \rangle \).

This shows that horizontally, the velocity is constant (\(1\)), while vertically, it varies with \( \cos t \), indicating the influence of the sine curve's rate of change.

Acceleration Function

The acceleration function, \( \vec{a}(t) \), provides information about how the velocity changes over time. In other words, it gives the rate of change of the velocity function and is found by differentiating the velocity function with respect to \( t \).

In this example, the velocity function \( \vec{v}(t) = \langle 1, \cos t \rangle \) is differentiated resulting in \( \frac{d}{dt} 1 = 0 \) for the horizontal component and \( \frac{d}{dt} \cos t = -\sin t \) for the vertical component. Therefore, the acceleration function is \( \vec{a}(t) = \langle 0, -\sin t \rangle \). This tells us there is no change in horizontal velocity (constant speed), while the vertical acceleration is constantly opposing the sine function, pointing downward as a result of the negative sign.

Derivative

Derivatives play a crucial role in calculus, particularly in understanding how functions change. In the context of vector calculus, deriving a vector function with respect to a parameter, like time \( t \), helps us understand how each component of the function changes.

For a position function \( \vec{r}(t) \), its derivative with respect to \( t \) gives us the velocity function \( \vec{v}(t) \). Similarly, deriving \( \vec{v}(t) \) gives us the acceleration function \( \vec{a}(t) \).

This cascade of derivatives provides detailed insights into motion:

• The first derivative tells us about speed and direction (velocity).
• The second derivative indicates the rate of change of that speed and direction (acceleration).

Parametric Curves

Parametric curves are a way to represent curves in the plane or in space, using one or more parameters to describe the coordinates. In the realm of vectors, they offer a flexible way to describe more complex paths that cannot be easily expressed with standard Cartesian equations.

The position function \( \vec{r}(t) = \langle t, \sin t \rangle \) is a typical example of a parametric curve. Here, \( t \) is the parameter, and as \( t \) changes, \( \langle t, \sin t \rangle \) traces out a path driven by the interaction between the linear \( t \) progression and the periodic behavior of \( \sin t \).

This use of parametric equations is valuable in describing motions that are not simple straight lines or geometric shapes, allowing the depiction of real-world complex movements and curves.