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 \(t_{0}\). $$ \vec{r}(t)=\left\langle t^{2}, \sin t^{2}\right\rangle \text { on }[0, \pi / 2] ; t_{0}=\sqrt{\pi / 4} $$

Short answer

\( \vec{v}(t) = \langle 2t, 2t \cos t^2 \rangle \), \( \vec{a}(t) = \langle 2, 2 \cos t^2 - 4t^2 \sin t^2 \rangle \).

Step-by-step solution

Determine the Velocity Function

To find the velocity function \( \vec{v}(t) \), differentiate the position function \( \vec{r}(t) = \langle t^2, \sin t^2 \rangle \) with respect to \( t \). Applying the derivative, we get:\[ \vec{v}(t) = \left\langle \frac{d}{dt}(t^2), \frac{d}{dt}(\sin t^2) \right\rangle = \left\langle 2t, \cos t^2 \cdot 2t \right\rangle = \langle 2t, 2t \cos t^2 \rangle \]

Determine the Acceleration Function

To find the acceleration function \( \vec{a}(t) \), differentiate the velocity function \( \vec{v}(t) = \langle 2t, 2t \cos t^2 \rangle \) with respect to \( t \). Using the product rule for the second component:\[ \vec{a}(t) = \left\langle \frac{d}{dt}(2t), \frac{d}{dt}(2t \cos t^2) \right\rangle \]\[ \vec{a}(t) = \langle 2, 2 \cos t^2 - 4t^2 \sin t^2 \rangle \]

Evaluate at \( t_{0} = \sqrt{\pi / 4} \)

First, calculate \( t_0 = \sqrt{\pi / 4} = \sqrt{\pi}/2 \). Now substitute \( t_0 \) into the functions to determine specific values. First, calculate \( \vec{r}(t_0) \):\[ \vec{r}(t_0) = \langle (\sqrt{\pi}/2)^2, \sin((\sqrt{\pi}/2)^2) \rangle = \langle \pi/4, \sin(\pi/4) \rangle = \langle \pi/4, \sqrt{2}/2 \rangle \]Next, calculate \( \vec{v}(t_0) \):\[ \vec{v}(t_0) = \langle 2(\sqrt{\pi}/2), 2(\sqrt{\pi}/2) \cos((\sqrt{\pi}/2)^2) \rangle = \langle \sqrt{\pi}, \sqrt{\pi} \cos(\pi/4) \rangle \]\[ = \langle \sqrt{\pi}, \sqrt{\pi}(\sqrt{2}/2) \rangle \]Finally, calculate \( \vec{a}(t_0) \):\[ \vec{a}(t_0) = \langle 2, 2 \cos(\pi/4) - 4(\sqrt{\pi}/2)^2 \sin(\pi/4) \rangle \]\[ = \langle 2, \sqrt{2} - \pi \sin(\pi/4) \rangle = \langle 2, \sqrt{2} - (\pi\sqrt{2}/4) \rangle \]

Sketch the Functions

Sketch the curve given by \( \vec{r}(t) = \langle t^2, \sin t^2 \rangle \) on the interval \([0, \pi/2]\). It is a parametric plot where the \( x \)-component is \( t^2 \) and the \( y \)-component is \( \sin t^2 \), which indicates that as \( t \) increases to \( \pi/2 \), the \( x \) values increase quadratically and the \( y \) values oscillate between 0 and 1. Highlight the position at \( t_0 = \sqrt{\pi}/2 \).Add the vector \( \vec{v}(t_0) = \langle \sqrt{\pi}, \sqrt{\pi}(\sqrt{2}/2) \rangle \) and vector \( \vec{a}(t_0) = \langle 2, \sqrt{2}-(\pi\sqrt{2}/4) \rangle \) with their initial points at \( \vec{r}(t_0) = \langle \pi/4, \sqrt{2}/2 \rangle \).

Key concepts

Position Function

In vector calculus, a position function describes the location of a point in space as a function of time, typically denoted as \( \vec{r}(t) \). It can be expressed in terms of components along different axes. For example, in a two-dimensional space, it can be represented as \( \vec{r}(t) = \langle x(t), y(t) \rangle \). The position function helps us track the trajectory of an object as it moves over time.

For the given problem, the position function is \( \vec{r}(t) = \langle t^2, \sin t^2 \rangle \), describing a path in a plane for \( t \) values within the interval \([0, \pi/2]\).

• The \( x \)-component, \( t^2 \), increases quadratically.
• The \( y \)-component, \( \sin t^2 \), oscillates between 0 and 1 as \( t \) changes.

Visualizing this path gives us an understanding of the object's movement relative to time, where the initial and final positions are key to understanding the journey.

Velocity Function

The velocity function represents the rate and direction of change of the position function over time. It is derived by differentiating the position function \( \vec{r}(t) \). If the position is given by \( \vec{r}(t) = \langle x(t), y(t) \rangle \), the velocity is given by \( \vec{v}(t) = \langle \frac{dx}{dt}, \frac{dy}{dt} \rangle \).

This yields the instantaneous rate of change of the object's position. For this exercise, differentiating \( \vec{r}(t) = \langle t^2, \sin t^2 \rangle \) gives:

• \( \vec{v}(t) = \langle 2t, 2t \cos t^2 \rangle \)

Here, \( 2t \) is the rate of change in the \( x \)-direction, and \( 2t \cos t^2 \) in the \( y \)-direction. When evaluated at a specific time \( t_0 = \sqrt{\pi/4} \), the velocity function explicitly tells us how quickly and in which direction the position is changing at that moment.

Acceleration Function

The acceleration function describes how the velocity of an object changes with time. It is derived by differentiating the velocity function with respect to time. In mathematical terms, if \( \vec{v}(t) = \langle v_x(t), v_y(t) \rangle \), then the acceleration function is \( \vec{a}(t) = \langle \frac{dv_x}{dt}, \frac{dv_y}{dt} \rangle \). This gives the rate of change of velocity, providing insight into the object's changing speed and direction.

For the problem at hand, differentiating the velocity function \( \vec{v}(t) = \langle 2t, 2t \cos t^2 \rangle \) results in:

• \( \vec{a}(t) = \langle 2, 2 \cos t^2 - 4t^2 \sin t^2 \rangle \)

Evaluating at \( t_0 = \sqrt{\pi/4} \) gives precise control over our understanding of how quickly the velocity is changing at that moment. The acceleration indicates whether the velocity is increasing or decreasing and in what direction.

Parametric Equations

Parametric equations define a set of related quantities as functions of one or more independent parameters. In the context of curves, these equations allow us to express a curve as a pair of equations, each representing one of the coordinates of a point on the curve.

In this exercise, two parametric equations \( x = t^2 \) and \( y = \sin t^2 \) are used to express the position function \( \vec{r}(t) = \langle t^2, \sin t^2 \rangle \).

• \( x = t^2 \) is the parametric equation for the x-component.
• \( y = \sin t^2 \) represents the y-component.

These equations allow the path to be described precisely and enable sketching the trajectory of the object in the plane. The use of parametric equations in such problems is crucial as it provides flexibility in tracking objects along complex, non-linear paths.