Question

Consider the vector-valued function \(\mathbf{r}(t)=t \mathbf{i}+\left(4-t^{2}\right) \mathbf{j}\) (a) Sketch the graph of \(\mathbf{r}(t)\). Use a graphing utility to verify your graph. (b) Sketch the vectors \(\mathbf{r}(1), \mathbf{r}(1.25)\), and \(\mathbf{r}(1.25)-\mathbf{r}(1)\) on the graph in part (a). (c) Compare the vector \(\mathbf{r}^{\prime}(1)\) with the vector \(\frac{\mathbf{r}(1.25)-\mathbf{r}(1)}{1.25-1}\).

Short answer

The graph of \(\mathbf{r}(t)\) is a parabola opening downwards with the vertex located at (0,4). The vectors \(\mathbf{r}(1)\), \(\mathbf{r}(1.25)\) are located on this graph and represented as arrows from the origin to the corresponding points. The vector \(\mathbf{r}^{\prime}(1)\) is tangent to the curve at \(\mathbf{r}(1)\), and the vector \(\frac{\mathbf{r}(1.25)-\mathbf{r}(1)}{1.25-1}\) points in the direction of the average rate of change of the function between \(\mathbf{r}(1)\) and \(\mathbf{r}(1.25)\). The two vectors are close in calculation assuming that the function changes linearly over small intervals like [1, 1.25].

Step-by-step solution

Sketch the graph of the vector-valued function

The vector-valued function \(\mathbf{r}(t)=t \mathbf{i}+(4-t^{2}) \mathbf{j}\) is a parametric function of time \(t\). The x-coordinate is simply \(t\) and the y-coordinate is \(4 - t^2\). To sketch the graph, plot these coordinates as values of \(t\) change.

Sketch the vectors \(\mathbf{r}(1), \mathbf{r}(1.25)\), and \(\mathbf{r}(1.25)-\mathbf{r}(1)\)

To sketch the vectors \(\mathbf{r}(1)\) and \(\mathbf{r}(1.25)\), substitute the corresponding values into the vector-valued function \(\mathbf{r}(t)\). The resulting vectors can then be represented as arrows on the graph from the origin to the relevant point. The vector \(\mathbf{r}(1.25)- \mathbf{r}(1)\) represents the change from \(\mathbf{r}(1)\) to \(\mathbf{r}(1.25)\) and can be sketched as an arrow between these two points in the graph.

Compare the vector \(\mathbf{r}^{\prime}(1)\) with the vector \(\frac{\mathbf{r}(1.25)-\mathbf{r}(1)}{1.25-1}\)

\(\mathbf{r}^{\prime}(1)\) is the derivative of the function \(\mathbf{r}(t)\) at \(t=1\), which gives the instantaneous rate of change of the function at this point. On the other hand, \(\frac{\mathbf{r}(1.25) - \mathbf{r}(1)}{1.25-1}\) represents an average rate of change of the function over the interval [1, 1.25]. To compare these vectors, first calculate \(\mathbf{r}^{\prime}(t) = \frac{d\mathbf{r}}{dt} = \mathbf{i} - 2t\mathbf{j}\). Then, substitute \(t=1\) to get \(\mathbf{r}^{\prime}(1)\). Calculate \(\frac{\mathbf{r}(1.25) - \mathbf{r}(1)}{1.25-1}\), and finally compare these two vectors.

Key concepts

Parametric Equations

When discussing vector-valued functions and their many attributes, one can't overlook the versatility of parametric equations. These equations express the coordinates of points that make up a curve or geometric object as functions of one or more parameters, typically denoted as independent variables, such as t in our exercise.

In the given problem, the vector function \(\textbf{r}(t)=t \textbf{i}+(4-t^2) \textbf{j}\) has separate equations for each of its components, which depend on t. The first component is related to the horizontal (x) axis by \(x(t)=t\), and the vertical (y) axis by \(y(t)=4-t^2\). These individual functions allow us to describe complex curves and movements in space, unique from standard line equations. Indeed, by understanding parametric equations, a student can approach geometric and physical problems with more tools at their disposal, as these equations efficiently encapsulate the essence of motion and change in space.

Derivatives of Vector Functions

To discuss the dynamics and physics of a situation, the concept of the derivative of a vector function is central. Just like derivatives in standard calculus provide a way of describing the rate of change of a function, the derivative of a vector function details the instantaneous rate of change of each component of a vector with respect to its parameter.

For our problem here, the vector function's derivative is found by differentiating each component with respect to t, resulting in a new vector function \(\textbf{r}^{\textbf{\prime}}(t) = \textbf{i} - 2t\textbf{j}\). This resulting vector essentially tells how quickly and in what direction the vector \(\textbf{r}(t)\) is changing at any given moment. For students, learning to find and interpret this derivative is key to understanding concepts like velocity and acceleration in physics, or the slope of a tangent line to the curve represented by the parametric equations in mathematics.

Graphing Vector Functions

Visual representation can be pivotal in conceptual learning, which is where the graphing of vector functions finds its importance. When one graphs a vector function like \(\textbf{r}(t)\), as demonstrated in this exercise, it involves plotting all possible vectors (position vectors in this context) denoted by the function for different parameters (time t in our problem).

Mapping this out on a coordinate plane, one would use the parametric equations to calculate specific points and then use these points to draw a curve. Moreover, individual vectors like \(\textbf{r}(1)\) and \(\textbf{r}(1.25)\) can be represented by arrows originating from the origin, pointing to the calculated coordinates. Such graphical depiction aids students in visualizing how a particle or object moves through space over time. It can also underscore the relationship between the graph of a function and its derivatives or how changes in the function’s parameters affect its position and orientation.

Instantaneous Rate of Change

Vector calculus becomes particularly gripping when referring to the instantaneous rate of change. Harkening back to the derivative, the instantaneous rate of change indicated by these derivatives shows us how fast a quantity represented by a vector function is changing at any precise moment in time.

By calculating \(\textbf{r}^{\textbf{\prime}}(1)\), we're determining this rate of change at t = 1 for our vector function \(\textbf{r}(t)\). It is akin to checking the speedometer of a car at a specific second of the journey; it tells us exactly how fast and in which direction we are moving at that point. Grasping this concept is crucial for students, as it expands beyond just vectors in physics or mathematics to inform almost any varying quantity, such as economics or biology, anytime something changes — we can analyze its behavior instantaneously with this handy tool.

Average Rate of Change

While instantaneous rate of change focuses on a single moment, average rate of change spans a specified interval. This concept is roughly the 'big picture' view in contrast to the zoomed-in, moment-by-moment scrutiny provided by the instantaneous version.

In our exercise context, we are instructed to look at the vector change over a period from t = 1 to t = 1.25, hence the calculation \(\frac{\textbf{r}(1.25) - \textbf{r}(1)}{1.25 - 1}\). This operation yields a vector that represents how much \(\textbf{r}(t)\) changes on average over that interval and serves as an approximation of the instantaneous rate of change at any point within the interval. For students, understanding average rates is imperative as they often serve as the stepping stones to grasp the deeper, more nuanced concept of instantaneous rates. Additionally, it is practically significant in real-world applications where precise rates at every moment may not be necessary or possible to obtain.