Question

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

Short answer

The process involves sketching the vector-valued function, plotting key vectors on the graph, and comparing the derivative of the vector function at a point with the slope of the line joining two points on the graph. The latter comparison enables us to understand how well the line approximates the tangent to the curve at that point.

Step-by-step solution

Sketching the graph

Start by sketching the graph of the function \( \mathbf{r}(t)=t \mathbf{i}+t^{2} \mathbf{j} \). You can do this by making a table of values for \( t \) and their corresponding \( t^{2} \), and then plotting these points in the cartesian plane. You may use a graphing calculator to verify your sketch.

Plotting the vectors

Next, calculate the values of \( \mathbf{r}(1 / 4), \mathbf{r}(1 / 2),\) and \( \mathbf{r}(1 / 2)-\mathbf{r}(1 / 4)\). These vectors will be plotted on the graph obtained from step 1.

Comparing vectors

Compute the derivative of \( \mathbf{r}(t) \), and find \( \mathbf{r}^{\prime}(1 / 4)\). This derivative gives the slope of the tangent line to the curve at \( t = 1/4 \). Now calculate \( \frac{\mathbf{r}(1 / 2)-\mathbf{r}(1 / 4)}{1 / 2-1 / 4} \), which represents the slope of the line from \( \mathbf{r}(1 / 4) \) to \( \mathbf{r}(1 / 2) \). Compare these two vectors to see how closely they approximate each other. The closer they are, the better is the approximation of the curve's tangent at \( t = 1/4 \) by the line from \( \mathbf{r}(1 / 4) \) to \( \mathbf{r}(1 / 2) \).

Key concepts

Graphing Vector Functions

Graphing vector-valued functions, like \( \mathbf{r}(t)=t \mathbf{i}+t^{2} \mathbf{j} \) is a visual way to understand the dynamics of such functions. Each vector function represents a position vector of a point in two-dimensional space (for planar vectors) or three-dimensional space (for spatial vectors). To graph this function, start by thinking of \( t \) as time and \( \mathbf{r}(t) \) as the position of an object at that time along its path.

For our example, create a table listing various time values (\( t \)) and calculate their associated positions (\( t \mathbf{i}+t^{2} \mathbf{j} \)). Plot these as points where the \( i \) and \( j \) components give the x and y coordinates, respectively. Connect these points to see the trajectory of the object. As \( t \) increases, the \( j \) (y-axis) movement grows quadratically, indicating an accelerating motion in that direction.

Using a graphing utility not only verifies your plot but also provides a more precise and smoother curve, especially valuable when working with complex vector functions.

Calculating Derivatives of Vector Functions

Understanding the process of finding derivatives of vector functions is essential for analyzing motion and other dynamic processes. For the given function

\( \mathbf{r}(t)=t \mathbf{i}+t^{2} \mathbf{j} \),

calculating the derivative will give us the velocity vector, which indicates the speed and direction of movement at any instant. The derivative \( \mathbf{r}^\prime(t) \) of this function is calculated component-wise, basically applying the usual rules of differentiation to each component of the vector. As derivatives represent the rate of change, here, you would get

\( \mathbf{r}^\prime(t) = \mathbf{i}+2t\mathbf{j} \).

This tells you how the object's position changes at each moment \( t \). For example, the velocity at \( t = 1/4 \) would be \( \mathbf{r}^\prime(1/4) = \mathbf{i}+2(1/4)\mathbf{j} \). Pay special attention in exams to apply the power rule correctly for the \( j \) component and don't forget the constant vector \( i \) component!

Tangent Vectors

Tangent vectors are pivotal in understanding motion along curves. For instance, the tangent vector to the curve of the vector-valued function \( \mathbf{r}(t) \) at a specific point provides the direction in which the curve is headed at that point. To visualize this, consider our function \( \mathbf{r}(t)=t \mathbf{i}+t^{2} \mathbf{j} \).The derivative \( \mathbf{r}^\prime(t) \) gives the tangent vector at any point \( t \). At \( t = 1/4 \), the tangent vector is \( \mathbf{r}^\prime(1/4) \), which indicates the direction of the object's instantaneous motion.

Moreover, comparing the tangent vector \( \mathbf{r}^\prime(1/4) \) to the difference quotient \( \frac{\mathbf{r}(1/2)-\mathbf{r}(1/4)}{1/2-1/4} \), which is a secant vector between points \( 1/4 \) and \( 1/2 \) on the curve, allows us to gauge how well the tangent vector approximates the curve at that point. The tangent vector is an exact directional arrow at a point, while the secant vector gives an average direction over an interval. Understanding this distinction is vital for problems relating to instantaneous rates of change and local linear approximations.