Chapter 3: Problem 27
Use a graphing utility to sketch each of the following vector-valued functions: $$ \mathbf{r}(t)=2 \cos t^{2} \mathbf{i}+(2-\sqrt{t}) \mathbf{j} $$
Short Answer
Expert verified
Use a graphing utility to plot the function with \( t \) in \( [0, 5] \) and observe its trajectory.
Step by step solution
01
Understand the Components of the Vector-Valued Function
The vector-valued function given is \( \mathbf{r}(t) = 2 \cos(t^2) \mathbf{i} + (2 - \sqrt{t}) \mathbf{j} \). It has an \( x \)-component \( x(t) = 2 \cos(t^2) \) and a \( y \)-component \( y(t) = 2 - \sqrt{t} \). We'll plot these components on a graph to visualize the trajectory of the function for values of \( t \).
02
Set the Range for Values of t
Decide a practical range for \( t \) based on the function components. Since \( \sqrt{t} \) is used, \( t \) must be non-negative. For \( t \) in the range \( [0, 5] \), both components are well-defined and capture the behavior of the function.
03
Plot the Vector Function's Trajectory
Using a graphing utility, plot the vector-valued function by calculating \( x(t) \) and \( y(t) \) for several points of \( t \) within the range \( [0, 5] \). Plot each resultant pair \( (x(t), y(t)) \) on a plane.
04
Analyze the Graph
Examine the graph plotted by the graphing utility. The curve will not resemble a typical function graph but rather the path traced by the vector-valued function. Note how the trigonometric and square root components affect the shape of the trajectory.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vector Components
In the realm of vector-valued functions, understanding the individual components is crucial. A vector-valued function consists of separate components that define its behavior with respect to each dimension. For the function \( \mathbf{r}(t) = 2 \cos(t^2) \mathbf{i} + (2 - \sqrt{t}) \mathbf{j} \), we identify two distinct parts:
\( x(t) = 2 \cos(t^2) \) and \( y(t) = 2 - \sqrt{t} \).
\( x(t) = 2 \cos(t^2) \) and \( y(t) = 2 - \sqrt{t} \).
- \( x(t) \) is the horizontal component and influences the left-right movement.
- \( y(t) \) is the vertical component and affects the up-down motion.
Graphing Utilities
Graphing utilities are powerful tools that aid us in visualizing mathematical functions, especially vector-valued functions which might not have simple analytical solutions. To use a graphing utility:
Identify your desired range for the parameter \( t \). Since \( t \) appears under a square root in \( y(t) \), it's limited to non-negative values. A typical choice might be \( 0 \leq t \leq 5 \).
Identify your desired range for the parameter \( t \). Since \( t \) appears under a square root in \( y(t) \), it's limited to non-negative values. A typical choice might be \( 0 \leq t \leq 5 \).
- Select your points strategically within this range.
- Calculate the corresponding \( x(t) \) and \( y(t) \) values.
- Input these pairs into your utility to plot.
Trigonometric Functions
Trigonometric functions play a significant role in shaping the behavior of vector components. In the function \( \mathbf{r}(t) = 2 \cos(t^2) \mathbf{i} + (2 - \sqrt{t}) \mathbf{j} \), the trigonometric component is \( 2 \cos(t^2) \). This contributes a periodic aspect to the function, influencing how \( x(t) \) oscillates as \( t \) changes.
- The cosine function typically oscillates between -1 and 1, but it's scaled by 2 here.
- As \( t^2 \) increases, the function displays rapid oscillations.
- This periodic behavior can result in multiple directional changes in the trajectory.
Function Trajectory
The trajectory of a vector-valued function is the path that its components trace on a coordinate plane as the parameter \( t \) varies. For \( \mathbf{r}(t) = 2 \cos(t^2) \mathbf{i} + (2 - \sqrt{t}) \mathbf{j} \), the trajectory reveals how the vector moves through space. This path provides a visual representation of the function's dynamism.
- The spiral or oscillating nature of \( x(t) = 2 \cos(t^2) \) may create wave-like paths horizontally.
- The descent aspect of \( 2 - \sqrt{t} \) usually causes a downward bending of the trajectory.
- Observing the trajectory helps in understanding how different components combine to form complex paths.