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Consider the following parametric equations. a. Make a brief table of values of \(t, x,\) and \(y.\) b. Plot the \((x, y)\) pairs in the table and the complete parametric curve, indicating the positive orientation (the direction of increasing \(t\)). c. Eliminate the parameter to obtain an equation in \(x\) and \(y.\) d. Describe the curve. $$x=-t+6, y=3 t-3 ;-5 \leq t \leq 5$$

Short Answer

Expert verified
Select any five values of \(t\), calculate the corresponding \(x\) and \(y\) values using the parametric equations, and organize the calculated values into a table. 2. How can we eliminate the parameter \(t\) and obtain an equation in \(x\) and \(y\)? Solve for \(t\) in one of the equations, for example, the \(x\) equation in this case: \(t = 6 - x\). Then, substitute this expression for \(t\) in the other equation, \(y = 3(6 - x) - 3\), and simplify. 3. What is the equation in \(x\) and \(y\) for the given parametric curve? The equation for the given parametric curve is \(y = -3x + 15\). 4. How would you describe the curve of the given parametric equations? The curve is a straight line with a negative slope of \(-3\) and a \(y\)-intercept at \((0, 15)\).

Step by step solution

01

Create the table of values for \(t\), \(x\), and \(y\).

For \(-5 \leq t \leq 5\), select any five values of \(t\) within this interval, calculate the corresponding \(x\) and \(y\) values using the parametric equations, and then organize the calculated values into a table. ###Step 2: Plotting the curve###
02

Plot the \((x, y)\) pairs from the table.

Graph the curve by plotting the \((x, y)\) pairs from the table of values, making sure to indicate the positive orientation (direction of increasing \(t\)). ###Step 3: Eliminate the parameter \(t\)###
03

Solve for \(t\) in one of the equations.

Solve for \(t\) in either the \(x\) or \(y\) parametric equation. In this case, solving for \(t\) in the \(x\) parametric equation is more straightforward: \(t = 6 - x\)
04

Substitute this expression for \(t\) in the other equation.

Now, substitute \((6 - x)\) for \(t\) in the \(y\) parametric equation: \(y = 3(6 - x) - 3\) ###Step 4: Obtain the equation in \(x\) and \(y\)###
05

Simplify the equation.

Simplify the equation obtained in the previous step to get the final equation in \(x\) and \(y\): \(y = -3x + 15\) ###Step 5: Describe the curve###
06

Analyze the equation.

With the equation \(y = -3x + 15\), observe that it is a linear function in slope-intercept form with a slope of \(-3\) and a \(y\)-intercept at \((0, 15)\). This describes a straight line with a negative slope and crossing the \(y\)-axis at \((0, 15)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Plotting Parametric Curves
Understanding parametric curves starts with recognizing that they represent a set of points in space, where each point is determined by parameters. In our case, the parameters are values of a variable, typically denoted as t.

When plotting parametric curves, we derive separate equations for x and y as functions of t. This can help in tracing the movement of a point in the coordinate plane over time. For students, creating a table of values is a practical approach to visualize this relationship. With our given parametric equations, x=-t+6, y=3t-3, we first select values for t within the specified interval and then compute the corresponding x and y values.

Graphing the Points

Once you have the table, each (x, y) pair can be plotted on a graph. If the t-values are chosen sequentially, and you plot the points and connect them in order, the 'positive orientation' or the 'direction of increasing t' appears naturally. This representation provides a visual understanding of how the position changes over time with t, enhancing the comprehension of parametric curves.
Eliminating the Parameter
Sometimes, it is useful to convert a parametric equation into a standard Cartesian equation to easily recognize the curve's form. This process is often called 'eliminating the parameter'.

To do this, we look for a way to express t in terms of x or y from one of the parametric equations and then substitute this expression into the other equation. By eliminating t, we derive an equation that relates x and y directly.

Conversion Process

In our example, once we have the value for t from the x equation, which is t=6-x, we can substitute it into the y equation leading to an equation solely in terms of x and y. This process demystifies the curve by transitioning from the parametric form to a more familiar Cartesian equation.
Linear Function
A linear function is one of the simplest forms of a function in algebra, usually described by a straight line on a graph. Each linear function is a first-degree polynomial, meaning it has the highest exponent of x as one.

Linear functions follow the formula y = mx + b, where m represents the slope or steepness of the line, and b represents the y-intercept, the point where the line crosses the y-axis.

Understanding Slope and Intercept

The slope, m, tells us how steep the line is and in which direction it 'runs'. A positive slope means the line ascends from left to right, while a negative slope indicates a descent. The y-intercept, b, gives us a starting point for the line on the graph. In our example, after we eliminate the parameter and rearrange, we recognize the linear function representing a line with these specific characteristics.
Slope-Intercept Form
The slope-intercept form is an equation of a straight line in the format y = mx + b. It is an easy way to graph linear equations because it provides clear information about the line's slope and where it intercepts the y-axis.

In our case, after eliminating the parameter t, we rearranged the equation to find that y = -3x + 15. This tells us that the line has a slope (m) of -3, indicating that for every unit increase in x, y decreases by 3 units. The intercept (b) is 15, indicating that the line crosses the y-axis at the point (0, 15).

Versatility of Slope-Intercept Form

The slope-intercept form is particularly helpful for plotting linear functions quickly and comprehending their characteristics at a glance. It makes the properties of the line – its steepness and its intersection with the y-axis – immediately evident, aiding in a deeper understanding of linear relationships.

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Most popular questions from this chapter

Find parametric equations that describe the circular path of the following objects. Assume \((x, y)\) denotes the position of the object relative to the origin at the center of the circle. Use the units of time specified in the problem. There is more than one way to describe any circle. A Ferris wheel has a radius of \(20 \mathrm{m}\) and completes a revolution in the clockwise direction at constant speed in 3 min. Assume that \(x\) and \(y\) measure the horizontal and vertical positions of a seat on the Ferris wheel relative to a coordinate system whose origin is at the low point of the wheel. Assume the seat begins moving at the origin.

Show that an equation of the line tangent to the ellipse \(x^{2} / a^{2}+y^{2} / b^{2}=1\) at the point \(\left(x_{0}, y_{0}\right)\) is $$ \frac{x x_{0}}{a^{2}}+\frac{y y_{0}}{b^{2}}=1 $$

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