Problem 14 - Calculus Early Transcendentals

Question

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. $$x=-t+6, y=3 t-3 ;-5 \leq t \leq 5$$

Step-by-Step Solution

Answer

Answer: The rectangular form of the given parametric equations is \(y = -3x + 15\).

Step 1: Create a table of values for t, x, and y

To make a table of values, evaluate the parametric equations at several values of \(t\) within the given range, \(-5 \leq t \leq 5\). Here, we choose to evaluate at \(t=-5, -3, 0, 3, 5\). | t | x | y | |----|----|----| | -5 | 11 | -18| | -3 | 9 | -12| | 0 | 6 | -3| | 3 | 3 | 6| | 5 | 1 | 12|

Step 2: Plot the points and the parametric curve

Using the values from the table, plot the \((x, y)\) pairs and then draw the parametric curve. Don't forget to indicate the positive orientation (direction of increasing \(t\)) with arrows. This step may need to be done by hand or using a graphing software like Desmos or Geogebra.

Step 3: Eliminate the parameter

Eliminating the parameter \(t\) means finding an equation that relates \(x\) and \(y\) without the need for a parameter. We can do this by solving one equation for the parameter, then substitute into the other equation. From the \(x\) equation: $$x = -t + 6$$ Solving for \(t\): $$t = 6 - x$$ Now, substitute this expression for \(t\) into the \(y\) equation: $$y = 3 (6 - x) - 3$$ $$y = 18 - 3x - 3$$ $$y = -3x + 15$$

Step 4: Describe the curve

The equation we obtained in step 3, \(y = -3x + 15\), is a linear equation representing a straight line with a slope of \(-3\) and a \(y\)-intercept at \((0, 15)\). This means that the parametric curve is a straight line and its direction is indicated by the positive orientation (increasing values of \(t\)) we mentioned in step 2.