a.
To create a table of x− and y− values using the t=0,1,2,3,4 for the given parametric equations.
a.

Answer to Problem 9E
t01234x01√2√32y210−1−2
Explanation of Solution
Given: The parametric equations x=√t and y=2−t .
Calculation:
When t=0 , x=√0=0,y=2−0=2 .
When t=1 , x=√1=1,y=2−1=1 .
When t=2 , x=√2,y=2−2=0 .
When t=3 , x=√3,y=2−3=−1 .
When t=4 , x=√4=2,y=2−4=−2 .
So, the table becomes as shown below.
t01234x01√2√32y210−1−2
b.
To plot the points (x,y) generated in part (a) and graph the parametric equation. Also, describe the orientation of the curve.
b.

Explanation of Solution
Given: The parametric equations x=√t and y=2−t .
Plot:
From part (a), the table is given below.
t01234x01√2√32y210−1−2
The points become (0,2),(1,1),(√2,0),(√3,−1),(2,−2) .
Now, plot these points on the coordinate axes as shown below.
Now, join the points to graph the parametric equations.
The curve is moving towards negative infinity as the value of x increases.
c.
To graph the parametric equations by using graphing utility.
c.

Explanation of Solution
Given: The parametric equations x=√t and y=2−t .
Graph:
The graph of the parametric equations using graphing utility can be given as shown below.
The curve is moving towards negative infinity as the value of x increases.
d.
To find the rectangular equation by eliminating the parameter t . Then, sketch its graph and describe how the graph differs from the graphs in part (b) and (c).
d.

Answer to Problem 9E
y=2−x2,x≥0
Explanation of Solution
Given: The parametric equations x=√t and y=2−t .
Calculation:
x=√t⇒t=x2 .
Then, y=2−t⇒y=2−x2 .
So, the rectangular equation for the given parametric equations is y=2−x2,x≥0 .
The graph of this equation can be given as shown below.
The curve is moving towards negative infinity as the value of x increases.
The graph is same as obtained in part (b) and (c).
Chapter 9 Solutions
PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)
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