Chapter 10: Q. 18. (page 694)

In problem graph the curve whose parametric equations are given and show its orientation. Find the rectangular equation of each curve. Verify your graph using a graphing utility.
$x = t^{\frac{3}{2}} + 1, y = \sqrt{t}; t \geq 0 .$

Short Answer

Expert verified

The answer is $x = y^{3} + 1 .$

Step by step solution

Step 1. Given information.

$x = t^{\frac{3}{2}} + 1, y = \sqrt{t}$ is given.

Step 2. Calculation.

Graph the curve defined by the parametric equations -

$t$	$x$	$y$	$(x, y)$
$0$	$0$	$0$	$(0, 0)$
$4$	$9$	$2$	$(9, 2)$

We find the rectangular equation by eliminating the parameter t from the parametric equations.

$y = \sqrt{t} t = y^{2}$

Put the value of t in equation-

role="math" localid="1648404939468" $x = t^{\frac{3}{2}} + 1 x = {(y^{2})}^{\frac{3}{2}} + 1 x = y^{3} + 1$

Thus, the equation isrole="math" localid="1648404969030" $x = y^{3} + 1 .$

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In problem graph the curve whose parametric equations are given and show its orientation. Find the rectangular equation of each curve. Verify your graph using a graphing utility.
$x = t^{\frac{3}{2}} + 1, y = \sqrt{t}; t \geq 0 .$

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Calculation.

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In problem graph the curve whose parametric equations are given and show its orientation. Find the rectangular equation of each curve. Verify your graph using a graphing utility.x=t32+1,y=t;t≥0.

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Calculation.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Applied Mathematics

Mechanics Maths

Geometry

Logic and Functions

Decision Maths

Discrete Mathematics

Study anywhere. Anytime. Across all devices.

In problem graph the curve whose parametric equations are given and show its orientation. Find the rectangular equation of each curve. Verify your graph using a graphing utility.
$x = t^{\frac{3}{2}} + 1, y = \sqrt{t}; t \geq 0 .$