Chapter 9: Q 53. (page 731)

In Exercises 48–55 convert the equations given in rectangular coordinates to equations in polar coordinates.
$y = - 3$

Short Answer

Expert verified

The required equation is $- 3 \csc θ$ .

Step by step solution

Step 1. Given information.

The given equation in rectangular coordinates is:

$y = - 3$

Step 2. Find the equation in rectangular coordinates.

$y = - 3$

As we know that in polar coordinates $y = r \sin θ$ , then substitute the value of y ,

$r \sin θ = - 3$

Divide both sides by role="math" localid="1649349326734" $\sin θ$ ,

$\frac{r \sin θ}{\sin θ} = \frac{- 3}{\sin θ} r = \frac{- 3}{\sin θ} r = - 3 \csc θ$

Therefore, the equation in polar coordinates is $- 3 \csc θ$ .

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In Exercises 48–55 convert the equations given in rectangular coordinates to equations in polar coordinates.
$y = - 3$

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Find the equation in rectangular coordinates.

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In Exercises 48–55 convert the equations given in rectangular coordinates to equations in polar coordinates.y=-3

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Find the equation in rectangular coordinates.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Statistics

Applied Mathematics

Pure Maths

Decision Maths

Discrete Mathematics

Mechanics Maths

Study anywhere. Anytime. Across all devices.

In Exercises 48–55 convert the equations given in rectangular coordinates to equations in polar coordinates.
$y = - 3$