# Thread: Parametric / polar equations

1. ## Parametric / polar equations

I have a few things im stuck on.. <.< I might be able to answere one through the example of others..

1) Find the polar equation for the curve x^2 + y^2 = 4

2)Find the equation in x and y for the curve with parametric equations x=(cost) -2 and y= (sin t) +3 for 0 is less or = to t and t is less or = to pie

3) Sketch the graph of the plar equation r=2-cos feta

4) Find an equation in x and y that has the same graph as r=2cos feta - 4sin feta

2. Originally Posted by 3deltat
I have a few things im stuck on.. <.< I might be able to answere one through the example of others..

1) Find the polar equation for the curve x^2 + y^2 = 4

2)Find the equation in x and y for the curve with parametric equations x=(cost) -2 and y= (sin t) +3 for 0 is less or = to t and t is less or = to pie

3) Sketch the graph of the plar equation r=2-cos feta

4) Find an equation in x and y that has the same graph as r=2cos feta - 4sin feta
x = r*cos(theta)
y = r*sin(theta)

r = sqrt(x^2 + y^2)
theta = tan^(-1)(y/x)
Two comments about the last equation. It's "theta" not "feta" and tan^(-1) is the inverse tangent function.

1) r^2 = x^2 + y^2 so,
r^2 = 4.

2) 0 <= t <= pi <== It's "pi", not "pie"
x=(cost) -2
y= (sin t) +3

Solve the x equation for cos(t):
x + 2 = cos(t)

Set up a right triangle. cos(t) is "adjacent" over hypotenuse so we want the side of the triangle adjecent to angle t to be x + 2 and the hypotenuse 1. Thus the "opposite" side will be sqrt{1 - (x + 2)^2}. So:
sin(t) = sqrt{1 - (x + 2)^2}/1 = sqrt{1 - (x + 2)^2}

So
y= (sin t) + 3 = sqrt{1 - (x + 2)^2} + 3

3) I can't do this one here. I'll leave it up to you.

4) r=2cos(theta) - 4sin(theta)
sqrt{x^2 + y^2} = 2*x/r - 4*y/r <== Square both sides
x^2 + y^2 = (2*x/r - 4*y/r)^2

x^2 + y^2 = 4x^2/r^2 + 16y^2/r^2 - 16xy/r^2

(x^2 + y^2)r^2 = 4x^2 + 16y^2 - 16xy

But r^2 = x^2 + y^2

(x^2 + y^2)^2 = 4(x^2 + 4y^2 - 4xy)

This will do. You can always solve this for y if you wish.

-Dan

3. Thanks for the help! One question though...
For problem 4, when it is " sqrt{x^2 + y^2} = 2*x/r - 4*y/r " Why is the x and y divided by r ?

Also, for number three; to start out at least, would this be a correct way?
r = 2-costheta
r^2 = 2r - rcostheta
X^2 +Y^2 = 2sqr(X^2 + Y^2) -x

4. Hello, 3deltat!

You're expected to know the polar-rectangular conversion formulas:
. . x = r·cosθ, . y = r·sinθ, . .= .x² + y²

By the way, you seem to have food on your mind.
"Pie" is a dessert; you're thinking of "pi": π
And "feta" is a Greek cheese; you mean "theta": θ

1) Find the polar equation for the curve: .x² + y² .= .4
This one is easy . . . replace x² + y² with r².

And we have: . .= .4 . . r .= .±2

2)Find the equation in x and y for the curve with parametric equations:
. . x .= .cosθ - 2 .and .y .= .sinθ + 3, .for 0 < t < π

This one is not easy (unless you know how) . . .

We have: .x + 2 .= .cosθ
. . . . . . . . y - 3 .= .sinθ

Square: .(x + 2)² .= .cos²θ
. . . . . . .(y - 3)² .= .sin²θ

Add: .(x + 2)² + (y - 3)² .= .sin²θ + cos²θ

Since sin²θ + cos²θ = 1, we have: .(x + 2)² + (y - 3)² .= .1

This is a circle with center at (-2, 3) and radius 1.

4) Find an equation in x and y that has the same graph as:
. . . r .= .2·cosθ - 4·sinθ

Multiply both sides by r: . .= .2·(r·cosθ) - 4·(r·sinθ)

Substitute: .x² + y² .= .2x - 4y

Basically, we're finished ... but we can show-off a bit.

We have: .x² - 2x + y² + 4y .= .0

. . Complete the square: .x² - 2x + 1 + y² + 4y + 4 .= .0 + 1 + 4

And we have: .(x - 1)² + (y + 2)² .= .5

This is a circle with center (1, -2) and radius √5.

5. Originally Posted by Soroban

By the way, you seem to have food on your mind.
"Pie" is a dessert; you're thinking of "pi": π
And "feta" is a Greek cheese; you mean "theta": θ
You da man, Soroban! I know he meant to write theta, but the fact that feta is a Greek cheese, is that common knowledge?

6. mm... food...

7. Originally Posted by 3deltat
mm... food...
which reminds me, i'm hungry. where's that feta!

8. Hello again, 3deltat!

I'll give #3 a try . . .

3) Sketch the graph of the polar equation: .r .= .2 - cosθ
You're expected to be familiar with the limacon . . . and the cardioid.

Plot the four compass-directions to get the orientation.

θ = 0: . r .= .2 - 1 .= .1

θ = π/2: . r .= .2 - 0 .= .2

θ = π: . r .= .2 - (-1) .= .3

θ = 3π/2: . r .= .2 - 0 .= .2
Code:
                    |
* * * |
*         |   *
*           |     *
*            |     *
|    *
*             |   *
- - * - - - - - - + - * - - - - - -
*             |   *
|    *
*            |     *
*           |     *
*         |   *
* * * |
|

9. Oh I see! For number three I was forgetting i could just graph it as it was. I was trying to get it into x^2 + y^2 form heh. Tyvm

10. Originally Posted by 3deltat
Thanks for the help! One question though...
For problem 4, when it is " sqrt{x^2 + y^2} = 2*x/r - 4*y/r " Why is the x and y divided by r ?
x = r*cos(theta)

Thus
cos(theta) = x/r

A similar equation holds for sin(theta).

-Dan