## Sunday, January 20, 2013

### Converting Polar Equations to Parametric Equations

Good Sunday!

This blog will show how to transform polar equations, in the form of r(θ) to a pair of parametric equations, x(t) and y(t).

Tools We Need

x = r * cos θ
y = r * sin θ

Let θ = t.

For each example, we will change each polar equation and display a graph for each form. The polar form, colored blue, is on top; the parametric form, in red, is on the bottom. (A TI nSpire-CX is used for the pictures)

Example (I):

r = 2*θ

Then θ = r/2 and let θ = t.

x = r * cos t
x = 2 * t * cos(r/2)
x = 2 * t * cos((2*t)/2)
x = 2 * t * cos t

Similarly,

y = r * sin t
y = 2 * t * sin(r/2)
y = 2 * t * sin t

To summarize:
r = 2*θ

is equivalent to

x = 2 * t * cos t
y = 2 * t * sin t

Example (II):

r = e^(2 * θ)

Then θ = 1/2 * ln r and let θ = t.

And...

x = r * cos t
x = e^(2*t) * cos(1/2 * ln r)
x = e^(2*t) * cos(1/2 * ln(e^(2*t)))
x = e^(2*t) * cos(1/2 * 2 * t)
x = e^(2*t) * cos t

y = r * sin t
y = e^(2*t) * sin(1/2 * ln r)
y = e^(2*t) * sin(1/2 * ln(e^(2*t)))
y = e^(2*t) * sin t

To summarize:
r = e^(2*θ)

is equivalent to:

x = e^(2*t) * cos t
y = e^(2*t) * sin t

Example (III):

r = 2 cos θ

Then:

θ = arccos(r/2) = cos⁻¹(r/2)

Then, with θ=t...

x = r * cos t
x = 2 * cos t * cos(cos⁻¹(r/2))
x = 2 * cos t * cos(cos⁻¹(2*cos(t)/2))
x = 2 * cos t * cos t
x = 2 * cos² t

And... (some trigonometric identities are required)

y = r * sin t
y = 2 * cos t * sin(cos⁻¹(r/2))

Note: sin(cos⁻¹ x) = √(1 - x²)

y = 2 * cos t * √((1 - (r/2)²)
y = 2 * cos t * √(1 - r²/4)
y = 2 * cos t * √(1 - (4 cos² t)/4)
y = 2 * cos t * √(1 - cos² t)

Note: sin² x + cos² x = 1

y = 2 * cos t * sin t

Note: sin(2*x) = 2 * cos x * sin x

y = sin (2*t)

To Summarize:
r = 2 cos θ

is equivalent to:

x = 2 * cos² t
y = sin (2*t)

Until next time, take care!

Eddie

This blog is property of Edward Shore. 2013

1. thanks for share...

1. You are quite welcome, Naveen and city!

2. Hope this blog isn't dead!
Struggling with this:
How can I modify the parametric equation of an archimedean spiral so it can be rotated by X degrees? I really can't figure this out!

1. Didn't explain myself correctly:
I want to find the intersection between each layer of the spiral. I thought rotating it would do the trick but it's not the case! Using parametric equations:
-Normal archimedean spiral:
x=r*t*(cos(t))
y=r*t*(sin(t))
-Rotated archimedean spiral:
x=r*(cos(t)-t*sin(t))
y=r*(sin(t)+t*cos(t))

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