How do you write the cartesian equation for #r=1-3cosx#?

Question

Christian Antunez · Accepted Answer

As below

#x = r cos theta, y = r sin theta, x^2+y^2=r^2#

#r = 1 - 3 cos x#

#sqrt(x^2 + y^2) = 1 - (3 x) / sqrt(x^2 + y^2)#

#x^2 + y^2 = sqrt(x^2 + y^2) - 3x#

Elijah Allen · Answer

undefined To write the Cartesian equation for $ r = 1 - 3 \cos(x) $, where $ r $ represents the distance from the origin to a point $ (x, y) $ in polar coordinates, follow these steps:

1. Use the relationship between polar coordinates $ (r, \theta) $ and Cartesian coordinates $ (x, y) $:
   $$ x = r \cos(\theta) $$
   $$ y = r \sin(\theta) $$

2. Substitute the given polar equation $ r = 1 - 3 \cos(x) $ into the expression for $ x $ and $ y $:
   $$ x = (1 - 3 \cos(x)) \cos(\theta) $$
   $$ y = (1 - 3 \cos(x)) \sin(\theta) $$

3. Since $ \theta $ is not specified in the equation, you may express $ x $ and $ y $ solely in terms of $ x $ and $ y $:
   $$ x = (1 - 3 \cos(x)) \frac{x}{r} $$
   $$ y = (1 - 3 \cos(x)) \frac{y}{r} $$

4. Substitute $ r = \sqrt{x^2 + y^2} $ into the expressions for $ x $ and $ y $:
   $$ x = (1 - 3 \cos(x)) \frac{x}{\sqrt{x^2 + y^2}} $$
   $$ y = (1 - 3 \cos(x)) \frac{y}{\sqrt{x^2 + y^2}} $$

5. Clear the denominators by multiplying both sides by $ \sqrt{x^2 + y^2} $:
   $$ x \sqrt{x^2 + y^2} = (1 - 3 \cos(x))x $$
   $$ y \sqrt{x^2 + y^2} = (1 - 3 \cos(x))y $$

6. Square both sides of the equations to eliminate the square roots:
   $$ x^2 (x^2 + y^2) = (1 - 3 \cos(x))^2 x^2 $$
   $$ y^2 (x^2 + y^2) = (1 - 3 \cos(x))^2 y^2 $$

7. Simplify the expressions to obtain the Cartesian equation in terms of $ x $ and $ y $.