Question 6.3.6: Converting Equations from Rectangular to Polar Coordinates C...
Converting Equations from Rectangular to Polar Coordinates
Convert each rectangular equation to a polar equation that expresses r in terms of θ:
a. x + y = 5 b. (x – 1)² + y² = 1.
Our goal is to obtain equations in which the variables are r and θ rather than x and y. We use x = r cos θ and y = r sin θ. We then solve the equations for r, obtaining equivalent equations that give r in terms of θ.
a. x + y = 5 This is the given equation in rectangular coordinates. The graph is a line passing through (5, 0) and (0, 5).
r cos θ + r sin θ = 5 Replace x with r cos θ and y with r sin θ.
r(cos θ + sin θ) = 5 Factor out r.
r=\frac{5}{\cos θ + \sin θ} Divide both sides of the equation by cos θ + sin θ and solve for r.
Thus, the polar equation for x + y = 5 is r=\frac{5}{\cos θ + \sin θ}.
b. 
This is the given equation in rectangular coordinates. The graph is a circle with radius 1 and center at (h, k) = (1, 0).
(r cos θ – 1)² + (r sin θ)² = 1 Replace x with r cos θ and y with r sin θ.
r² cos² θ – 2r cos θ + 1 + r² sin² θ = 1 Use (A – B)² = A² – 2AB + B² to square r cos θ – 1.
r² cos² θ + r² sin² θ – 2r cos θ = 0 Subtract 1 from both sides and rearrange terms.
r² – 2r cos θ = 0 Simplify: r 2 cos² θ + r² sin² θ = r 2(cos² θ + sin² θ) = r² · 1 = r².
r(r – 2 cos θ) = 0 Factor out r.
r = 0 or r – 2 cos θ = 0 Set each factor equal to 0.
r = 2 cos θ Solve for r.
The graph of r = 0 is a single point, the pole. Because the pole also satisfies the equation r = 2 cos θ (for θ = \frac{\pi}{2}, r = 0), it is not necessary to include the equation r = 0. Thus, the polar equation for (x – 1)² + y² = 1 is r = 2 cos θ.