Converting Equations from Polar to Rectangular Form Convert each polar equation to a rectangular equation in x and y: a. r = 5 b. θ = π/4 c. r = 3 csc θ d. r = -6 cos θ.

Question

Converting Equations from Polar to Rectangular Form

Convert each polar equation to a rectangular equation in x and y:

a. r = 5 b. θ = [latex]\frac{π}{4}[/latex] c. r = 3 csc θ d. r = -6 cos θ.

Accepted Answer

In each case, let’s express the rectangular equation in a form that enables us to recognize its graph.

a. We use r² = x² + y² to convert the polar equation r = 5 to a rectangular equation.

r = 5 This is the given polar equation.

r² = 25 Square both sides.

x² + y² = 25 Use r² = x² + y² on the left side.

The rectangular equation for r = 5 is x² + y² = 25. The graph is a circle with center at (0, 0) and radius 5.

b. We use [latex] an θ =\frac{y}{x}[/latex] to convert the polar equation [latex]θ =\frac{\pi}{4}[/latex] to a rectangular equation in x and y.

[latex]θ =\frac{\pi}{4}[/latex] This is the given polar equation.

[latex] an θ = an\frac{\pi}{4}[/latex] Take the tangent of both sides.

tan θ = 1 [latex] an\frac{\pi}{4}=1[/latex]

[latex]\frac{y}{x}=1[/latex] Use [latex] an θ=\frac{y}{x}[/latex] on the left side.

y = x Multiply both sides by x.

The rectangular equation for [latex]θ =\frac{\pi}{4}[/latex] is y = x. The graph is a line that bisects quadrants I and III. Figure 6.28 shows the line drawn in a polar coordinate system.

c. We use r sin θ = y to convert the polar equation r = 3 csc θ to a rectangular equation. To do this, we express the cosecant in terms of the sine.

r = 3 csc θ This is the given polar equation.

[latex]r=\frac{3}{\sin θ}[/latex] [latex]\csc θ=\frac{1}{\sin θ}[/latex]

r sin θ = 3 Multiply both sides by sin θ.

y = 3 Use r sin θ = y on the left side.

The rectangular equation for r = 3 csc θ is y = 3. The graph is a horizontal line three units above the x-axis. Figure 6.29 shows the line drawn in a polar coordinate system.

d. To convert r = -6 cos θ to rectangular coordinates, we multiply both sides by r. Then we use r² = x² + y² on the left side and r cos θ = x on the right side.

r = -6 cos θ This is the given polar equation.

r² = -6r cos θ Multiply both sides by r.

x² + y² = -6x Convert to rectangular coordinates: r² = x² + y² and r cos θ = x.

x² + 6x + y² = 0 Add 6x to both sides.

x² + 6x + 9 + y² = 9 Complete the square on x: [latex]\frac{1}{2}·6=3[/latex] and 3² =9.

(x + 3)² + y² = 9 Factor.

The rectangular equation for r = -6 cos θ is (x + 3)² + y² = 9. This last equation is the standard form of the equation of a circle, (x - h)² + (y - k)² = r², with radius r and center at (h, k). Thus, the graph of (x + 3)² + y² = 9 is a circle with center at (-3, 0) and radius 3.

Question 6.3.7: Converting Equations from Polar to Rectangular Form Convert ...

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