Find the polar equations of the lines or curves whose Cartesian equations are (a) y =2x. (b) xy = 4 (c) y =4 (d) 3x-y+2=0

Answer 1

x=rcosθ, y=rsinθ

y=2x

rsinθ=2(rcosθ)

divide both sides by r

sinθ=2cosθ

sinθ/cosθ=2

tanθ=2

tan^-1θtanθ=tan^-12

θ=tan^-12

θ=0.63.435° (also, r=0)

xy=4

rcosθrsinθ=4

r^2sinθcosθ=4

r^2=4/sinθcosθ

r=2/√(sinθcosθ)

r=2√(sinθcosθ)/(sinθcosθ)

y=4

rsinθ=4

r=4/sinθ

r=4cscθ

3x-y+2=0

3x-y=-2

3(rcosθ)-rsinθ=-2

r(3cosθ-sinθ)=-2

r=-2/(3cosθ-sinθ)

r=2/(sinθ-3cosθ)

Answer 2

To convert between polar and rectangular equations we have to remember the following identities/formulas

1) r² = x² + y²

2) x = r • cos(θ)

3) y = r • sin(θ)

4) tan(θ) = y/x

A) cartesian equation: y = 2x

Notice that we have a polar equivalent for y/x. So we divide both sides of the equation by x to get

y/x = 2

We then make the substitution using identity #4 to get

tan(θ) = 2. This is the polar equation.

Just like the cartesian equation y = -4(a horizontal line), polar equations don't need both r and θ.

B) cartesian equation: xy = 4

Notice that we have polar equivalents for both x and y. Using identities #2 and #3, we know that

xy = [r • cos(θ)][r • sin(θ)]

We simplify that to

xy = r² cos(θ) sin(θ)

We then make a final substitution into the original cartesian equation to get

r² cos(θ) sin(θ) = 4. This is the polar equation.

C) cartesian equation: y = 4

Notice that we have a polar equivalent for y. We directly substitute into the original cartesian equation to get

r • sin(θ) = 4. This is the polar equation.

Some people would rearrange it to r = 4/sin(θ).

D) cartesian equation: 3x - y + 2 = 0

Notice that we have polar equivalents for x and y. We substitute those into the original cartesian to get

3[r • cos(θ)] - [r • sin(θ)] + 2 = 0

We simplify to get

3r • cos(θ) - r • sin(θ) + 2 = 0. This is the polar equation.

Some people would notice that the first two terms have a common factor and write

r[3cos(θ) - sin(θ)] + 2 = 0

r[3cos(θ) - sin(θ)] = -2

r = -2/[3cos(θ) - sin(θ)]