Polar Functions Flashcards

The polar point (6, 2π/3) is converted to rectangular coordinates. Which is the result?

Answer: (-3, 3√3)

x = 6cos(2π/3) = 6(-1/2) = -3 and y = 6sin(2π/3) = 6(√3/2) = 3√3.

The rectangular point (-2, 2) is written in polar form (r, θ) with r > 0 and 0 ≤ θ < 2π. Which is correct?

Answer: (2√2, 3π/4)

r = √((-2)² + 2²) = 2√2. The point lies in Quadrant II and tan θ = -1, so θ = 3π/4.

For the polar function r = 2 + 4 sin θ, which values of θ in [0, 2π) give r = 0, so the graph passes through the pole?

Answer: 7π/6 and 11π/6

Setting 2 + 4 sin θ = 0 gives sin θ = -1/2, which occurs at θ = 7π/6 and 11π/6.

How many petals does the graph of r = 5 sin(4θ) have?

Answer: 8

For r = a sin(nθ) with n even, the graph is a rose with 2n petals, so n = 4 gives 8 petals.

For r = 1 + 2 sin θ, consider the interval 7π/6 < θ < 3π/2. On this interval, r is negative and decreasing as θ increases. What is happening to the distance of the point (r, θ) from the pole?

Answer: It is increasing, because |r| is increasing.

Distance from the pole is |r|. When r is negative and decreasing, it becomes more negative, so |r| increases.

What is the average rate of change of r = 3 cos θ with respect to θ on the interval 0 ≤ θ ≤ π/2?

Answer: -6/π

r changes from 3 at θ = 0 to 0 at θ = π/2, so the average rate is (0 - 3)/(π/2 - 0) = -6/π.

Which polar function has a graph that is symmetric about the y-axis (the line θ = π/2) but not about the x-axis?

Answer: r = 4 + 3 sin θ

Because sin(π - θ) = sin θ, a function of the form r = a + b sin θ gives the same r for θ and π - θ, so its graph is symmetric about the y-axis.

Which rectangular equation is equivalent to the polar equation r = 6 cos θ?

Answer: (x - 3)² + y² = 9

Multiplying by r gives r² = 6r cos θ, so x² + y² = 6x. Completing the square gives (x - 3)² + y² = 9.

The graphs of r = 2 and r = 4 cos θ intersect at two points. Which values of θ in [0, 2π) give these intersection points?

Answer: π/3 and 5π/3

Setting 4 cos θ = 2 gives cos θ = 1/2, so θ = π/3 and θ = 5π/3 in [0, 2π).

What is the period, as a function of θ, of the polar function r = 5 + 2 sin(3θ)?

Answer: 2π/3

For a sinusoid with argument bθ, the period is 2π/|b|. Here b = 3, so the period is 2π/3.