AP Precalculus Polar Functions

Question 1

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

Accepted Answer

(-3, 3√3)

Answer

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

Question 2

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

Accepted Answer

(2√2, 3π/4)

Answer

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

Question 3

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?

Accepted Answer

7π/6 and 11π/6

Answer

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

Question 4

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

Accepted Answer

8

Answer

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

Question 5

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?

Accepted Answer

It is increasing, because |r| is increasing.

Answer

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

AP Precalculus Practice Test

AP Precalculus Practice Test

AP Precalculus Polar Functions