AP Precalculus Polar Functions — Questions and Answers
Question 1: The polar point (6, 2π/3) is converted to rectangular coordinates. Which is the result?
- (-3, 3√3) (Correct answer)
- (3, 3√3)
- (-3√3, 3)
- (3√3, -3)
Correct answer: (-3, 3√3)
x = 6cos(2π/3) = 6(-1/2) = -3 and y = 6sin(2π/3) = 6(√3/2) = 3√3.
To convert from polar to rectangular form, use x = r cos θ and y = r sin θ. The angle 2π/3 is in Quadrant II, where cosine is negative and sine is positive. Its reference angle is π/3, so cos(2π/3) = -1/2 and sin(2π/3) = √3/2. Multiplying by r = 6 gives (-3, 3√3). Swapping the cosine and sine values is a common error, and it gives the wrong point (-3√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?
- (2√2, π/4)
- (2√2, 5π/4)
- (2√2, 3π/4) (Correct answer)
- (2, 3π/4)
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.
The radius is r = √(x² + y²) = √8 = 2√2. The angle satisfies tan θ = y/x = -1, which holds at both 3π/4 and 7π/4. Since x < 0 and y > 0, the point is in Quadrant II, so θ = 3π/4. Choosing π/4 or 5π/4 puts the point in the wrong quadrant. Using r = 2 forgets the square root of the sum of squares.
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?
- π/6 and 5π/6
- 7π/6 and 11π/6 (Correct answer)
- 2π/3 and 4π/3
- 5π/6 and 7π/6
Correct 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.
The graph reaches the pole when r = 0. That gives 4 sin θ = -2, so sin θ = -1/2. Sine is negative in Quadrants III and IV, and the reference angle is π/6. The solutions are therefore θ = π + π/6 = 7π/6 and θ = 2π - π/6 = 11π/6. This limaçon has an inner loop because r takes negative values between these two angles.
Question 4: How many petals does the graph of r = 5 sin(4θ) have?
Correct answer: 8
For r = a sin(nθ) with n even, the graph is a rose with 2n petals, so n = 4 gives 8 petals.
A rose curve r = a sin(nθ) or r = a cos(nθ) has n petals when n is odd and 2n petals when n is even. Here n = 4 is even, so there are 8 petals. Each petal extends a maximum distance of |a| = 5 from the pole. If n were odd, such as r = 5 sin(3θ), the graph would have only 3 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?
- It is decreasing, because r is decreasing.
- It is decreasing, because r is negative.
- It is constant, because the angle is in Quadrant III.
- It is increasing, because |r| is increasing. (Correct answer)
Correct 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.
On this interval, sin θ falls from -1/2 to -1, so r falls from 0 to -1. Since r is negative and getting more negative, its absolute value grows from 0 to 1. The distance from the pole is |r|, so the point moves farther from the pole. Whenever r is negative, a decreasing r means increasing distance, which is the opposite of the positive case.
Question 6: What is the average rate of change of r = 3 cos θ with respect to θ on the interval 0 ≤ θ ≤ π/2?
- -6/π (Correct answer)
- -3/π
- -3
- -6π
Correct 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.
The average rate of change of r with respect to θ is Δr/Δθ. At θ = 0, r = 3cos 0 = 3. At θ = π/2, r = 3cos(π/2) = 0. So Δr = -3 and Δθ = π/2, giving -3 ÷ (π/2) = -6/π. The negative sign means r decreases on average as θ increases, so the graph moves toward the pole.
Question 7: Which polar function has a graph that is symmetric about the y-axis (the line θ = π/2) but not about the x-axis?
- r = 3 + 2 cos θ
- r = 5 cos θ
- r = 2 - 2 cos θ
- r = 4 + 3 sin θ (Correct answer)
Correct 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.
Symmetry about the y-axis holds when replacing θ with π - θ leaves r unchanged. Sine satisfies sin(π - θ) = sin θ, so r = 4 + 3 sin θ is symmetric about the y-axis. Cosine satisfies cos(-θ) = cos θ, so the three cosine functions are symmetric about the x-axis instead. The limaçons and the cardioid among them are not symmetric about the y-axis, and neither is the circle r = 5 cos θ.
Question 8: Which rectangular equation is equivalent to the polar equation r = 6 cos θ?
- x² + (y - 3)² = 9
- (x - 3)² + y² = 9 (Correct answer)
- (x - 6)² + y² = 36
- (x - 3)² + y² = 3
Correct 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.
Start with r = 6 cos θ and multiply both sides by r to get r² = 6r cos θ. Substituting r² = x² + y² and r cos θ = x gives x² + y² = 6x. Moving 6x to the left and completing the square gives (x - 3)² + y² = 9. This is a circle of radius 3 centered at (3, 0), which passes through the pole and reaches r = 6 at θ = 0.
Question 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?
- π/6 and 11π/6
- π/3 and 2π/3
- π/3 and 5π/3 (Correct answer)
- π/4 and 7π/4
Correct 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).
At an intersection point with the same θ, the r-values are equal, so 4 cos θ = 2 and cos θ = 1/2. Cosine equals 1/2 at θ = π/3 in Quadrant I and θ = 5π/3 in Quadrant IV. The circle r = 2 is centered at the pole, and r = 4 cos θ is a circle through the pole that lies to the right of it. They cross at these two symmetric points, each at distance 2 from the pole.
Question 10: What is the period, as a function of θ, of the polar function r = 5 + 2 sin(3θ)?
- 2π/3 (Correct answer)
- 3π
- π/3
- 6π
Correct answer: 2π/3
For a sinusoid with argument bθ, the period is 2π/|b|. Here b = 3, so the period is 2π/3.
The function r = 5 + 2 sin(3θ) is a sinusoidal function of θ with amplitude 2, midline r = 5, and frequency factor b = 3. Its period is 2π/b = 2π/3, so the pattern of r-values repeats every 2π/3 radians. Multiplying θ by 3 makes the cycle shorter than 2π. A period of 3π or 6π would come from mistakenly multiplying by b instead of dividing.
