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AP Precalculus -3.15 Rates of Change in Polar Functions- MCQ Exam Style Questions - Effective Fall 2023

AP Precalculus -3.15 Rates of Change in Polar Functions- MCQ Exam Style Questions – Effective Fall 2023

AP Precalculus -3.15 Rates of Change in Polar Functions- MCQ Exam Style Questions – AP Precalculus- per latest AP Precalculus Syllabus.

AP Precalculus – MCQ Exam Style Questions- All Topics

Question

Consider the graph of the polar function \( r = f(\theta) \), where \( f(\theta) = 4\sin(2\theta) \), in the polar coordinate system. On the interval \( 0 \leq \theta \leq 2\pi \), which of the following is true about the graph of \( r = f(\theta) \)?
(A) For the input values \( \theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} \), the function \( r = f(\theta) \) has extrema that correspond to points that are farthest from the origin.
(B) For the input values \( \theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} \), the function \( r = f(\theta) \) has extrema. However, only the points corresponding to \( \theta = \frac{\pi}{4} \) and \( \theta = \frac{5\pi}{4} \) are farthest from the origin.
(C) For the input values \( \theta = \frac{\pi}{2} \) and \( \theta = \frac{3\pi}{2} \), the function \( r = f(\theta) \) has extrema that correspond to points that are farthest from the origin.
(D) For the input values \( \theta = \frac{\pi}{2} \) and \( \theta = \frac{3\pi}{2} \), the function \( r = f(\theta) \) has extrema. However, only the point corresponding to \( \theta = \frac{\pi}{2} \) is farthest from the origin.
▶️ Answer/Explanation
Detailed solution

For \( r = 4\sin(2\theta) \), the extrema occur where \( \sin(2\theta) = \pm 1 \).
This happens when \( 2\theta = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2} \), so \( \theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} \).
At these angles, \( r = 4 \) or \( r = -4 \), but the distance from the origin is \( |r| = 4 \) in all four cases, so they are all farthest from the origin.
✅ Answer: (A)

Question

Consider the graph of the polar function \( r = f(\theta) \), where \( f(\theta) = 1 + 2 \sin \theta \), in the polar coordinate system for \( 0 \leq \theta \leq 2\pi \). Which of the following statements is true about the distance between the point with polar coordinates \( (f(\theta), \theta) \) and the origin?
(A) The distance is increasing for \( 0 \leq \theta \leq \frac{\pi}{2} \), because \( f(\theta) \) is positive and increasing on the interval.
(B) The distance is increasing for \( \frac{3\pi}{2} \leq \theta \leq \frac{11\pi}{6} \), because \( f(\theta) \) is negative and increasing on the interval.
(C) The distance is decreasing for \( 0 \leq \theta \leq \frac{\pi}{2} \), because \( f(\theta) \) is positive and decreasing on the interval.
(D) The distance is decreasing for \( \frac{3\pi}{2} \leq \theta \leq \frac{11\pi}{6} \), because \( f(\theta) \) is negative and decreasing on the interval.
▶️ Answer/Explanation
Detailed solution

In polar coordinates, the distance from the origin is given by \( r = f(\theta) \). However, distance must be non-negative. When \( f(\theta) \) is negative, the point is plotted \( |f(\theta)| \) units in the opposite direction, but the distance from the origin is \( |f(\theta)| \).

We check each option:

  • On \( 0 \leq \theta \leq \frac{\pi}{2} \): \( f'(\theta) = 2 \cos \theta > 0 \) (increasing), and \( f(\theta) > 0 \), so distance \( = f(\theta) \) is increasing. ✅ (A) is correct.
  • On \( \frac{3\pi}{2} \leq \theta \leq \frac{11\pi}{6} \): \( f(\theta) = 1 + 2 \sin \theta \). For \( \theta = \frac{3\pi}{2} \), \( f(\theta) = -1 \); for \( \theta = \frac{11\pi}{6} \), \( f(\theta) = 0 \). Since \( f(\theta) \) is negative and increasing toward 0, \( |f(\theta)| \) is decreasing. So the distance is decreasing, not increasing. (B) is false.
  • (C) contradicts (A), so false.
  • (D) states “because \( f(\theta) \) is negative and decreasing” — but \( f(\theta) \) is increasing here, not decreasing, so false.

✅ Answer: (A)

Question

A polar function is given by \( r = f(\theta) = -1 + \sin \theta \). As \(\theta\) increases on the interval \(0 < \theta < \frac{\pi}{2}\), which of the following is true about the points on the graph of \(r = f(\theta)\) in the \(xy\)-plane?
(A) The points on the graph are above the \(x\)-axis and are getting closer to the origin.
(B) The points on the graph are above the \(x\)-axis and are getting farther from the origin.
(C) The points on the graph are below the \(x\)-axis and are getting closer to the origin.
(D) The points on the graph are below the \(x\)-axis and are getting farther from the origin.
▶️ Answer/Explanation
Detailed solution

For \(0 < \theta < \frac{\pi}{2}\), \(\sin \theta\) increases from 0 to 1.
Thus \(f(\theta) = -1 + \sin \theta\) increases from \(-1\) to \(0\).
Since \(f(\theta)\) is negative, points are in the direction opposite to terminal ray \(\theta\); i.e., for \(0 < \theta < \frac{\pi}{2}\), the corresponding Cartesian points lie in Quadrant IV (below the \(x\)-axis).
The distance from origin is \( |f(\theta)| = 1 – \sin \theta\), which decreases as \(\theta\) increases.
✅ Answer: (C)

Question

Consider the graph of the polar function \( r = f(\theta) \), where \( f(\theta) = \theta (\theta – 2)(\theta – 4) \), in the polar coordinate system for \( 0 \leq \theta \leq 4 \). Which of the following statements is true?
(A) On the interval \( 2 < \theta < 2.1 \), the distance between \( (f(\theta), \theta) \) and the origin is increasing because the values of \( f(\theta) \) are negative and decreasing.
(B) On the interval \( 2 < \theta < 2.1 \), the distance between \( (f(\theta), \theta) \) and the origin is decreasing because the values of \( f(\theta) \) are negative and decreasing.
(C) On the interval \( 2 < \theta < 2.1 \), the distance between \( (f(\theta), \theta) \) and the origin is increasing because the values of \( f(\theta) \) are negative and increasing.
(D) On the interval \( 2 < \theta < 2.1 \), the distance between \( (f(\theta), \theta) \) and the origin is decreasing because the values of \( f(\theta) \) are negative and increasing.
▶️ Answer/Explanation
Detailed solution

For \( 2 < \theta < 2.1 \):
\( \theta > 0 \), \( \theta – 2 > 0 \), \( \theta – 4 < 0 \) ⇒ \( f(\theta) = \theta(\theta-2)(\theta-4) < 0 \).

Check derivative: \( f'(\theta) = (\theta-2)(\theta-4) + \theta(\theta-4) + \theta(\theta-2) \).
At \( \theta = 2 \), \( f'(2) = -4 < 0 \), so \( f \) is decreasing near \( \theta = 2 \).
Thus on \( 2 < \theta < 2.1 \), \( f(\theta) \) is negative and decreasing.

Distance from origin = \( |f(\theta)| = -f(\theta) \).
Since \( f \) is decreasing (more negative), \( |f| \) is increasing.
Therefore, distance is increasing.
✅ Answer: (A)

Question

Consider the graph of the polar function \( r = f(\theta) \), where \( f(\theta) = 2 \sin \theta – 1 \), in the polar coordinate system. Which of the following descriptions is true?
(A) As \( \theta \) increases from 0 to \( \frac{\pi}{6} \), the polar function \( r = f(\theta) \) is increasing, and the distance between the point \( (f(\theta), \theta) \) on the curve and the origin is increasing.
(B) As \( \theta \) increases from 0 to \( \frac{\pi}{6} \), the polar function \( r = f(\theta) \) is increasing, and the distance between the point \( (f(\theta), \theta) \) on the curve and the origin is decreasing.
(C) As \( \theta \) increases from 0 to \( \frac{\pi}{6} \), the polar function \( r = f(\theta) \) is decreasing, and the distance between the point \( (f(\theta), \theta) \) on the curve and the origin is increasing.
(D) As \( \theta \) increases from 0 to \( \frac{\pi}{6} \), the polar function \( r = f(\theta) \) is decreasing, and the distance between the point \( (f(\theta), \theta) \) on the curve and the origin is decreasing.
▶️ Answer/Explanation
Detailed solution

On \( 0 \leq \theta \leq \frac{\pi}{6} \):
\( f(\theta) = 2\sin\theta – 1 \).
Since \( \sin\theta \) increases from 0 to \( \frac12 \), \( f(\theta) \) increases from \( -1 \) to \( 0 \).
So \( f \) is increasing.

Also \( f(\theta) \) is negative (or zero at \( \theta = \frac{\pi}{6} \)).
Distance from origin = \( |f(\theta)| = 1 – 2\sin\theta \).
As \( \theta \) increases, \( 1 – 2\sin\theta \) decreases.
Thus the distance is decreasing.

Conclusion: \( f(\theta) \) is increasing, distance is decreasing.
✅ Answer: (B)

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