Unit 3 · Topic 3.15

Rates of Change in Polar Functions

For a polar function, the average rate of change of r on [a, b] is (r(b) − r(a))/(b − a) units per radian, the slope of a secant line on the graph of r against θ .

Practice 2: Multiple RepresentationsPractice 3: Communication and Reasoning

Part 1 · Hook

Why this matters

A garden sprinkler turns slowly, and the water lands farther in some directions than others. As the nozzle turns, is the spray reaching farther or falling short, and how quickly? That is a question about how rr changes as θ\theta changes.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. What is the average rate of change of ff on [1,4][1, 4] if f(1)=2f(1) = 2 and f(4)=11f(4) = 11?

  1. 33
  2. 99
  3. 13\frac{1}{3}
  4. 4.54.5
Show the answer

f(4)−f(1)4−1\frac{f(4) - f(1)}{4 - 1}=93= \frac{9}{3}=3= 3, the slope of the secant line.

  • Correct: 33:
  • 99:
  • 13\frac{1}{3}:
  • 4.54.5:

2. Where is the point of r=1+2cos⁡θr = 1 + 2\cos\theta at θ=π\theta = \pi?

  1. 11 unit right of the pole
  2. 11 unit left of the pole
  3. 33 units left of the pole
  4. At the pole
Show the answer

r=1+2cos⁡π=−1r = 1 + 2\cos\pi = -1; the angle points left and the negative rr reverses it, so the point is (1,0)(1, 0).

  • Correct: 11 unit right of the pole:
  • 11 unit left of the pole:
  • 33 units left of the pole:
  • At the pole:

Part 4 · See it

See it first

The polar graph of r = 1 + 2 sin theta: a limacon whose large outer loop reaches highest straight above the origin and crosses the horizontal axis 1 unit on each side of the origin, with a small inner loop above the origin.
r=1+2sin⁡θr = 1 + 2\sin\theta. The big loop moves away from the origin while rr grows; the inner loop is drawn while r<0r < 0. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. Pick two angles aa and bbr(b)−r(a)b−a\frac{r(b) - r(a)}{b - a} is the average rate of change of rr, in units per radian
  2. The distance from the origin is |r||r|it increases where r>0r > 0 and increasing, or r<0r < 0 and decreasing
  3. r>0r > 0 and decreasing, or r<0r < 0 and increasingthe distance from the origin decreases
  4. |r||r| reaches a peaka point of the curve farthest from the origin nearby

Part 6 · Key ideas

Key ideas

  • The average rate of change of rr on [a,b][a, b] is r(b)−r(a)b−a\frac{r(b) - r(a)}{b - a}: the slope of the secant line on the graph of rr against θ\theta.
  • Whether the curve moves toward or away from the origin depends on the sign of rr and on whether rr is increasing.
  • Positive and increasing, or negative and decreasing: moving away. The other two cases: moving closer.
  • The largest |r||r| is the farthest point; r=0r = 0 is a visit to the origin.

Part 7 · Misconception

A common mistake

The wrong idea: When rr is decreasing, the curve is getting closer to the origin.

What actually happens: Only when r>0r > 0. If rr is negative and decreasing, such as from −1-1 to −3-3, then |r||r| grows and the curve moves away from the origin.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

The polar function r=1+2sin⁡θr = 1 + 2\sin\theta and its graph of rr against θ\theta

The first graph shows r=1+2sin⁡θr = 1 + 2\sin\theta against θ\theta on rectangular axes, for 0≤θ≤2π0 \le \theta \le 2\pi. The second shows the same function in the polar plane.

0π/2π3π/22π−1123θr
The polar graph of r = 1 + 2 sin theta: a limacon whose large outer loop reaches highest straight above the origin and crosses the horizontal axis 1 unit on each side of the origin, with a small inner loop above the origin.
LevlPrep original diagram.

1. Find the average rate of change of r=1+2sin⁡θr = 1 + 2\sin\theta with respect to θ\theta on [0,π2]\left[0, \frac{\pi}{2}\right]. Give an exact value.

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

r(π/2)−r(0)π/2−0\frac{r(\pi/2) - r(0)}{\pi/2 - 0}=3−1π/2= \frac{3 - 1}{\pi/2}=4π= \frac{4}{\pi}, the slope of the secant line on the graph of rr against θ\theta. On average rr grows 4π≈1.27\frac{4}{\pi} \approx 1.27 units per radian of turning.

  • Answer: 4π\frac{4}{\pi} ≈ 1.273

2. On the interval π2<θ<7π6\frac{\pi}{2} < \theta < \frac{7\pi}{6}, how is the distance from the origin to the curve changing, and why?

  1. Decreasing, because rr is positive and decreasing
  2. Increasing, because rr is positive
  3. Decreasing, because rr is negative and increasing
  4. Increasing, because rr is decreasing
Show the answer

On the graph of rr, the values fall from 33 at π2\frac{\pi}{2} to 00 at 7π6\frac{7\pi}{6} and stay positive. The distance from the origin is |r|=r|r| = r here, so it decreases: the curve moves back toward the pole.

  • Correct: Decreasing, because rr is positive and decreasing: Right: positive and decreasing rr means a shrinking distance.
  • Increasing, because rr is positive: Positive tells you the direction of the point, not whether rr is growing; here it is shrinking.
  • Decreasing, because rr is negative and increasing: rr is still positive on this interval; it reaches 00 only at 7π6\frac{7\pi}{6}.
  • Increasing, because rr is decreasing: When rr is positive, a decreasing rr means the points get closer to the pole.

3. On which interval is rr increasing while the distance from the origin to the curve is decreasing?

  1. (3π2,11π6)\left(\frac{3\pi}{2}, \frac{11\pi}{6}\right)
  2. (11π6,2π)\left(\frac{11\pi}{6}, 2\pi\right)
  3. (0,π2)\left(0, \frac{\pi}{2}\right)
  4. (7π6,3π2)\left(\frac{7\pi}{6}, \frac{3\pi}{2}\right)
Show the answer

On (3π2,11π6)\left(\frac{3\pi}{2}, \frac{11\pi}{6}\right), rr rises from −1-1 to 00: increasing but negative, so |r||r| shrinks from 11 to 00 and the curve heads back to the pole.

  • Correct: (3π2,11π6)\left(\frac{3\pi}{2}, \frac{11\pi}{6}\right): Right: negative and increasing rr means a shrinking distance.
  • (11π6,2π)\left(\frac{11\pi}{6}, 2\pi\right): There rr rises from 00 to 11 and is positive, so the distance increases.
  • (0,π2)\left(0, \frac{\pi}{2}\right): There rr rises from 11 to 33 and is positive, so the distance increases.
  • (7π6,3π2)\left(\frac{7\pi}{6}, \frac{3\pi}{2}\right): There rr is decreasing, from 00 to −1-1.

Table

The reach of a garden sprinkler

A sprinkler turns counterclockwise. When its nozzle points in the direction θ\theta radians, the water lands r(θ)r(\theta) meters away. Use a calculator.

Distance r(θ) the water reaches, in meters, at selected angles θ, in radians
θ\theta (radians)00π6\frac{\pi}{6}π3\frac{\pi}{3}π2\frac{\pi}{2}2π3\frac{2\pi}{3}
r(θ)r(\theta) (meters)4.04.05.55.56.26.26.46.46.06.0

4. Find the average rate of change of rr on [0,π3]\left[0, \frac{\pi}{3}\right], correct to three decimal places.

Calculator allowed. Give the answer correct to three decimal places, in meters per radian.

Show the answer

r(π/3)−r(0)π/3−0\frac{r(\pi/3) - r(0)}{\pi/3 - 0}=6.2−4.0π/3= \frac{6.2 - 4.0}{\pi/3}=6.6π= \frac{6.6}{\pi}≈2.100845\approx 2.100845 meters per radian, the slope of the secant line through (0,4.0)(0, 4.0) and (π3,6.2)\left(\frac{\pi}{3}, 6.2\right).

  • Answer: 2.101 meters per radian

5. The average rate of change of rr is about 2.8652.865 meters per radian on [0,π6]\left[0, \frac{\pi}{6}\right] and about 1.3371.337 meters per radian on [π6,π3]\left[\frac{\pi}{6}, \frac{\pi}{3}\right]. Which statement is supported?

  1. On [0,π3]\left[0, \frac{\pi}{3}\right] the reach is increasing, and the rate of change of the reach is decreasing
  2. On [0,π3]\left[0, \frac{\pi}{3}\right] the reach is decreasing, because the rates are getting smaller
  3. On [0,π3]\left[0, \frac{\pi}{3}\right] the reach is increasing, and the rate of change of the reach is increasing
  4. On [0,π3]\left[0, \frac{\pi}{3}\right] the reach is constant, because both rates are positive
Show the answer

5.5−4.0π/6\frac{5.5 - 4.0}{\pi/6}=9π= \frac{9}{\pi}≈2.865\approx 2.865 and 6.2−5.5π/6\frac{6.2 - 5.5}{\pi/6}=4.2π= \frac{4.2}{\pi}≈1.337\approx 1.337. Both are positive, so the reach (the distance from the sprinkler, since r>0r > 0) increases. Over equal intervals the rate drops, so the rate of change of the reach is decreasing.

  • Correct: On [0,π3]\left[0, \frac{\pi}{3}\right] the reach is increasing, and the rate of change of the reach is decreasing: Right: positive rates, getting smaller over equal intervals.
  • On [0,π3]\left[0, \frac{\pi}{3}\right] the reach is decreasing, because the rates are getting smaller: Smaller positive rates still mean the reach grows, just more slowly.
  • On [0,π3]\left[0, \frac{\pi}{3}\right] the reach is increasing, and the rate of change of the reach is increasing: The later rate is smaller, so the rate of change is decreasing.
  • On [0,π3]\left[0, \frac{\pi}{3}\right] the reach is constant, because both rates are positive: Positive rates mean the reach changes; it grows.

6. Which point of r=3+cos⁡θr = 3 + \cos\theta is closest to the origin?

  1. The point at θ=π\theta = \pi, 22 units from the origin
  2. The point at θ=0\theta = 0, 44 units from the origin
  3. The point at θ=π2\theta = \frac{\pi}{2}, 33 units from the origin
  4. The origin itself, at θ=π\theta = \pi
Show the answer

r=3+cos⁡θr = 3 + \cos\theta is always positive, so the distance is rr itself. It is smallest where cos⁡θ=−1\cos\theta = -1: θ=π\theta = \pi, r=2r = 2.

  • Correct: The point at θ=π\theta = \pi, 22 units from the origin: Right: the least value of rr is 22.
  • The point at θ=0\theta = 0, 44 units from the origin: r=4r = 4 at θ=0\theta = 0 is the farthest point.
  • The point at θ=π2\theta = \frac{\pi}{2}, 33 units from the origin: r=3r = 3 at π2\frac{\pi}{2} is not the least value.
  • The origin itself, at θ=π\theta = \pi: r=2r = 2 at θ=π\theta = \pi, so the curve never reaches the origin.

Part 9 · Summary

Summary

For a polar function, the average rate of change of rr on [a,b][a, b] is r(b)−r(a)b−a\frac{r(b) - r(a)}{b - a} units per radian, the slope of a secant line on the graph of rr against θ\theta. The distance from the origin is |r||r|: it increases where rr is positive and increasing or negative and decreasing, and decreases in the other two cases. Extremes of |r||r| give the farthest and closest points.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections

Question set for this topic