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Graphs of polar equations

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55115912
Consider the polar equation \(r=2\). Without graphing, describe the curve in Cartesian geometric terms by giving its center and radius.

Hints

- In polar coordinates, \(r\) measures distance from the pole. - Ask what set of points has one fixed distance from a single center.

Solution

1. The equation \(r=2\) says every point on the curve is exactly \(2\) units from the pole. 2. Therefore, the curve is a circle centered at the origin with radius \(2\).

Answer

A circle centered at \((0, 0)\) with radius \(2\)
55113412
Convert the polar equation \(r=4\cos\theta\) to a Cartesian equation. Then identify the graph by giving its center and radius.

Hints

- Multiplying the polar equation by \(r\) can create expressions that have standard Cartesian replacements. - Look for \(r^2\) and \(r\cos\theta\) before converting. - After conversion, rewrite the quadratic equation in a form that displays its geometric features.

Solution

1. Multiply by \(r\): \(r^2=4r\cos\theta\). 2. Use \(r^2=x^2+y^2\) and \(r\cos\theta=x\) to get \(x^2+y^2=4x\). 3. Complete the square: \((x-2)^2+y^2=4\). 4. The graph is a circle centered at \((2, 0)\) with radius \(2\).

Answer

\((x-2)^2+y^2=4\); a circle centered at \((2, 0)\) with radius \(2\)
55113512
Consider the polar equation \(r=3\sin(2\theta)\). a) How many petals does its rose graph have? b) Give the four directions \(\theta\) along which the petals are centered, using angles in \([0,2\pi)\).

Hints

- For an even multiplier of \(\theta\), positive and negative radius values produce distinct petals. - Petal centers occur in directions where the magnitude of \(r\) reaches its maximum. - Account for the fact that a negative radius points in the opposite direction.

Solution

1. For a rose \(r=a\sin(n\theta)\) with even \(n\), the graph has \(2n\) petals. Here \(n=2\), so there are \(4\) petals. 2. Maximum distance occurs when \(|r|=3\). The resulting petal directions are the diagonal directions \(\theta=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\).

Answer

a) \(4\) petals b) \(\theta=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\)
55116012
Analyze the polar equation \(r=2+2\cos\theta\) algebraically. a) State the line of symmetry. b) Find the angle at which the curve reaches the pole. c) Find the point farthest from the pole.

Hints

- Compare the equation at \(\theta\) and \(-\theta\) to test symmetry. - A point is at the pole exactly when its radius is zero. - The farthest point occurs where the radius is as large as possible.

Solution

1. Because replacing \(\theta\) by \(-\theta\) leaves \(\cos\theta\) unchanged, the graph is symmetric about the polar axis. 2. The graph reaches the pole when \(r=0\): \(2+2\cos\theta=0\), so \(\theta=\pi\). 3. The largest radius occurs when \(\cos\theta=1\), at \(\theta=0\). Then \(r=4\), giving the point \((4, 0)\).

Answer

a) The polar axis b) \(\theta=\pi\) c) \((4, 0)\)
55116312
Which equation matches the graph shown? Justify your choice using the number of petals and their orientation. A. \(r=2\cos(2\theta)\) B. \(r=2\sin(2\theta)\) C. \(r=2\cos\theta\) D. \(r=2+2\cos\theta\)
Figure for problem 551163

Hints

- Use the visible number of petals to narrow the equation family. - Then use the visible directions of the petals to distinguish sine from cosine. - Test where each remaining equation attains a maximum radius.

Solution

1. The graph has four petals, so equations C and D do not match the curve family. 2. The petals are centered on the coordinate-axis directions. For \(r=2\cos(2\theta)\), the radius is maximal at \(\theta=0\), producing a petal on the positive x-axis and, by symmetry, petals on the other coordinate-axis directions. 3. For \(r=2\sin(2\theta)\), the petals are centered on diagonal directions instead. Therefore, A matches the graph.

Answer

A. \(r=2\cos(2\theta)\)
55582712
A polar curve has the following sampled values. The distinct corresponding points are plotted and labeled but not connected; the row at \(2\pi\) returns to the same plotted point as the row at \(0\). <table><tr><th>\(\theta\)</th><th>\(r\)</th></tr><tr><td>\(0\)</td><td>\(4\)</td></tr><tr><td>\(\frac{\pi}{2}\)</td><td>\(2\)</td></tr><tr><td>\(\pi\)</td><td>\(0\)</td></tr><tr><td>\(\frac{3\pi}{2}\)</td><td>\(2\)</td></tr><tr><td>\(2\pi\)</td><td>\(4\)</td></tr></table> a) Which labeled plotted point corresponds to the row \(\theta=\frac{\pi}{2}\)? b) State the sampled angle at which the curve reaches the pole. c) Which equation is consistent with all five rows? A. \(r=2+2\cos\theta\) B. \(r=2+2\sin\theta\) C. \(r=2\cos(2\theta)\) D. \(r=2\)
Figure for problem 555827

Hints

- Convert the polar pair in part a to a plotted location, then match that location to a label in the diagram. - A radius of zero places a point at the pole. - Test the candidate equations against more than one row of the table.

Solution

1. At \(\theta=\frac{\pi}{2}\), \(r=2\), which plots at \((0, 2)\). The diagram labels this point \(Q\). 2. The radius is \(0\) at \(\theta=\pi\), so the curve reaches the pole there. 3. Substituting the sampled angles into \(r=2+2\cos\theta\) gives \(4,2,0,2,4\), matching every row. The other choices fail at one or more sampled angles.

Answer

a) Point \(Q\) b) \(\theta=\pi\) c) A. \(r=2+2\cos\theta\)
55583012
Use a graphing table, not conversion to a Cartesian equation, to analyze the polar equation \(r=2\sin\theta\) on \(0\le\theta\le\pi\). a) For \(\theta=0,\frac{\pi}{4},\frac{\pi}{2},\frac{3\pi}{4},\pi\), find \(r\) and the corresponding rectangular point using \(x=r\cos\theta\) and \(y=r\sin\theta\). b) Using the generated points and their symmetry about the y-axis, identify the curve and give its defining geometric data. c) Explain what the first and last table rows tell you about how the curve is traced over \(0\le\theta\le\pi\).

Hints

- Build the requested table row by row before deciding what the whole graph is. - After finding each radius, use \(x=r\cos\theta\) and \(y=r\sin\theta\) to locate the plotted point. - Compare the two points at angles equally far from \(\frac{\pi}{2}\). - Use the four distinct plotted points to identify a center that is the same distance from each one.

Solution

1. At \(\theta=0\), \(r=0\), so the point is \((0, 0)\). 2. At \(\theta=\frac{\pi}{4}\), \(r=\sqrt{2}\), so the point is \((1, 1)\). 3. At \(\theta=\frac{\pi}{2}\), \(r=2\), so the point is \((0, 2)\). 4. At \(\theta=\frac{3\pi}{4}\), \(r=\sqrt{2}\), so the point is \((-1, 1)\). At \(\theta=\pi\), \(r=0\), so the point is again \((0, 0)\). 5. The points \((0, 0)\), \((1, 1)\), \((0, 2)\), and \((-1, 1)\) are the bottom, right, top, and left points of a circle centered at \((0, 1)\) with radius \(1\). 6. The rows at \(\theta=0\) and \(\theta=\pi\) both give the pole, while the intermediate rows move around the right side, top, and left side. Thus the interval returns to the starting point after one complete trace.

Answer

a) \(\theta=0\): \(r=0\), point \((0, 0)\) \(\theta=\frac{\pi}{4}\): \(r=\sqrt{2}\), point \((1, 1)\) \(\theta=\frac{\pi}{2}\): \(r=2\), point \((0, 2)\) \(\theta=\frac{3\pi}{4}\): \(r=\sqrt{2}\), point \((-1, 1)\) \(\theta=\pi\): \(r=0\), point \((0, 0)\) b) A circle centered at \((0, 1)\) with radius \(1\) c) Both endpoint rows give the pole; the intermediate symmetric points trace one loop and return to the starting point at \(\theta=\pi\).
55113612
Use substitution tests to determine whether the polar graph \(r=3\cos(2\theta)\) is symmetric about a) the polar axis, b) the line \(\theta=\frac{\pi}{2}\), and c) the pole.

Hints

- Each symmetry has a corresponding angle substitution that can leave a polar equation unchanged. - Simplify the cosine expressions using evenness and periodicity rather than trying to sketch first. - Treat the three symmetry tests independently.

Solution

1. For the polar axis, replace \(\theta\) by \(-\theta\): \(3\cos(-2\theta)=3\cos(2\theta)\), so the equation is unchanged. 2. For the line \(\theta=\frac{\pi}{2}\), replace \(\theta\) by \(\pi-\theta\): \(3\cos(2\pi-2\theta)=3\cos(2\theta)\), so the equation is unchanged. 3. For the pole, replace \(\theta\) by \(\theta+\pi\): \(3\cos(2\theta+2\pi)=3\cos(2\theta)\), so the equation is unchanged.

Answer

a) Yes b) Yes c) Yes
55116112
The displayed graph is generated by \(r=1+2\cos\theta\). Its pole is labeled \(O\), and its two nonzero x-axis intersections are labeled \(A\) and \(B\). a) Find the two angles in \([0,2\pi)\) at which the curve passes through \(O\). b) At \(\theta=\pi\), compute \(r\). Use the sign of \(r\) and the displayed labels to decide whether the resulting point is \(A\) or \(B\). c) Determine the interval of angles between the two pole crossings for which \(r<0\), and explain how those negative-radius values produce the inner loop visible in the graph.
Figure for problem 551161

Hints

- Passing through the pole means the radius is zero. - A negative radius places the point in the direction opposite the stated angle. - Compare the sign of \(1+2\cos\theta\) between its two zeros, then connect that interval to the displayed loop.

Solution

1. At the pole, \(r=0\). Thus \(1+2\cos\theta=0\), so \(\cos\theta=-\frac{1}{2}\). The angles are \(\theta=\frac{2\pi}{3}\) and \(\theta=\frac{4\pi}{3}\). 2. At \(\theta=\pi\), \(r=1+2(-1)=-1\). A negative radius reverses the direction by \(\pi\), so \((r,\theta)=(-1,\pi)\) is the Cartesian point \((1,0)\). On the displayed graph, that point is labeled \(B\). 3. Between \(\frac{2\pi}{3}\) and \(\frac{4\pi}{3}\), \(\cos\theta<-\frac{1}{2}\), so \(r<0\). Plotting those radii in the opposite direction produces the small loop between the two pole crossings.

Answer

a) \(\theta=\frac{2\pi}{3}\) and \(\theta=\frac{4\pi}{3}\) b) \(r=-1\), giving \((1,0)\), which is point \(B\). c) \(\frac{2\pi}{3}<\theta<\frac{4\pi}{3}\); the negative radii reverse direction and trace the inner loop.
55116212
A polar curve is given by \(r=\frac{\theta}{\pi}\) for \(0\le\theta\le4\pi\). a) How many complete revolutions does the curve make? b) What is the radius after the first complete revolution? c) Find the rectangular coordinates of the final point.

Hints

- Compare the total angular change with the angle in one full revolution. - Substitute the angle reached after one revolution into the radius equation. - The final angle is coterminal with the positive x-axis.

Solution

1. The angle increases from \(0\) to \(4\pi\). Since one revolution is \(2\pi\), the curve makes \(2\) complete revolutions. 2. After the first revolution, \(\theta=2\pi\), so \(r=\frac{2\pi}{\pi}=2\). 3. At the final value \(\theta=4\pi\), \(r=4\). Thus \(x=4\cos4\pi=4\) and \(y=4\sin4\pi=0\), giving \((4, 0)\).

Answer

a) \(2\) b) \(2\) c) \((4, 0)\)
55582812
Consider the polar equation \(r^2=9\cos(2\theta)\). a) State the directions of the two lobes of the graph. b) Find all angles in \([0,2\pi)\) at which the graph passes through the pole. c) Explain why the equation has no real point with \(\theta=\frac{\pi}{2}\). d) Identify the standard polar-curve family.

Hints

- The lobe directions occur where the right side makes \(r^2\) as large as possible. - To find pole crossings, set the radius equal to zero. - Check the sign of the right side at the specified angle before trying to take a square root.

Solution

1. The largest possible value of \(r^2\) is \(9\), which occurs when \(\cos(2\theta)=1\). This aligns the lobes with the polar axis, so they extend along the positive and negative \(x\)-axis directions. 2. At the pole, \(r=0\), so \(\cos(2\theta)=0\). Thus \(2\theta=\frac{\pi}{2}+k\pi\), giving \(\theta=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\) in \([0,2\pi)\). 3. At \(\theta=\frac{\pi}{2}\), \(\cos(2\theta)=\cos\pi=-1\), so the equation would require \(r^2=-9\), which has no real radius. 4. An equation of the form \(r^2=a^2\cos(2\theta)\) is a lemniscate.

Answer

a) Along the positive and negative \(x\)-axis directions b) \(\theta=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\) c) It would require \(r^2=-9\), so there is no real radius. d) A lemniscate
55582912
Panels a) and b) show the two polar curves \(r=2+\cos\theta\) and \(r=2-\cos\theta\), but the equations are not assigned to the panels. a) Compute the radii of both equations at \(\theta=0\) and \(\theta=\pi\). Use those values together with the displayed graphs to decide which equation belongs to panel a) and which belongs to panel b). b) Show algebraically that each curve is symmetric about the polar axis. c) Describe the reflection that maps one curve to the other, and justify it from the equations.
Figure for problem 555829

Hints

- The values at \(\theta=0\) and \(\theta=\pi\) tell you how far each curve reaches to the right and left. - Match those directional radii to the actual panels rather than relying on the equation signs alone. - For the reflection, look for an angle substitution that changes the sign of cosine.

Solution

1. For \(r=2+\cos\theta\), \(r(0)=3\) and \(r(\pi)=1\), so the curve extends farther to the right than to the left. The displayed panel a) has that feature, so panel a) is \(r=2+\cos\theta\). 2. For \(r=2-\cos\theta\), \(r(0)=1\) and \(r(\pi)=3\), so the curve extends farther to the left. Thus panel b) is \(r=2-\cos\theta\). 3. Replacing \(\theta\) by \(-\theta\) leaves \(\cos\theta\) unchanged, so each equation is unchanged. Both curves are symmetric about the polar axis. 4. Replacing \(\theta\) by \(\pi-\theta\) in \(r=2+\cos\theta\) gives \(r=2-\cos\theta\). Therefore, the curves are reflections of each other across the y-axis.

Answer

a) Panel a): \(r=2+\cos\theta\); panel b): \(r=2-\cos\theta\) b) Both are symmetric about the polar axis. c) They are reflections of each other across the y-axis.

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