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

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 \(2\).

A circle centered at \((0, 0)\) with radius \(2\)

- Begin by evaluating the radius rule at the stated angle. - Decide which direction that angle points, then consider the effect of the radius sign.

1. At \(\theta=\frac{\pi}{2}\), the radius is \(r=-2\sin\frac{\pi}{2}=-2\). 2. The angle points straight up, but the negative radius reverses the direction. The plotted point is \((x,y)=(-2\cos\frac{\pi}{2},-2\sin\frac{\pi}{2})=(0,-2)\).

\(r=-2\), plotted point \((0,-2)\); the negative radius places it opposite the upward ray.

- What stays the same when a polar equation has a constant positive radius? - Follow the angle from its first value through its middle value to its last value.

1. The radius remains \(2\), so every point lies on the circle centered at the origin with radius \(2\). 2. At \(\theta=0\), the point is \((2,0)\). At \(\theta=\frac{\pi}{2}\), it is \((0,2)\). At \(\theta=\pi\), it is \((-2,0)\). 3. The increasing angle traces the upper semicircle counterclockwise.

Upper semicircle, counterclockwise; \((2,0)\), \((0,2)\), \((-2,0)\).

- 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.

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 \(2\).

\((x-2)^2+y^2=4\); a circle centered at \((2, 0)\) with radius \(2\)

- 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.

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 \(\theta=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\).

a) \(4\) petals b) \(\theta=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\)

- 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.

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 \((4, 0)\).

a) The polar axis b) \(\theta=\pi\) c) \((4, 0)\)

- 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.

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.

A. \(r=2\cos(2\theta)\)

- Convert the polar pair in part a to a plotted location, then match that location to a label in the diagram. - Which sampled point has zero distance from the origin? - Which row lets you find one coefficient without knowing the other? - Use another row to find the remaining coefficient, then check the other rows.

1. At \(\theta=\frac{\pi}{2}\), \(r=2\), which gives \((0, 2)\). The diagram labels this point \(Q\). 2. The radius is \(0\) at \(\theta=\pi\), so the curve reaches the pole there. 3. At \(\theta=\frac{\pi}{2}\), the cosine term is zero, so \(a=2\). At \(\theta=0\), \(a+b=4\), so \(b=2\). 4. The equation \(r=2+2\cos\theta\) gives \(4,2,0,2,4\) at the five listed angles, as required.

a) Point \(Q\) b) \(\theta=\pi\) c) \(a=2\), \(b=2\), so \(r=2+2\cos\theta\). It matches all five rows.

- 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.

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.

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 \(\theta=\pi\).

- Find the radius for each angle before placing its point. - The angles lie along the coordinate axes; which way does each ray point? - What happens to a plotted point when its radius is zero?

1. At \(\theta=0\), \(r=2\), so the point is \((2,0)\). 2. At \(\theta=\frac{\pi}{2}\), \(r=1\), so the point is \((0,1)\). 3. At \(\theta=\pi\), \(r=0\), so the point is the origin \((0,0)\).

At \(\theta=0\): \(r=2\), point \((2,0)\). At \(\theta=\frac{\pi}{2}\): \(r=1\), point \((0,1)\). At \(\theta=\pi\): \(r=0\), point \((0,0)\).

- Find a radius for each chosen angle, then convert that polar point to rectangular coordinates. - Keep the plotted points in the order of increasing angle. - Can you turn the polar equation into a Cartesian equation whose shape and center are easier to recognize?

1. Substitution and \(x=r\cos\theta,\ y=r\sin\theta\) give the ordered points \((0,0)\), \((1,-1)\), \((2,0)\), \((1,1)\), \((0,0)\). 2. Since \(r=2\cos\theta\), we have \(r^2=2r\cos\theta\), or \(x^2+y^2=2x\). 3. Rearranging gives \((x-1)^2+y^2=1\): a full circle centered at \((1,0)\) with radius \(1\), traced once in the order listed.

Ordered points: \((0,0)\), \((1,-1)\), \((2,0)\), \((1,1)\), \((0,0)\). Full circle centered at \((1,0)\), radius \(1\).

- Begin by evaluating the given radius expression at each of the five angles. - What does a zero radius mean on the trace? - Could the sign of the radius place a sample at a tip you already found?

1. The listed radii are \(2,0,-2,0,2\). The zeros mark two returns to the pole, and the negative value must be plotted on the opposite ray. 2. At \(\theta=\frac{\pi}{3}\), \(r=-2\) plots on the \(\frac{4\pi}{3}\) ray, at distance \(2\) from the pole. 3. The three petal-tip directions are \(0,\frac{2\pi}{3},\frac{4\pi}{3}\), each at distance \(2\). The negative radius gives an existing tip, not an extra petal.

Radii in order: \(2,0,-2,0,2\). At \(\theta=\frac{\pi}{3}\) the plotted direction is \(\frac{4\pi}{3}\). Three petal tips are at \(0,\frac{2\pi}{3},\frac{4\pi}{3}\), each distance \(2\) from the pole.

- Find the radius at each listed angle before converting to coordinates. - Think about what happens at each quarter-turn as the radius increases. - Compare the first and last positions, not only their directions.

1. The corresponding radii are \(0,\frac{1}{2},1,\frac{3}{2},2,4\). 2. Converting to rectangular coordinates gives \((0,0)\), \((0,\frac{1}{2})\), \((-1,0)\), \((0,-\frac{3}{2})\), \((2,0)\), and \((4,0)\). 3. The radius grows as the angle increases. The path is an outward spiral that makes two full turns and does not return to its start.

\((0,0)\), \((0,\frac{1}{2})\), \((-1,0)\), \((0,-\frac{3}{2})\), \((2,0)\), \((4,0)\); outward spiral, two full turns, does not return to the start.

- Start by finding each radius, then turn both polar points into rectangular coordinates. - What changes about the coordinates when a point is reflected across the y-axis? - Compare the sine and cosine of an angle with those of its supplement.

1. Both angles give \(r=1\). Their rectangular points are \((\frac{\sqrt{3}}{2},\frac{1}{2})\) and \((-\frac{\sqrt{3}}{2},\frac{1}{2})\), which are reflections across the y-axis. 2. In general, \(\sin(\pi-\theta)=\sin\theta\), so the radius is unchanged. The cosines have opposite signs, and the sines agree, so the two rectangular points have opposite x-coordinates and equal y-coordinates.

Points: \((\frac{\sqrt{3}}{2},\frac{1}{2})\) and \((-\frac{\sqrt{3}}{2},\frac{1}{2})\). They mirror across the y-axis; the same radius and opposite x-coordinates show the symmetry for any \(\theta\).

- Pick an easy angle and imagine reflecting its plotted point in each axis. - One matching pair is not enough to prove symmetry. What identity could show that the matching points always exist? - Check the origin separately: would a half-turn change the set of points?

1. For A, \(\cos(-\theta)=\cos\theta\), so the x-axis is a symmetry axis. The right-facing cardioid is not symmetric about the y-axis or origin. 2. For B, \(\sin(\pi-\theta)=\sin\theta\), so the y-axis is a symmetry axis. The upward-facing cardioid is not symmetric about the x-axis or origin. 3. For C, \(\cos(-2\theta)=\cos(2\theta)\), \(\cos(2(\pi-\theta))=\cos(2\theta)\), and \(\cos(2(\theta+\pi))=\cos(2\theta)\). Thus all three symmetries hold.

A. x-axis only. B. y-axis only. C. x-axis, y-axis, and origin.

- Evaluate the sine factor at each stated angle. - Does the sign of the computed radius change the plotted direction?

1. At \(\theta=\frac{\pi}{4}\), \(r=2\sin\frac{\pi}{2}=2\). The point is distance \(2\) in direction \(\frac{\pi}{4}\). 2. At \(\theta=\frac{3\pi}{4}\), \(r=2\sin\frac{3\pi}{2}=-2\). Its plotted direction is \(\frac{3\pi}{4}+\pi=\frac{7\pi}{4}\), also at distance \(2\). 3. Treating the negative radius as positive would put that petal tip on the wrong ray.

At \(\theta=\frac{\pi}{4}\): distance \(2\), direction \(\frac{\pi}{4}\). At \(\theta=\frac{3\pi}{4}\): distance \(2\), direction \(\frac{7\pi}{4}\). A negative radius reverses the ray.

- Try each listed angle in the equation before deciding whether a point exists. - Which real numbers, if any, could have the resulting square? Check all possibilities.

1. At \(\theta=0\), the right side is \(0\), so \(r=0\). 2. At \(\theta=\frac{\pi}{2}\), the right side is \(4\), so \(r=2\) or \(r=-2\). 3. At \(\theta=\frac{3\pi}{2}\), the right side is \(-4\), so no real radius exists.

\(\theta=0:\ r=0\); \(\theta=\frac{\pi}{2}:\ r=\pm2\); \(\theta=\frac{3\pi}{2}\): no real \(r\).

- Find where the curve begins on the positive x-axis. - Look at its distance from the origin where it crosses the negative x-axis. - Compare those two distances with the formulas in the choices.

1. The plotted spiral begins at \((1,0)\), so its radius at \(\theta=0\) is \(1\). This rules out A and D. 2. At \(\theta=\pi\), the curve passes through \((-2,0)\), so its radius is \(2\). Choice B gives \(r=2\) there, while C gives \(r=3\). 3. Therefore only B matches both positions and the outward movement of the curve.

B. \(r=1+\frac{\theta}{\pi}\), with radii \(1\) at \(\theta=0\) and \(2\) at \(\theta=\pi\).

- Replace the polar angle with the parameter, then project the radius horizontally and vertically. - Ask whether a negative radius means an invalid point or a point in the opposite direction.

1. Use \(x=r\cos\theta\) and \(y=r\sin\theta\). 2. Substituting \(r=2\cos(2t)\) gives \(x(t)=2\cos(2t)\cos t\), \(y(t)=2\cos(2t)\sin t\), for \(0\le t\le2\pi\). 3. The cosine factor sometimes becomes negative. Those points lie on rays opposite their parameter angles and complete the petals, so no interval should be removed.

\(x(t)=2\cos(2t)\cos t,\ y(t)=2\cos(2t)\sin t,\ 0\le t\le2\pi\); negative radii are included because they trace parts of the rose.

- Compare the two radii at the same angle before plotting their points. - How does each radius change as the angle increases? - Do the two angle rules turn in different directions, or only change the distance differently?

1. At \(\theta=\frac{\pi}{2}\), the radii are \(r_1=\frac{5}{2}\) and \(r_2=\frac{3}{2}\), so the plotted points are \((0,\frac{5}{2})\) and \((0,\frac{3}{2})\). 2. Both radii are positive throughout the interval. The first radius increases from \(2\) to \(3\), so its point moves away from the origin. The second radius decreases from \(2\) to \(1\), so its point moves toward the origin. 3. Both angles increase from \(0\) to \(\pi\), so both traces turn counterclockwise through a half-turn.

At \(\theta=\frac{\pi}{2}\): \(r_1\) gives \((0,\frac{5}{2})\), and \(r_2\) gives \((0,\frac{3}{2})\). The first moves away; the second moves toward the origin. Both turn counterclockwise.

- What happens to the plotted direction when the radius is negative? - Could two equal-size radii at different angles lead to points on different rays? - Compare the locations farthest from the pole before deciding how many petal tips there are.

1. At \(\theta=0\), the radius is \(2\), giving \((2,0)\). 2. At \(\theta=\frac{\pi}{2}\), the radius is \(-2\), giving \((0,-2)\); a negative radius reverses the plotted direction. 3. At \(\theta=\pi\), the radius is \(2\), giving \((-2,0)\). At \(\theta=\frac{3\pi}{2}\), the radius is \(-2\), giving \((0,2)\). 4. These four different points are the tips of four petals. Ignoring negative radii would miss two petals.

In order: \((2,0)\), \((0,-2)\), \((-2,0)\), \((0,2)\). They are four distinct petal tips; negative radii are plotted on opposite rays, not discarded.

- 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.

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.

a) Yes b) Yes c) Yes

- 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.

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.

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.

- 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.

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 \((4, 0)\).

a) \(2\) b) \(2\) c) \((4, 0)\)

- 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.

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.

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

- 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.

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.

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.

- Make a short radius table before converting the points. - Look for where the radius is zero and where it is greatest. - Compare the locations for angles above and below the x-axis.

1. The radii at the listed angles are \(0,1,2,1,0\). 2. Converting gives \((0,0)\), \((0,1)\), \((-2,0)\), \((0,-1)\), \((0,0)\). 3. This is a left-facing cardioid with a cusp at the pole, farthest point \((-2,0)\), and symmetry across the x-axis.

Ordered points: \((0,0)\), \((0,1)\), \((-2,0)\), \((0,-1)\), \((0,0)\). Pole at \((0,0)\), farthest point \((-2,0)\); symmetric about the x-axis.

- The pole is reached wherever the radius becomes zero. - What happens to the plotted position when the radius becomes negative? - Separate the parameter interval where the radius is negative from the intervals where it is positive.

1. The radius is zero when \(\cos\theta=-\frac{1}{2}\), at \(\theta=\frac{2\pi}{3}\) and \(\theta=\frac{4\pi}{3}\). 2. Between those angles, the cosine is below \(-\frac{1}{2}\), so \(r<0\); these angles trace the inner loop. 3. At \(\theta=\pi\), \(r=1-2=-1\). The negative radius on the left-pointing ray plots at \((1,0)\). 4. The outer loop reaches \((3,0)\) at \(\theta=0\). The full curve is a limaçon with a smaller inner loop on the right.

Pole at \(\theta=\frac{2\pi}{3}\) and \(\theta=\frac{4\pi}{3}\); \(r<0\) for \(\frac{2\pi}{3}<\theta<\frac{4\pi}{3}\); point at \(\theta=\pi\): \((1,0)\). A large outer loop with a smaller inner loop to the right of the pole.

- Before plotting, ask when the right side of an equation for \(r^2\) can be negative. - Check where the radius reaches zero and where its magnitude is largest. - When the right side of the equation is positive, how many real radii can it give?

1. Real radii require \(\cos(2\theta)\ge0\). In the stated interval this means \(0\le\theta\le\frac{\pi}{4}\), \(\frac{3\pi}{4}\le\theta\le\frac{5\pi}{4}\), or \(\frac{7\pi}{4}\le\theta<2\pi\). 2. The radius is zero at \(\theta=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\). 3. The largest magnitude of the radius is \(2\), reached on the horizontal axis; both signs of \(r\) give the farthest points \((-2,0)\) and \((2,0)\). 4. Connecting valid angles with both signs forms a horizontal figure-eight curve with two loops meeting at the pole.

Real radii for \(\theta\in[0,\frac{\pi}{4}]\cup[\frac{3\pi}{4},\frac{5\pi}{4}]\cup[\frac{7\pi}{4},2\pi)\). The four pole angles are \(\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\). Farthest points: \((-2,0)\), \((2,0)\). Two horizontal loops meeting at the pole.

- Choose the sampled angle that makes the sine term disappear. - Use another sample to find the remaining coefficient. - After finding the missing radius, consider what its sign means for the plotted curve.

1. At \(\theta=0\), \(\sin\theta=0\), so \(a=2\). 2. At \(\theta=\frac{\pi}{2}\), \(\sin\theta=1\), so \(a+b=5\) and \(b=3\). 3. At \(\theta=\frac{3\pi}{2}\), \(r=a-b=-1\). 4. The radius changes sign, so the limaçon crosses the pole and has an inner loop.

\(a=2\), \(b=3\); \(r=2+3\sin\theta\). At \(\theta=\frac{3\pi}{2}\), \(r=-1\), so the curve has an inner loop.

- Both coordinates have the same positive factor. What does that tell you about the distance from the origin? - Compare the direction of the point with the parameter \(t\). - Which values of the trigonometric factor make the distance greatest?

1. Since \(1+\sin^2 t>0\), the distance from the origin is \(r=\sqrt{x^2+y^2}=1+\sin^2 t\). 2. The point's direction is \(\theta=t\) for \(0\le t\le2\pi\). Therefore, the full curve has polar equation \(r=1+\sin^2\theta\), for \(0\le\theta\le2\pi\). 3. The radius is greatest when \(\sin^2\theta=1\), at \(\theta=\frac{\pi}{2}\) and \(\theta=\frac{3\pi}{2}\). The greatest radius is \(2\).

\(r=1+\sin^2\theta\), for \(0\le\theta\le2\pi\). Greatest radius \(2\) at \(\theta=\frac{\pi}{2}\) and \(\theta=\frac{3\pi}{2}\).

- Which angles make the sine easy to evaluate without a calculator? - Think about the range of the sine function while choosing the samples. - Compare points on opposite sides of the y-axis. What must match for the whole curve to have that symmetry?

1. One choice is \(\theta=0,\frac{\pi}{2},\pi,\frac{3\pi}{2}\). The corresponding radii are \(2,1,2,3\). 2. The plotted points in that order are \((2,0)\), \((0,1)\), \((-2,0)\), and \((0,-3)\). 3. Since \(-1\le\sin\theta\le1\), the radius ranges from \(1\) at \(\frac{\pi}{2}\) to \(3\) at \(\frac{3\pi}{2}\). The curve extends farther below the x-axis. 4. Reflecting an angle \(\theta\) to \(\pi-\theta\), modulo \(2\pi\), keeps \(\sin\theta\) unchanged. The reflected points therefore have equal radii, so the curve is symmetric about the y-axis. Other valid angle choices are possible. 5. The radius never reaches zero, so the curve does not pass through the origin. After one full turn, it returns to \((2,0)\), forming a single closed loop.

One valid set is \(\theta=0,\frac{\pi}{2},\pi,\frac{3\pi}{2}\), with radii \(2,1,2,3\) and points \((2,0)\), \((0,1)\), \((-2,0)\), \((0,-3)\). The greatest radius is \(3\) below the x-axis, and the smallest is \(1\) above it. The curve is symmetric about the y-axis, forms one closed loop, and never passes through the origin. Other choices of the four angles are valid if they include both extrema and the matching points are correct.