Polar Coordinates and Parametric Equations: Precalculus Math Worksheet Answer Key – PDF Download

Precalculus Polar Parametric Worksheet Answer Key: problems 1–5

Problem 1: Read points on a polar grid

Count rings for \(r\) and read the spoke for \(\theta\).

1. A: \((3,30^{\circ})\)
2. B: \((4,120^{\circ})\)
3. C: \((2,225^{\circ})\)
4. D: \((5,300^{\circ})\)

Problem 2: Convert polar to rectangular

1. \(x=4\cos\frac{\pi}{3}=2\), \(y=4\sin\frac{\pi}{3}=2\sqrt{3}\): \((2,2\sqrt{3})\)
2. \(x=6\cdot(-\frac{\sqrt{2}}{2})=-3\sqrt{2}\), \(y=6\cdot\frac{\sqrt{2}}{2}=3\sqrt{2}\): \((-3\sqrt{2},3\sqrt{2})\)
3. \(x=5\cos\pi=-5\), \(y=5\sin\pi=0\): \((-5,0)\)
4. \(x=2\cos(-\frac{\pi}{2})=0\), \(y=2\sin(-\frac{\pi}{2})=-2\): \((0,-2)\)

Problem 3: Convert rectangular to polar

1. \(r=\sqrt{9+27}=6\); \(\tan\theta=\sqrt{3}\) in Quadrant I, so \(\theta=\frac{\pi}{3}\): \((6,\frac{\pi}{3})\)
2. The point lies on the negative x-axis, so \(r=4\) and \(\theta=\pi\): \((4,\pi)\)
3. \(r=\sqrt{25+25}=5\sqrt{2}\); \(\tan\theta=-1\) in Quadrant II, so \(\theta=\frac{3\pi}{4}\): \((5\sqrt{2},\frac{3\pi}{4})\)
4. \(r=\sqrt{4+12}=4\); \(\tan\theta=-\sqrt{3}\) in Quadrant IV, so \(\theta=\frac{5\pi}{3}\): \((4,\frac{5\pi}{3})\)

Problem 4: Rewrite equivalent polar pairs

1. Change the sign of \(r\) and add \(180^{\circ}\): \((2,250^{\circ})\)
2. Add \(360^{\circ}\): \(-45^{\circ}+360^{\circ}=315^{\circ}\), so \((5,315^{\circ})\)
3. Change the sign of \(r\) and add \(\pi\): \(\frac{\pi}{6}+\pi=\frac{7\pi}{6}\), so \((4,\frac{7\pi}{6})\)
4. Subtract \(4\pi\) (two full turns): \(5\pi-4\pi=\pi\), so \((3,\pi)\)

Problem 5: Distance between polar points

1. The angle between the rays is \(60^{\circ}\). \(d^{2}=16+36-2(4)(6)\cos 60^{\circ}=52-24=28\), so \(d=2\sqrt{7}\approx 5.29\)
2. The angle between the rays is \(120^{\circ}\). \(d^{2}=25+25-2(5)(5)\cos 120^{\circ}=50+25=75\), so \(d=5\sqrt{3}\approx 8.66\)

Precalculus Polar Parametric Worksheet Answer Key: problems 6–10

Problem 6: Rectangular equations to polar form

1. Replace \(x\) with \(r\cos\theta\): \(r\cos\theta=4\)
2. \(r\sin\theta=r\cos\theta\) gives \(\tan\theta=1\), so \(\theta=\frac{\pi}{4}\)
3. \(r^{2}=25\), so \(r=5\)
4. \(r^{2}=6r\sin\theta\), so \(r=6\sin\theta\) (the pole is included at \(\theta=0\))

Problem 7: Polar equations to rectangular form

1. \(r^{2}=49\), so \(x^{2}+y^{2}=49\)
2. \(\tan\theta=\sqrt{3}\) gives \(y=\sqrt{3}\,x\)
3. Multiply by \(r\): \(x^{2}+y^{2}=8y\). Complete the square: \(x^{2}+(y-4)^{2}=16\)
4. Multiply by \(r\): \(x^{2}+y^{2}=2x+2y\). Complete both squares: \((x-1)^{2}+(y-1)^{2}=2\)

Problem 8: Identify a polar circle

The circle passes through the pole and reaches \(r=10\) at \(\theta=0\), so its diameter is 10.

Center \((5,0)\), radius 5. Algebraically, \(r^{2}=10r\cos\theta\) gives \(x^{2}+y^{2}=10x\), so \((x-5)^{2}+y^{2}=25\).

Problem 9: Complete a cardioid table

\(\theta\)	0	\(\frac{\pi}{3}\)	\(\frac{\pi}{2}\)	\(\frac{2\pi}{3}\)	\(\pi\)	\(\frac{4\pi}{3}\)	\(\frac{3\pi}{2}\)	\(\frac{5\pi}{3}\)
\(r\)	4	3	2	1	0	1	2	3

Because \(a=b=2\), the graph is a cardioid. The greatest value is \(r=4\) at \(\theta=0\), and the curve passes through the pole at \(\theta=\pi\).

Problem 10: Classify polar graphs

1. \(a=b=3\): cardioid
2. \(a=2< b=4\): limaçon with an inner loop
3. \(\frac{a}{b}=1.5\), between 1 and 2: dimpled limaçon
4. \(\frac{a}{b}=2.5\ge 2\): convex limaçon (no loop, no dimple)
5. \(n=3\) is odd: rose with 3 petals
6. \(n=2\) is even: rose with 4 petals

Precalculus Polar Parametric Worksheet Answer Key: problems 11–15

Problem 11: Find the tips of a rose

Since \(n=3\) is odd there are 3 petals, each of length 5.

Solve \(\cos 3\theta=1\): \(3\theta=0,2\pi,4\pi\), so \(\theta=0,\frac{2\pi}{3},\frac{4\pi}{3}\). These angles point to the three petal tips.

Problem 12: Pole crossings of a rose

Solve \(4\cos 2\theta=0\): \(2\theta=\frac{\pi}{2}+\pi k\), so \(\theta=\frac{\pi}{4}+\frac{\pi k}{2}\). In \([0,2\pi)\): \(\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\).

\(n=2\) is even, so there are 4 petals. \(r=4\) when \(\cos 2\theta=1\): \(2\theta=0,2\pi\), so \(\theta=0,\pi\).

Problem 13: Limaçon with an inner loop

The greatest value is \(2+4=\) 6 at \(\theta=0\).

\(2+4\cos\theta=0\) gives \(\cos\theta=-\frac{1}{2}\), so \(\theta=\frac{2\pi}{3},\frac{4\pi}{3}\).

At \(\theta=\pi\), \(r=2-4=-2\). The point \((-2,\pi)\) is \((x,y)=(-2\cos\pi,-2\sin\pi)=\) \((2,0)\), on the inner loop.

Problem 14: Make a table of parametric points

Substitute each \(t\): \((-3,1)\), \((-1,-2)\), \((1,-3)\), \((3,-2)\), \((5,1)\). They lie on a parabola opening upward with its lowest point at \((1,-3)\).

Problem 15: Eliminate the parameter, linear

1. From \(y=t+4\), \(t=y-4\). Then \(x=3(y-4)-2=3y-14\), so \(y=\frac{x+14}{3}\).
2. From \(x=5-2t\), \(t=\frac{5-x}{2}\). Then \(y=4\cdot\frac{5-x}{2}+1=10-2x+1\), so \(y=11-2x\).

Precalculus Polar Parametric Worksheet Answer Key: problems 16–20

Problem 16: Eliminate the parameter, quadratic

1. From \(y=2t-1\), \(t=\frac{y+1}{2}\). Then \(x=\frac{(y+1)^{2}}{4}\), a parabola opening to the right.
2. From \(x=t+3\), \(t=x-3\). Then \(y=(x-3)^{2}-6(x-3)+5=x^{2}-6x+9-6x+18+5\), so \(y=x^{2}-12x+32\), a parabola opening upward. Check: \(t=1\) gives \((4,0)\), and \(16-48+32=0\).

Problem 17: Eliminate the parameter, ellipse

Write \(\cos t=\frac{x}{4}\) and \(\sin t=\frac{y}{3}\). Since \(\cos^{2}t+\sin^{2}t=1\): \(\frac{x^{2}}{16}+\frac{y^{2}}{9}=1\), an ellipse.

The points are \((4,0)\), \((0,3)\), \((-4,0)\), \((0,-3)\), so the curve is traced counterclockwise.

Problem 18: Shifted circle from parametric equations

Rewrite as \(x-2=3\cos t\) and \(y+1=3\sin t\). Squaring and adding gives \((x-2)^{2}+(y+1)^{2}=9\).

Center \((2,-1)\), radius 3. At \(t=0\) the point is \((5,-1)\); at \(t=\pi\) it is \((-1,-1)\). The motion is counterclockwise.

Problem 19: Soccer kick

1. \(y=40-16=\) 24 ft
2. \(40t-16t^{2}=0\) gives \(8t(5-2t)=0\), so \(t=2.5\) s
3. \(x=24(2.5)=\) 60 ft
4. The vertex is at \(t=\frac{40}{32}=1.25\), where \(y=50-25=\) 25 ft
5. \(t=\frac{x}{24}\), so \(y=\frac{40x}{24}-\frac{16x^{2}}{576}=\) \(\frac{5x}{3}-\frac{x^{2}}{36}\). Check: \(x=60\) gives \(100-100=0\).

Problem 20: Ferris wheel

1. One turn needs \(\frac{\pi t}{10}=2\pi\), so \(t=20\) s
2. \(\cos\frac{\pi}{2}=0\), \(\sin\frac{\pi}{2}=1\): \((0,65)\), the top
3. The lowest point is \(35-30=\) 5 ft, at \(t=15\)
4. \(x^{2}+(y-35)^{2}=900\cos^{2}+900\sin^{2}=\) \(x^{2}+(y-35)^{2}=900\)
5. \(35+30\sin\frac{\pi t}{10}=50\) gives \(\sin\frac{\pi t}{10}=\frac{1}{2}\), so \(\frac{\pi t}{10}=\frac{\pi}{6}\) and \(t=\frac{5}{3}\approx 1.67\) s

Precalculus Polar Parametric Worksheet Answer Key: problems 21–22

Problem 21: Radar stations

For boat 1: \(x=12\cos 40^{\circ}\approx\) 9.19 and \(y=12\sin 40^{\circ}\approx\) 7.71.

The angle between the rays is \(60^{\circ}\). \(d^{2}=144+81-2(12)(9)\cos 60^{\circ}=225-108=117\), so \(d=3\sqrt{13}\approx 10.82\) mi.

Problem 22: Rose flower bed

1. \(n=3\) is odd: 3 petals
2. \(\sin 3\theta=1\) gives \(3\theta=\frac{\pi}{2}+2\pi k\), so \(\theta=\frac{\pi}{6},\frac{5\pi}{6},\frac{3\pi}{2}\)
3. Tips are 8 ft from the pole and \(120^{\circ}\) apart: \(d^{2}=64+64-128\cos 120^{\circ}=192\), so \(d=8\sqrt{3}\approx 13.86\) ft
4. The three tips form an equilateral triangle: \(3\cdot 8\sqrt{3}=24\sqrt{3}\approx\) 41.57 ft

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