Polar coordinates – further maths worksheet answers to download as PDF

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1. Answer to Question 1

   Method: use \(x=r\cos\theta\), \(y=r\sin\theta\) to go from polar to Cartesian, and \(r=\sqrt{x^2+y^2}\), \(\tan\theta=\dfrac yx\) to go back. Always plot the point to choose the correct quadrant for \(\theta\).

   1. \(x=4\cos\tfrac\pi3=4\times\tfrac12=2\) and \(y=4\sin\tfrac\pi3=4\times\tfrac{\sqrt3}2=2\sqrt3\): \((2,2\sqrt3)\) This point is in the first quadrant, as expected for \(\theta=\tfrac\pi3\).
   2. \(x=2\sqrt2\cos\tfrac{3\pi}4=2\sqrt2\times(-\tfrac{\sqrt2}2)=-2\) and \(y=2\sqrt2\times\tfrac{\sqrt2}2=2\): \((-2,2)\) This point is in the second quadrant.
   3. \(r=\sqrt{9+27}=\sqrt{36}=6\). The point is in the second quadrant, so \(\theta=\pi-\arctan\dfrac{3\sqrt3}3=\pi-\tfrac\pi3=\tfrac{2\pi}3\): \((6,\tfrac{2\pi}3)\) Check: \(6\cos\tfrac{2\pi}3=-3\) and \(6\sin\tfrac{2\pi}3=3\sqrt3\).
   4. \(r=\sqrt{25+144}=13\). The point is in the fourth quadrant, so \(\theta=-\arctan\tfrac{12}5=-1.176\): \((13,\,-1.176)\) A calculator gives \(\arctan\tfrac{-12}5\) directly in the fourth quadrant here.

2. Answer to Question 2

   1. Multiply by \(r\): \(r^2=4r\cos\theta\), so \(x^2+y^2=4x\), i.e. \((x-2)^2+y^2=4\): a circle with centre \((2,0)\) and radius 2.
   2. \(r\cos\theta=2\), so \(x=2\): a vertical straight line.
   3. \(\sin2\theta=2\sin\theta\cos\theta\), so \(2(r\sin\theta)(r\cos\theta)=8\), i.e. \(2xy=8\): \(xy=4\), a rectangular hyperbola.
   4. \(r-r\cos\theta=2\), so \(r=x+2\). Square: \(x^2+y^2=x^2+4x+4\), so \(y^2=4x+4\), a parabola with vertex \((-1,0)\).

3. Answer to Question 3

   Method: replace \(x^2+y^2\) by \(r^2\), \(x\) by \(r\cos\theta\) and \(y\) by \(r\sin\theta\), then divide by any common power of \(r\) (the pole is usually still on the curve).

   1. \(r^2=9\), so \(r=3\): a circle centred at the pole.
   2. \(\tan\theta=1\) with \(x>0\): \(\theta=\tfrac\pi4\), a half-line from the pole.
   3. \(r^2=6r\sin\theta\), so \(r=6\sin\theta\) (a circle through the pole with centre \((0,3)\)).
   4. \(r^4=4r^2\sin\theta\cos\theta=2r^2\sin2\theta\), so \(r^2=2\sin2\theta\).

4. Answer to Question 4

   1. \(r=4,\;3,\;2,\;1,\;0\)
   2. \(\cos(-\theta)=\cos\theta\), so the points \((r,\theta)\) and \((r,-\theta)\) are both on the curve: the curve is its own reflection in the initial line.
   3. \(\cos\theta\le1\), so \(r_{\max}=4\) at \(\theta=0\); \(r=0\) when \(\cos\theta=-1\), i.e. \(\theta=\pi\).
   4. The curve is a cardioid (heart shape) with its cusp at the pole:

5. Answer to Question 5

   Method: the area swept out by a polar curve is \(A=\tfrac12\displaystyle\int_\alpha^\beta r^2\,d\theta\). Square \(r\) first, replace \(\cos^2\theta\) or \(\sin^2\theta\) by double angle expressions, then integrate term by term.

   1. Here \(r^2=9\) is constant and a full turn is \(0\le\theta\le2\pi\). \(A=\tfrac12\displaystyle\int_0^{2\pi}9\,d\theta=\tfrac12\times9\times2\pi=\) \(9\pi\) \(=\pi\times3^2\) ✓ The formula agrees with \(\pi r^2\).
   2. The cardioid is traced once as \(\theta\) goes from 0 to \(2\pi\). Use \(\cos^2\theta=\tfrac12(1+\cos2\theta)\). \(r^2=4(1+2\cos\theta+\cos^2\theta)=4\left(\tfrac32+2\cos\theta+\tfrac12\cos2\theta\right)\). Integrating from 0 to \(2\pi\) the cosine terms give 0, so \(A=\tfrac12\times4\times\tfrac32\times2\pi=\) \(6\pi\). So the cardioid encloses \(6\pi\approx18.8\) square units.
   3. The spiral and the initial line enclose the region swept as \(\theta\) goes from 0 to \(\pi\). \(A=\tfrac12\displaystyle\int_0^\pi\theta^2\,d\theta=\tfrac12\left[\tfrac{\theta^3}3\right]_0^\pi=\) \(\dfrac{\pi^3}6\) \(\approx5.17\) So the area is \(\tfrac{\pi^3}6\).

6. Answer to Question 6

   1. \(r\sin\theta=\sin\theta+\sin\theta\cos\theta=\sin\theta+\tfrac12\sin2\theta\). Differentiate: \(\cos\theta+\cos2\theta=\cos\theta+2\cos^2\theta-1\) ✓
   2. \(2\cos^2\theta+\cos\theta-1=(2\cos\theta-1)(\cos\theta+1)=0\), so \(\cos\theta=\tfrac12\) (or \(\cos\theta=-1\), which is the pole). \(\theta=\pm\tfrac\pi3\) and \(r=\tfrac32\): \((\tfrac32,\tfrac\pi3)\) and \((\tfrac32,-\tfrac\pi3)\).
   3. \(r\cos\theta=\cos\theta+\cos^2\theta\), derivative \(-\sin\theta-2\sin\theta\cos\theta=-\sin\theta(1+2\cos\theta)=0\). \(\sin\theta=0\): \(\theta=0\), \(r=2\). \(\cos\theta=-\tfrac12\): \(\theta=\pm\tfrac{2\pi}3\), \(r=\tfrac12\). \((2,0)\), \((\tfrac12,\tfrac{2\pi}3)\) and \((\tfrac12,-\tfrac{2\pi}3)\) (\(\theta=\pi\) gives the pole, where the curve has a cusp).

7. Answer to Question 7

   1. \(P\): \((40\cos\tfrac\pi6,\;40\sin\tfrac\pi6)=\) \((34.64,\;20)\). \(Q\): \((30\cos\tfrac{2\pi}3,\;30\sin\tfrac{2\pi}3)=\) \((-15,\;25.98)\).
   2. \(\tfrac{2\pi}3-\tfrac\pi6=\) \(\tfrac\pi2\), a right angle.
   3. By the cosine rule, \(PQ^2=40^2+30^2-2\times40\times30\cos\tfrac\pi2=1600+900-0=2500\), so \(PQ=50\) km (or use Pythagoras since the angle is \(90^\circ\)).
   4. The aircraft are 50 km apart, well above 5 km, so they are safely separated at the moment.

8. Answer to Question 8

   1. \(\cos\theta\) ranges from \(-1\) to \(1\), so greatest 6 m (at \(\theta=0\)) and least 2 m (at \(\theta=\pi\)).
   2. \(A=\tfrac12\displaystyle\int_0^{2\pi}(16+16\cos\theta+4\cos^2\theta)\,d\theta\). With \(4\cos^2\theta=2+2\cos2\theta\): \(A=\tfrac12\displaystyle\int_0^{2\pi}(18+16\cos\theta+2\cos2\theta)\,d\theta=\tfrac12\times18\times2\pi=\) \(18\pi\) ✓ (about 56.5 m²).
   3. \(\dfrac{18\pi}{80}\times100=\) 70.7%
   4. \(r=4+2\cos\theta\ge4-2=2>0\) for every \(\theta\), so \(r\) is never zero.

9. Answer to Question 9

   1. \(r=\tfrac\pi4,\;\tfrac\pi2,\;\pi,\;2\pi\)
   2. Along the initial line the spiral crosses at \(\theta=2\pi,4\pi,\dots\), where \(r=\pi,2\pi,\dots\). The gap is \(\pi\) dm \(\approx3.14\) dm.
   3. \(A=\tfrac12\displaystyle\int_0^{2\pi}\tfrac14\theta^2\,d\theta=\tfrac18\left[\tfrac{\theta^3}3\right]_0^{2\pi}=\dfrac{8\pi^3}{24}=\dfrac{\pi^3}3\approx\) 10.3 dm².
   4. The gap between turns is \(\pi\) dm \(\approx31.4\) cm, which is much more than the 3 cm width, so the turns do not touch. For \(r=k\theta\) the gap is \(2\pi k\) dm. Setting \(2\pi k=0.3\) dm gives \(k=\dfrac{0.3}{2\pi}\approx0.0477\) dm per radian.

10. Answer to Question 10

    1. \(\sin2\theta\ge0\) for \(0\le\theta\le\tfrac\pi2\), starting and ending at \(r=0\): one petal. The greatest value \(r=3\) occurs at \(\theta=\tfrac\pi4\), so each petal is 3 cm long.
    2. \(A=\tfrac12\displaystyle\int_0^{\pi/2}9\sin^22\theta\,d\theta=\tfrac94\displaystyle\int_0^{\pi/2}(1-\cos4\theta)\,d\theta=\tfrac94\times\tfrac\pi2=\) \(\dfrac{9\pi}8\) cm² \(\approx3.53\) cm²
    3. \(4\times\dfrac{9\pi}8=\) \(\dfrac{9\pi}2\approx14.1\) cm²
    4. \(14.137\times2=28.3\)p, so about 28p.

11. Answer to Question 11

    Method: the area swept out by a polar curve is \(A=\tfrac12\displaystyle\int_\alpha^\beta r^2\,d\theta\). Square \(r\) first, replace \(\cos^2\theta\) or \(\sin^2\theta\) by double angle expressions, then integrate term by term.

    1. Solve the equation \(1+\cos\theta=1.5\) for \(-180^\circ\lt \theta\le180^\circ\). \(1+\cos\theta=1.5\) gives \(\cos\theta=0.5\): \(\theta=\pm60^\circ\). So the sensitivity is 1.5 at \(60^\circ\) either side of the singer.
    2. Expand \(r^2\) and use \(\int_0^{2\pi}\cos^2\theta\,d\theta=\pi\). \(A=\tfrac12\displaystyle\int_0^{2\pi}(1+2\cos\theta+\cos^2\theta)\,d\theta=\tfrac12(2\pi+0+\pi)=\) \(\dfrac{3\pi}2\) So the whole curve encloses \(\tfrac{3\pi}2\approx4.71\) square units.
    3. \(\tfrac12\displaystyle\int_{-\pi/2}^{\pi/2}\left(\tfrac32+2\cos\theta+\tfrac12\cos2\theta\right)d\theta=\tfrac12\left[\tfrac32\theta+2\sin\theta+\tfrac14\sin2\theta\right]_{-\pi/2}^{\pi/2}=\tfrac12\left(\tfrac{3\pi}2+4+0\right)=\dfrac{3\pi}4+2\) ✓ This is the required result, about 4.36.
    4. \(\dfrac{3\pi/4+2}{3\pi/2}=\dfrac12+\dfrac4{3\pi}=0.924\), so about 92% lies in front: the microphone mostly ignores sound from the speakers behind it, which reduces feedback.

12. Answer to Question 12

    1. \(\theta=0\): \(r=8\); \(\theta=\pi\): \(r=2\), on the other side of the hut. Total \(8+2=\) 10, i.e. 1000 m.
    2. \(A=\tfrac12\displaystyle\int_0^{2\pi}(25+30\cos\theta+9\cos^2\theta)\,d\theta=\tfrac12(50\pi+0+9\pi)=\dfrac{59\pi}2\approx92.7\) (hundreds of m)². Each unit is \(100\times100=10\,000\) m², so the area is about 927 000 m².
    3. \(a=5\), \(2b=6\), and \(5<6\), so the course is not convex: it has a slight dent (dimple) near \(\theta=\pi\), where the runner turns slightly the other way.

13. Answer to Question 13

    1. The coast is the line \(x=4\), i.e. \(r\cos\theta=4\), so \(r=\dfrac4{\cos\theta}=4\sec\theta\) ✓
    2. \(r=4\sec\tfrac\pi3=4\times2=\) 8 km; the point is \((8\cos\tfrac\pi3,\;8\sin\tfrac\pi3)=\) \((4,\;4\sqrt3)\).
    3. \(A=\tfrac12\displaystyle\int_0^{\pi/3}16\sec^2\theta\,d\theta=8\left[\tan\theta\right]_0^{\pi/3}=\) \(8\sqrt3\approx13.9\) km²
    4. The region is a right-angled triangle with vertices \((0,0)\), \((4,0)\) and \((4,4\sqrt3)\): area \(\tfrac12\times4\times4\sqrt3=8\sqrt3\) ✓

14. Answer to Question 14

    1. \(r\) is least when \(\cos\theta=1\): \(r=\dfrac{8000}{1.2}=\) 6667 km (perigee). \(r\) is greatest when \(\cos\theta=-1\): \(r=\dfrac{8000}{0.8}=\) 10 000 km (apogee).
    2. \(\cos\tfrac\pi2=0\), so \(r=8000\) km.
    3. \(6667-6370=\) about 297 km.
    4. \(\cos(-\theta)=\cos\theta\), so \((r,\theta)\) and \((r,-\theta)\) both lie on the orbit: reflection in the initial line maps the orbit to itself.

15. Answer to Question 15

    1. \(3\cos\theta=1+\cos\theta\) gives \(\cos\theta=\tfrac12\), so \(\theta=\pm\tfrac\pi3\) and \(r=\tfrac32\): \((\tfrac32,\tfrac\pi3)\) and \((\tfrac32,-\tfrac\pi3)\).
    2. \(C_1\) passes through the pole at \(\theta=\pm\tfrac\pi2\) and \(C_2\) at \(\theta=\pi\). Equating \(r\)-values only finds intersections at the same \(\theta\), so the pole must be checked separately.
    3. \(R\) is the crescent between the two intersection points, for \(-\tfrac\pi3\le\theta\le\tfrac\pi3\), where the circle is outside the cardioid.
    4. By symmetry \(A=2\times\tfrac12\displaystyle\int_0^{\pi/3}\left(9\cos^2\theta-(1+\cos\theta)^2\right)d\theta=\displaystyle\int_0^{\pi/3}(8\cos^2\theta-2\cos\theta-1)\,d\theta\). With \(8\cos^2\theta=4+4\cos2\theta\): \(\displaystyle\int_0^{\pi/3}(3+4\cos2\theta-2\cos\theta)\,d\theta=\left[3\theta+2\sin2\theta-2\sin\theta\right]_0^{\pi/3}=\pi+\sqrt3-\sqrt3=\) \(\pi\) ✓
    5. \(r^2=3r\cos\theta\) gives \(x^2+y^2=3x\), i.e. \((x-\tfrac32)^2+y^2=\tfrac94\): radius \(\tfrac32\), area \(\tfrac94\pi\). Check: \(\tfrac12\displaystyle\int_{-\pi/2}^{\pi/2}9\cos^2\theta\,d\theta=\tfrac92\times\tfrac\pi2=\tfrac{9\pi}4\) ✓

16. Answer to Question 16

    1. \(r\sin\theta=3\sin\theta+\sin2\theta\), so the derivative is \(3\cos\theta+2\cos2\theta=3\cos\theta+4\cos^2\theta-2\) ✓
    2. \(\cos\theta=\dfrac{-3+\sqrt{9+32}}8=\dfrac{-3+\sqrt{41}}8=0.4254\) (the other root is less than \(-1\)). \(\theta=1.131\). Then \(r=3+2(0.4254)=3.851\) and the tangent is \(y=r\sin\theta=3.851\times0.9050=\) 3.485 units from the initial line.
    3. \(r\cos\theta=3\cos\theta+2\cos^2\theta\), derivative \(-3\sin\theta-4\sin\theta\cos\theta=-\sin\theta(3+4\cos\theta)\) ✓
    4. \(\sin\theta=0\): \(\theta=0\) (\(r=5\)) and \(\theta=\pi\) (\(r=1\)). \(\cos\theta=-\tfrac34\): \(\theta=\pm2.419\), \(r=3-\tfrac32=1.5\) and \(x=1.5\times(-\tfrac34)=\) \(-1.125\). Points: \((5,0)\), \((1,\pi)\), \((1.5,\pm2.419)\).
    5. At \(\theta=\pi\) the curve is at \(x=-1\), but at \(\theta=\pm2.419\) it reaches further left, \(x=-1.125\). So the leftmost part of the curve bends inwards at \(\theta=\pi\): a dimple (this agrees with \(a=3<2b=4\)).

17. Answer to Question 17

    1. \(\cos2\theta\ge0\) when \(-\tfrac\pi4\le\theta\le\tfrac\pi4\) or \(\tfrac{3\pi}4\le\theta\le\pi\), \(-\pi\lt \theta\le-\tfrac{3\pi}4\). \(r_{\max}=2\) at \(\theta=0\) and \(\theta=\pi\).
    2. One loop: \(\tfrac12\displaystyle\int_{-\pi/4}^{\pi/4}4\cos2\theta\,d\theta=\left[\sin2\theta\right]_{-\pi/4}^{\pi/4}=1-(-1)=\) 2. Total area 4.
    3. \(y=r\sin\theta\), so \(y^2=r^2\sin^2\theta=4\cos2\theta\sin^2\theta\) ✓. Product rule: \(-2\sin2\theta\sin^2\theta+\cos2\theta\cdot2\sin\theta\cos\theta=\sin2\theta(\cos2\theta-2\sin^2\theta)=\sin2\theta(\cos2\theta-(1-\cos2\theta))=\sin2\theta(2\cos2\theta-1)\) ✓
    4. The maximum of \(y\) in \((0,\tfrac\pi4)\) needs \(\cos2\theta=\tfrac12\), so \(\theta=\tfrac\pi6\) and \(r^2=2\), \(r=\sqrt2\). Height \(y=\sqrt2\sin\tfrac\pi6=\dfrac{\sqrt2}2\approx0.707\).
    5. \(r^2=4(\cos^2\theta-\sin^2\theta)\). Multiply by \(r^2\): \(r^4=4(r^2\cos^2\theta-r^2\sin^2\theta)\), so \((x^2+y^2)^2=4(x^2-y^2)\) ✓

18. Answer to Question 18

    1. \(r=2\pi\) and \(r=4\pi\)
    2. \(\tfrac12\displaystyle\int_0^{2\pi}\theta^2\,d\theta=\tfrac16(2\pi)^3=\) \(\dfrac{4\pi^3}3\)
    3. \(\tfrac12\displaystyle\int_{2\pi}^{4\pi}\theta^2\,d\theta=\tfrac16(64\pi^3-8\pi^3)=\) \(\dfrac{28\pi^3}3\)
    4. The second sweep covers the region inside the second turn, which includes the region inside the first turn; removing that leaves the strip. \(\dfrac{28\pi^3}3-\dfrac{4\pi^3}3=\) \(8\pi^3\approx248\) mm² ✓
    5. The sweep for turn \(n+1\) is \(\tfrac16\left((2n+2)^3-(2n)^3\right)\pi^3\) and for turn \(n\) is \(\tfrac16\left((2n)^3-(2n-2)^3\right)\pi^3\). Their difference is \(\tfrac{8\pi^3}6\left((n+1)^3-2n^3+(n-1)^3\right)=\tfrac{8\pi^3}6\times6n=\) \(8n\pi^3\), which increases by \(8\pi^3\) each time: an arithmetic sequence.

19. Answer to Question 19

    1. \(r=0,\;\sqrt2,\;2,\;\sqrt2,\;0\). \(r\) starts and ends at 0 and is positive in between: one petal, centred on \(\theta=0\).
    2. \(r=2\) when \(\cos3\theta=1\): \(3\theta=0,\pm2\pi\), so \(\theta=0\), \(\pm\tfrac{2\pi}3\).
    3. \(\tfrac12\displaystyle\int_{-\pi/6}^{\pi/6}4\cos^23\theta\,d\theta=\displaystyle\int_{-\pi/6}^{\pi/6}(1+\cos6\theta)\,d\theta=\tfrac\pi3+0=\) \(\dfrac\pi3\). Total \(\pi\).
    4. \(\dfrac{\pi}{4\pi}=\) \(\dfrac14\)
    5. The area of a petal is proportional to \(a^2\): \(\dfrac{a^2}4\times\dfrac\pi3=3\), so \(a^2=\dfrac{36}\pi\) and \(a=\dfrac6{\sqrt\pi}\approx3.39\) cm.

20. Answer to Question 20

    1. \(A=\tfrac12a^2\displaystyle\int_0^{2\pi}\left(\tfrac32+2\cos\theta+\tfrac12\cos2\theta\right)d\theta=\tfrac12a^2\times3\pi=\tfrac32\pi a^2\). The circle \(r=a\) has area \(\pi a^2\), so the ratio is \(\tfrac32\) ✓
    2. \(r=a+a\cos\theta\); multiply by \(r\): \(r^2=ar+ax\), so \(r^2-ax=ar\). Square: \((x^2+y^2-ax)^2=a^2r^2=\) \(a^2(x^2+y^2)\) ✓
    3. \(y=a(1+\cos\theta)\sin\theta\), \(\dfrac{dy}{d\theta}=a(\cos\theta+\cos2\theta)=a(2\cos\theta-1)(\cos\theta+1)=0\) gives \(\theta=\tfrac\pi3\) and \(y=a\cdot\tfrac32\cdot\tfrac{\sqrt3}2=\) \(\dfrac{3\sqrt3}4a\).
    4. The curves meet where \(\cos\theta=0\), \(\theta=\pm\tfrac\pi2\). For \(|\theta|\lt \tfrac\pi2\) the cardioid is outside the circle. Area \(=2\times\tfrac12\displaystyle\int_0^{\pi/2}a^2\left((1+\cos\theta)^2-1\right)d\theta=a^2\displaystyle\int_0^{\pi/2}\left(2\cos\theta+\tfrac12+\tfrac12\cos2\theta\right)d\theta=a^2\left(2+\tfrac\pi4\right)\): \(\left(2+\dfrac\pi4\right)a^2\).