Polar coordinates – further maths test answers to download as PDF

← Back to the test

1. Answer to Question 1 (4 marks)

   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 \(\theta\). Mark scheme as shown in brackets.

   1. \(r=\sqrt{12+4}=4\) [1]; second quadrant, \(\theta=\pi-\tfrac\pi6=\tfrac{5\pi}6\): \((4,\tfrac{5\pi}6)\) [1] Check: \(4\cos\tfrac{5\pi}6=-2\sqrt3\) and \(4\sin\tfrac{5\pi}6=2\).
   2. Multiply both sides by \(r\) so that \(r^2\) and \(r\sin\theta\) appear, then complete the square in \(y\). \(r^2=6r\sin\theta\): \(x^2+y^2=6y\) [1], i.e. \(x^2+(y-3)^2=9\): circle, centre \((0,3)\), radius 3 [1] The circle passes through the pole.

2. Answer to Question 2 (4 marks)

   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. Mark scheme as shown in brackets.

   1. \(\sin\theta\) takes values from \(-1\) to \(1\). \(r_{\max}=5\) at \(\theta=\tfrac\pi2\) [1]; \(r_{\min}=1\) at \(\theta=\tfrac{3\pi}2\) [1] Since \(r\ge1\gt 0\), the curve never passes through the pole.
   2. Use \(4\sin^2\theta=2-2\cos2\theta\); the sine and cosine terms integrate to 0 over a full turn. \(\tfrac12\displaystyle\int_0^{2\pi}(9+12\sin\theta+4\sin^2\theta)\,d\theta=\tfrac12\displaystyle\int_0^{2\pi}(11+12\sin\theta-2\cos2\theta)\,d\theta\) [1] \(=\tfrac12\times22\pi=\) \(11\pi\) [1] So the area is \(11\pi\approx34.6\) square units.

3. Answer to Question 3 (4 marks)

   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. Mark scheme as shown in brackets.

   1. The tip is where \(r\) is largest. \(\sin2\theta=1\) at \(\theta=\tfrac\pi4\): \((4,\tfrac\pi4)\) [1]
   2. Use \(\sin^22\theta=\tfrac12(1-\cos4\theta)\). \(A=\tfrac12\displaystyle\int_0^{\pi/2}16\sin^22\theta\,d\theta\) [1] \(=4\displaystyle\int_0^{\pi/2}(1-\cos4\theta)\,d\theta\) [1] \(=4\left[\theta-\tfrac14\sin4\theta\right]_0^{\pi/2}=4\times\tfrac\pi2=2\pi\) ✓ [1] So the petal has area \(2\pi\approx6.28\).

4. Answer to Question 4 (4 marks)

   Method: a tangent is parallel to the initial line where \(\dfrac{dy}{d\theta}=0\), with \(y=r\sin\theta\). Write \(y\) in terms of \(\theta\), differentiate, and factorise using \(\cos2\theta=2\cos^2\theta-1\). Mark scheme as shown in brackets.

   1. \(y=4(\sin\theta+\tfrac12\sin2\theta)\), \(\dfrac{dy}{d\theta}=4(\cos\theta+\cos2\theta)\) [1] \(=4(2\cos\theta-1)(\cos\theta+1)=0\) [1], so \(\theta=\pm\tfrac\pi3\) [1]
   2. \(r=6\): \((6\cos\tfrac\pi3,\,6\sin\tfrac\pi3)=\) \((3,\,3\sqrt3)\) [1]

   At \(\theta=\pi\), \(r=0\): this is the cusp at the pole, which is why it is excluded.

   Check: at \(\theta=\tfrac\pi3\), \(\cos\tfrac\pi3+\cos\tfrac{2\pi}3=\tfrac12-\tfrac12=0\) ✓.

5. Answer to Question 5 (4 marks)

   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. For the region between two curves, use \(\tfrac12\displaystyle\int(r_1^2-r_2^2)\,d\theta\) with \(r_1\) the outer curve. Mark scheme as shown in brackets.

   1. \(4=4(1-\cos\theta)\) gives \(\cos\theta=0\): \(\theta=\pm\tfrac\pi2\) [1]
   2. \(A=2\times\tfrac12\displaystyle\int_0^{\pi/2}\left(16-16(1-\cos\theta)^2\right)d\theta=16\displaystyle\int_0^{\pi/2}(2\cos\theta-\cos^2\theta)\,d\theta\) [1]; \(=16\left[2\sin\theta-\tfrac12\theta-\tfrac14\sin2\theta\right]_0^{\pi/2}\) [1] \(=16\left(2-\tfrac\pi4\right)=\) \(32-4\pi\approx19.4\) m² ✓ [1]

   By symmetry about the initial line, we integrate from 0 to \(\tfrac\pi2\) and double.

Total : 20 marks