Further Maths Polar coordinates Practice Questions

Free Further Maths Polar coordinates practice questions with full step-by-step worked solutions. Covers polar-coordinates, conversion, polar-to-cartesian, cartesian-to-polar. Practise exam-style problems and check your method.

polar-coordinatesconversionpolar-to-cartesiancartesian-to-polarpolar-equationspolar-curves
Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
The point PP has polar coordinates (4, 5π6)\left(4,\ \frac{5 \pi}{6}\right). Find the exact Cartesian coordinates of PP.
Show worked solution

Worked solution

  1. Write down the conversion formulae

    x=rcos⁡θ,y=rsin⁡θx=r\cos\theta,\qquad y=r\sin\theta

    The Cartesian coordinates are the components of the radius vector.

  2. Substitute r=4r=4 and θ=5π6\theta=\frac{5 \pi}{6}

    x=4cos⁡(5π6),y=4sin⁡(5π6)x=4\cos\left(\frac{5 \pi}{6}\right),\qquad y=4\sin\left(\frac{5 \pi}{6}\right)

    Both coordinates use the same rr and the same θ\theta.

  3. State the Cartesian coordinates

    P(−23, 2)P\left(- 2 \sqrt{3},\ 2\right)

    These are the exact Cartesian coordinates of PP.

Answer
(−23, 2)\left(- 2 \sqrt{3},\ 2\right)
Question 2
2 markseasy
Which of the following gives the polar coordinates, with r≥0r\ge0 and −π<θ≤π-\pi<\theta\le\pi, of the point with Cartesian coordinates (−4, 4)\left(-4,\ 4\right)?
Show worked solution

Worked solution

  1. Find rr

    r=(−4)2+(4)2=42r=\sqrt{\left(-4\right)^{2}+\left(4\right)^{2}}=4 \sqrt{2}

    The polar radius is the distance from the pole.

  2. Identify the quadrant

    (−4, 4)\left(-4,\ 4\right)

    The signs of xx and yy fix the quadrant, and hence the sign of θ\theta.

  3. Find θ\theta in the principal range

    θ=3π4\theta=\frac{3 \pi}{4}

    The angle must satisfy −π<θ≤π-\pi<\theta\le\pi.

  4. Select the correct polar coordinates

    (42, 3π4)\left(4 \sqrt{2},\ \frac{3 \pi}{4}\right)

    These are the polar coordinates of the point with r≥0r\ge0 and −π<θ≤π-\pi<\theta\le\pi.

Answer
(42, 3π4)\left(4 \sqrt{2},\ \frac{3 \pi}{4}\right)
Question 3
4 marksintermediate
Which of the following best describes the curve with polar equation r=5+3cos⁡θr=5+3\cos\theta?
Show worked solution

Worked solution

  1. Compare the polar equation with the standard form

    r=a+bcos⁡θ with a=5, b=3r=a+b\cos\theta\ \text{with}\ a=5,\ b=3

    The relative sizes of aa and bb decide which curve this is.

  2. Test the symmetry

    r(−θ)=r(θ)r\left(-\theta\right)=r\left(\theta\right)

    The curve is unchanged by this reflection, so it is symmetrical about the initial line.

  3. Find the greatest value of rr

    rmax⁡=8r_{\max}=8

    This happens when the cosine term takes its extreme value.

  4. Decide whether the curve reaches the pole

    r=0 has no solutionr=0\ \text{has no solution}

    The curve passes through the pole exactly when r=0r=0 can be solved.

  5. Recall the classification of r=a+bcos⁡θr=a+b\cos\theta

    a=b ⇒ cardioid,a<b ⇒ inner loop,a>b ⇒ no inner loopa=b\ \Rightarrow\ \text{cardioid},\qquad a<b\ \Rightarrow\ \text{inner loop},\qquad a>b\ \Rightarrow\ \text{no inner loop}

    This is the standard classification of the limacon family.

  6. Check the value of rr at θ=0\theta=0

    r(0)=8r\left(0\right)=8

    This is where the curve crosses the initial line.

  7. Select the correct description

    r=5+3cos⁡θr=5+3\cos\theta

    This option gets the family, the axis of symmetry, the greatest value of rr and the behaviour at the pole all correct.

Answer
A limaçon with no inner loop, symmetrical about the initial line, with greatest value r=8r=8, not passing through the pole
Question 4
6 markshard
Which of the following is a Cartesian equation of the curve with polar equation rcos⁡(θ−π6)=2r\cos\left(\theta-\frac{\pi}{6}\right)=2?
Show worked solution

Worked solution

  1. Write down the conversion relations

    x=rcos⁡θ,y=rsin⁡θ,r2=x2+y2x=r\cos\theta,\quad y=r\sin\theta,\quad r^{2}=x^{2}+y^{2}

    These are the only tools needed for the conversion.

  2. Rearrange the polar equation to expose those combinations

    rcos⁡(θ−π6)=2r\cos\left(\theta-\frac{\pi}{6}\right)=2

    Multiplying by rr produces r2r^{2}, rcos⁡θr\cos\theta and rsin⁡θr\sin\theta.

  3. Replace each combination by its Cartesian form

    3x+y=4\sqrt{3} x + y=4

    Every rr and θ\theta has now been eliminated.

  4. Test a point of the curve in each option

    θ=0 gives a point of C\theta=0\ \text{gives a point of }C

    Substituting a known point rules out the wrong options at once.

  5. Note that the pole may satisfy a wrong option too

    (0, 0)\left(0,\ 0\right)

    A single test point is not always enough; a second test point is safer.

  6. Recall the conversion formulae

    x=rcos⁡θ,y=rsin⁡θx=r\cos\theta,\qquad y=r\sin\theta

    These take a point from polar form to Cartesian form.

  7. Recall the reverse conversion formulae

    r2=x2+y2,tan⁡θ=yxr^{2}=x^{2}+y^{2},\qquad\tan\theta=\frac{y}{x}

    These take a point from Cartesian form back to polar form.

  8. Recall the polar area formula

    A=12∫αβr2 dθA=\frac{1}{2}\int_{\alpha}^{\beta}r^{2}\,\mathrm{d}\theta

    The area is swept out by the radius vector between the two half-lines.

  9. Recall the double-angle form of cos⁡2θ\cos^{2}\theta

    cos⁡2θ=12(1+cos⁡2θ)\cos^{2}\theta=\frac{1}{2}\left(1+\cos2\theta\right)

    A squared cosine must be reduced to a multiple angle before it can be integrated.

  10. Recall the double-angle form of sin⁡2θ\sin^{2}\theta

    sin⁡2θ=12(1−cos⁡2θ)\sin^{2}\theta=\frac{1}{2}\left(1-\cos2\theta\right)

    A squared sine must be reduced to a multiple angle before it can be integrated.

  11. Select the matching Cartesian equation

    3x+y=4\sqrt{3} x + y=4

    This is the only equation satisfied by every point of CC.

Answer
3x+y=4\sqrt{3} x + y=4
Question 5
9 markschallenging
Which of the following gives the tangents at the pole to the curve with polar equation r=6cos⁡2θr=6\cos 2\theta for 0≤θ<2π0\le\theta<2 \pi?
Show worked solution

Worked solution

  1. State the condition for a tangent at the pole

    r=0r=0

    The tangents at the pole are the half-lines on which rr vanishes.

  2. Solve r=0r=0

    cos⁡2θ=0\cos 2\theta=0

    The zeros of the cosine give every direction in which the curve meets the pole.

  3. List the solutions in the interval

    θ=π4, θ=3π4, θ=5π4, θ=7π4\theta=\frac{\pi}{4},\ \theta=\frac{3 \pi}{4},\ \theta=\frac{5 \pi}{4},\ \theta=\frac{7 \pi}{4}

    These are all the solutions with 0≤θ<2π0\le\theta<2 \pi.

  4. Reject the stationary values of rr

    drdθ=0 gives the tips of the petals\frac{\mathrm{d}r}{\mathrm{d}\theta}=0\ \text{gives the tips of the petals}

    The tips are where rr is greatest, not where it is zero.

  5. Check one of the values

    r(π4)=0r\left(\frac{\pi}{4}\right)=0

    This confirms that the half-line really is a tangent at the pole.

  6. Count the tangents

    4 half-lines4\ \text{half-lines}

    Each solution of r=0r=0 contributes one tangent at the pole.

  7. Recall the conversion formulae

    x=rcos⁡θ,y=rsin⁡θx=r\cos\theta,\qquad y=r\sin\theta

    These take a point from polar form to Cartesian form.

  8. Recall the reverse conversion formulae

    r2=x2+y2,tan⁡θ=yxr^{2}=x^{2}+y^{2},\qquad\tan\theta=\frac{y}{x}

    These take a point from Cartesian form back to polar form.

  9. Recall the polar area formula

    A=12∫αβr2 dθA=\frac{1}{2}\int_{\alpha}^{\beta}r^{2}\,\mathrm{d}\theta

    The area is swept out by the radius vector between the two half-lines.

  10. Recall the double-angle form of cos⁡2θ\cos^{2}\theta

    cos⁡2θ=12(1+cos⁡2θ)\cos^{2}\theta=\frac{1}{2}\left(1+\cos2\theta\right)

    A squared cosine must be reduced to a multiple angle before it can be integrated.

  11. Recall the double-angle form of sin⁡2θ\sin^{2}\theta

    sin⁡2θ=12(1−cos⁡2θ)\sin^{2}\theta=\frac{1}{2}\left(1-\cos2\theta\right)

    A squared sine must be reduced to a multiple angle before it can be integrated.

  12. Recall the condition for a tangent parallel to the initial line

    dydθ=0wherey=rsin⁡θ\frac{\mathrm{d}y}{\mathrm{d}\theta}=0\quad\text{where}\quad y=r\sin\theta

    A horizontal tangent means yy is stationary as θ\theta varies.

  13. Recall the condition for a tangent perpendicular to the initial line

    dxdθ=0wherex=rcos⁡θ\frac{\mathrm{d}x}{\mathrm{d}\theta}=0\quad\text{where}\quad x=r\cos\theta

    A vertical tangent means xx is stationary as θ\theta varies.

  14. Recall the condition for a tangent at the pole

    r=0 ⇒ θ=α is a tangent at the poler=0\ \Rightarrow\ \theta=\alpha\ \text{is a tangent at the pole}

    The curve reaches the pole along the half-line whose angle makes rr vanish.

  15. Recall the principal range for the polar angle

    −π<θ≤π-\pi<\theta\le\pi

    Every polar angle in this bank is given in the principal range.

  16. Select the correct set of tangents

    θ=π4, θ=3π4, θ=5π4, θ=7π4\theta=\frac{\pi}{4},\ \theta=\frac{3 \pi}{4},\ \theta=\frac{5 \pi}{4},\ \theta=\frac{7 \pi}{4}

    These half-lines are exactly the tangents to CC at the pole.

Answer
θ=π4, θ=3π4, θ=5π4, θ=7π4\theta=\frac{\pi}{4},\ \theta=\frac{3 \pi}{4},\ \theta=\frac{5 \pi}{4},\ \theta=\frac{7 \pi}{4}

Unlock 65 more Polar coordinates questions

Create a free account to work through every Further Maths Polar coordinates question with instant step-by-step worked solutions, progress tracking and interactive lessons.

  • Full worked solutions for every question
  • Interactive lessons and instant feedback
  • Track your mastery across every topic
Create a Free Account

No card required · Free forever

More Polar coordinates practice

Hard QuestionsWorked Solutions

Related Pure Maths topics

Argand diagramsComplex numbers: arithmeticModelling with differential equationsComplex numbers: de MoivreFirst-order differential equationsFurther calculus