Polar coordinates Worked Solutions — Further Maths Maths

Fully worked, step-by-step solutions to Further Maths Polar coordinates questions. See exactly how to solve problems on polar-coordinates, conversion, polar-to-cartesian, cartesian-to-polar.

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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.

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
The point PP has polar coordinates (6, −π3)\left(6,\ - \frac{\pi}{3}\right). Find the exact Cartesian coordinates of PP.

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=6r=6 and θ=−π3\theta=- \frac{\pi}{3}

    x=6cos⁡(−π3),y=6sin⁡(−π3)x=6\cos\left(- \frac{\pi}{3}\right),\qquad y=6\sin\left(- \frac{\pi}{3}\right)

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

  3. Use the exact trigonometric values

    cos⁡(−π3)=12,sin⁡(−π3)=−32\cos\left(- \frac{\pi}{3}\right)=\frac{1}{2},\qquad\sin\left(- \frac{\pi}{3}\right)=- \frac{\sqrt{3}}{2}

    The angle is a special angle, so exact surd values are available.

  4. State the Cartesian coordinates

    P(3, −33)P\left(3,\ - 3 \sqrt{3}\right)

    These are the exact Cartesian coordinates of PP.

Answer
(3, −33)\left(3,\ - 3 \sqrt{3}\right)
Question 3
2 markseasy
The point PP has polar coordinates (2, 3π4)\left(2,\ \frac{3 \pi}{4}\right). Find the exact Cartesian coordinates of PP.

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=2r=2 and θ=3π4\theta=\frac{3 \pi}{4}

    x=2cos⁡(3π4),y=2sin⁡(3π4)x=2\cos\left(\frac{3 \pi}{4}\right),\qquad y=2\sin\left(\frac{3 \pi}{4}\right)

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

  3. State the Cartesian coordinates

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

    These are the exact Cartesian coordinates of PP.

Answer
(−2, 2)\left(- \sqrt{2},\ \sqrt{2}\right)
Question 4
2 markseasy
The point PP has Cartesian coordinates (−1, 3)\left(-1,\ \sqrt{3}\right). Find the polar coordinates of PP, where r≥0r\ge0 and −π<θ≤π-\pi<\theta\le\pi.

Worked solution

  1. Find rr from r2=x2+y2r^{2}=x^{2}+y^{2}

    r=(−1)2+(3)2=2r=\sqrt{\left(-1\right)^{2}+\left(\sqrt{3}\right)^{2}}=2

    The polar radius is the distance of the point from the pole.

  2. Find the acute angle the radius makes with the xx-axis

    tan⁡α=∣3−1∣ ⇒ α=π3\tan\alpha=\left|\frac{\sqrt{3}}{-1}\right|\ \Rightarrow\ \alpha=\frac{\pi}{3}

    Working with the acute angle first avoids sign errors.

  3. Place the angle in the correct quadrant

    (−1, 3) ⇒ θ=2π3\left(-1,\ \sqrt{3}\right)\ \Rightarrow\ \theta=\frac{2 \pi}{3}

    The signs of xx and yy decide which quadrant the point is in.

  4. State the polar coordinates

    P(2, 2π3)P\left(2,\ \frac{2 \pi}{3}\right)

    This is the point in polar form with r≥0r\ge0 and −π<θ≤π-\pi<\theta\le\pi.

Answer
(2, 2π3)\left(2,\ \frac{2 \pi}{3}\right)
Question 5
2 markseasy
The point PP has Cartesian coordinates (3, −3)\left(3,\ -3\right). Find the polar coordinates of PP, where r≥0r\ge0 and −π<θ≤π-\pi<\theta\le\pi.

Worked solution

  1. Find rr from r2=x2+y2r^{2}=x^{2}+y^{2}

    r=(3)2+(−3)2=32r=\sqrt{\left(3\right)^{2}+\left(-3\right)^{2}}=3 \sqrt{2}

    The polar radius is the distance of the point from the pole.

  2. Find the acute angle the radius makes with the xx-axis

    tan⁡α=∣−33∣ ⇒ α=0.785398163397448\tan\alpha=\left|\frac{-3}{3}\right|\ \Rightarrow\ \alpha=0.785398163397448

    Working with the acute angle first avoids sign errors.

  3. State the polar coordinates

    P(32, −π4)P\left(3 \sqrt{2},\ - \frac{\pi}{4}\right)

    This is the point in polar form with r≥0r\ge0 and −π<θ≤π-\pi<\theta\le\pi.

Answer
(32, −π4)\left(3 \sqrt{2},\ - \frac{\pi}{4}\right)

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