Hard Further Maths Polar coordinates Questions

Challenging, exam-style Further Maths Polar coordinates questions with worked solutions. Stretch yourself on the hardest polar-coordinates, polar-equations, conversion, polar-to-cartesian problems.

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Further Maths34 questionsStep-by-step solutions
Question 1
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}
Question 2
9 markschallenging
Which of the following is the value of θ\theta at which the tangent to the curve with polar equation r=2(1−cos⁡θ)r=2\left(1-\cos\theta\right), 0<θ<π0<\theta<\pi, is parallel to the initial line?
Show worked solution

Worked solution

  1. State the condition for a tangent parallel to the initial line

    dydθ=0,y=rsin⁡θ\frac{\mathrm{d}y}{\mathrm{d}\theta}=0,\qquad y=r\sin\theta

    A tangent parallel to the initial line is a stationary value of yy.

  2. Write yy in terms of θ\theta

    y=2(1−cos⁡(θ))sin⁡(θ)y=2 \left(1 - \cos{\left(\theta \right)}\right) \sin{\left(\theta \right)}

    The polar equation has been substituted for rr.

  3. Differentiate and factorise

    dydθ=−2(cos⁡(θ)−1)(2cos⁡(θ)+1)\frac{\mathrm{d}y}{\mathrm{d}\theta}=- 2 \left(\cos{\left(\theta \right)} - 1\right) \left(2 \cos{\left(\theta \right)} + 1\right)

    Factorising exposes the roots of the derivative.

  4. Solve in the given interval

    θ=2π3\theta=\frac{2 \pi}{3}

    This is the only root with 0<θ<π0<\theta<\pi.

  5. Check the derivative at the chosen angle

    dydθ∣θ=2π3=0\left.\frac{\mathrm{d}y}{\mathrm{d}\theta}\right|_{\theta=\frac{2 \pi}{3}}=0

    The derivative really is zero there.

  6. Reject the angles that fail the test

    dydθ∣θ=π2=2\left.\frac{\mathrm{d}y}{\mathrm{d}\theta}\right|_{\theta=\frac{\pi}{2}}=2

    A non-zero derivative means the tangent is not parallel to the initial line.

  7. Note the corresponding value of rr

    r=3r=3

    The point of contact is (3, 2π3)\left(3,\ \frac{2 \pi}{3}\right).

  8. 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.

  9. 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.

  10. 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.

  11. 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.

  12. 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.

  13. 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.

  14. 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.

  15. Select the correct value of θ\theta

    θ=2π3\theta=\frac{2 \pi}{3}

    At this angle the tangent to CC is parallel to the initial line.

Answer
2π3\frac{2 \pi}{3}
Question 3
9 markschallenging
Which of the following is the exact area of the region bounded by the curve with polar equation r=1+2cos⁡θr=1+2\cos\theta and the half-lines θ=2π3\theta=\frac{2 \pi}{3} and θ=4π3\theta=\frac{4 \pi}{3}?
Show worked solution

Worked solution

  1. Quote the polar area formula

    A=12∫2π34π3r2 dθA=\frac{1}{2}\int_{\frac{2 \pi}{3}}^{\frac{4 \pi}{3}}r^{2}\,\mathrm{d}\theta

    The factor 12\frac{1}{2} and the square are what distinguish it from a Cartesian integral.

  2. Substitute the polar equation and reduce the integrand

    r2=4cos⁡(θ)+2cos⁡(2θ)+3r^{2}=4 \cos{\left(\theta \right)} + 2 \cos{\left(2 \theta \right)} + 3

    Squared trigonometric terms must be written as multiple angles first.

  3. Integrate and apply the limits

    [3θ2+2sin⁡(θ)+sin⁡(2θ)2]2π34π3=π−332\left[\frac{3 \theta}{2} + 2 \sin{\left(\theta \right)} + \frac{\sin{\left(2 \theta \right)}}{2}\right]_{\frac{2 \pi}{3}}^{\frac{4 \pi}{3}}=\pi - \frac{3 \sqrt{3}}{2}

    This is the exact area of the region.

  4. Reject the option that omits the factor 12\frac{1}{2}

    ∫2π34π3r2 dθ=−33+2π\int_{\frac{2 \pi}{3}}^{\frac{4 \pi}{3}}r^{2}\,\mathrm{d}\theta=- 3 \sqrt{3} + 2 \pi

    Forgetting the 12\frac{1}{2} doubles the answer.

  5. Reject the option that forgets to square rr

    12∫2π34π3r dθ=−3+π3\frac{1}{2}\int_{\frac{2 \pi}{3}}^{\frac{4 \pi}{3}}r\,\mathrm{d}\theta=- \sqrt{3} + \frac{\pi}{3}

    The integrand is r2r^{2}, not rr.

  6. Check the answer numerically

    A≈0.543516A\approx 0.543516

    The decimal value identifies the correct option beyond doubt.

  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. Recall the product rule

    ddθ(uv)=udvdθ+vdudθ\frac{\mathrm{d}}{\mathrm{d}\theta}\left(uv\right)=u\frac{\mathrm{d}v}{\mathrm{d}\theta}+v\frac{\mathrm{d}u}{\mathrm{d}\theta}

    Both rsin⁡θr\sin\theta and rcos⁡θr\cos\theta are products, so the product rule is needed.

  17. Select the correct area

    A=π−332A=\pi - \frac{3 \sqrt{3}}{2}

    This is the exact area of the region.

Answer
π−332\pi - \frac{3 \sqrt{3}}{2}
Question 4
9 markschallenging
Which of the following best describes the curve with polar equation r=3(1−cos⁡θ)r=3\left(1-\cos\theta\right)?
Show worked solution

Worked solution

  1. Compare the polar equation with the standard form

    r=a+bcos⁡θ with a=3, b=−3r=a+b\cos\theta\ \text{with}\ a=3,\ 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⁡=6r_{\max}=6

    This happens when the cosine term takes its extreme value.

  4. Decide whether the curve reaches the pole

    r=0 has a solutionr=0\ \text{has a 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)=0r\left(0\right)=0

    This is where the curve crosses the initial line.

  7. Reject the options with the wrong greatest value of rr

    rmax⁡=6r_{\max}=6

    Only one option quotes the correct maximum.

  8. 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.

  9. 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.

  10. 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.

  11. 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.

  12. 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.

  13. 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.

  14. 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.

  15. 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.

  16. Select the correct description

    r=3(1−cos⁡θ)r=3\left(1-\cos\theta\right)

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

Answer
A cardioid, symmetrical about the initial line, with greatest value r=6r=6, passing through the pole
Question 5
9 markschallenging
The curve CC has Cartesian equation (x2−x+y2)2=x2+y2\left(x^{2} - x + y^{2}\right)^{2}=x^{2} + y^{2}. Find a polar equation of CC in the form r=f(θ)r=f\left(\theta\right).
Show worked solution

Worked solution

  1. Substitute x=rcos⁡θx=r\cos\theta and y=rsin⁡θy=r\sin\theta

    (x2−x+y2)2=x2+y2\left(x^{2} - x + y^{2}\right)^{2}=x^{2} + y^{2}

    This turns the Cartesian equation into an equation in rr and θ\theta.

  2. Use x2+y2=r2x^{2}+y^{2}=r^{2} wherever it appears

    x2+y2=r2x^{2}+y^{2}=r^{2}

    Recognising this combination is what keeps the algebra short.

  3. Substitute x=rcos⁡θx=r\cos\theta, y=rsin⁡θy=r\sin\theta

    (r2−rcos⁡θ)2=r2\left(r^{2}-r\cos\theta\right)^{2}=r^{2}

    Every Cartesian coordinate becomes a polar expression.

  4. Take the square root of both sides

    r2−rcos⁡θ=rr^{2}-r\cos\theta=r

    The positive root is the one with r≥0r\ge0.

  5. Divide by rr

    r−cos⁡θ=1r-\cos\theta=1

    The factor r=0r=0 is only the pole, which lies on the curve anyway.

  6. Divide through by rr and discard the pole r=0r=0

    r=1+cos⁡θr=1+\cos\theta

    The factor r=0r=0 is only the pole, which already lies on the curve.

  7. Check the result at θ=0\theta=0

    r=2r=2

    This value satisfies the original Cartesian equation.

  8. Note that the polar form is not unique

    r=1+cos⁡θr=1+\cos\theta

    Any equation with the same solution set is acceptable.

  9. 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.

  10. 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.

  11. 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.

  12. 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.

  13. 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.

  14. 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.

  15. State the polar equation

    r=1+cos⁡θr=1+\cos\theta

    This is a polar equation of CC in the form r=f(θ)r=f\left(\theta\right).

Answer
r=1+cos⁡θr=1+\cos\theta

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