1.5 - Sketching polar curves

Symmetry of a polar curve

A polar curve is symmetric about a line when reflection in that line preserves its set of points. For an equation $r=f(\theta)$, compare the radii at reflected angles and check that the angle domain contains the corresponding directions, allowing complete turns.

Worked example 1: a full angle interval

Question. State the symmetry of $r=3+\cos\theta$, $0\le\theta<2\pi$, and locate the points on the horizontal axis.

Plan. Compare reflected directions, then use angles $0$ and $\pi$.

Since $\cos(2\pi-\theta)=\cos\theta$, reflection in the horizontal axis preserves the radius and domain. But $\cos(\pi-\theta)=-\cos\theta$, so the vertical reflection generally changes it.

Answer. Symmetric about the horizontal axis, with horizontal-axis points $(r,\theta)=(4,0),(2,\pi)$. It is not symmetric about the vertical axis or under a half-turn.

Check. The corresponding Cartesian points are $(4,0)$ and $(-2,0)$. Their unequal distances from the pole rule out the latter two symmetries.

Worked example 2: a restricted angle interval

Question. Describe which part of $r=3+\cos\theta$ is traced for $0\le\theta\le\pi$. Is this restricted curve symmetric about the horizontal axis?

Plan. Use the signs of $r$ and $\sin\theta$, then test a reflected interior point.

Here $2\le r\le4$, and $\sin\theta\ge0$, so $y=r\sin\theta\ge0$. As the ray turns, the curve goes from $(4,0)$, through $(0,3)$, to $(-2,0)$, in Cartesian coordinates.

Answer. Only the upper half of the full curve is traced. It is not symmetric about the horizontal axis.

Check. The point $(0,3)$ occurs at $\theta=\pi/2$, but its reflection $(0,-3)$ would require $\theta=3\pi/2$, outside the interval.

Problem-solving

Find where $r=0$, and find the least and greatest permitted values of $r$. If the equation gives $r^2$, take its nonnegative square root before stating a radius.

Worked example 3: a cusp at the pole

Question. For $r=3(1+\cos\theta)$, $-\pi<\theta\le\pi$, find the extreme radii and describe how the curve meets the pole.

Plan. Use the cosine bounds, then follow the curve on either side of its zero radius, identifying $-\pi$ and $\pi$ as the same ray.

Near either end of the interval, $r$ is small and positive, $x=r\cos\theta<0$, and $y=r\sin\theta$ has opposite signs on the two branches. Also $y/x=\tan\theta\to0$.

Answer. The furthest point is $(6,0)$ in polar coordinates. The curve has a cusp at the pole, with both branches on its left and tangent line $y=0$.

Check. Writing $\theta=\pi-\delta$ gives $r=3(1-\cos\delta)>0$ for small positive $\delta$. Thus the upper branch approaches from the left; its reflection supplies the lower branch on the same side, confirming a cusp rather than a crossing.

Worked example 4: an equation for the squared radius

Question. For $r^2=3(1+\cos\theta)$, $-\pi<\theta\le\pi$, find the extreme radii and the tangent at the pole.

Plan. Bound $r^2$, take the nonnegative square root, then use the limiting direction.

The maximum is at $\theta=0$, and the minimum zero at $\theta=\pi$. Near the pole, $x<0$ on both branches and $y/x=\tan\theta\to0$.

Answer. The furthest point is $(\sqrt6,0)$ in polar coordinates, and the pole tangent is $y=0$. The branches again approach on the left with a common tangent.

Check. At $\theta=\pi/2$, the radius is $\sqrt3<\sqrt6$. Substituting the claimed maximum $\sqrt6$ into the equation gives $6=3(1+1)$; radius $6$ would fail it.

Watch out

If $r^2\le k$, the distance bound is $r\le\sqrt{k}$, not $r\le k$. Give the furthest point as a polar pair, with its angle, when asked for polar coordinates.

Worked example 5: two horizontal lobes

Question. Sketch $r^2=16\cos2\theta$, $0\le\theta<2\pi$, stating its allowed angle intervals and significant features.

Plan. Require the squared radius to be nonnegative, then locate its zeros, maxima and symmetries.

On these intervals, $r=4\sqrt{\cos2\theta}$. The largest radius is $4$, at $\theta=0,\pi$, and the radius is zero at $\pi/4,3\pi/4,5\pi/4,7\pi/4$. Reflection in either axis preserves the equation and allowed directions.

Answer. Two equal lobes lie to the right and left of the pole, symmetric about both axes. Their tips are Cartesian points $(4,0)$, $(-4,0)$; the positive tip is the nonzero initial-line intersection. The lobes meet at the pole with tangent lines $y=x$ and $y=-x$.

Check. At $\theta=\pi/2$, the right side is $-16$, so no vertical-axis point is allowed except the pole represented at other angles. At $\theta=\pi/8$, $r=4\sqrt{\sqrt2/2}$, placing the upper right branch strictly between its tip and its pole endpoint.

Worked example 6: rotated lobes

Question. Sketch $r^2=16\sin2\theta$, $0\le\theta<2\pi$, stating its significant features.

Plan. Solve $\sin2\theta\ge0$, and use its zero and maximum values.

The allowed intervals are $[0,\pi/2]\cup[\pi,3\pi/2]$. The radius is zero at $0,\pi/2,\pi,3\pi/2$. Its maximum $4$ occurs at $\theta=\pi/4,5\pi/4$.

The identity $\cos(2(\theta-\pi/4))=\sin2\theta$ shows that this is the preceding sketch rotated anticlockwise through $\pi/4$.

Answer. Two equal lobes lie in the first and third quadrants, symmetric about $y=x$ and $y=-x$, with a half-turn symmetry. Their polar tips are $(4,\pi/4)$, $(4,5\pi/4)$; their pole tangents are the two coordinate axes. The initial line contains only the pole.

Check. Converting the first tip gives $(2\sqrt2,2\sqrt2)$, which lies on the stated diagonal. At $\theta=3\pi/4$, the squared-radius expression is negative, confirming that no lobe belongs in quadrant two.

Watch out

Do not plot a negative radius on the opposite ray. Under our convention, exclude that angle. For $r^2=f(\theta)$, retain only $f(\theta)\ge0$ and use the nonnegative square root.