Lemniscates in Polar Form

Introduction

A lemniscate is a curve with two loops meeting at the pole. Two standard polar equations that produce lemniscates are \(r^2=a^2\cos(2\theta)\) and \(r^2=a^2\sin(2\theta)\). Their shapes are similar, but their loops point in different directions.

To sketch either curve, first find the angles for which the right-hand side is nonnegative. Those are the angles where a real radius exists. Then track the nonnegative radius as \(\theta\) increases: the radius starts at zero, grows to a maximum, and returns to zero, tracing one loop.

Finding the Real-Valued Intervals

Solving either equation for a nonnegative radius requires taking a square root. For the cosine equation, \(r=a\sqrt{\cos(2\theta)}\) is real exactly when \(\cos(2\theta)\geq 0\). Cosine is nonnegative on intervals centred at integer multiples of \(2\pi\). Accounting for the doubled angle gives

For the sine equation, \(r=a\sqrt{\sin(2\theta)}\) is real exactly when \(\sin(2\theta)\geq 0\). Sine is nonnegative from \(2k\pi\) to \((2k+1)\pi\), so

These intervals repeat as the angle increases. Each interval has endpoints where \(r=0\), so the corresponding trace begins and ends at the pole. A full interval between consecutive zeros traces one loop using nonnegative \(r\).

Sketching the Cosine Lemniscate

Consider \(r^2=a^2\cos(2\theta)\). One interval is \([-\pi/4,\pi/4]\). At its endpoints, the radius is zero. At its midpoint, \(\theta=0\), the cosine equals one, giving \(r=a\). The loop therefore lies on the right side of the pole: as the angle increases from \(-\pi/4\) to \(\pi/4\), the trace leaves the pole, reaches its greatest distance at \(\theta=0\), and returns to the pole.

The next interval, \([3\pi/4,5\pi/4]\), produces the left loop. Its midpoint is \(\theta=\pi\), where the radius is again \(a\). With increasing angle, this trace also leaves the pole, reaches its farthest point, and returns. Thus the cosine lemniscate has loops directed along the horizontal axis.

Sketching the Sine Lemniscate

For \(r^2=a^2\sin(2\theta)\), the interval \([0,\pi/2]\) traces a loop in the first quadrant. The radius is zero at both endpoints and greatest at the midpoint, \(\theta=\pi/4\), where \(\sin(2\theta)=1\). The next interval, \([\pi,3\pi/2]\), traces the other loop in the third quadrant, with its maximum at \(\theta=5\pi/4\). The loops therefore point along the diagonal line through those directions.

On the first interval, increasing \(\theta\) takes the trace from the pole into the first quadrant, out to the farthest point, and back to the pole. On the second interval it does the same in the third quadrant. For comparison, \(\sin(2\theta)=\cos(2(\theta-\pi/4))\): this is the cosine lemniscate with its directions turned by \(\pi/4\).

Why the Maximum Radius Is \(a\)

Both equations have the form \(r^2=a^2 T(\theta)\), where \(T(\theta)\) is either \(\cos(2\theta)\) or \(\sin(2\theta)\). On the real-valued intervals, \(T(\theta)\geq 0\), and for every angle \(T(\theta)\leq 1\). Therefore \(r^2\leq a^2\). Since \(r\geq 0\) and \(a>0\), this implies \(r\leq a\).

The bound is attained: cosine equals \(1\) when \(2\theta=2k\pi\), and sine equals \(1\) when \(2\theta=\pi/2+2k\pi\). Thus \(r=a\) is not just an upper bound; it is the maximum radius. This also shows why changing \(a\) scales the whole curve: the loop directions stay the same, while the greatest distance from the pole changes to \(a\).

Coefficient and Angle Checks

When a coefficient is given, distinguish \(a^2\) from \(a\). For example, in \(r^2=12\cos(2\theta)\), the scale parameter is \(a=\sqrt{12}=2\sqrt{3}\), not \(12\). The real-valued intervals are unchanged by the positive coefficient: they are still \([-\pi/4+k\pi,\pi/4+k\pi]\). The maximum radius is \(2\sqrt{3}\).

A Common Mistake

A frequent error is to forget that the angle in the trigonometric function is \(2\theta\). For example, one might incorrectly claim that \(r^2=a^2\cos(2\theta)\) is real for \(-\pi/2\leq\theta\leq\pi/2\), reasoning from where \(\cos(\theta)\) is nonnegative. That substitutes the sign pattern of \(\cos(\theta)\) for the sign pattern of \(\cos(2\theta)\).

Check \(\theta=\pi/3\): it lies in the claimed interval, but \(\cos(2\theta)=\cos(2\pi/3)=-1/2\), so \(r^2=-a^2/2<0\). No real radius exists there. The correction is to set \(u=2\theta\), find where \(\cos(u)\geq 0\), and then divide the resulting angle bounds by \(2\). For the first interval, \(-\pi/2\leq u\leq\pi/2\) becomes \(-\pi/4\leq\theta\leq\pi/4\).