From a Polar Equation to a Sketch
A polar equation \(r=f(\theta)\) assigns a signed radius to each chosen angle. To sketch its graph, calculate several values of \(r\), plot the corresponding points, and follow them in the order in which the angle increases. The table is not just a list of distances: the sign of each radius and the order of the angles both affect the curve.
A few carefully chosen values can reveal where the curve reaches farthest from the pole, where it passes through the pole, and how it travels between those points. But a table only samples the graph. A reliable sketch also takes account of continuity and uses enough intermediate angles to show the path without connecting the plotted points as a series of straight segments.
Choosing Angles for a Value Table
Begin by selecting angles across the interval you intend to sketch. For equations involving sine or cosine, multiples of \(\frac{\pi}{2}\) and \(\frac{\pi}{3}\), as well as angles where the radius is zero, are often useful. These angles make trigonometric values manageable and can expose important changes in the graph.
For a complete trace, choose an interval that covers the part of the angle range you intend to examine. The angle interval depends on the equation and what portion of the curve is being sketched; do not assume that every polar equation is fully represented by the same interval. Subdivide the interval when the radius changes substantially, especially near a zero or where its sign changes.
Each row should be evaluated from the equation itself. If \(r\) is negative, do not replace it with its absolute value and keep the same angle. A negative radius places the point in the direction opposite the angle. Equivalently, the same point can be represented by a positive radius at an angle shifted by \(\pi\), since \((r,\theta)\) and \((-r,\theta+\pi)\) represent the same point.
Worked Example: A Table with Positive Radii
Worked Example: A Table with Positive Radii
Build a value table for \(r=1+\frac{1}{2}\sin(2\theta)\) on \(0\leq\theta\leq\pi\), and plot its selected points. Choose angles at intervals of \(\frac{\pi}{4}\).
Evaluate the radius at each angle:
All five radii are positive, so each point lies along the ray specified by its angle. Convert each one using \(x=r\cos\theta\) and \(y=r\sin\theta\).
| \(\theta\) | \(r\) | \((x,y)\) |
|---|---|---|
| \(0\) | \(1\) | \((1,0)\) |
| \(\frac{\pi}{4}\) | \(\frac32\) | \((\frac{3\sqrt2}{4},\frac{3\sqrt2}{4})\) |
| \(\frac{\pi}{2}\) | \(1\) | \((0,1)\) |
| \(\frac{3\pi}{4}\) | \(\frac12\) | \((-\frac{\sqrt2}{4},\frac{\sqrt2}{4})\) |
| \(\pi\) | \(1\) | \((-1,0)\) |
For example, at \(\theta=\frac{\pi}{4}\), the coordinates are \(x=\frac32\cos(\frac{\pi}{4})=\frac{3\sqrt2}{4}\) and \(y=\frac32\sin(\frac{\pi}{4})=\frac{3\sqrt2}{4}\). At \(\theta=\frac{3\pi}{4}\), they are \(x=\frac12(-\frac{\sqrt2}{2})=-\frac{\sqrt2}{4}\) and \(y=\frac12(\frac{\sqrt2}{2})=\frac{\sqrt2}{4}\).
Check: At \(\theta=\frac{\pi}{4}\), the plotted point has distance \(\sqrt{(\frac{3\sqrt2}{4})^2+(\frac{3\sqrt2}{4})^2}=\frac32\), the calculated radius. Its coordinates also have equal positive components, as expected for an angle of \(\frac{\pi}{4}\). At \(\theta=\pi\), \(r=1\) gives \((x,y)=(1\cdot(-1),1\cdot0)=(-1,0)\), matching the table.
Plot the points in increasing angle order and sketch a smooth path guided by the equation. The table gives landmarks; it does not claim that the curve between two successive landmarks is a straight line.
Worked Example: Sketching \(r=1+2\cos(\theta)\)
Worked Example: Sketching \(r=1+2\cos(\theta)\)
We will build a table across \(0\leq\theta\leq2\pi\). Values at multiples of \(\frac{\pi}{6}\) provide useful intermediate points, while the zeros of the radius identify where the curve reaches the pole. Solve for those zeros first:
In this interval, the solutions are \(\theta=\frac{2\pi}{3}\) and \(\theta=\frac{4\pi}{3}\). Now evaluate \(r\) at selected angles and convert the signed radii into points.
| \(\theta\) | \(r=1+2\cos\theta\) | Plotted point \((r\cos\theta,r\sin\theta)\) |
|---|---|---|
| \(0\) | \(3\) | \((3,0)\) |
| \(\frac{\pi}{6}\) | \(1+\sqrt3\) | \((\frac{3+\sqrt3}{2},\frac{1+\sqrt3}{2})\) |
| \(\frac{\pi}{3}\) | \(2\) | \((1,\sqrt3)\) |
| \(\frac{\pi}{2}\) | \(1\) | \((0,1)\) |
| \(\frac{2\pi}{3}\) | \(0\) | \((0,0)\) |
| \(\frac{5\pi}{6}\) | \(1-\sqrt3\) | \((\frac{3-\sqrt3}{2},\frac{1-\sqrt3}{2})\) |
| \(\pi\) | \(-1\) | \((1,0)\) |
| \(\frac{7\pi}{6}\) | \(1-\sqrt3\) | \((\frac{3-\sqrt3}{2},\frac{\sqrt3-1}{2})\) |
| \(\frac{4\pi}{3}\) | \(0\) | \((0,0)\) |
| \(\frac{3\pi}{2}\) | \(1\) | \((0,-1)\) |
| \(\frac{5\pi}{3}\) | \(2\) | \((1,-\sqrt3)\) |
| \(\frac{11\pi}{6}\) | \(1+\sqrt3\) | \((\frac{3+\sqrt3}{2},-\frac{1+\sqrt3}{2})\) |
| \(2\pi\) | \(3\) | \((3,0)\) |
Here is the sign change in detail. At \(\theta=\frac{5\pi}{6}\), \(r=1-\sqrt3<0\). Therefore the point is not on the ray at angle \(\frac{5\pi}{6}\). Its coordinates are
The negative \(y\)-coordinate places this point below the horizontal axis. At \(\theta=\pi\), the radius is also negative: \(r=-1\). Thus \((r,\theta)=(-1,\pi)\) plots at \(((-1)(-1),(-1)(0))=(1,0)\), not at \((-1,0)\).
Check: At \(\theta=\frac{5\pi}{6}\), the distance of the listed point from the pole is \(\sqrt{(\frac{3-\sqrt3}{2})^2+(\frac{1-\sqrt3}{2})^2}=\sqrt{4-2\sqrt3}=\sqrt3-1\). Since \(\sqrt3-1=|1-\sqrt3|\), this agrees with the magnitude of the signed radius. The point is in the direction opposite \(\frac{5\pi}{6}\), as required for a negative radius. At both zero-radius angles, the coordinates are \((0,0)\).
To sketch, plot the table points and follow them as \(\theta\) increases: start at \((3,0)\), pass through the listed upper points, reach the pole at \(\frac{2\pi}{3}\), then continue through the points produced by negative radii. The curve reaches the pole again at \(\frac{4\pi}{3}\) and continues through the remaining points until it returns to \((3,0)\). The angle labels determine the order of travel; the signs determine where the points lie. Use the function between sample angles to guide a smooth connection rather than drawing straight segments between the dots.
Why the Points Must Be Followed in Order
The coordinate formulas explain why the plotting procedure works. At each angle, \(r\cos\theta\) gives the horizontal coordinate and \(r\sin\theta\) gives the vertical coordinate. As \(\theta\) varies, these coordinates vary with it whenever \(r=f(\theta)\) is continuous. Thus the equation describes a continuous path, and the table records selected points along that path. The angle order tells you which point comes next.
For each selected angle, calculate \(r=f(\theta)\), keeping its sign.
Use \(x=r\cos\theta\) and \(y=r\sin\theta\), or plot the signed radius along its angle's ray.
Plot them in increasing \(\theta\), and mark any zero-radius values at the pole.
Use intermediate samples and the behavior of the equation to draw a smooth curve through the points.
A zero of \(r\) places the plotted point at the pole regardless of the angle. If \(f\) is continuous and its values change sign between two angles, the Intermediate Value Theorem guarantees that \(f\) is zero at some angle between them. Consequently, the curve reaches the pole somewhere in that interval. A sign change alone does not tell you the exact angle; solve \(f(\theta)=0\) when possible to locate it precisely.
Worked Example: Tracking a Sign Change
Worked Example: Tracking a Sign Change
For \(r=1-2\sin\theta\), make a table at selected angles from \(0\) to \(2\pi\), including the zeros, and account for the negative radii. Solving \(1-2\sin\theta=0\) gives \(\sin\theta=\frac12\), so the zeros in this interval are \(\frac{\pi}{6}\) and \(\frac{5\pi}{6}\).
| \(\theta\) | \(r=1-2\sin\theta\) | Plotted point \((r\cos\theta,r\sin\theta)\) |
|---|---|---|
| \(0\) | \(1\) | \((1,0)\) |
| \(\frac{\pi}{6}\) | \(0\) | \((0,0)\) |
| \(\frac{\pi}{3}\) | \(1-\sqrt3\) | \((\frac{1-\sqrt3}{2},\frac{\sqrt3-3}{2})\) |
| \(\frac{\pi}{2}\) | \(-1\) | \((0,-1)\) |
| \(\frac{2\pi}{3}\) | \(1-\sqrt3\) | \((\frac{\sqrt3-1}{2},\frac{\sqrt3-3}{2})\) |
| \(\frac{5\pi}{6}\) | \(0\) | \((0,0)\) |
| \(\pi\) | \(1\) | \((-1,0)\) |
| \(\frac{7\pi}{6}\) | \(2\) | \((-\sqrt3,-1)\) |
| \(\frac{4\pi}{3}\) | \(1+\sqrt3\) | \((-\frac{1+\sqrt3}{2},-\frac{3+\sqrt3}{2})\) |
| \(\frac{3\pi}{2}\) | \(3\) | \((0,-3)\) |
| \(2\pi\) | \(1\) | \((1,0)\) |
For example, at \(\theta=\frac{\pi}{3}\), the radius is \(1-\sqrt3<0\), and the coordinates are \(x=(1-\sqrt3)\frac12=\frac{1-\sqrt3}{2}\) and \(y=(1-\sqrt3)\frac{\sqrt3}{2}=\frac{\sqrt3-3}{2}\). At \(\theta=\frac{7\pi}{6}\), \(r=2\), so \(x=2(-\frac{\sqrt3}{2})=-\sqrt3\) and \(y=2(-\frac12)=-1\).
Check: For the point at \(\theta=\frac{\pi}{3}\), the distance from the pole is \(\sqrt{(\frac{1-\sqrt3}{2})^2+(\frac{\sqrt3-3}{2})^2}=\sqrt{4-2\sqrt3}=\sqrt3-1=|1-\sqrt3|\). The coordinates therefore have the correct distance for the signed radius, and their signs place the point opposite the angle's ray. At \(\frac{7\pi}{6}\), \(r=2\) produces distance \(\sqrt{(-\sqrt3)^2+(-1)^2}=2\), as expected.
A Common Plotting Mistake
A frequent error is to treat a negative radius as a positive distance on the same ray. For \(r=1+2\cos\theta\) at \(\theta=\pi\), the incorrect calculation is \(r=-1\), followed by “plot distance \(1\) along angle \(\pi\),” which gives \((-1,0)\). That point is wrong: the radius is signed, not an instruction to take its absolute value and retain the angle.
The correction is to use the coordinate formulas: \(x=(-1)\cos\pi=1\) and \(y=(-1)\sin\pi=0\), giving \((1,0)\). Equivalently, the same point is represented by positive radius \(1\) and angle \(0\). Always preserve the sign until the point has been placed.
Making a Sketch More Reliable
A useful table includes more than the angles where sine or cosine is easiest to evaluate. Include zeros of the radius, angles on both sides of each zero, and additional intermediate angles when the radius changes quickly. If neighboring sampled radii have opposite signs, investigate the interval: for a continuous radius function, the graph passes through the pole somewhere within it.
Finally, remember what the table can and cannot establish. It provides exact landmarks and, with continuity, a basis for tracing the curve between them. It does not determine the precise shape between widely separated samples by itself. If the sketch looks uncertain, add rows rather than guessing how the curve travels. The essential routine is to evaluate, preserve the sign, plot in angular order, and refine the samples where the path needs clarification.
Check Your Understanding
Answer these questions using the value-table and plotting method.
- For a polar value with \(r<0\), what direction should you use to plot the point?
- Why does a zero value of \(r\) place the point at the pole?
- For \(r=1+2\cos\theta\), find the radius at \(\theta=\pi\) and give the plotted point.
- What does continuity guarantee if \(r=f(\theta)\) changes sign between two angles?
- Why should the points in a polar table be connected in increasing angle order?
