Polar Function Grapher

The plot is a polar grid of rings around the pole with spokes every 30°; the ring spacing adjusts to fit the curve and is labeled once (for example "r=1"). The traced curve is coral, and a dashed teal segment runs from the pole to a white dot at the current angle.

The readout under the equation gives the current angle, the value of r and the rectangular point, for example θ = 60° → r = -3 point (-1.50, -2.60). Drag the trace slider slowly: when r is negative, the dot sits on the opposite side of the pole from the direction θ points. Note that the slider is labeled "petals k", but k is only the petal count when k is odd.

A polar function gives the distance r as a function of the angle $\theta$. Every point converts to rectangular coordinates with $$x = r\cos\theta, \qquad y = r\sin\theta, \qquad x^2 + y^2 = r^2.$$ If $r < 0$, the point $(r, \theta)$ lies $|r|$ units from the pole in the direction $\theta + 180^\circ$.

The curve passes through the pole exactly where $r = 0$. Between zeros, AP Precalculus Topic 3.15 asks how r is changing: if r is positive and increasing, the point is moving away from the pole; if r is positive and decreasing, it is moving toward the pole. When r is negative, the distance from the pole is $|r|$, so the reasoning flips.

Graph $r = 3\cos(3\theta)$ and locate the points at $\theta = 20^\circ$ and $\theta = 60^\circ$. This is the simulator's default setting.

Step 1: petals. $k = 3$ is odd, so the rose has 3 petals, each of length 3. One petal is centered on $\theta = 0^\circ$, where $r = 3$.

Step 2: θ = 20°. $r = 3\cos 60^\circ = 1.5$. Then $x = 1.5\cos 20^\circ \approx 1.41$ and $y = 1.5\sin 20^\circ \approx 0.51$. Set the trace slider to 20° and the readout shows r = 1.50 and the point (1.41, 0.51).

Step 3: θ = 60°. $r = 3\cos 180^\circ = -3$. Because r is negative, the point lies 3 units in the direction $240^\circ$: $x = -3\cos 60^\circ = -1.5$, $y = -3\sin 60^\circ \approx -2.60$. That is the tip of the petal in Quadrant III, not a point in Quadrant I.

Step 4: zeros. $r = 0$ when $3\theta = 90^\circ$, so $\theta = 30^\circ$; the curve returns to the pole there between petals.

Step 5: watch it trace. Press Animate trace. The three petals are complete by 180°; from 180° to 360° the dot retraces the same petals. With k = 4 instead, nothing repeats, and all 8 petals need the full 360°.

Follow-up: choose Limaçon with a = 1 and b = 2. Since $b > a$, there is an inner loop. $r = 0$ when $\cos\theta = -\tfrac12$, at 120° and 240°, and at 180° the readout shows r = -1, the far edge of the inner loop.

Polar functions appear at the end of Unit 3 (Topics 3.13 to 3.15). Typical questions give a polar equation or its graph and ask for the point at a given $\theta$, for intervals where the distance from the origin is increasing or decreasing, or for the average rate of change of r over an interval, $\frac{r(\theta_2) - r(\theta_1)}{\theta_2 - \theta_1}$. Practise matching a graph of r against $\theta$ to the polar picture.