Slope of a polar curve

When you reach for it

Tsering has a nautilus shell sawn in half. The cut shows a spiral wall winding out from a center point, and its outline is close to r = e^{\theta/3}, with r in centimeters measured from the center (the pole) and \theta in radians. She wants the direction of the wall at a given point. The curve is given in polar form (Polar coordinates); it winds around the pole and is not the graph of any function y = F(x).

But every polar curve is a parametric curve with the angle as its parameter, and Derivatives of parametric curves already gives the slope of a parametric curve: \displaystyle \frac{dy}{dx} = \frac{dy/dt}{dx/dt}, horizontal where only the top is 0, vertical where only the bottom is 0, and a limit to take where both are 0. This lesson applies that to r = f(\theta). Reach for it whenever a polar curve needs a slope, a tangent line, its level and upright points, its tangent lines at the pole, or the angle it makes with the radius.

The procedure

A polar curve is a parametric curve. The point with polar coordinates (r, \theta) has rectangular coordinates x = r\cos\theta, y = r\sin\theta (Converting between polar and rectangular). On the curve r = f(\theta), substitute:

The angle \theta plays the part of the parameter t. A negative value of f needs no special care: that point is plotted on the opposite ray, and the formulas put it there, since both coordinates change sign.

The two rates. The product rule on each coordinate gives

The slope. Where f is differentiable and \displaystyle \frac{dx}{d\theta} \ne 0, the parametric result gives

The tangent line at \theta_0 passes through \big(f(\theta_0)\cos\theta_0,\ f(\theta_0)\sin\theta_0\big) with this slope, built as in Tangent lines and normal lines. The tangent is horizontal where \displaystyle \frac{dy}{d\theta} = 0 and \displaystyle \frac{dx}{d\theta} \ne 0, vertical where \displaystyle \frac{dx}{d\theta} = 0 and \displaystyle \frac{dy}{d\theta} \ne 0, and undecided by the ratio where both are 0.

What the two rates mean. Let \mathbf{u} = (\cos\theta, \sin\theta), the unit vector pointing out along the ray at angle \theta, and \mathbf{v} = (-\sin\theta, \cos\theta), the same vector turned a quarter turn counterclockwise. Reading the two rates off term by term,

The tangent direction has a part f'(\theta) along the radius and a part f(\theta) across it. Four facts follow.

• \displaystyle \frac{dr}{d\theta} = f'(\theta) is only the radial part. It says how fast the point moves away from the pole, not how steep the curve is. The slope \displaystyle \frac{dy}{dx} measures the whole direction against the x axis, so it uses both parts and the angle \theta as well. On the circle r = c, \displaystyle \frac{dr}{d\theta} = 0 everywhere, yet the slope -\cot\theta takes every real value.
• Where f(\theta_0) \ne 0 and f'(\theta_0) = 0, the tangent is perpendicular to the radius: for that instant the curve neither approaches nor leaves the pole.
• At the pole: if f(\theta_0) = 0 and f'(\theta_0) \ne 0, the tangent direction is f'(\theta_0)\,\mathbf{u}, along the ray itself. When \cos\theta_0 \ne 0, the slope is \displaystyle \frac{dy}{dx} = \frac{f'(\theta_0)\sin\theta_0}{f'(\theta_0)\cos\theta_0} = \tan\theta_0, and when \cos\theta_0 = 0 the tangent is vertical. Either way the tangent line at the pole is the line \theta = \theta_0. A curve that reaches the pole at several angles has a tangent line for each.
• The angle with the radius: where f(\theta) > 0, let \psi be the angle from the outward radius \mathbf{u} to the tangent direction of increasing \theta. Its cosine and sine are the two parts divided by \displaystyle \sqrt{f'^2 + f^2}, so where f'(\theta) \ne 0, \displaystyle \tan\psi = \frac{f(\theta)}{f'(\theta)} = \frac{r}{dr/d\theta}, and \displaystyle \psi = \frac{\pi}{2} where f'(\theta) = 0.

Worked examples

Boundaries

• The formula needs f differentiable at \theta_0 and \displaystyle \frac{dx}{d\theta} \ne 0 there. Where \displaystyle \frac{dx}{d\theta} = 0 and \displaystyle \frac{dy}{d\theta} \ne 0 the tangent is vertical and has no slope as a number.
• Where both rates are 0, the ratio is \displaystyle \frac{0}{0} and decides nothing; take the limit of \displaystyle \frac{dy}{dx} as \theta \to \theta_0, which may or may not exist. At the pole both rates are 0 exactly when f(\theta_0) = f'(\theta_0) = 0, as at the cusp of a cardioid.
• The rule “the tangent at the pole is the line \theta = \theta_0” assumes f'(\theta_0) \ne 0.
• A slope belongs to a value of \theta, not to a point. The pole, and any point the curve passes more than once, can carry several tangent lines.
• \displaystyle \tan\psi = \frac{r}{dr/d\theta} as derived assumes r > 0; at the pole \psi is not defined.
• \displaystyle \frac{dr}{d\theta} is not a slope. It is 0 all around a circle centered at the pole, whose tangents point in every direction.

Where this goes next

The decomposition f'(\theta)\,\mathbf{u} + f(\theta)\,\mathbf{v} also gives the speed at which a polar curve is traced: \mathbf{u} and \mathbf{v} are perpendicular unit vectors, so the length of that vector is \displaystyle \sqrt{f'(\theta)^2 + f(\theta)^2}, the integrand of polar arc length in Arc length and speed on parametric and polar curves. Area inside a polar curve, which uses the angles where it meets the pole found here, is in Area in polar coordinates. Circles through the pole, such as those in Polar circles and symmetry tests, meet the pole at a single angle each, and their tangent there is the line through the pole at that angle. In Vector valued functions and motion in the plane the pair \displaystyle \left(\frac{dx}{d\theta}, \frac{dy}{d\theta}\right) becomes a velocity vector.

Exercises

Slopes and tangent lines

1. Find the tangent line to r = 3 + 2\cos\theta at \displaystyle \theta = \frac{\pi}{2}.
2. Find the tangent line to the spiral \displaystyle r = \frac{4}{\theta} at \displaystyle \theta = \frac{\pi}{2}.
3. Find the tangent line to r = \theta^2 at \theta = \pi.
4. Find the tangent line to Dorje’s rose r = 7\sin 2\theta at \displaystyle \theta = \frac{\pi}{3}.
5. Find the tangent line to r = 1 + 4\cos\theta at \displaystyle \theta = \frac{2\pi}{3}, where r is negative.

Level and upright points

6. Find every point of r = 4 + 2\sin\theta, 0 \le \theta < 2\pi, with a horizontal tangent and every point with a vertical tangent.
7. Find the highest point of the petal of r = 7\sin 2\theta traced for \displaystyle 0 \le \theta \le \frac{\pi}{2}.
8. Find the point of the same petal farthest to the right.
9. Show that r = 2\csc\theta, 0 < \theta < \pi, has a horizontal tangent at every point, and name the curve.

Tangents at the pole

10. Find the tangent lines at the pole of \displaystyle r = \sqrt{3} + 2\sin\theta.
11. Find the tangent lines at the pole of r = 3\sin 4\theta, 0 \le \theta < \pi.
12. The spiral r = \theta - 1 passes through the pole at \theta = 1. Find its tangent line there.
13. For r = \theta^2, both f(0) and f'(0) are 0. Find the limit of \displaystyle \frac{dy}{dx} as \theta \to 0 and the tangent line at the pole.

The rate of r and the angle with the radius

14. On the circle r = 4, find \displaystyle \frac{dr}{d\theta} and \displaystyle \frac{dy}{dx} at \displaystyle \theta = \frac{\pi}{3}.
15. For \displaystyle r = e^{\sqrt{3}\,\theta}, find the angle \psi between the radius and the tangent, and the slope at \displaystyle \theta = \frac{\pi}{2}.
16. For the spiral r = 2\theta, \theta > 0, find \psi at \theta = 1 and at \theta = 10. What happens to \psi as \theta grows?

Reasoning

17. Prove that if f(\theta_0) \ne 0 and f'(\theta_0) = 0, the tangent line at \theta_0 is perpendicular to the radius.
18. Refute the claim: “If \displaystyle \frac{dr}{d\theta} = 0 at a point, the tangent line there is horizontal.”
19. Suppose f > 0 and the angle \psi between radius and tangent is the same at every point, with \displaystyle \tan\psi = \frac{1}{k}. Show that f(\theta) = Ce^{k\theta} for a constant C > 0.
20. The curve in Exercise 10 has two tangent lines at the pole, yet its slope is a single formula in \theta. Explain why there is no contradiction.