LINES AND TANGENTS IN POLAR COORDINATES

ROGER ALEXANDER DEPARTMENT OF

1. Polar-coordinate equations for lines A polar in the is determined by a P , called the pole, and a half- known as the polar axis, shown extending from P to the right in Figure 1 below. In polar coordinates, lines occur in two species. A line through the pole, making θ0 with the polar axis, has an equation

(1.1) θ = constant = θ0. If a line ` does not pass through the pole, the normal line from the pole to ` intersects ` at its point nearest the pole, say Q(d, θ0) with d > 0. So the equation of ` is (see Figure 1)

(1.2) r = d/ cos(θ − θ0) = d sec(θ − θ0).

Figure 1. Equation of a Line in Polar Coordinates

Since d is the from the pole P to the foot Q of the normal, we call d the pedal1 distance.

1pedal: of or relating to the foot. 1 2 ROGER ALEXANDER

1.1. Polar-coordinate equation for the line through given points. Given points A(r1, θ1) and B(r2, θ2) we see that the line AB passes through the Pole if and only if θ2 − θ1 is an integral multiple of π; and then its equation is θ = θ1. If AB does not pass through the Pole, we want to find parameters θ0 and d so that the line through A and B is represented by Equation (1.2). We may suppose (1) radii r1 and r2 have the same sign (if sgn(r2) 6= sgn(r1) replace (r2, θ2) by (−r2, θ2+π); and (2) the satisfy |θ2−θ1| < π (if not, replace θ2 by θ2 + 2kπ for suitable k.)

Figure 2. Parameters for the Line Through Given Points.

Figure 2 shows that the pedal length d is the altitude on the base AB in 4P AB. It follows that 2 · (4P AB) (1.3) d = . AB In 4P AB the altitude on the base PA (not shown in Figure 2) is

(1.4) h = |r2||sin(θ2 − θ1)|, so that, using

(1.5) σ = sgn(sin(θ2 − θ1)), we have

2 · Area (4P AB) = |r1||r2||sin(θ2 − θ1)|

(1.6) = σr1r2 sin(θ2 − θ1), since r1 and r2 have the same sign. LINES AND TANGENTS IN POLAR COORDINATES 3

By the Law of Cosines in 4P AB we have q 2 2 (1.7) AB = r1 + r2 − 2r1r2 cos(θ2 − θ1). Inserting Equations (1.6) and (1.7) into the formula (1.3) for d gives σr r sin(θ − θ ) (1.8) d = 1 2 2 1 p 2 2 r1 + r2 − 2r1r2 cos(θ2 − θ1)

It is convenient to express θ0 in the form (see Figure 2)

(1.9) θ0 = θ1 + ϕ.

Then the equation r1 cos(θ1 − θ0) = d becomes

d r2 (1.10) cos ϕ = = σ sin(θ2 − θ1). r1 AB

To find sin ϕ we expand the equation r2 cos (θ2 − (θ1 + ϕ)) = d and simplify, leading to −σ (1.11) sin ϕ = (r2 cos(θ2 − θ1) − r1) . AB Finally, using Equation (1.9) and the addition formulas for and cosine, we readily find that σ(r sin θ − r sin θ ) (1.12) cos θ = 2 2 1 1 0 p 2 2 r1 + r2 − 2r1r2 cos(θ2 − θ1) −σ(r cos θ − r cos θ ) (1.13) sin θ = 2 2 1 1 0 p 2 2 r1 + r2 − 2r1r2 cos(θ2 − θ1)

2. Exercises for Section 1 1.1 For each pair of points given in polar coordinates, find a polar- coordinate equation for the line determined by the points. √ (a) (r1, θ1) = ( 3, 0), (r2, θ2) = (1,√ π/2). (b) (r1, θ1) = (6, π/2), (r2, θ2) = (3 2, 3π/4). (c) (r1, θ1) = (4, π/2), (r2, θ2) =√ (8, 3π/2). (d) (r1, θ1) = (5, 0), (r2, θ2) = (5 2, π/4). 1.2 (a) Explain why every line in the x-y plane that does not pass through the has an equation of the form ax + by = c with c > 0, and a, b not both zero. (Use the fact that any line not passing through the origin either is vertical or has a -intercept equation with a nonzero intercept.) 4 ROGER ALEXANDER

(b) Substitute x = r cos θ, y = r sin θ into the equation for the line and solve for r. Then show that if θ0 and d are defined by a b c cos θ0 = √ , sin θ0 = √ , d = √ , a2 + b2 a2 + b2 a2 + b2 the equation for the line takes the standard form (1.2). 1.3 The Cartesian coordinates of A and B in Figure 2 are

(xi, yi) = (ri cos θi, ri sin θi), i = 1, 2.

Show that if the angle α is defined by x − x y − y cos α = 2 1 , sin α = 2 1 , AB AB then the angle θ0 = θ1 + ϕ satisfies θ0 = α − π/2.

3. Polar Equation for the Tangent Line Suppose that a polar curve is defined by r = f(θ) with a continuously differentiable f defined on some open θ-, and that θ1 is an interior point of this interval. Set r1 = f(θ1); we seek the equation of the tangent line to the curve at (r1, θ1). What the equation for the tangent line is depends on whether r1 = 0. We consider first the case r1 6= 0 and then the case r1 = 0.

3.1. Tangent at a point where r1 6= 0. Let ∆θ be an increment and write θ2 = θ1 + ∆θ, and r2 = f(θ2). Since f is continuous, r1 and r2 will have the same sign for all sufficiently small increments ∆θ. We find the parameters θ0, d for the tangent line as limits of the corresponding parameters for chords joining A(r1, θ1) and B(r2, θ2) as ∆θ → 0; see Figure 3. Since f is differentiable there are (see the Appendix, Section 5) wobble functions wf , wc and ws of ∆θ, all with limit 0 as ∆θ → 0, so that (writing 0 0 r1 = f (θ1)) 0  (3.1) f(θ1 + ∆θ) = r1 + ∆θ r1 + wf (∆θ) , (3.2) cos(θ1 + ∆θ) = cos θ1 + ∆θ (− sin θ1 + wc(∆θ)) ,

(3.3) sin(θ1 + ∆θ) = sin θ1 + ∆θ (cos θ1 + ws(∆θ)) .

First we determine the limit of the pedal distance d. In our notation the quantity inside the in the denominator in Equation (1.8) is

2 2 2 2 1 r2 + r1 − 2r1r2 + 2r1r2(1 − cos ∆θ) = (r2 − r1) + 4r1r2 sin 2 ∆θ 2 0 2 2 1 (3.4) = (∆θ) (r1 + wf ) + 4r1r2 sin 2 ∆θ. LINES AND TANGENTS IN POLAR COORDINATES 5

Figure 3. ABC and Tangent AT .

Taking d from Equation (1.8), dividing numerator and denominator by ∆θ and taking the limit ∆θ → 0 we find (using σ∆θ = |∆θ| = p(∆θ)2)

sin(∆θ) r1r2 lim d = lim ∆θ r 1 ∆θ→0 ∆θ→0 sin2 ∆θ 0 2 2 (r1 + wf ) + 4r1r2 (∆θ)2 r2 (3.5) = 1 . p 2 0 2 r1 + (r1)

To find the angle θ0 for the tangent line, represent θ0 = θ1 + ϕ as we did in Section 1. We calculate the limit of cos ϕ from Equation (1.10) to be

σr sin(∆θ) lim cos ϕ = lim 2 . ∆θ→0 ∆θ→0 AB 6 ROGER ALEXANDER

Divide numerator and denominator by ∆θ and use Equation (3.4) to get

sin(∆θ) r2 (3.6) lim cos ϕ = lim ∆θ ∆θ→0 ∆θ→0 1 σ∆θ AB r (3.7) = 1 . p 2 0 2 r1 + (r1) A similar calculation for the limit of sin ϕ (Equation (1.11)) leads to −σ(r cos(θ − θ ) − r ) lim sin ϕ = lim 2 2 1 1 ∆θ→0 ∆θ→0 AB   σ (r1 − r2) + r2 1 − cos(θ2 − θ1) = lim , ∆θ→0 AB and, dividing numerator and denominator by σ(∆θ),

2 1 r2 − r1 2 sin 2 ∆θ − − r2 = lim ∆θ ∆θ 1 ∆θ→0 AB σ∆θ −r0 (3.8) = 1 . p 2 0 2 r1 + (r1) Combining the equations for d, cos ϕ and sin ϕ we find that the parameters of the tangent line at a point (r1, θ1) with r1 6= 0 are simply expressed by

(3.9) d = r1 cos ϕ, θ0 = θ1 + ϕ. 3.2. Smooth curves; arc length in polar coordinates. A free dividend of our derivation is the formula for the differential of arc length in polar coordinates. The denominator in the formula (1.8) for d is just the length of the chord from (r1, θ1) to (r2, θ2), or ∆s. Hence the quantity under the square root in the denominator of (3.5) gives ds2, the squared differential of arc length: p (3.10) ds = r2 + (r0)2 dθ. Recall that a curve is smooth if the differential form ds never vanishes. We see that the condition for a polar-coordinate curve to be smooth is that r and r0 are never simultaneously zero.

3.3. Tangent at the pole. Suppose now that a smooth curve given by r = f(θ) passes through the pole at a certain parameter value θ = θ1: 0 f(θ1) = 0, f (θ1) 6= 0.

When ∆θ is small, the point B(f(θ1 + ∆θ), θ1 + ∆θ) is given by

 0   0  f(θ1) + (f (θ1) + wf )∆θ, θ1 + ∆θ = (f (θ1) + wf )∆θ, θ1 + ∆θ LINES AND TANGENTS IN POLAR COORDINATES 7 and the chord PB has the direction θ1 + ∆θ. In the limit ∆θ → 0 the direction is obviously θ1. The tangent line to a smooth curve at a point where f(θ1) = 0 is the line θ = θ1. 4. Exercises for Section 3 3.1 A with diameter 2a lying along the polar axis with one end at the pole has the equation r = 2a cos θ, 0 ≤ θ ≤ π. Show that the parameters ϕ and d of the tangent line at a point (r, θ) on the circle are given by ϕ = θ, d = 2a cos2 θ. Find all values of θ where the tangent line is (a) horizontal; (b) vertical. a 3.2 For the r = , show that the parameters ϕ and d 1 + cos θ of the tangent line at a point (r, θ) are given by a ϕ = − 1 θ, d = sec 1 θ. 2 2 2 2 1 (Use 1 + cos θ = 2 cos 2 θ.) 3.3 For the r = a(1 + cos θ), show that the parameters ϕ and d of the tangent line at a point (r, θ) are given by 1 3 1 ϕ = 2 θ, d = 2a cos 2 θ. 5. Appendix: Differentiability and the Wobble Function Suppose a function f is defined on some open interval containing a point a, and is differentiable at a. Then f(a + h) − f(a) (5.1) f 0(a) = lim . h→0 h Subtracting the constant f 0(a) from both sides and rearranging, we get f(a + h) − f(a)  (5.2) lim − f 0(a) = 0. h→0 h We define the wobble function w(h) to be the parenthesized expression on the left hand side. We call it the wobble function because w(h) is the difference between of the chord and the tangent line. Now take the equation defining the wobble function, f(a + h) − f(a) w(h) = − f 0(a), h and solve for f(a + h). The result is (5.3) f(a + h) = f(a) + f 0(a) + w(h) · h. This simple calculation shows that if f is differentiable at a, then there is a wobble function w defined for small h so that Equation (5.3) holds. (The converse is also true, but it is not needed here.)