Defining Polar Coordinates and Differentiating in Polar Form: Study Guide (Calculus BC 9.7)

BC-only material. Polar coordinates are part of Calculus BC only. Calculus AB students do not need this topic.

Before you start: prerequisites

This topic reuses the parametric tools of Topics 9.1 and 9.2 with a new parameter, the angle θ. If any row is shaky, revisit it from the Calculus BC roadmap first.

Prerequisite	Topic	What you use it for here
dy/dx for parametric curves	9.1	dy/dx = (dy/dθ) ÷ (dx/dθ)
Second derivatives of parametric curves	9.2	d²y/dx² for a polar curve
Product rule and chain rule	2.8, 3.1	Differentiating x = f(θ) cos θ and y = f(θ) sin θ
Unit-circle values in radians	Precalculus	Exact values such as cos(π/3) = ½

Notation on this page: r′ or f′(θ) means dr/dθ. Angles are in radians. The pole is the origin.

What polar coordinates are

Rectangular coordinates (x, y) describe a point by how far to go across and up. Polar coordinates (r, θ) describe it by a direction and a distance:

• θ is the angle measured anticlockwise from the positive x-axis (the polar axis);
• r is the signed distance from the pole along the ray at angle θ.

Right-angled triangle trigonometry links the two systems:

x = r cos θ, y = r sin θ, r² = x² + y², tan θ = y/x (when x ≠ 0)

Example. The polar point (4, 5π/6) has x = 4 cos(5π/6) = −2√3 and y = 4 sin(5π/6) = 2. So it is (−2√3, 2).

Going the other way. For the rectangular point (−3, 3), r = √(9 + 9) = 3√2. The point is in the second quadrant, so θ = 3π/4, not the calculator’s tan⁻¹(−1) = −π/4. Always check the quadrant.

One point, many names. Adding 2π to θ gives the same point, so (2, π/6) and (2, 13π/6) coincide. A negative r means “face along the ray at angle θ, then walk backwards”. So (−2, π/4) is the point (−√2, −√2), the same point as (2, 5π/4). Rectangular coordinates are unique; polar coordinates are not.

Polar curves and equations

A polar equation r = f(θ) gives one distance for each angle. As θ increases, the point (f(θ), θ) sweeps round the pole and traces a curve. Some shapes you will meet often:

Equation	Shape
r = a	Circle of radius a centred at the pole
θ = α	Straight line through the pole
r = 2a cos θ or r = 2a sin θ	Circle through the pole
r = a + b cos θ (or sin θ)	Limaçon; a cardioid when a = b
r = a sin(nθ) or a cos(nθ)	Rose with petals

You do not have to memorise this table. You can always sketch a curve from a table of values: choose θ = 0, π/6, π/4, … and plot each (r, θ).

Converting equations. Use r² = x² + y² and r cos θ = x. For r = 6 cos θ, multiply both sides by r: r² = 6r cos θ, so x² + y² = 6x, which is (x − 3)² + y² = 9. That is a circle of radius 3 centred at (3, 0).

A polar curve is a parametric curve

Here is the key idea of the topic. For the curve r = f(θ), every point has

x = f(θ) cos θ, y = f(θ) sin θ

These are parametric equations with parameter θ. So everything you learned in Topics 9.1 and 9.2 applies, with θ in the place of t.

Differentiate each with the product rule:

dx/dθ = f′(θ) cos θ − f(θ) sin θ dy/dθ = f′(θ) sin θ + f(θ) cos θ

and then, as for any parametric curve,

dy/dx = (dy/dθ) ÷ (dx/dθ), provided dx/dθ ≠ 0

Second derivative. As in Topic 9.2, dy/dx is a function of θ, so differentiate it with respect to θ and divide by dx/dθ again:

d²y/dx² = [d/dθ (dy/dx)] ÷ (dx/dθ)

Horizontal and vertical tangents.

• Horizontal tangent: dy/dθ = 0 and dx/dθ ≠ 0.
• Vertical tangent: dx/dθ = 0 and dy/dθ ≠ 0.
• Both zero: the formula gives 0/0 and tells you nothing on its own. Investigate further.

A tangent at the pole. If f(α) = 0 and f′(α) ≠ 0, the curve passes through the pole at θ = α, and the formula simplifies to dy/dx = tan α. The tangent there is the line θ = α. For example, r = 2 sin 3θ passes through the pole at θ = π/3 with r′ = 6 cos π = −6 ≠ 0, so the tangent there is the line θ = π/3.

Three derivatives, three meanings

For a polar curve you can be asked about several rates. Keep them apart.

Derivative	What it measures	How to read its sign
dr/dθ	How r changes as θ increases	If r > 0: positive means moving away from the pole. If r < 0, the reverse.
dx/dθ	How the x-coordinate changes as θ increases	Positive: the point moves right
dy/dθ	How the y-coordinate changes as θ increases	Positive: the point moves up
dy/dx	Slope of the tangent line in the xy-plane	Ordinary slope

The rule for distance comes from |r|: the distance from the pole is increasing when r and dr/dθ have the same sign, and decreasing when they have opposite signs.

Worked example 1: a limaçon by hand

Question. The curve r = 3 − 2 cos θ for 0 ≤ θ < 2π is shown in Figure 1. Without a calculator:

(a) find dr/dθ at θ = π/2 and say what it tells you; (b) find the slope of the curve at θ = π/2 and the equation of the tangent line there; (c) find d²y/dx² at θ = π/2 and state the concavity; (d) find every point where the curve has a vertical tangent.

(a) dr/dθ = 2 sin θ, so at θ = π/2, dr/dθ = 2. Also r(π/2) = 3 > 0. Since r and dr/dθ are both positive, the point is moving away from the pole as θ increases through π/2.

(b) Write the parametric form: x = (3 − 2 cos θ) cos θ, y = (3 − 2 cos θ) sin θ.

1. At θ = π/2: r = 3, r′ = 2, cos θ = 0, sin θ = 1.
2. dx/dθ = r′ cos θ − r sin θ = 2(0) − 3(1) = −3.
3. dy/dθ = r′ sin θ + r cos θ = 2(1) + 3(0) = 2.
4. dy/dx = 2 ÷ (−3) = −2/3.
5. The point is x = 3 cos(π/2) = 0, y = 3 sin(π/2) = 3, so Q(0, 3).
6. Tangent line: y = 3 − (2/3)x.

Check with the picture: at Q the curve is near its top, sloping gently down to the right. A small negative slope fits.

(c) It helps to simplify first. Expanding and using double-angle identities:

• dy/dθ = 2 sin²θ + 3 cos θ − 2 cos²θ = 3 cos θ − 2 cos 2θ
• dx/dθ = 2 sin θ cos θ − 3 sin θ + 2 sin θ cos θ = 2 sin 2θ − 3 sin θ

So dy/dx = N/D with N = 3 cos θ − 2 cos 2θ and D = 2 sin 2θ − 3 sin θ.

1. At θ = π/2: N = 2, D = −3, N′ = −3 sin θ + 4 sin 2θ = −3, D′ = 4 cos 2θ − 3 cos θ = −4.
2. Quotient rule: d/dθ (dy/dx) = (N′D − ND′)/D² = ((−3)(−3) − (2)(−4))/9 = 17/9.
3. Divide by dx/dθ = −3: d²y/dx² = (17/9) ÷ (−3) = −17/27.

Negative, so the curve is concave down at Q. That matches the figure: near Q the curve lies below its tangent line.

(d) Vertical tangents need dx/dθ = 0. Factorise: dx/dθ = sin θ (4 cos θ − 3).

• sin θ = 0 gives θ = 0 or π.
• cos θ = 3/4 gives θ = cos⁻¹(3/4) (about 0.723) or 2π − cos⁻¹(3/4).

Now check dy/dθ = 3 cos θ − 2 cos 2θ at each:

θ	dy/dθ	r	Point (x, y)
0	3 − 2 = 1	1	(1, 0)
π	−3 − 2 = −5	5	(−5, 0)
cos θ = 3/4	9/4 − 2(1/8) = 2	3/2	(9/8, ±3√7/8)

For cos θ = 3/4, cos 2θ = 2(9/16) − 1 = 1/8, and sin θ = ±√7/4. All four dy/dθ values are non-zero, so there are four vertical tangents: at (1, 0), (−5, 0) and (9/8, ±3√7/8). The two at x = 9/8 sit either side of the “dimple” on the right of the curve, slightly further right than (1, 0).

Worked example 2: a pond edge, with a calculator

Context. The edge of a fictional garden pond is the polar curve r = 3 + sin 2θ for 0 ≤ θ < 2π. The fountain is at the pole, r is in metres, the positive x-axis points east and the positive y-axis points north. A robot cleaner travels round the edge as θ increases. Use a calculator in radian mode.

(a) Find dr/dθ at θ = 2 and interpret it. dr/dθ = 2 cos 2θ, so at θ = 2, dr/dθ = 2 cos 4 ≈ −1.307. Since r(2) = 3 + sin 4 ≈ 2.243 > 0 and dr/dθ < 0, the robot is getting closer to the fountain, at about 1.307 metres per radian of turning.

(b) Is the robot moving north or south at θ = 2? dy/dθ = r′ sin θ + r cos θ ≈ (−1.307)(0.909) + (2.243)(−0.416) ≈ −2.122. Negative, so y is decreasing: the robot is moving south as θ increases.

(c) Find dy/dx and d²y/dx² at θ = 2. dx/dθ = r′ cos θ − r sin θ ≈ (−1.307)(−0.416) − (2.243)(0.909) ≈ −1.496. dy/dx ≈ (−2.122) ÷ (−1.496) ≈ 1.419. A calculator (differentiating dy/dx numerically with respect to θ, then dividing by dx/dθ) gives d²y/dx² ≈ −0.496, so the edge is concave down there.

(d) How far from the fountain does the edge get in the first quadrant? No calculator needed. On 0 ≤ θ ≤ π/2, dr/dθ = 2 cos 2θ = 0 at θ = π/4, where r = 3 + sin(π/2) = 4. At the endpoints r(0) = r(π/2) = 3. So the greatest distance is 4 m, at θ = π/4.

(e) Is that also the easternmost point? No. The easternmost point has the largest x, which needs dx/dθ = 0, not dr/dθ = 0. Solving dx/dθ = 0 numerically gives θ ≈ 0.377, where x ≈ 3.426 m (and r ≈ 3.684). At θ = π/4, x = 4 cos(π/4) ≈ 2.828. Being furthest from the pole is a different question from being furthest east.

Checks. Units: r in metres, θ in radians, so dr/dθ is in metres per radian. The slope dy/dx has no units. Signs: in the second quadrant with r shrinking, moving south and towards the fountain is consistent.

Common misconceptions

• “dr/dθ is the slope.” It is not. The slope of the curve is dy/dx = (dy/dθ) ÷ (dx/dθ). In Worked example 1, dr/dθ = 2 but the slope is −2/3.
• “The slope at angle θ is tan θ.” tan θ is the slope of the ray from the pole to the point. It equals the tangent slope only in special cases, such as a curve passing through the pole.
• Forgetting the product rule. dx/dθ is not −f(θ) sin θ. Both factors of f(θ) cos θ depend on θ.
• “d²y/dx² = (d²y/dθ²) ÷ (d²x/dθ²).” Differentiate dy/dx with respect to θ, then divide by dx/dθ.
• “dr/dθ > 0 always means moving away from the pole.” Only when r > 0. If r < 0, a positive dr/dθ means |r| is shrinking.
• Using degree mode. Every derivative formula here assumes radians.
• Declaring a horizontal tangent from dy/dθ = 0 alone. Check dx/dθ ≠ 0; if both are 0, investigate.
• Taking tan⁻¹(y/x) without checking the quadrant when converting to polar form.

Where this leads

Topic 9.8 uses polar coordinates to find areas. The building block is a thin sector of area ½r²Δθ, and adding them gives ½ ∫ r² dθ. Continue with Topic 9.8, Finding the Area of a Polar Region or the Area Bounded by a Single Polar Curve, or see the order on the Calculus BC roadmap.

Try the practice questions now, then use the revision notes and the checklist to consolidate.