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Unit 9 · Topic 9.7

BC only

9.7 Defining Polar Coordinates and Differentiating in Polar Form

BC only. Polar coordinates locate a point by its distance r from the origin and its angle θ. To find slopes of a polar curve r = f(θ), write x = r cos θ and y = r sin θ, then use dy/dx = (dy/dθ)/(dx/dθ). The derivative dr/dθ is different: it tells you whether the curve is moving toward or away from the origin.

Key terms

  • polar coordinates
  • x = r cos θ, y = r sin θ
  • dy/dx
  • dr/dθ
  • pole (origin)

Polar coordinates

This whole unit is BC only. A point in polar form is (r, θ): go out a distance r from the origin (called the pole) in the direction of the angle θ, measured from the positive x-axis. A negative r means go the opposite direction, so (−2, π/4) is the same point as (2, 5π/4).

Converting between forms:

  • x = r cos θ and y = r sin θ
  • r² = x² + y² and tan θ = y/x (for x ≠ 0)
  • A polar curve r = f(θ) becomes the parametric curve x = f(θ) cos θ, y = f(θ) sin θ, with θ as the parameter.

Slope of a polar curve

Since a polar curve is a parametric curve in θ, use the 9.1 formula:

dy/dx = (dy/dθ)/(dx/dθ), where dx/dθ = r′ cos θ − r sin θ and dy/dθ = r′ sin θ + r cos θ, with r′ = dr/dθ.

Those come from the product rule on x = r cos θ and y = r sin θ. You don't need to memorize them; just differentiate. Horizontal tangents occur where dy/dθ = 0 and dx/dθ ≠ 0, and vertical tangents where dx/dθ = 0 and dy/dθ ≠ 0.

What dr/dθ tells you

dr/dθ is the rate at which the distance from the origin changes as θ increases. It is not the slope of the curve. Whether the curve moves toward or away from the origin depends on the signs of both r and dr/dθ:

rdr/dθThe curve is…
positivepositivemoving away from the origin
positivenegativemoving toward the origin
negativenegativemoving away from the origin (the size of r is growing)
negativepositivemoving toward the origin (the size of r is shrinking)

Rates of change of x and y

Read the question's wording carefully. “Distance from the origin” is about r, so use dr/dθ. “Distance from the x-axis” or “height” is about y, so use dy/dθ. dx/dθ and dy/dθ tell you how the x- and y-coordinates change as θ increases. A question might ask whether the curve is getting closer to the x-axis at some θ. That's about whether |y| is decreasing: look at the signs of y and dy/dθ together. Always compute these with y = r sin θ and x = r cos θ, not with r alone.

Worked examples

Try each one yourself first, then open the solution.

  1. Example 1

    Slope of a cardioid

    Find the slope of the curve r = 2 + 2 cos θ at θ = π/2, and give the point.

    Show the solution
    1. Step 1: r(π/2) = 2 + 0 = 2. The point is x = 2 cos(π/2) = 0, y = 2 sin(π/2) = 2, so (0, 2).
    2. Step 2: r′ = −2 sin θ, so r′(π/2) = −2.
    3. Step 3: dx/dθ = r′ cos θ − r sin θ = (−2)(0) − (2)(1) = −2.
    4. Step 4: dy/dθ = r′ sin θ + r cos θ = (−2)(1) + (2)(0) = −2.
    5. Step 5: dy/dx = (−2)/(−2) = 1.

    Answer: Slope 1 at the point (0, 2)

  2. Example 2

    Trap: dr/dθ with negative r

    For r = 1 + 2 cos θ, is the curve moving toward or away from the origin at θ = π/3? At θ = 3π/4? Also find dy/dθ at θ = π/2.

    Show the solution
    1. Step 1: dr/dθ = −2 sin θ.
    2. Step 2: At θ = π/3: r = 1 + 2(½) = 2 > 0 and dr/dθ = −√3 < 0. r is positive and shrinking, so the curve moves toward the origin.
    3. Step 3: At θ = 3π/4: r = 1 + 2(−√2/2) = 1 − √2 ≈ −0.414 < 0 and dr/dθ = −2(√2/2) = −√2 < 0. r is negative and getting more negative, so |r| is growing: the curve moves away from the origin.
    4. Step 4: A student who only looks at dr/dθ < 0 would wrongly say “toward” both times.
    5. Step 5: For dy/dθ: y = r sin θ = sin θ + 2 sin θ cos θ = sin θ + sin 2θ, so dy/dθ = cos θ + 2 cos 2θ. At θ = π/2: 0 + 2(−1) = −2.

    Answer: Toward the origin at θ = π/3; away from it at θ = 3π/4. dy/dθ = −2 at θ = π/2.

Common mistakes

  • Using dr/dθ as the slope of the curve. The slope is dy/dx.
  • Forgetting the product rule when differentiating r cos θ and r sin θ.
  • Deciding toward/away from dr/dθ alone when r is negative.
  • Working in degree mode on the calculator.

On the exam

  • BC free-response polar questions often ask for the slope at a given θ, the meaning of dr/dθ in context, and an area (9.8, 9.9).
  • When explaining dr/dθ, mention both r's sign and dr/dθ's sign: “Since r > 0 and dr/dθ < 0, the curve is getting closer to the origin.”

Connected topics

Videos

  • Calculus BC – 9.7 Defining Polar Coordinates and Differentiating in Polar Form

    The AlgebrosWatch on YouTube (opens in a new tab)

  • Polar functions derivatives | Advanced derivatives | AP Calculus BC | Khan Academy

    Khan AcademyWatch on YouTube (opens in a new tab)

  • Tangent Line Equations, Slope, & Derivatives In Polar Form | Calculus 2

    The Organic Chemistry TutorWatch on YouTube (opens in a new tab)

  • Finding dy/dx of a polar curve (the formula and an example)

    bprp calculus basicsWatch on YouTube (opens in a new tab)

  • The Slope of Tangent Lines to Polar Curves

    Mathispower4uWatch on YouTube (opens in a new tab)

  • Worked example: differentiating polar functions | AP Calculus BC | Khan Academy

    Khan AcademyWatch on YouTube (opens in a new tab)

Check yourself

4 questions on 9.7 Defining Polar Coordinates and Differentiating in Polar Form. Pick an answer to see if you got it, and why.

Question 1 of 4

What is the slope of the line tangent to the polar curve r = 1 + 2sin θ at the point where θ = π/6?

Question 2 of 4

For the polar curve r = 1 + 2cos θ, which of the following is true at θ = 3π/4?

Question 3 of 4Calculator allowed

A particle moves along the polar curve r = 1 + θ sin θ. What is the rate of change of the particle's y-coordinate with respect to θ when θ = 2?

Question 4 of 4

A point has polar coordinates (r, θ) = (−2, π/3). What are its rectangular coordinates (x, y)?

0 of 4 answered