AP Calculus BC Unit 9 Notes: Polar Curves, Slopes, and Areas

Defining Polar Coordinates and Differentiating in Polar Form

What polar coordinates are (and why you use them)

In the usual Cartesian plane, you locate a point by walking horizontally and vertically: $x$ units left or right and $y$ units up or down. Polar coordinates describe location differently: you start at the origin and describe a point by (1) how far away it is from the origin and (2) the direction you travel.

A point in polar form is written as $(r,\theta)$ where:

• $r$ is the directed distance from the origin (the pole) to the point.
• $\theta$ is the angle (measured from the positive $x$-axis) that tells you the direction.

Polar coordinates matter because many curves that look complicated in Cartesian form become simple in polar form. Curves with circular symmetry, petals, loops, spirals, and “distance-from-the-origin” rules are often easiest to express and analyze using $r$ as a function of $\theta$.

Multiple representations of the same point

A key idea in polar is that the same point can be written in many ways. That happens for two reasons:

1. Angles repeat every full rotation:

$(r,\theta) = (r,\theta + 2\pi k)$

for any integer $k$.

2. A negative radius points you in the opposite direction:

$(r,\theta) = (-r,\theta + \pi)$

This “negative $r$” fact is extremely important for graphing and for solving intersection problems—if you forget it, you can miss where curves overlap.

Converting between polar and Cartesian

Polar and Cartesian are connected through right-triangle geometry:

$x = r\cos\theta$

$y = r\sin\theta$

From these, you also get:

$r^2 = x^2 + y^2$

and (with quadrant awareness):

$\tan\theta = \frac{y}{x}$

It’s common in AP problems to switch forms depending on what you’re asked to compute. For example:

• Slopes and areas are often cleanest if you keep $r$ as a function of $\theta$.
• Interpreting a curve in the plane might be easier if you convert to an $x$-$y$ equation.

How polar graphs work (intuition for sketching)

A polar equation typically looks like $r = f(\theta)$. You can think of it as a “radar sweep”: as $\theta$ changes, the function tells you how far out from the origin the point is.

A few sketching instincts help a lot:

• If $r = c$ (a constant), the graph is a circle centered at the origin with radius $|c|$.
• If $r$ becomes negative for some angles, the plotted point jumps to the opposite direction (rotate $\theta$ by $\pi$).
• Symmetry checks save time. Common symmetry tests:
  • If replacing $\theta$ by $-\theta$ leaves the equation unchanged, the graph is symmetric about the $x$-axis.
  • If replacing $\theta$ by $\pi - \theta$ leaves it unchanged, symmetric about the $y$-axis.
  • If replacing $\theta$ by $\theta + \pi$ leaves it unchanged, symmetric about the origin.

(These symmetry patterns often appear in AP problems when choosing integration bounds for area.)

Differentiating in polar form: finding $\frac{dy}{dx}$

Polar curves are naturally parameterized by $\theta$. You should treat $x$ and $y$ as parametric functions of $\theta$:

$x(\theta) = r(\theta)\cos\theta$

$y(\theta) = r(\theta)\sin\theta$

Once you have a parametric setup, the slope of the tangent line is

$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$

This matters because AP questions often ask for the slope of the tangent line to a polar curve at a specified angle, or they ask you to find where tangents are horizontal or vertical.

Computing $\frac{dx}{d\theta}$ and $\frac{dy}{d\theta}$

Let $r = r(\theta)$ and define

$r' = \frac{dr}{d\theta}$

Differentiate using the product rule:

$\frac{dx}{d\theta} = r'\cos\theta - r\sin\theta$

$\frac{dy}{d\theta} = r'\sin\theta + r\cos\theta$

So the standard polar slope formula is

$\frac{dy}{dx} = \frac{r'\sin\theta + r\cos\theta}{r'\cos\theta - r\sin\theta}$

A big conceptual point: this is not a new kind of derivative rule—it’s the parametric derivative rule applied to the polar-to-Cartesian conversion.

Horizontal and vertical tangents (a common AP skill)

Because $\theta$ is the parameter, you use parametric tangent logic:

• Horizontal tangent when $\frac{dy}{d\theta} = 0$ and $\frac{dx}{d\theta} \ne 0$.
• Vertical tangent when $\frac{dx}{d\theta} = 0$ and $\frac{dy}{d\theta} \ne 0$.

A common mistake is to set the numerator or denominator of $\frac{dy}{dx}$ to zero without checking the other derivative—this can accidentally include cusps or undefined points.

Worked example 1: slope of a tangent line to a polar curve

Find $\frac{dy}{dx}$ for the curve $r = 2\cos\theta$ at $\theta = \frac{\pi}{3}$.

Step 1: Compute $r$ and $r'$.

$r = 2\cos\theta$

$r' = -2\sin\theta$

At $\theta = \frac{\pi}{3}$:

$r = 2\cos\left(\frac{\pi}{3}\right) = 1$

$r' = -2\sin\left(\frac{\pi}{3}\right) = -\sqrt{3}$

Step 2: Use the polar slope formula.

$\frac{dy}{dx} = \frac{r'\sin\theta + r\cos\theta}{r'\cos\theta - r\sin\theta}$

Compute numerator:

$r'\sin\theta + r\cos\theta = (-\sqrt{3})\left(\frac{\sqrt{3}}{2}\right) + 1\left(\frac{1}{2}\right) = -\frac{3}{2} + \frac{1}{2} = -1$

Compute denominator:

$r'\cos\theta - r\sin\theta = (-\sqrt{3})\left(\frac{1}{2}\right) - 1\left(\frac{\sqrt{3}}{2}\right) = -\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} = -\sqrt{3}$

So

$\frac{dy}{dx} = \frac{-1}{-\sqrt{3}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$

Worked example 2: horizontal tangents on a polar curve

Consider $r = 1 + \cos\theta$. Find where the curve has horizontal tangents.

Step 1: Compute $r'$.

$r' = -\sin\theta$

Step 2: Set $\frac{dy}{d\theta} = 0$ and check $\frac{dx}{d\theta} \ne 0$.

$\frac{dy}{d\theta} = r'\sin\theta + r\cos\theta = (-\sin\theta)\sin\theta + (1 + \cos\theta)\cos\theta$

Simplify:

$\frac{dy}{d\theta} = -\sin^2\theta + \cos\theta + \cos^2\theta = \cos\theta + (\cos^2\theta - \sin^2\theta)$

Use $\cos^2\theta - \sin^2\theta = \cos(2\theta)$:

$\frac{dy}{d\theta} = \cos\theta + \cos(2\theta)$

Solve

$\cos\theta + \cos(2\theta) = 0$

At this point you’d typically use trig identities or a calculator (depending on what the problem allows) to find solutions in the interval of interest, then verify $\frac{dx}{d\theta} \ne 0$ at those angles. The key learning is the method: horizontal tangents come from $\frac{dy}{d\theta} = 0$, not from setting $\frac{dy}{dx} = 0$ directly.

Exam Focus

• Typical question patterns:
  • “Given $r=f(\theta)$, find $\frac{dy}{dx}$ at a specific $\theta$.”
  • “Find all values of $\theta$ where the tangent line is horizontal or vertical.”
  • “Write parametric equations $x(\theta)$ and $y(\theta)$ for the polar curve and use them for a derivative-related quantity.”
• Common mistakes:
  • Forgetting to use $\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$ and instead differentiating $r$ as if it were $y$.
  • Missing points because $r$ can be negative, so the same location might occur at a different $\theta$ value.
  • Declaring a horizontal tangent when $\frac{dy}{d\theta} = 0$ but $\frac{dx}{d\theta}$ is also $0$ (a cusp or undefined tangent needs special care).

Finding the Area of a Polar Region

What “area in polar” means

In Cartesian coordinates, you learn area under a curve using rectangles: $\int y\,dx$ accumulates thin vertical slices. In polar, the natural “thin slice” is not a rectangle—it’s a thin sector (a slice of a circle).

If you fix an angle interval of width $d\theta$ and go out to radius $r$, the small region looks like a sector with central angle $d\theta$ and radius $r$.

The area of a sector with radius $r$ and angle $\Delta\theta$ (in radians) is

$A = \frac{1}{2}r^2\Delta\theta$

So an “infinitesimal sector” has area

$dA = \frac{1}{2}r^2\,d\theta$

This is the key idea behind the polar area formula.

The polar area formula

If a region is traced by $r=f(\theta)$ from $\theta=a$ to $\theta=b$, and the curve does not retrace itself over that interval, the area enclosed is

$A = \frac{1}{2}\int_a^b \left(r(\theta)\right)^2\,d\theta$

Why the square? Because the sector area depends on $r^2$. A common conceptual pitfall is to think “area should be proportional to $r$,” but in two dimensions, scaling distance scales area quadratically.

Choosing correct bounds: the real skill

The formula is simple; the challenge is picking $a$ and $b$ so you cover exactly the region you want.

You typically choose bounds by:

1. Identifying where the curve starts and ends a full “loop” or a single petal.
2. Finding where $r=0$ (often indicates a pole crossing) or where the curve intersects itself.
3. Using symmetry to reduce work (then multiplying), but only after you’re sure the curve doesn’t double-trace.

A classic mistake is integrating over $0$ to $2\pi$ automatically. Some polar curves trace the same region multiple times over $0$ to $2\pi$, which would make your computed area too large.

Worked example 1: area enclosed by a cardioid

Find the area enclosed by the curve

$r = 1 + \cos\theta$

Step 1: Decide bounds.
A cardioid like $1+\cos\theta$ traces exactly once as $\theta$ goes from $0$ to $2\pi$. So you can use

$a=0$

$b=2\pi$

Step 2: Apply the polar area formula.

$A = \frac{1}{2}\int_0^{2\pi} (1+\cos\theta)^2\,d\theta$

Expand:

$(1+\cos\theta)^2 = 1 + 2\cos\theta + \cos^2\theta$

So

$A = \frac{1}{2}\int_0^{2\pi} \left(1 + 2\cos\theta + \cos^2\theta\right)d\theta$

Use the identity

$\cos^2\theta = \frac{1+\cos(2\theta)}{2}$

Integrate term-by-term:

• $\int_0^{2\pi} 1\,d\theta = 2\pi$
• $\int_0^{2\pi} 2\cos\theta\,d\theta = 0$
• $\int_0^{2\pi} \cos^2\theta\,d\theta = \int_0^{2\pi} \frac{1+\cos(2\theta)}{2}\,d\theta = \pi$

So

$A = \frac{1}{2}(2\pi + 0 + \pi) = \frac{3\pi}{2}$

Worked example 2: area of one petal of a rose curve

Find the area of one petal of

$r = 2\sin(3\theta)$

Step 1: Identify the interval for one petal.
One petal is traced between consecutive zeros of $r$ where the curve goes out and back to the pole.

Solve

$2\sin(3\theta)=0$

This happens when

$3\theta = 0,\pi,2\pi,\dots$

So consecutive zeros occur at

$\theta=0$

and

$\theta=\frac{\pi}{3}$

Over $\left[0,\frac{\pi}{3}\right]$, $\sin(3\theta)$ is nonnegative, so you get a single outward petal without relying on negative $r$.

Step 2: Compute the area.

$A = \frac{1}{2}\int_0^{\pi/3} (2\sin(3\theta))^2\,d\theta = \frac{1}{2}\int_0^{\pi/3} 4\sin^2(3\theta)\,d\theta$

$A = 2\int_0^{\pi/3} \sin^2(3\theta)\,d\theta$

Use

$\sin^2(u) = \frac{1-\cos(2u)}{2}$

So

$\sin^2(3\theta)=\frac{1-\cos(6\theta)}{2}$

Then

$A = 2\int_0^{\pi/3} \frac{1-\cos(6\theta)}{2}\,d\theta = \int_0^{\pi/3} (1-\cos(6\theta))\,d\theta$

Integrate:

$\int (1-\cos(6\theta))\,d\theta = \theta - \frac{1}{6}\sin(6\theta)$

Evaluate:

$A = \left[\theta - \frac{1}{6}\sin(6\theta)\right]_0^{\pi/3} = \frac{\pi}{3} - \frac{1}{6}\sin(2\pi) - 0 = \frac{\pi}{3}$

So one petal has area $\frac{\pi}{3}$.

Exam Focus

• Typical question patterns:
  • “Find the area enclosed by $r=f(\theta)$ over a specified interval.”
  • “Find the area of one loop/petal” (you must determine bounds where the loop is traced once).
  • “Use symmetry to find total area” (often: compute a fraction of the curve and multiply).
• Common mistakes:
  • Integrating over an interval that traces the region more than once (double-counting area).
  • Using degrees in trig integrals or sector reasoning; the area formula assumes $\theta$ is in radians.
  • Forgetting the square: using $\int r\,d\theta$ instead of $\frac{1}{2}\int r^2\,d\theta$.

Finding the Area of Regions Bounded by Polar Curves

What “bounded by polar curves” means

Many problems don’t ask for the area inside a single polar curve. Instead, they describe a region trapped between two polar graphs—an “outer boundary” and an “inner boundary.”

In Cartesian coordinates, you’d compute area between curves with

$\int (\text{top} - \text{bottom})\,dx$

In polar, you do the same idea but with sectors:

• Outer boundary: $r = r_{\text{outer}}(\theta)$
• Inner boundary: $r = r_{\text{inner}}(\theta)$

For each small angle $d\theta$, the region looks like a “sector ring” (a washer sector). Its area is the difference of two sector areas:

$dA = \frac{1}{2}\left(r_{\text{outer}}(\theta)^2 - r_{\text{inner}}(\theta)^2\right)d\theta$

So the area between polar curves from $\theta=a$ to $\theta=b$ is

$A = \frac{1}{2}\int_a^b \left(r_{\text{outer}}(\theta)^2 - r_{\text{inner}}(\theta)^2\right)d\theta$

How to decide “outer” vs “inner”

Unlike Cartesian “top minus bottom,” in polar you compare distances from the origin.

For a fixed angle $\theta$:

• The curve with larger $r$ value is farther from the origin along that ray.
• That curve is the outer boundary for that angle.

However, be careful: if one curve has negative $r$ values, “larger $r$” is not automatically “farther out” in the picture the way you might expect. In AP problems, the bounded-region intervals are usually chosen so the relevant radii are nonnegative, but you still need to check.

A reliable approach is: sketch or test a sample angle in the interval and compare the actual plotted distances.

Finding the bounds: intersections in polar

The boundary angles often come from intersection points, found by solving

$r_1(\theta)=r_2(\theta)$

over the interval of interest.

But polar intersections can be subtle because the same point can satisfy

$r_1(\theta)=r_2(\theta)$

or can occur via the “negative $r$” equivalence. For most standard AP bounded-area questions, solving $r_1=r_2$ in the intended interval is enough, but you should still sanity-check with a sketch.

Worked example: area between two polar curves

Find the area of the region inside $r = 2\cos\theta$ and outside $r = 1$.

Step 1: Understand the curves.

• $r=1$ is the unit circle centered at the origin.
• $r=2\cos\theta$ is a circle of radius $1$ centered at $(1,0)$ (you can confirm by converting: $r=2\cos\theta$ gives $r^2=2r\cos\theta$, so $x^2+y^2=2x$, which becomes $(x-1)^2+y^2=1$).

The region “inside $r=2\cos\theta$ and outside $r=1$” is the part of the right-shifted circle that lies outside the unit circle.

Step 2: Find intersection angles.
Solve

$2\cos\theta = 1$

So

$\cos\theta = \frac{1}{2}$

Thus

$\theta = \pm \frac{\pi}{3}$

Step 3: Decide outer and inner on that interval.
On $\left[-\frac{\pi}{3},\frac{\pi}{3}\right]$, $\cos\theta$ is at least $\frac{1}{2}$, so $2\cos\theta \ge 1$. That means:

• Outer radius: $r_{\text{outer}} = 2\cos\theta$
• Inner radius: $r_{\text{inner}} = 1$

Step 4: Set up and compute the area integral.

$A = \frac{1}{2}\int_{-\pi/3}^{\pi/3} \left((2\cos\theta)^2 - 1^2\right)d\theta$

Simplify:

$A = \frac{1}{2}\int_{-\pi/3}^{\pi/3} (4\cos^2\theta - 1)\,d\theta$

The integrand is even, so

$A = \int_0^{\pi/3} (4\cos^2\theta - 1)\,d\theta$

Use

$\cos^2\theta = \frac{1+\cos(2\theta)}{2}$

Then

$4\cos^2\theta - 1 = 4\cdot\frac{1+\cos(2\theta)}{2} - 1 = 2(1+\cos(2\theta)) - 1 = 1 + 2\cos(2\theta)$

So

$A = \int_0^{\pi/3} (1 + 2\cos(2\theta))\,d\theta$

Integrate:

$\int (1 + 2\cos(2\theta))\,d\theta = \theta + \sin(2\theta)$

Evaluate:

$A = \left[\theta + \sin(2\theta)\right]_0^{\pi/3} = \frac{\pi}{3} + \sin\left(\frac{2\pi}{3}\right)$

And

$\sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2}$

So the area is

$A = \frac{\pi}{3} + \frac{\sqrt{3}}{2}$

A note on “bounded by” vs “enclosed by” language

AP problems may say “area of the region bounded by the curves” without explicitly saying which is inside/outside. In that case, you usually:

1. Find intersection angles.
2. Determine which curve is farther from the origin (outer) on each sub-interval.
3. If the outer/inner relationship switches, split the integral.

Not splitting when the outer curve changes is a common way students get the wrong sign or the wrong region.

Exam Focus

• Typical question patterns:
  • “Find the area of the region inside $r=f(\theta)$ and outside $r=g(\theta)$.”
  • “Find the area of the region bounded by two polar curves” (you must identify intersection angles and which curve is outer).
  • “Compute area of overlap” (often requires careful interval choice and sometimes symmetry).
• Common mistakes:
  • Using $\frac{1}{2}\int (r_{\text{outer}}-r_{\text{inner}})^2 d\theta$ instead of $\frac{1}{2}\int (r_{\text{outer}}^2-r_{\text{inner}}^2)d\theta$.
  • Failing to solve for the correct intersection angles (or forgetting to check for additional intersections due to periodicity).
  • Not splitting the integral when the “outer” curve changes across the interval, leading to subtracting in the wrong order for part of the region.