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Calculus BC · Unit 9 · 9.7
AP Calculus BC · Parametric, Polar & Vector-Valued Functions
Polar Coordinates and Differentiating in Polar Form
r is a distance, not a height — and dr/dθ is not the slope of anything in the plane.
key idea$x=r\cos\theta,\quad y=r\sin\theta\quad\Rightarrow\quad\frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}$
What it says
Polar coordinates
\[(r,\theta):\quad x=r\cos\theta,\qquad y=r\sin\theta\]
A point is located by a distance $r$ from the origin and an angle $\theta$ measured from the positive $x$-axis. A polar equation $r=f(\theta)$ says how far out the curve sits in each direction — it is a rule for a radius, not for a height.
Differentiating a polar curve
\[\frac{dy}{dx}=\frac{\dfrac{d}{d\theta}\big(r\sin\theta\big)}{\dfrac{d}{d\theta}\big(r\cos\theta\big)}\]
The slope in the $xy$-plane is not $\dfrac{dr}{d\theta}$. Convert to $x$ and $y$ first, treating $\theta$ as a parameter, then use the parametric rule. Both numerator and denominator require the product rule, since $r$ itself depends on $\theta$.
Reading the notation
| Polar fact | Meaning | Watch for |
|---|
| $r=f(\theta)$ | distance from the pole at angle $\theta$ | $r$ is a radius, not a height |
| $r=0$ | the curve passes through the pole | often the start or end of a petal |
| $\dfrac{dr}{d\theta}=0$ | the radius stops changing | not a horizontal tangent |
| $\dfrac{dy}{d\theta}=0$ | $y$ stops changing | a horizontal tangent |
| $\dfrac{dx}{d\theta}=0$ | $x$ stops changing | a vertical tangent |
| $r\lt 0$ | plot in the opposite direction | $\left(-2,\tfrac{\pi}{3}\right)$ sits where $\left(2,\tfrac{4\pi}{3}\right)$ does |
Row 3 is the trap this topic is built around. $\dfrac{dr}{d\theta}$ measures how fast the radius grows, which is a completely different question from how the curve is tilted in the plane.
PoleThe origin. A polar curve reaches it exactly when $r=0$.
Polar axisThe ray $\theta=0$, matching the positive $x$-axis.
$\dfrac{dr}{d\theta}$The rate at which the curve moves away from or toward the pole. Not a slope.
$\dfrac{dy}{dx}$The ordinary slope in the plane, obtained through the parametric rule after converting.
Non-uniqueness$(r,\theta)$ and $(r,\theta+2\pi)$ are the same point, and so is $(-r,\theta+\pi)$. One point has infinitely many polar names.
Worked examples
EX 1
Convert a point
polar to CartesianConvert $\left(4,\dfrac{2\pi}{3}\right)$ to Cartesian coordinates.
1
Apply the conversion
$x=r\cos\theta$ and $y=r\sin\theta$.
2
Evaluate the cosine
$\cos\dfrac{2\pi}{3}=-\dfrac12$, so $x=4\left(-\dfrac12\right)=-2$.
3
Evaluate the sine
$\sin\dfrac{2\pi}{3}=\dfrac{\sqrt3}{2}$, so $y=4\cdot\dfrac{\sqrt3}{2}=2\sqrt3$.
4
State the point
$\left(-2,\ 2\sqrt3\right)$.
5
Check the quadrant
$\dfrac{2\pi}{3}$ lies in the second quadrant, and indeed $x\lt 0$ while $y\gt 0$. ✓
EX 2
The slope of a polar curve
product rule twiceFind $\dfrac{dy}{dx}$ for $r=2+2\cos\theta$ at $\theta=\dfrac{\pi}{2}$.
1
Write $x$ and $y$
$x=\left(2+2\cos\theta\right)\cos\theta$ and $y=\left(2+2\cos\theta\right)\sin\theta$.
2
Differentiate $y$
$\dfrac{dy}{d\theta}=-2\sin^{2}\theta+\left(2+2\cos\theta\right)\cos\theta$; at $\theta=\dfrac{\pi}{2}$ this is $-2+0=-2$.
3
Differentiate $x$
$\dfrac{dx}{d\theta}=-2\sin\theta\cos\theta-\left(2+2\cos\theta\right)\sin\theta$; at $\theta=\dfrac{\pi}{2}$ this is $0-2=-2$.
4
Form the ratio
$\dfrac{dy}{dx}=\dfrac{-2}{-2}=1$.
5
Note what was not used
$\dfrac{dr}{d\theta}=-2\sin\theta=-2$ at this angle — a different number entirely, and not the slope.
EX 3
Horizontal and vertical tangents
set each rate to zeroFor $r=2\cos\theta$, find where the curve has a vertical tangent.
1
Write $x$
$x=2\cos^{2}\theta$.
2
Differentiate
$\dfrac{dx}{d\theta}=-4\cos\theta\sin\theta=-2\sin2\theta$.
3
Solve
$\sin2\theta=0$ gives $\theta=0,\ \dfrac{\pi}{2},\ \pi$ within one revolution.
4
Check the other rate
$y=2\cos\theta\sin\theta=\sin2\theta$, so $\dfrac{dy}{d\theta}=2\cos2\theta$, which is $2$ at $\theta=0$ and $-2$ at $\theta=\dfrac{\pi}{2}$ — both non-zero, so these are genuine vertical tangents.
5
Locate them
At $\theta=0$: the point $(2,0)$. At $\theta=\dfrac{\pi}{2}$: $r=0$, the pole. The curve is the circle of radius $1$ centred at $(1,0)$.
EX 4
Reaching the pole
$r=0$Find every $\theta$ in $[0,2\pi)$ for which $r=1+2\cos\theta$ passes through the pole.
1
Set $r=0$
$1+2\cos\theta=0$, so $\cos\theta=-\dfrac12$.
2
Solve on the interval
$\theta=\dfrac{2\pi}{3}$ and $\theta=\dfrac{4\pi}{3}$.
3
Interpret
The curve passes through the origin at these two angles — the boundaries of its inner loop.
4
Note the tangent there
At a point where $r=0$ and $\dfrac{dr}{d\theta}\neq0$, the line $\theta=$ that angle is tangent to the curve.
5
Explain why this matters
Pole crossings are exactly the limits you need when integrating a single loop — which is why they are worth finding first.
Boundary cases
$\dfrac{dr}{d\theta}$ is not a slope
It measures how fast the curve leaves the pole. The slope in the plane requires converting to $x$ and $y$ first.
Both derivatives need the product rule
$x=r(\theta)\cos\theta$ is a product of two functions of $\theta$. Differentiating only one factor is the most common error here.
Negative $r$ plots backwards
$\left(-2,\tfrac{\pi}{3}\right)$ lands where $\left(2,\tfrac{4\pi}{3}\right)$ does. Curves like $r=1+2\cos\theta$ trace an inner loop for exactly this reason.
Polar names are not unique
Adding $2\pi$ to $\theta$, or negating $r$ and adding $\pi$, gives the same point. Two polar curves can intersect at a point without ever sharing a $\theta$.