1. Parametric curves and their derivatives
The equation $y = f(x)$ forces a single $y$ value for each $x$. That rules out circles, figure-eights, any path that loops back on itself, and any motion where the natural variable is time rather than horizontal position. The fix is to describe both coordinates as functions of a third variable, the parameter:
$$ x = x(t), \qquad y = y(t), \qquad t \in [a, b]. $$As $t$ sweeps from $a$ to $b$, the point $(x(t), y(t))$ traces a curve. The parameter is usually pictured as time, and the curve is the path of a particle.
If $x(t)$ and $y(t)$ are differentiable and $dx/dt \neq 0$, then
$$ \frac{dy}{dx} = \frac{dy/dt}{dx/dt}. $$This is the slope of the tangent line at the point $(x(t), y(t))$.
The formula falls out of the chain rule. Treat $y$ as a function of $x$ and $x$ as a function of $t$; then $dy/dt = (dy/dx)(dx/dt)$, and dividing both sides by $dx/dt$ gives the result. The same idea, used a second time, yields the second derivative:
$$ \frac{d^2 y}{dx^2} = \frac{d}{dx}\!\left(\frac{dy}{dx}\right) = \frac{\dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right)}{\dfrac{dx}{dt}}. $$Notice that you must differentiate the quotient $dy/dx$ with respect to $t$ and then divide by $dx/dt$ again — not just take $d^2y/dt^2$ over $d^2x/dt^2$, which is wrong.
Two big payoffs. First, every curve in the plane can be parameterized — even ones that fail the vertical line test. Second, parametric form carries orientation: the parameter tells you the direction of travel along the curve, which matters for physics (velocity vs. just position) and for signed area.
2. Arc length of a parametric curve
How long is a curve? The trick is to chop it into tiny line segments, sum their lengths, and let the chopping get infinitely fine. Between $t$ and $t + dt$, the point moves by $(dx, dy)$ where $dx = (dx/dt)\,dt$ and $dy = (dy/dt)\,dt$. By the Pythagorean theorem, the length of that segment is
$$ ds = \sqrt{(dx)^2 + (dy)^2} = \sqrt{\!\left(\tfrac{dx}{dt}\right)^2 + \!\left(\tfrac{dy}{dt}\right)^2}\,dt. $$Integrating from $t = a$ to $t = b$ gives the arc length:
$$ L = \int_a^b \sqrt{\!\left(\tfrac{dx}{dt}\right)^2 + \!\left(\tfrac{dy}{dt}\right)^2}\,dt. $$The integrand $\sqrt{(dx/dt)^2 + (dy/dt)^2}$ is the speed of the parametrization — the magnitude of the velocity vector. Arc length is just the total distance a particle traveling along the curve covers.
If the parametrization traces the curve more than once over $[a, b]$, the integral counts those traversals. Arc length depends on the parametrization range, not the curve alone — pick $a$ and $b$ so the curve is traced exactly once.
3. Area under a parametric curve
For a function $y = f(x)$ between $x = a$ and $x = b$, the area under the graph is $\int_a^b f(x)\,dx$. For a parametric curve, $x$ is itself a function of $t$, so the substitution $dx = (dx/dt)\,dt$ gives
$$ A = \int_a^b y \,dx = \int_{t_1}^{t_2} y(t)\,\frac{dx}{dt}\,dt, $$where $t = t_1$ corresponds to $x = a$ and $t = t_2$ to $x = b$. As with any integral computed via substitution, the bounds change when the variable does.
The integral is signed in two ways. First, $y(t)$ can be negative — then the contribution is below the $x$-axis. Second, $dx/dt$ can be negative — if the parametrization sweeps right-to-left, you collect negative width. For a closed curve traced counterclockwise, $\oint y\,dx$ yields the (negative of the) enclosed area; clockwise gives positive area. Reverse-engineering the right sign is the kind of detail that bites in practice.
The formula $\int y\,dx$ slices the region into vertical rectangles. For parametric curves, you're still slicing vertically — you just have to express both the height $y$ and the slice width $dx$ in terms of the same parameter $t$.
4. Polar coordinates and polar derivatives
In polar coordinates a point is described not by $(x, y)$ but by $(r, \theta)$: distance from the origin and angle from the positive $x$-axis. The conversion is
$$ x = r\cos\theta, \qquad y = r\sin\theta. $$A polar curve is a function $r = f(\theta)$ — for each angle, a radius. You can read off any such curve as a parametric curve in $\theta$ by substituting:
$$ x(\theta) = f(\theta)\cos\theta, \qquad y(\theta) = f(\theta)\sin\theta. $$Then the parametric machinery applies immediately. Differentiating using the product rule:
$$ \frac{dx}{d\theta} = f'(\theta)\cos\theta - f(\theta)\sin\theta, \qquad \frac{dy}{d\theta} = f'(\theta)\sin\theta + f(\theta)\cos\theta. $$So the slope in polar form is
$$ \frac{dy}{dx} = \frac{f'(\theta)\sin\theta + f(\theta)\cos\theta}{f'(\theta)\cos\theta - f(\theta)\sin\theta}. $$Don't memorize the closed form. Memorize the recipe: convert to parametric in $\theta$, then differentiate componentwise. The closed form will reassemble itself every time.
5. Polar area from the wedge approximation
Cartesian area is a sum of vertical rectangles. Polar area is a sum of thin wedges radiating from the origin — pie slices.
A wedge from angle $\theta$ to $\theta + d\theta$, with outer radius $r$, is a sector of a circle. The exact area of a circular sector with radius $r$ and angle $\Delta\theta$ is $\tfrac{1}{2} r^2 \Delta\theta$. Letting $\Delta\theta \to d\theta$ and summing all the wedges between $\theta = \alpha$ and $\theta = \beta$:
$$ A = \frac{1}{2}\int_\alpha^\beta r(\theta)^2\,d\theta. $$That factor of $\tfrac{1}{2}$ is the part everyone forgets. It is not optional. It comes from the geometry of the sector — the wedge is a triangle with base $r\,d\theta$ and height $r$, so its area is $\tfrac{1}{2}(r\,d\theta)(r)$.
For curves like the cardioid $r = 1 + \cos\theta$, sweeping $\theta$ from $0$ to $2\pi$ gives the area enclosed once. But for a four-petaled rose $r = \cos(2\theta)$, each petal is traced when $r \geq 0$ over an arc of length $\pi/2$, and the same petals are retraced over the next $2\pi$. Always check the range of $\theta$ over which $r \geq 0$ and the curve is traced exactly once.
6. Polar arc length
Substitute $x = r\cos\theta$, $y = r\sin\theta$ into the parametric arc-length formula. Compute $dx/d\theta$ and $dy/d\theta$:
$$ \frac{dx}{d\theta} = r'\cos\theta - r\sin\theta, \qquad \frac{dy}{d\theta} = r'\sin\theta + r\cos\theta. $$Square and add — the cross-terms cancel and the trig identities collapse cleanly:
$$ \!\left(\tfrac{dx}{d\theta}\right)^2 + \!\left(\tfrac{dy}{d\theta}\right)^2 = (r')^2 + r^2. $$So the polar arc length formula is
$$ L = \int_\alpha^\beta \sqrt{r^2 + \!\left(\tfrac{dr}{d\theta}\right)^2}\,d\theta. $$Compare with the polar area formula $\tfrac{1}{2}\int r^2\,d\theta$. Area cares only about $r$; arc length picks up an extra $(dr/d\theta)^2$ because the curve doesn't just sit at radius $r$ — it's also drifting in or out, and that drift contributes to length.
7. The classic curves: cardioid, rose, lemniscate
Three named polar curves show up everywhere, and their calculus reveals their shape.
| Curve | Equation | Period | Key feature |
|---|---|---|---|
| Cardioid | $r = 1 + \cos\theta$ | $\theta \in [0, 2\pi]$ | Heart shape; $r = 0$ only at $\theta = \pi$ (the cusp). |
| Rose (4-petal) | $r = \cos(2\theta)$ | $\theta \in [0, 2\pi]$ | Four petals; $r = 0$ at $\theta = \pi/4, 3\pi/4, \ldots$ (between petals). |
| Lemniscate | $r^2 = \cos(2\theta)$ | $\theta \in [-\pi/4, \pi/4] \cup [3\pi/4, 5\pi/4]$ | Figure-eight; only exists where $\cos(2\theta) \geq 0$. |
For each, the calculus answers slightly different questions. The cardioid: what's its total area, and how long is the heart's outline? The rose: what's the area of one petal, and where do tangent lines go vertical or horizontal? The lemniscate: when is $r$ even defined?
The number of petals on the rose $r = \cos(n\theta)$ depends on the parity of $n$ in an annoying way: even $n$ gives $2n$ petals over $[0, 2\pi]$; odd $n$ gives $n$ petals. The reason is that for odd $n$, $r$ goes negative for half the period and re-traces the existing petals from the other side.
8. Common pitfalls
The integrand is $\tfrac{1}{2}r^2$, not $r^2$. The factor of one-half comes from the area of a sector and is non-negotiable. Forgetting it doubles every answer.
Naively integrating from $0$ to $2\pi$ for a multi-petal rose double-, triple-, or quadruple-counts the petals. For each curve, find the smallest interval over which it is traced exactly once, and use that as the bounds (or compute one petal and multiply).
For a parametric curve, the second derivative is not $\dfrac{d^2y/dt^2}{d^2x/dt^2}$. It's $\dfrac{d/dt(dy/dx)}{dx/dt}$. The shortcut is wrong; the chain rule is non-negotiable.
In polar form, $r$ is sometimes allowed to be negative, and conventions vary. Always check whether the curve $r = f(\theta)$ is being interpreted as "skip the region where $r < 0$" or "plot the point on the opposite ray." It changes which curve you have.
The integral $\int y\,dx$ comes out negative for counterclockwise-traced closed curves and positive for clockwise. If you want a positive enclosed area, take the absolute value (or use Green's theorem, which makes the sign explicit).
9. Worked examples
Example 1 · Arc length of one arch of the cycloid
The cycloid (the path traced by a point on the rim of a rolling wheel of radius $1$) is parameterized by
$$ x(t) = t - \sin t, \qquad y(t) = 1 - \cos t. $$One arch corresponds to $t \in [0, 2\pi]$.
Step 1. Derivatives:
$$ \frac{dx}{dt} = 1 - \cos t, \qquad \frac{dy}{dt} = \sin t. $$Step 2. Speed squared:
$$ (1 - \cos t)^2 + \sin^2 t = 1 - 2\cos t + \cos^2 t + \sin^2 t = 2 - 2\cos t = 2(1 - \cos t). $$Step 3. Use the identity $1 - \cos t = 2\sin^2(t/2)$:
$$ \sqrt{2(1 - \cos t)} = \sqrt{4\sin^2(t/2)} = 2\left|\sin(t/2)\right|. $$On $[0, 2\pi]$, $\sin(t/2) \geq 0$, so the absolute value drops.
Step 4. Integrate:
$$ L = \int_0^{2\pi} 2\sin(t/2)\,dt = \left[-4\cos(t/2)\right]_0^{2\pi} = -4(-1) - (-4)(1) = 8. $$The arch is $8$ units long — exactly four diameters of the rolling wheel, which is the kind of clean answer that makes calculus feel like a magic trick.
Example 2 · Area inside the cardioid $r = 1 + \cos\theta$
The cardioid is traced exactly once as $\theta$ ranges over $[0, 2\pi]$.
Step 1. Apply the polar area formula:
$$ A = \frac{1}{2}\int_0^{2\pi} (1 + \cos\theta)^2\,d\theta. $$Step 2. Expand:
$$ (1 + \cos\theta)^2 = 1 + 2\cos\theta + \cos^2\theta. $$Step 3. Use $\cos^2\theta = \tfrac{1 + \cos(2\theta)}{2}$:
$$ 1 + 2\cos\theta + \tfrac{1}{2} + \tfrac{1}{2}\cos(2\theta) = \tfrac{3}{2} + 2\cos\theta + \tfrac{1}{2}\cos(2\theta). $$Step 4. Integrate term by term over $[0, 2\pi]$. Both $\cos\theta$ and $\cos(2\theta)$ integrate to $0$ over a full period, leaving
$$ A = \tfrac{1}{2} \cdot \tfrac{3}{2} \cdot 2\pi = \frac{3\pi}{2}. $$Example 3 · Intersection points of $r = 1$ and $r = 2\sin\theta$
The first is a unit circle centered at the origin. The second is a circle of radius $1$ centered at $(0, 1)$ — that's a fact worth remembering: $r = 2a\sin\theta$ is a circle of radius $a$ centered at $(0, a)$.
Step 1. Set the radii equal:
$$ 1 = 2\sin\theta \quad\Longrightarrow\quad \sin\theta = \tfrac{1}{2}, $$which gives $\theta = \pi/6$ and $\theta = 5\pi/6$.
Step 2. Translate to Cartesian coordinates using $x = r\cos\theta$ and $y = r\sin\theta$:
- At $\theta = \pi/6$: $(x, y) = (\cos(\pi/6), \sin(\pi/6)) = (\tfrac{\sqrt{3}}{2}, \tfrac{1}{2})$.
- At $\theta = 5\pi/6$: $(x, y) = (-\tfrac{\sqrt{3}}{2}, \tfrac{1}{2})$.
Step 3. Check for a third intersection at the origin (a common surprise — polar intersections can occur at the origin even when there's no shared $\theta$). The first curve reaches the origin... never; $r = 1$ means $r$ is always $1$. So the origin isn't on the first circle and is not an intersection.
The two curves meet at exactly the two points above.
Example 4 · Slope at a point on a parametric curve
Find the slope of the tangent to $x(t) = t^2$, $y(t) = t^3 - 3t$ at $t = 2$.
Step 1. Differentiate componentwise:
$$ \frac{dx}{dt} = 2t, \qquad \frac{dy}{dt} = 3t^2 - 3. $$Step 2. Form the parametric derivative:
$$ \frac{dy}{dx} = \frac{3t^2 - 3}{2t}. $$Step 3. Evaluate at $t = 2$:
$$ \frac{dy}{dx}\bigg|_{t=2} = \frac{3(4) - 3}{2(2)} = \frac{9}{4}. $$Sanity check: $dx/dt = 4 \neq 0$ at $t = 2$, so the formula is valid (no vertical tangent).
Example 5 · Area of one petal of the rose $r = \cos(2\theta)$
One petal of $r = \cos(2\theta)$ is traced as $\theta$ ranges over the interval where $r \geq 0$ centered on $\theta = 0$ — that is, from $-\pi/4$ to $\pi/4$.
Step 1. Set up:
$$ A = \frac{1}{2}\int_{-\pi/4}^{\pi/4} \cos^2(2\theta)\,d\theta. $$Step 2. Use $\cos^2 u = \tfrac{1 + \cos(2u)}{2}$ with $u = 2\theta$:
$$ \cos^2(2\theta) = \tfrac{1 + \cos(4\theta)}{2}. $$Step 3. Integrate:
$$ A = \tfrac{1}{2}\int_{-\pi/4}^{\pi/4}\!\!\left[\tfrac{1}{2} + \tfrac{1}{2}\cos(4\theta)\right]d\theta = \tfrac{1}{4}\!\left[\theta + \tfrac{1}{4}\sin(4\theta)\right]_{-\pi/4}^{\pi/4}. $$$\sin(4 \cdot \pi/4) = \sin(\pi) = 0$ and same at the lower bound, so the sine term vanishes:
$$ A = \tfrac{1}{4}\!\left[\tfrac{\pi}{4} - \!\left(-\tfrac{\pi}{4}\right)\right] = \tfrac{1}{4} \cdot \tfrac{\pi}{2} = \frac{\pi}{8}. $$The total area of all four petals is $4 \cdot \tfrac{\pi}{8} = \tfrac{\pi}{2}$.