# Parametric Equations, Polar Coordinates, and Vector-Valued Functions

Up to now, curves have mostly been described as $y = f(x)$ — one output for each input. That description breaks down for a fly looping around a room, a planet on an elliptical orbit, or a projectile drifting sideways in the wind: these paths can double back on themselves, cross their own tracks, or simply depend on time in a way a single equation $y = f(x)$ cannot capture. This unit develops three connected tools that handle these situations — parametric equations, vector-valued functions, and polar coordinates — and shows how the calculus you already know (derivatives, integrals, arc length) carries over into each one.

# Parametric Equations and Calculus

# Defining a curve parametrically

A parametric curve is given by a pair of functions of a third variable, usually $t$:

$$x = x(t), \qquad y = y(t)$$

As $t$ increases, the point $(x(t), y(t))$ traces a path. Because $x$ and $y$ are each free to do their own thing, parametric curves can loop, cross themselves, and move backward — none of which a single $y = f(x)$ can do.

title: A parametric curve — x = t^2, y = t^3 - 3t
xlabel: x
ylabel: y
bounds: -1, 8, -5, 5
x=t^2
y=t^3-3t

• eliminating the parameter (solving one equation for $t$ and substituting into the other) converts a parametric pair into a single rectangular equation $y = f(x)$, when that's possible.
• not every parametric curve can be written this way — many are only convenient in parametric form.

# The chain rule for parametrics

The slope of a parametric curve at a given $t$ comes from the chain rule, treating $x$ and $y$ as both depending on $t$:

$$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}, \quad \text{provided } \frac{dx}{dt} \neq 0$$

example. Find the slope of $x = t^2$, $y = t^3 - 3t$ at $t = 2$.

$\dfrac{dx}{dt} = 2t$, $\dfrac{dy}{dt} = 3t^2 - 3$. At $t = 2$: $\dfrac{dx}{dt} = 4$, $\dfrac{dy}{dt} = 9$.

$$\frac{dy}{dx} = \frac{9}{4}$$

example. Find where $x = \cos t$, $y = \sin(2t)$ has horizontal tangents, for $0 \le t < 2\pi$.

Horizontal tangents need $dy/dt = 0$ with $dx/dt \neq 0$.

$$\frac{dy}{dt} = 2\cos(2t) = 0 \implies t = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$$

At each of these, $dx/dt = -\sin t \neq 0$, so all four are valid. The points are $\left(\tfrac{\sqrt2}{2}, 1\right)$, $\left(-\tfrac{\sqrt2}{2}, -1\right)$, $\left(-\tfrac{\sqrt2}{2}, 1\right)$, and $\left(\tfrac{\sqrt2}{2}, -1\right)$.

# Second derivatives and concavity

To find concavity, you need $\dfrac{d^2y}{dx^2}$ — but you're still differentiating with respect to $t$, so a second division by $dx/dt$ is required:

$$\frac{d^2y}{dx^2} = \frac{\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)}{\dfrac{dx}{dt}}$$

crucial note. the most common error here is forgetting the second division by $dx/dt$ — treating $\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)$ itself as the answer.

example. Find $\dfrac{d^2y}{dx^2}$ for $x = e^t$, $y = te^{-t}$.

$\dfrac{dx}{dt} = e^t$, $\dfrac{dy}{dt} = e^{-t}(1-t)$, so $\dfrac{dy}{dx} = e^{-2t}(1-t)$.

Differentiating that with respect to $t$:

$$\frac{d}{dt}\left[e^{-2t}(1-t)\right] = -2e^{-2t}(1-t) - e^{-2t} = e^{-2t}(2t-3)$$

Dividing by $dx/dt = e^t$:

$$\frac{d^2y}{dx^2} = \frac{e^{-2t}(2t-3)}{e^t} = e^{-3t}(2t-3)$$

example. Find $t$ where $x = t^3 - 3t$, $y = t^2$ has a vertical tangent.

Vertical tangents need $dx/dt = 0$ with $dy/dt \neq 0$: $3t^2 - 3 = 0 \implies t = \pm 1$. At both values $dy/dt = 2t \neq 0$, so both give vertical tangents.

# Arc length of a parametric curve

If a curve is traced exactly once as $t$ runs from $a$ to $b$:

$$L = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}, dt$$

example. Find the length of $x = \sin(3t)$, $y = \cos(3t)$ from $t = 0$ to $t = \pi$.

$x' = 3\cos(3t)$, $y' = -3\sin(3t)$, so $(x')^2 + (y')^2 = 9$.

$$L = \int_0^\pi \sqrt{9}, dt = \int_0^\pi 3, dt = 3\pi$$

# Vector-Valued Functions and Motion in the Plane

# Position, velocity, and acceleration

A vector-valued function $\mathbf{r}(t) = \langle x(t), y(t) \rangle$ describes a particle's position at time $t$. Differentiating component-by-component gives velocity and acceleration:

$$\mathbf{v}(t) = \mathbf{r}'(t) = \langle x'(t), y'(t) \rangle \qquad \mathbf{a}(t) = \mathbf{v}'(t) = \langle x''(t), y''(t) \rangle$$

The velocity vector is always tangent to the path — it points in the direction of motion at that instant.

example. For $\mathbf{r}(t) = \langle t^3, t^2 \rangle$, find $\mathbf{v}(1)$ and $\mathbf{a}(1)$.

$\mathbf{v}(t) = \langle 3t^2, 2t \rangle \implies \mathbf{v}(1) = \langle 3, 2 \rangle$.

$\mathbf{a}(t) = \langle 6t, 2 \rangle \implies \mathbf{a}(1) = \langle 6, 2 \rangle$.

example. A particle moves along $\mathbf{r}(t) = \langle \cos(t^2), \sin(t^2) \rangle$. Does its acceleration always point toward the origin?

$$\mathbf{v}(t) = \langle -2t\sin(t^2),\ 2t\cos(t^2) \rangle$$

$$\mathbf{a}(t) = \langle -2\sin(t^2) - 4t^2\cos(t^2),\ 2\cos(t^2) - 4t^2\sin(t^2) \rangle$$

Since $\mathbf{a}(t)$ isn't a pure scalar multiple of $\mathbf{r}(t)$ (the extra $-2\sin(t^2)$ and $2\cos(t^2)$ terms don't cancel), the acceleration is not purely centripetal — it doesn't point straight at the origin at every instant, even though the path is a circle.

# Speed and distance traveled

Speed is a scalar — the magnitude of velocity:

$$\text{speed} = |\mathbf{v}(t)| = \sqrt{(x'(t))^2 + (y'(t))^2}$$

Total distance traveled over $[a,b]$ is the integral of speed (identical in form to parametric arc length):

$$\text{distance} = \int_a^b \sqrt{(x'(t))^2 + (y'(t))^2}, dt$$

example. Find the speed at $t=2$ for $\mathbf{r}(t) = \langle \ln t, t^2 \rangle$.

$\mathbf{v}(t) = \langle 1/t, 2t \rangle \implies \mathbf{v}(2) = \langle 1/2, 4 \rangle$.

$$\text{speed} = \sqrt{(1/2)^2 + 4^2} = \sqrt{16.25} = \frac{\sqrt{65}}{2}$$

example (calculator active). $x'(t) = \sqrt{1+t^2}$, $y'(t) = \sin(t^2)$. Find the distance traveled from $t=0$ to $t=2$.

$$\text{distance} = \int_0^2 \sqrt{1+t^2+\sin^2(t^2)}, dt \approx 2.731$$

example. A particle has $\mathbf{r}(t) = \langle R\cos(\omega t), R\sin(\omega t) \rangle$ — uniform circular motion of radius $R$ and angular speed $\omega$. Find its speed.

$$\mathbf{v}(t) = \langle -R\omega\sin(\omega t),\ R\omega\cos(\omega t) \rangle$$

$$\text{speed} = \sqrt{R^2\omega^2\sin^2(\omega t) + R^2\omega^2\cos^2(\omega t)} = |R\omega|$$

the speed is constant — a hallmark of uniform circular motion, even though velocity's direction is always changing.

# Recovering position from velocity or acceleration

Since $\mathbf{r}(t) = \mathbf{r}(0) + \displaystyle\int_0^t \mathbf{v}(u), du$, integrating each component (and using an initial condition to pin down constants) recovers position from velocity, or velocity from acceleration.

example. $\mathbf{v}(t) = \langle 2t, 3 \rangle$, $\mathbf{r}(0) = \langle 1, 5 \rangle$. Find $\mathbf{r}(2)$.

$$x(2) = 1 + \int_0^2 2t, dt = 1 + 4 = 5 \qquad y(2) = 5 + \int_0^2 3, dt = 5 + 6 = 11$$

$$\mathbf{r}(2) = \langle 5, 11 \rangle$$

example. $\mathbf{a}(t) = \langle \cos t, \sin t \rangle$, $\mathbf{v}(0) = \langle 0, 1 \rangle$. Find $\mathbf{v}(t)$.

$$\mathbf{v}(t) = \langle \sin t + C_1,\ -\cos t + C_2 \rangle$$

At $t=0$: $0 + C_1 = 0 \implies C_1 = 0$, and $-1 + C_2 = 1 \implies C_2 = 2$. So:

$$\mathbf{v}(t) = \langle \sin t,\ 2 - \cos t \rangle$$

example. Given $\mathbf{v}(t) = \langle \sin t, \cos t \rangle$, find the displacement from $t=0$ to $t = \pi$.

$$\Delta x = \int_0^\pi \sin t, dt = 2 \qquad \Delta y = \int_0^\pi \cos t, dt = 0$$

displacement $= \langle 2, 0 \rangle$ — note this only captures net change in position, not total distance traveled, which would require integrating speed instead.

# Polar Coordinates

# Defining and converting polar coordinates

A polar point $(r, \theta)$ is located by a directed distance $r$ from the origin (the pole), measured along a ray at angle $\theta$ from the positive $x$-axis (the polar axis). Unlike rectangular coordinates, $r$ can be negative — if $r < 0$, the point sits $|r|$ units out along the direction $\theta + \pi$ instead.

$$x = r\cos\theta, \qquad y = r\sin\theta$$

$$r^2 = x^2 + y^2, \qquad \tan\theta = \frac{y}{x}$$

example. Convert $(x,y) = (-1, 1)$ to polar, with $r > 0$, $0 \le \theta < 2\pi$.

$r^2 = 1 + 1 = 2 \implies r = \sqrt2$. Since $\tan\theta = -1$ and the point is in Quadrant II, $\theta = 3\pi/4$. Polar: $(\sqrt2,\ 3\pi/4)$.

example. Convert $r = 4\sin\theta$ to rectangular form.

Multiply both sides by $r$: $r^2 = 4r\sin\theta$. Substituting $r^2 = x^2+y^2$ and $y = r\sin\theta$:

$$x^2 + y^2 = 4y \implies x^2 + (y-2)^2 = 4$$

a circle of radius $2$ centered at $(0,2)$.

example. Find the polar equation of the vertical line $x = 3$.

$r\cos\theta = 3 \implies r = 3\sec\theta$.

# Slope of a polar curve

Given $r = f(\theta)$, write $x = f(\theta)\cos\theta$ and $y = f(\theta)\sin\theta$, then apply the parametric slope formula with $\theta$ playing the role of $t$:

$$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{f'(\theta)\sin\theta + f(\theta)\cos\theta}{f'(\theta)\cos\theta - f(\theta)\sin\theta}$$

example. Find the slope of $r = 2$ (a circle) at $\theta = \pi/4$.

$x = 2\cos\theta \implies dx/d\theta = -2\sin\theta$; $y = 2\sin\theta \implies dy/d\theta = 2\cos\theta$.

$$\frac{dy}{dx} = \frac{2\cos(\pi/4)}{-2\sin(\pi/4)} = \frac{\sqrt2}{-\sqrt2} = -1$$

this makes sense geometrically — the tangent to a circle is always perpendicular to the radius, and the radius at $\theta = \pi/4$ has slope $1$.

example. Find the slope of $r = 1 + \cos\theta$ (a cardioid) as it passes through the pole.

$r = 0 \implies 1+\cos\theta = 0 \implies \theta = \pi$. At the pole, the tangent line's slope is simply $\tan\theta$ evaluated where $r=0$:

$$\text{slope} = \tan(\pi) = 0$$

# Area in polar coordinates

The area of a thin circular sector of angle $d\theta$ is $\tfrac12 r^2, d\theta$. Summing these over an interval gives the area swept by a polar curve:

$$A = \frac{1}{2}\int_\alpha^\beta [r(\theta)]^2, d\theta$$

title: A four-petal rose, r = cos(2 theta)
bounds: -1.2, 1.2, -1.2, 1.2
r=\cos(2\theta)

example. Find the area of $r = 3$ (a full circle) and confirm it matches $\pi r^2$.

$$A = \frac12\int_0^{2\pi} 9, d\theta = \frac{18\pi}{2} = 9\pi \quad \checkmark$$

example. Find the area of one petal of the rose $r = \cos(2\theta)$.

One petal is traced as $r$ goes from $0$ up to a peak and back to $0$; solving $\cos(2\theta) = 0$ nearest the origin gives $\theta = -\pi/4, \pi/4$.

$$A = \frac12\int_{-\pi/4}^{\pi/4}\cos^2(2\theta), d\theta = \frac14\int_{-\pi/4}^{\pi/4}\left(1+\cos(4\theta)\right)d\theta = \frac14\left[\theta + \frac{\sin(4\theta)}{4}\right]_{-\pi/4}^{\pi/4} = \frac{\pi}{8}$$

title: A cardioid, r = 1 + sin(theta)
bounds: -2.2, 2.2, -0.5, 2.5
r=1+\sin(\theta)

example. Find the total area enclosed by the cardioid $r = 1 + \sin\theta$.

The whole cardioid is traced for $0 \le \theta \le 2\pi$.

$$A = \frac12\int_0^{2\pi}(1+\sin\theta)^2,d\theta = \frac12\int_0^{2\pi}\left(1+2\sin\theta + \frac{1-\cos2\theta}{2}\right)d\theta = \frac{3\pi}{2}$$

example. Find the area inside $r = 3\sin\theta$ and outside $r = 1+\sin\theta$.

Setting the curves equal: $3\sin\theta = 1+\sin\theta \implies \sin\theta = \tfrac12 \implies \theta = \pi/6,\ 5\pi/6$.

$$A = \frac12\int_{\pi/6}^{5\pi/6}\left[(3\sin\theta)^2 - (1+\sin\theta)^2\right]d\theta = \frac12\int_{\pi/6}^{5\pi/6}(8\sin^2\theta - 2\sin\theta - 1), d\theta = \pi$$

title: An inner loop — r = 1 - 2 sin(theta)
bounds: -2, 2, -3.2, 1.2
r=1-2\sin(\theta)

example. Find the area of the inner loop of the limaçon $r = 1 - 2\sin\theta$.

An inner loop appears where $r < 0$. Solving $1-2\sin\theta = 0$ gives $\theta = \pi/6,\ 5\pi/6$, and $r$ is negative between them, tracing the inner loop.

$$A = \frac12\int_{\pi/6}^{5\pi/6}(1-2\sin\theta)^2, d\theta = \frac12\left[3\theta + 4\cos\theta - \sin(2\theta)\right]_{\pi/6}^{5\pi/6} = \pi - \frac{3\sqrt3}{2}$$

# Arc length in polar coordinates

Treating $\theta$ itself as the parameter in the parametric arc length formula gives:

$$L = \int_\alpha^\beta \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2}, d\theta$$

example. Find the length of $r = e^\theta$ from $\theta=0$ to $\theta=1$.

$r' = e^\theta$, so:

$$L = \int_0^1 \sqrt{e^{2\theta}+e^{2\theta}}, d\theta = \int_0^1 \sqrt2, e^\theta, d\theta = \sqrt2(e-1)$$

example. Find the length of the spiral $r = \theta^2$ for $0 \le \theta \le \sqrt5$.

$r' = 2\theta$, so $r^2 + (r')^2 = \theta^4 + 4\theta^2$, and:

$$L = \int_0^{\sqrt5} \theta\sqrt{\theta^2+4}, d\theta$$

Substituting $u = \theta^2+4$, $du = 2\theta, d\theta$:

$$L = \frac12\int_4^9 \sqrt{u}, du = \frac13\left[u^{3/2}\right]_4^9 = \frac13(27-8) = \frac{19}{3}$$

# Symmetry and intersections

Testing whether a polar equation is unchanged under $\theta \to -\theta$, $\theta \to \pi - \theta$, or $r \to -r$ reveals symmetry about the polar axis, the line $\theta = \pi/2$, or the pole, respectively — often shortening area and arc length computations by letting you integrate over half the interval and double the result.

example. Show $r = 2\cos(3\theta)$ is symmetric about the polar axis.

Replacing $\theta$ with $-\theta$: $r = 2\cos(-3\theta) = 2\cos(3\theta)$, unchanged — so the curve is symmetric about the $x$-axis.

example. Where do $r = 2\sin\theta$ and $r = 2\cos\theta$ intersect, other than at the pole?

$$2\sin\theta = 2\cos\theta \implies \tan\theta = 1 \implies \theta = \pi/4$$

$r = 2\sin(\pi/4) = \sqrt2$, giving the point $(\sqrt2,\ \pi/4)$.

# Exercises

1. Find the slope of $x = t^2 - 1$, $y = t^3 + t$ at $t = 1$.
2. Find $\dfrac{d^2y}{dx^2}$ for $x = t^2$, $y = t^4$.
3. Find the arc length of $x = 2\cos t$, $y = 2\sin t$ from $t=0$ to $t=\pi/2$.
4. A particle has $\mathbf{r}(t) = \langle t^2 - 4, t^3 \rangle$. Find $\mathbf{v}(t)$ and $\mathbf{a}(t)$.
5. Find the speed at $t=1$ for $\mathbf{r}(t) = \langle e^t, e^{-t} \rangle$.
6. If $\mathbf{v}(t) = \langle 4t, 6t^2 \rangle$ and $\mathbf{r}(0) = \langle 0, 1 \rangle$, find $\mathbf{r}(1)$.
7. Convert $(x, y) = (2, -2)$ to polar with $r > 0$, $0 \le \theta < 2\pi$.
8. Convert $r = 6\cos\theta$ to rectangular form and identify the curve.
9. Find the area enclosed by $r = 2 + 2\cos\theta$.
10. Find the area of one petal of $r = \sin(3\theta)$.
11. Find the length of $r = 3\theta$ from $\theta = 0$ to $\theta = 2$.
12. Show that $r = 3 + 3\sin\theta$ is symmetric about the line $\theta = \pi/2$.