Math Core

A parametric curve gives you two functions, $x(t)$ and $y(t)$ . Bundling them into a single object, a vector that changes with $t$ , turns out to be the cleanest language for motion: position, velocity and acceleration all become vectors, and calculus works on them one component at a time.

Vectors that depend on t

Definition

Vector-valued function

A vector-valued function assigns a vector to each value of $t$ :

\vec{r}(t) = \langle x(t),\, y(t) \rangle.

The functions $x(t)$ and $y(t)$ are its components. When the vector is drawn from the origin, its tip is at the point $(x(t), y(t))$ , so as $t$ changes the tip traces the same curve as the parametric equations $x = x(t)$ , $y = y(t)$ .

You may also see this written as $\vec{r}(t) = x(t)\,\mathbf{i} + y(t)\,\mathbf{j}$ , where $\mathbf{i} = \langle 1, 0 \rangle$ and $\mathbf{j} = \langle 0, 1 \rangle$ . Both notations mean the same thing, and AP exams use both.

For example, $\vec{r}(t) = \langle \cos t, \sin t \rangle$ is a position vector whose tip goes counterclockwise around the unit circle, and $\vec{r}(t) = \langle t, t^2 \rangle$ traces the parabola $y = x^2$ .

Derivatives, one component at a time

The derivative of a vector-valued function is defined by the same limit as always:

\vec{r}\,'(t) = \lim_{h \to 0} \frac{\vec{r}(t + h) - \vec{r}(t)}{h}.

Subtracting vectors and dividing by $h$ both happen component by component, so the limit does too.

Calculus on vectors is componentwise

If $\vec{r}(t) = \langle x(t), y(t) \rangle$ , then

\vec{r}\,'(t) = \langle x'(t),\, y'(t) \rangle, \qquad \vec{r}\,''(t) = \langle x''(t),\, y''(t) \rangle,

\int \vec{r}(t)\, dt = \left\langle \int x(t)\, dt,\; \int y(t)\, dt \right\rangle.

The derivative $\vec{r}\,'(t)$ points in the direction the tip is moving, so it is tangent to the curve. Its slope is $\dfrac{y'(t)}{x'(t)}$ , which is exactly the $\dfrac{dy}{dx}$ you computed for parametric curves.

Worked example: Derivative at a point

Let $\vec{r}(t) = \langle t^2,\, 3t - 1 \rangle$ . Find $\vec{r}\,'(2)$ and the slope of the curve at $t = 2$ .

Differentiate each component: $\vec{r}\,'(t) = \langle 2t,\, 3 \rangle$ , so $\vec{r}\,'(2) = \langle 4, 3 \rangle$ .

The slope of the curve is $\dfrac{3}{4}$ . The tangent vector $\langle 4, 3 \rangle$ says "4 right, 3 up," which is a slope of $\dfrac{3}{4}$ .

For a picture, take $\vec{r}(t) = \langle 2\cos t,\, 2\sin t \rangle$ , whose tip runs counterclockwise around a circle of radius 2. Its derivative is $\vec{r}\,'(t) = \langle -2\sin t,\, 2\cos t \rangle$ . At $t = \dfrac{\pi}{4}$ the tip is at $(\sqrt{2}, \sqrt{2})$ and $\vec{r}\,'\left(\dfrac{\pi}{4}\right) = \langle -\sqrt{2}, \sqrt{2} \rangle$ : it points up and to the left, along the tangent line, which is the direction of travel. Notice that the derivative vector is perpendicular to the position vector here. That's special to circles centered at the origin, but it's a good sanity check that "derivative = tangent direction."

The circle traced by r(t) = ⟨2 cos t, 2 sin t⟩. At t = π/4 the derivative ⟨−√2, √2⟩ points along the dashed tangent line, up and to the left.Open in grapher →

The length of $\vec{r}\,'(t)$ matters too. For this circle, $|\vec{r}\,'(t)| = \sqrt{4\sin^2 t + 4\cos^2 t} = 2$ for every $t$ , so the tip moves at a constant rate of 2 units per unit of $t$ . In the next lesson, that length becomes the particle's speed.

Every derivative rule you know applies inside each component: chain rule, product rule, and so on.

Worked example: Chain rule inside the components

Let $\vec{r}(t) = \langle e^{2t},\, \sin(\pi t) \rangle$ . Find $\vec{r}\,'(0)$ and $\vec{r}\,''(t)$ .

$\vec{r}\,'(t) = \langle 2e^{2t},\, \pi\cos(\pi t) \rangle$ , so $\vec{r}\,'(0) = \langle 2, \pi \rangle$ .

Differentiating again: $\vec{r}\,''(t) = \langle 4e^{2t},\, -\pi^2\sin(\pi t) \rangle$ .

Integrals and initial conditions

Integrating a vector-valued function also works componentwise. An indefinite integral picks up a constant vector $\vec{C} = \langle C_1, C_2 \rangle$ , one constant per component. An initial condition like $\vec{r}(0) = \langle 1, 3 \rangle$ gives you both constants at once.

Worked example: Recovering a function from its derivative

Suppose $\vec{r}\,'(t) = \langle 6t,\, 2\cos t \rangle$ and $\vec{r}(0) = \langle 1, 3 \rangle$ . Find $\vec{r}(t)$ .

Integrate each component:

\vec{r}(t) = \langle 3t^2 + C_1,\; 2\sin t + C_2 \rangle.

At $t = 0$ : $\langle C_1, C_2 \rangle = \langle 1, 3 \rangle$ . So

\vec{r}(t) = \langle 3t^2 + 1,\; 2\sin t + 3 \rangle.

For instance, $\vec{r}\left(\dfrac{\pi}{2}\right) = \left\langle \dfrac{3\pi^2}{4} + 1,\; 5 \right\rangle$ .

A definite integral of a vector function is a vector whose components are definite integrals:

\int_0^2 \langle 3t^2,\, 4t \rangle\, dt = \left\langle t^3\Big|_0^2,\; 2t^2\Big|_0^2 \right\rangle = \langle 8, 8 \rangle.

By the Fundamental Theorem, $\displaystyle\int_a^b \vec{r}\,'(t)\, dt = \vec{r}(b) - \vec{r}(a)$ . That is the change in position between $t = a$ and $t = b$ , which will be important in the next lesson.

Common mistake

Don't forget that each component has its own constant. Writing $\vec{r}(t) = \langle 3t^2, 2\sin t \rangle + C$ with a single number $C$ added to both components is wrong unless the initial condition happens to make both constants equal. Solve for $C_1$ and $C_2$ separately.

Entering vector answers

On this site, type a vector like $\langle 4, 3 \rangle$ as an ordered pair: (4, 3).

Practice

Practice 1

Let $\vec{r}(t) = \langle t^3,\, 2t^2 - t \rangle$ . Find $\vec{r}\,'(1)$ . Enter the vector as $(a, b)$ .

Enter a point like (2, -3)

Practice 2

Let $\vec{r}(t) = \langle \ln t,\, \sqrt{t} \rangle$ for $t > 0$ . Find $\vec{r}\,'(4)$ . Enter the vector as $(a, b)$ .

Enter a point like (2, -3)

Practice 3

If $\vec{r}(t) = \langle \sin t,\, t^2 \rangle$ , what is $\vec{r}\,''(t)$ ?

Practice 4

Evaluate $\displaystyle\int_0^1 \langle 4t^3,\, 6t^2 + 1 \rangle\, dt$ . Enter the vector as $(a, b)$ .

Enter a point like (2, -3)

Practice 5

The curve traced by $\vec{r}(t) = \langle t^2 + 1,\, t^3 - 4t \rangle$ passes through $(5, 0)$ at $t = 2$ . What is the slope of the curve there?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 6

Suppose $\vec{r}\,'(t) = \langle 2t,\, \pi\cos(\pi t) \rangle$ and $\vec{r}(0) = \langle 3, 1 \rangle$ . Find $\vec{r}(2)$ . Enter the vector as $(a, b)$ .

Enter a point like (2, -3)

Practice 7

Suppose $\vec{r}\,''(t) = \langle 6t,\, -2 \rangle$ , $\vec{r}\,'(0) = \langle 1, 0 \rangle$ and $\vec{r}(0) = \langle 0, 5 \rangle$ . Find $\vec{r}(1)$ . Enter the vector as $(a, b)$ .

Enter a point like (2, -3)