Vector-Valued Functions

A vector-valued function packs a pair of parametric equations into one object: a vector whose tip traces the curve as $t$ changes. The good news is that the calculus is no harder than what you already know. You differentiate and integrate one component at a time. This notation is the language of motion in the plane, which comes next in motion with vectors.

Key ideas

Vectors and vector-valued functions

A vector in the plane has two components, written with angle brackets: $\langle 3, -2 \rangle$ means “3 right, 2 down”. A vector-valued function assigns a vector to each $t$:

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

You’ll also see it written $\vec{r}(t) = x(t)\,\mathbf{i} + y(t)\,\mathbf{j}$, or with a bold $\mathbf{r}(t)$. They all mean the same thing. AP questions usually use the angle-bracket form, so we’ll use $\langle\ ,\ \rangle$ for components and $\vec{r}(t)$ for the function’s name.

Think of $\vec{r}(t)$ as an arrow from the origin to the point $(x(t), y(t))$. As $t$ changes, the arrow’s tip traces the same curve as the parametric equations $x = x(t)$, $y = y(t)$.

Derivatives, component by component

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

All the usual rules (power, product, chain, and so on) apply to each component separately.

What $\vec{r}\,'(t)$ means. Drawn starting at the point $(x(t), y(t))$, the vector $\vec{r}\,'(t)$ is tangent to the curve and points in the direction of increasing $t$. Its slope is $\dfrac{y'(t)}{x'(t)}$, which is the same $\dfrac{dy}{dx}$ you found for parametric curves.

For $\vec{r}(t) = \langle 3\cos t, 2\sin t \rangle$ (Example 2): the position vector $\vec{r}\left( \frac{\pi}{3} \right)$ and the tangent vector $\vec{r}\,'\left( \frac{\pi}{3} \right)$.

Integrals, component by component

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

An indefinite integral gets a constant in each component. You can write them as a constant vector: $+\ \vec{C}$, where $\vec{C} = \langle C_1, C_2 \rangle$.

A definite integral gives a vector of numbers:

$$\int_a^b \vec{r}\,'(t)\, dt = \vec{r}(b) - \vec{r}(a)$$

This is the Fundamental Theorem of Calculus, applied to each component. It’s the net change in $\vec{r}$.

Initial conditions

If you know $\vec{r}\,'(t)$ and one value such as $\vec{r}(0)$, you can find $\vec{r}(t)$: integrate each component, then use the initial condition to find each constant. Equivalently,

$$\vec{r}(t) = \vec{r}(0) + \int_0^t \vec{r}\,'(u)\, du$$

Trig functions on this page use radians, as all AP calculus does.

Worked examples

Example 1: Differentiating

Let $\vec{r}(t) = \langle t^3 - 2t,\ e^{2t} \rangle$. Find $\vec{r}\,'(t)$, $\vec{r}\,''(t)$ and $\vec{r}\,'(0)$.

Solution. Differentiate each component:

$$\vec{r}\,'(t) = \langle 3t^2 - 2,\ 2e^{2t} \rangle, \qquad \vec{r}\,''(t) = \langle 6t,\ 4e^{2t} \rangle$$ $$\vec{r}\,'(0) = \langle 0 - 2,\ 2e^0 \rangle = \langle -2, 2 \rangle$$

Example 2: The tangent vector on an ellipse

Let $\vec{r}(t) = \langle 3\cos t,\ 2\sin t \rangle$. Find $\vec{r}\left( \dfrac{\pi}{3} \right)$, $\vec{r}\,'\left( \dfrac{\pi}{3} \right)$, and the slope of the tangent line at $t = \dfrac{\pi}{3}$.

Solution. Using $\cos\dfrac{\pi}{3} = \dfrac{1}{2}$ and $\sin\dfrac{\pi}{3} = \dfrac{\sqrt{3}}{2}$:

$$\vec{r}\left( \frac{\pi}{3} \right) = \left\langle \frac{3}{2},\ \sqrt{3} \right\rangle$$ $$\vec{r}\,'(t) = \langle -3\sin t,\ 2\cos t \rangle, \qquad \vec{r}\,'\left( \frac{\pi}{3} \right) = \left\langle -\frac{3\sqrt{3}}{2},\ 1 \right\rangle$$

The slope of the tangent line is

$$\frac{y'}{x'} = \frac{1}{-3\sqrt{3}/2} = -\frac{2}{3\sqrt{3}} = -\frac{2\sqrt{3}}{9}$$

The figure above shows this: at $\left( \tfrac{3}{2}, \sqrt{3} \right)$ the tangent vector points up and to the left, because the ellipse is traced counterclockwise.

Example 3: Finding r(t) from r′(t)

Given $\vec{r}\,'(t) = \left\langle 6t^2,\ \dfrac{1}{t + 1} \right\rangle$ for $t \gt -1$ and $\vec{r}(0) = \langle 1, -3 \rangle$, find $\vec{r}(t)$.

Solution. Integrate each component:

$$\vec{r}(t) = \big\langle 2t^3 + C_1,\ \ln(t + 1) + C_2 \big\rangle$$

(No absolute value is needed, since $t + 1 \gt 0$.) Use $\vec{r}(0) = \langle 1, -3 \rangle$:

$$0 + C_1 = 1 \ \Rightarrow\ C_1 = 1, \qquad \ln 1 + C_2 = -3 \ \Rightarrow\ C_2 = -3$$ $$\vec{r}(t) = \big\langle 2t^3 + 1,\ \ln(t + 1) - 3 \big\rangle$$

Check: $\vec{r}(0) = \langle 1, 0 - 3 \rangle = \langle 1, -3 \rangle$.

Example 4: A definite integral and net change

Evaluate $\displaystyle\int_0^{\pi} \langle \sin t,\ 2t \rangle\, dt$. If $\vec{r}\,'(t) = \langle \sin t,\ 2t \rangle$ and $\vec{r}(0) = \langle 1, 0 \rangle$, find $\vec{r}(\pi)$.

Solution.

$$\int_0^{\pi} \langle \sin t,\ 2t \rangle\, dt = \left\langle \Big[ -\cos t \Big]_0^{\pi},\ \Big[ t^2 \Big]_0^{\pi} \right\rangle = \langle 1 + 1,\ \pi^2 \rangle = \langle 2,\ \pi^2 \rangle$$

This is the net change $\vec{r}(\pi) - \vec{r}(0)$, so

$$\vec{r}(\pi) = \langle 1, 0 \rangle + \langle 2, \pi^2 \rangle = \langle 3,\ \pi^2 \rangle$$

Common mistakes

Only one constant of integration. Each component gets its own constant. Writing $\langle 2t^3, \ln(t + 1) \rangle + C$ with a single number $C$ added to both is wrong unless $C$ is a vector $\langle C_1, C_2 \rangle$.

Treating the vector as a single number. You can’t “multiply out” or divide the components by each other. $\vec{r}\,'(t)$ is a vector; the slope $\dfrac{y'}{x'}$ is a number. Make sure your answer has the form the question asks for.

Forgetting the initial position. $\int_0^t \vec{r}\,'(u)\, du$ is the change in $\vec{r}$, not $\vec{r}(t)$ itself. Add $\vec{r}(0)$.

Mixing up the components. Keep $x$ first and $y$ second all the way through. A swapped answer like $\langle \pi^2, 2 \rangle$ in Example 4 is a different vector.

Drawing the tangent vector from the origin. The position vector starts at the origin. The derivative $\vec{r}\,'(t)$ is drawn starting at the point on the curve, where it shows the direction of motion.

Practice

1. (Warm-up) Let $\vec{r}(t) = \langle t^2,\ 5t - 1 \rangle$. Find $\vec{r}\,'(t)$ and $\vec{r}\,'(2)$.

Solution

$$\vec{r}\,'(t) = \langle 2t,\ 5 \rangle, \qquad \vec{r}\,'(2) = \langle 4,\ 5 \rangle$$

2. (Warm-up) Find $\displaystyle\int \langle 2t,\ 3 \rangle\, dt$.

Solution

$$\int \langle 2t,\ 3 \rangle\, dt = \langle t^2 + C_1,\ 3t + C_2 \rangle$$

3. (Warm-up) Let $\vec{r}(t) = \langle e^t,\ \ln t \rangle$ for $t \gt 0$. Find $\vec{r}\,''(t)$.

Solution

$$\vec{r}\,'(t) = \left\langle e^t,\ \frac{1}{t} \right\rangle, \qquad \vec{r}\,''(t) = \left\langle e^t,\ -\frac{1}{t^2} \right\rangle$$

4. (Core) Let $\vec{r}(t) = \langle t^2 - 1,\ t^3 - t \rangle$. Find $\vec{r}\,'(t)$, and the equation of the tangent line to the curve at $t = 2$.

Solution

$\vec{r}\,'(t) = \langle 2t,\ 3t^2 - 1 \rangle$, so $\vec{r}\,'(2) = \langle 4, 11 \rangle$ and the slope is $\dfrac{11}{4}$.

The point is $\vec{r}(2) = \langle 3, 6 \rangle$. Tangent line:

$$y - 6 = \frac{11}{4}(x - 3)$$

5. (Core) Given $\vec{r}\,'(t) = \langle \cos t,\ -2\sin(2t) \rangle$ and $\vec{r}(0) = \langle 3, 1 \rangle$, find $\vec{r}(t)$ and $\vec{r}\left( \dfrac{\pi}{2} \right)$.

Solution

$$\vec{r}(t) = \langle \sin t + C_1,\ \cos(2t) + C_2 \rangle$$

At $t = 0$: $0 + C_1 = 3$ and $1 + C_2 = 1$, so $C_1 = 3$ and $C_2 = 0$.

$$\vec{r}(t) = \langle 3 + \sin t,\ \cos(2t) \rangle, \qquad \vec{r}\left( \frac{\pi}{2} \right) = \langle 3 + 1,\ \cos\pi \rangle = \langle 4,\ -1 \rangle$$

6. (Core) Evaluate $\displaystyle\int_1^4 \left\langle \sqrt{t},\ \frac{1}{t} \right\rangle dt$.

Solution

$$\left\langle \left[ \frac{2}{3}t^{3/2} \right]_1^4,\ \Big[ \ln t \Big]_1^4 \right\rangle = \left\langle \frac{2}{3}(8 - 1),\ \ln 4 \right\rangle = \left\langle \frac{14}{3},\ \ln 4 \right\rangle$$

7. (Core) A ball is thrown from a point $2$ m above the ground. Its acceleration is $\vec{r}\,''(t) = \langle 0, -9.8 \rangle$ m/s², its initial velocity is $\vec{r}\,'(0) = \langle 20, 15 \rangle$ m/s, and $\vec{r}(0) = \langle 0, 2 \rangle$ (in metres, with $x$ horizontal and $y$ the height). Find $\vec{r}(t)$, and the height of the ball when it is $30$ m away horizontally.

Solution

Integrate once and use $\vec{r}\,'(0)$:

$$\vec{r}\,'(t) = \langle 0 + C_1,\ -9.8t + C_2 \rangle = \langle 20,\ 15 - 9.8t \rangle$$

Integrate again and use $\vec{r}(0)$:

$$\vec{r}(t) = \langle 20t,\ 2 + 15t - 4.9t^2 \rangle$$

$x = 30$ when $20t = 30$, so $t = 1.5$ s. The height is

$$y(1.5) = 2 + 22.5 - 11.025 = 13.475 \text{ m}$$

8. (Challenge) Given $\vec{r}\,'(t) = \left\langle 2te^{t^2},\ \dfrac{1}{1 + t^2} \right\rangle$ and $\vec{r}(0) = \langle -1, 2 \rangle$, find $\vec{r}(1)$ exactly.

Solution

Use $\vec{r}(1) = \vec{r}(0) + \displaystyle\int_0^1 \vec{r}\,'(t)\, dt$.

First component (substitute $u = t^2$, $du = 2t\, dt$):

$$\int_0^1 2te^{t^2}\, dt = \Big[ e^{t^2} \Big]_0^1 = e - 1$$

Second component:

$$\int_0^1 \frac{1}{1 + t^2}\, dt = \Big[ \arctan t \Big]_0^1 = \frac{\pi}{4}$$$$\vec{r}(1) = \left\langle -1 + e - 1,\ 2 + \frac{\pi}{4} \right\rangle = \left\langle e - 2,\ 2 + \frac{\pi}{4} \right\rangle$$

9. (Challenge) (Calculator active.) Given $\vec{r}\,'(t) = \left\langle \sqrt{1 + t^3},\ \cos(t^2) \right\rangle$ and $\vec{r}(0) = \langle 2, -1 \rangle$, find $\vec{r}(2)$. Explain why a calculator is needed.

Solution

Neither $\sqrt{1 + t^3}$ nor $\cos(t^2)$ has an antiderivative you can write with ordinary functions, so evaluate the definite integrals numerically (radian mode):

$$\vec{r}(2) = \left\langle 2 + \int_0^2 \sqrt{1 + t^3}\, dt,\ -1 + \int_0^2 \cos(t^2)\, dt \right\rangle \approx \langle 2 + 3.241,\ -1 + 0.461 \rangle = \langle 5.241,\ -0.539 \rangle$$