Defining and Differentiating Vector-Valued Functions

AP Calculus BC · Parametric, Polar & Vector-Valued Functions

Defining and Differentiating Vector-Valued Functions

One function, two components, and differentiation that acts on each independently.

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

What it says

Vector-valued function

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

A single function whose output is a vector rather than a number. Each input $t$ produces a point in the plane, so $\vec{r}$ describes a moving position. It is the same information a pair of parametric equations carries, packaged into one object.

Componentwise differentiation

\[\vec{r}\,'(t)=\big\langle x'(t),\,y'(t)\big\rangle\]

Differentiation acts on each component independently. There is no interaction between them and no new rule to learn — if you can differentiate $x(t)$ and $y(t)$ separately, you can differentiate $\vec{r}$.

the three objects

$\vec{r}\,'(t)$ is the velocity vector: it points along the direction of motion.
Its length $\left\lVert\vec{r}\,'(t)\right\rVert=\sqrt{\left(x'\right)^{2}+\left(y'\right)^{2}}$ is the speed — a scalar.
$\vec{r}\,''(t)$ is the acceleration vector, again computed componentwise.
Speed is a number; velocity is a vector. A question asking ‘how fast’ wants the first.

Quantity	Formula	Vector or scalar?
position	$\vec{r}(t)=\langle x,y\rangle$	vector
velocity	$\vec{v}(t)=\langle x',y'\rangle$	vector
speed	$\left\lVert\vec{v}(t)\right\rVert=\sqrt{\left(x'\right)^{2}+\left(y'\right)^{2}}$	scalar
acceleration	$\vec{a}(t)=\langle x'',y''\rangle$	vector
slope of the path	$\dfrac{y'}{x'}$	scalar

Only speed and slope are numbers; the other three are vectors. Answering a vector question with a scalar — or the reverse — loses the point even when the arithmetic is right.

$\langle a,b\rangle$Angle-bracket notation for a vector with components $a$ and $b$. Equivalent to $a\vec{i}+b\vec{j}$.

Magnitude $\lVert\cdot\rVert$The length of a vector, $\sqrt{a^{2}+b^{2}}$. Always a non-negative number.

VelocityThe derivative of position. Its direction is the direction of travel; its length is the speed.

AccelerationThe derivative of velocity. It need not point along the path — a turning object accelerates even at constant speed.

At rest$\vec{v}(t)=\vec{0}$, meaning both components vanish at the same $t$. One zero component is not enough.

Reading the notation

$\vec{r}\,'$velocity — a vector

$\left\lVert\vec{r}\,'\right\rVert$speed — a number

$\vec{r}\,''$acceleration — a vector

Worked examples

EX 1

Velocity, speed and acceleration

one function, three answers

For $\vec{r}(t)=\left\langle t^{2},\,t^{3}-3t\right\rangle$, find $\vec{v}(2)$, the speed at $t=2$, and $\vec{a}(2)$.

Differentiate componentwise

$\vec{v}(t)=\left\langle 2t,\,3t^{2}-3\right\rangle$.

Evaluate the velocity

$\vec{v}(2)=\langle4,\,9\rangle$.

Take its magnitude

Speed $=\sqrt{16+81}=\sqrt{97}\approx9.85$.

Differentiate again

$\vec{a}(t)=\langle2,\,6t\rangle$, so $\vec{a}(2)=\langle2,12\rangle$.

Check the forms

Two vectors and one number — matching what each question asked for.

EX 2

When is the object at rest?

both components at once

For $\vec{r}(t)=\left\langle t^{2}-4t,\,t^{2}-4\right\rangle$, find every $t$ at which the object is at rest.

Find the velocity

$\vec{v}(t)=\langle2t-4,\,2t\rangle$.

Set the first component to zero

$2t-4=0$ gives $t=2$.

Set the second to zero

$2t=0$ gives $t=0$.

Compare

No value of $t$ makes both zero at once.

Conclude

The object is never at rest. At $t=2$ it moves purely vertically; at $t=0$, purely horizontally — but it never stops.

EX 3

Constant speed, non-zero acceleration

turning is accelerating

For $\vec{r}(t)=\langle\cos t,\,\sin t\rangle$, show the speed is constant while the acceleration is not zero.

Find the velocity

$\vec{v}(t)=\langle-\sin t,\,\cos t\rangle$.

Compute the speed

$\sqrt{\sin^{2}t+\cos^{2}t}=1$ for every $t$ — constant.

Find the acceleration

$\vec{a}(t)=\langle-\cos t,\,-\sin t\rangle$, whose magnitude is also $1$.

Reconcile the two

The speed never changes, but the direction does, and changing direction is itself acceleration.

Note the geometry

$\vec{a}=-\vec{r}$ here: the acceleration points straight at the centre, which is exactly what keeps the motion circular.

EX 4

Direction of motion

reading the velocity vector

For $\vec{r}(t)=\left\langle e^{t},\,e^{-t}\right\rangle$, describe the direction of motion at $t=0$ and the slope of the path there.

Differentiate

$\vec{v}(t)=\left\langle e^{t},\,-e^{-t}\right\rangle$.

Evaluate at $t=0$

$\vec{v}(0)=\langle1,\,-1\rangle$.

Describe the direction

Rightward and downward, at $45^{\circ}$ below the horizontal.

Find the slope

$\dfrac{dy}{dx}=\dfrac{y'}{x'}=\dfrac{-e^{-t}}{e^{t}}=-e^{-2t}$, which is $-1$ at $t=0$. ✓

Note the two answers

The direction is a vector, the slope is a number. The slope alone cannot distinguish rightward-and-down from leftward-and-up.

Boundary cases

Differentiate each component separately

There is no product or chain rule between the components. $\vec{r}\,'=\langle x',y'\rangle$, always.

Speed is the magnitude, not a component

$\left\lVert\vec{v}\right\rVert=\sqrt{\left(x'\right)^{2}+\left(y'\right)^{2}}$. Reporting $x'$ or $y'$ alone answers a different question.

At rest requires both components to vanish

$\vec{v}=\vec{0}$ means $x'=0$ and $y'=0$ at the same $t$. Solving the two equations separately and reporting both roots is wrong.

Constant speed does not mean zero acceleration

Circular motion has constant speed and non-zero acceleration throughout. Acceleration responds to changes in direction as well as in magnitude.

Course map. Topic 9.4 · LO CHA-3.H — calculate derivatives of vector-valued functions. BC-only.