Solving Motion Problems Using Parametric and Vector-Valued Functions

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

Solving Motion Problems in the Plane

Vector, scalar, or time — decide what kind of answer the question wants before computing.

key idea$\vec{r}\ \xrightarrow{\ \frac{d}{dt}\ }\ \vec{v}\ \xrightarrow{\ \frac{d}{dt}\ }\ \vec{a}\qquad\text{speed}=\lVert\vec{v}\rVert$

Setting up the model

Planar motion is the one-dimensional story told twice, once per coordinate, with a magnitude tying them together. Read the question for whether it wants a vector, a scalar, or a time — almost every lost point on these questions is a mismatch of type.

Answering a planar-motion question

Write down which of $\vec{r}$, $\vec{v}$, $\vec{a}$ you are given and which you need. Differentiating moves right along the chain, integrating moves left.
For anything involving how fast, compute $\left\lVert\vec{v}\right\rVert=\sqrt{\left(x'\right)^{2}+\left(y'\right)^{2}}$ — a single number.
For anything involving where or which way, keep both components and report a vector or a point.
For total distance, integrate the speed: $\displaystyle\int_{t_1}^{t_2}\left\lVert\vec{v}\right\rVert dt$. Never integrate the components and take the magnitude afterwards.
For extremes — highest point, slowest moment — differentiate the right quantity and set it to zero. The highest point needs $y'=0$; the slowest moment needs $\dfrac{d}{dt}\left\lVert\vec{v}\right\rVert=0$.

A projectile path $x=2t$, $y=6t-t^{2}$. The arrows show the direction of travel and the amber vector is the velocity at $t=1.5$. Notice it points up and to the right early on and will tilt downward past the apex — the horizontal component never changes.

Question	What to compute	Type of answer
Where is the particle at $t=3$?	$\vec{r}(3)$	a point
How fast is it moving?	$\left\lVert\vec{v}(t)\right\rVert$	a number
In what direction?	$\vec{v}(t)$	a vector
How far has it travelled?	$\displaystyle\int\left\lVert\vec{v}\right\rVert dt$	a number
What is its net change in position?	$\displaystyle\int\vec{v}\,dt$	a vector
When is it highest?	solve $y'(t)=0$	a time
When is it slowest?	solve $\dfrac{d}{dt}\left\lVert\vec{v}\right\rVert=0$	a time

Six different questions, six different computations, from one $\vec{r}(t)$. Identify the type of the answer before starting — it tells you immediately whether a magnitude belongs in the calculation.

Worked examples

EX 1

Speed at an instant

magnitude of velocity

A particle has $\vec{r}(t)=\left\langle 2t,\ 6t-t^{2}\right\rangle$. Find its speed at $t=1$.

Differentiate

$\vec{v}(t)=\langle2,\ 6-2t\rangle$.

Evaluate

$\vec{v}(1)=\langle2,\,4\rangle$.

Take the magnitude

$\sqrt{4+16}=\sqrt{20}=2\sqrt5$.

State the answer

Speed $=2\sqrt5\approx4.47$ units per unit time.

Check the type

A single number, as ‘how fast’ requires. The vector $\langle2,4\rangle$ would have answered ‘in what direction’.

EX 2

The highest point

vertical rate only

For the same particle, when is it highest, and how high?

Isolate the vertical motion

Height is the second component, $y=6t-t^{2}$.

Set its rate to zero

$y'=6-2t=0$ gives $t=3$.

Confirm it is a maximum

$y''=-2\lt 0$ throughout, so $t=3$ is the maximum.

Evaluate the height

$y(3)=18-9=9$.

Note what was ignored

The horizontal component played no part. Highest is a question about $y$ alone, and the speed is not zero there.

EX 3

Total distance travelled

integrate the speed

Find the distance the particle travels from $t=0$ to $t=3$.

Write the speed

$\left\lVert\vec{v}\right\rVert=\sqrt{4+(6-2t)^{2}}$.

Set up the integral

$\displaystyle\int_{0}^{3}\sqrt{4+(6-2t)^{2}}\,dt$.

Recognise the situation

This needs a trigonometric substitution, so it is a calculator integral.

Evaluate numerically

$\approx10.75$.

Sanity-check

The straight-line displacement is from $(0,0)$ to $(6,9)$, length $\sqrt{36+81}\approx10.8$. The path is nearly straight here, so the two agree closely. ✓

EX 4

The slowest moment

differentiate the speed

For $\vec{r}(t)=\left\langle t^{2}-4t,\ t^{2}\right\rangle$, find when the particle is moving slowest.

Find the velocity

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

Write the speed squared

$\left\lVert\vec{v}\right\rVert^{2}=(2t-4)^{2}+4t^{2}=8t^{2}-16t+16$.

Minimise the square instead

Speed is non-negative, so it is smallest exactly where its square is. Differentiate: $16t-16=0$, giving $t=1$.

Compute the minimum speed

$\left\lVert\vec{v}(1)\right\rVert=\sqrt{8-16+16}=\sqrt8=2\sqrt2$.

Note the tactic

Minimising $\left\lVert\vec{v}\right\rVert^{2}$ avoids differentiating a square root, and gives the same $t$.

Units & interpretation

matching the question to the computation

$\vec{v}$ carries units of length per time in each component; $\left\lVert\vec{v}\right\rVert$ carries the same units as a single number.
‘Highest’ and ‘farthest right’ are questions about one component, so set that component's derivative to zero.
‘Slowest’ and ‘fastest’ are questions about the magnitude, so differentiate the speed — or its square.
The particle is at rest only when both components of $\vec{v}$ vanish at the same instant.

Watch out

Speed is a magnitude, not a component

$\left\lVert\vec{v}\right\rVert=\sqrt{\left(x'\right)^{2}+\left(y'\right)^{2}}$. Using $x'$ or $y'$ alone answers a different question and is rarely worth partial credit.

Distance integrates the speed

$\int\left\lVert\vec{v}\right\rVert dt$. Integrating the components first and taking the magnitude afterwards gives the displacement, which can be zero for a long trip.

The highest point is not the slowest point

At the apex the vertical rate is zero but the horizontal rate usually is not. Speed at the apex equals $\lvert x'\rvert$, which need not be minimal.

Minimise the square when hunting extremes of speed

Differentiating $\sqrt{\;\cdot\;}$ invites algebra errors. The square has its minimum at the same $t$, so work with that instead.

Course map. Topic 9.6 · LO FUN-8.B — determine values for positions and rates of change in problems involving planar motion. BC-only.