Introduction

Welcome to this comprehensive guide on solving motion problems using parametric and vector-valued functions! This topic represents one of the most practical applications of calculus, bridging the gap between mathematical theory and real-world physics. When particles move along curves in the plane, we need more sophisticated tools than simple position functions to describe their motion. Parametric and vector-valued functions provide the perfect framework for analyzing complex planar motion, allowing us to determine positions, velocities, speeds, accelerations, displacements, and total distances traveled.

This topic builds directly on your knowledge of parametric equations, vector-valued functions, and their derivatives and integrals. You'll learn to interpret the physical meaning of mathematical operations and solve comprehensive motion problems that mirror real-world scenarios in physics and engineering.

Understanding Planar Motion with Parametric Functions

When a particle moves along a curve in the plane, its position can be described using parametric equations: $x=f(t)$, $y=g(t)$

Here, $t$ represents time (or another parameter), and the functions $f(t)$ and $g(t)$ describe the horizontal and vertical positions respectively. This parametric representation allows us to track motion along any curve, not just simple straight lines or basic functions.

The same motion can be expressed using a vector-valued function: $\vec{r}(t)=⟨f(t),g(t)⟩=f(t)\vec{i}+g(t)\vec{j}$

Both representations are equivalent and describe the particle's position at any time $t$.

Velocity, Speed, and Acceleration in Planar Motion

The essential knowledge tells us that derivatives can be used to determine velocity, speed, and acceleration for a particle moving along a curve in the plane defined using parametric or vector-valued functions.

Velocity Vector

The velocity vector represents both the direction and rate of change of position:

For parametric equations: $\vec{v}(t)=⟨\frac{dx}{dt},\frac{dy}{dt}⟩$

For vector-valued functions: $\vec{v}(t)={\vec{r}}^{\prime }(t)=⟨f^{\prime }(t),g^{\prime }(t)⟩$

The velocity vector is tangent to the path of motion and points in the direction of travel.

Speed

Speed is the magnitude of the velocity vector and represents how fast the particle is moving: $\text{Speed}=|\vec{v}(t)|=\sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2}$

Speed is always non-negative and gives the instantaneous rate of change of distance along the curve.

Acceleration Vector

The acceleration vector represents the rate of change of velocity: $\vec{a}(t)={\vec{v}}^{\prime }(t)={\vec{r}}^{\prime \prime }(t)=⟨\frac{d^{2}x}{d{t}^2},\frac{d^{2}y}{d{t}^2}⟩$

Acceleration indicates how the velocity vector is changing in both magnitude and direction.

Basic Motion Analysis Examples

Example 1: Position Analysis
A particle moves along a curve with parametric equations $x(t)=3t^2+1$ and $y(t)=2t-4$ for $t\ge 0$. Find the position, velocity, speed, and acceleration at $t=2$.

Position at $t=2$:

$x(2)=3(2{)}^2+1=3(4)+1=13$

$y(2)=2(2)-4=4-4=0$

Position: $(13,0)$

Velocity:

$\frac{dx}{dt}=6t$

$\frac{dy}{dt}=2$

$\vec{v}(t)=⟨6t,2⟩$

At $t=2$: $\vec{v}(2)=⟨12,2⟩$

Speed at $t=2$: $|\vec{v}(2)|=\sqrt{12^2+2^2}=\sqrt{144+4}=\sqrt{148}=2\sqrt{37}$

Acceleration:

$\frac{d^{2}x}{d{t}^2}=6$

$\frac{d^{2}y}{d{t}^2}=0$

$\vec{a}(t)=⟨6,0⟩$

At $t=2$: $\vec{a}(2)=⟨6,0⟩$

Example 2: Trigonometric Motion
A particle moves with position vector $\vec{r}(t)=⟨4\cos (t),3\sin (t)⟩$. Find the velocity and speed functions.

Velocity: $\vec{v}(t)={\vec{r}}^{\prime }(t)=⟨-4\sin (t),3\cos (t)⟩$

Speed: $|\vec{v}(t)|=\sqrt{16{\sin }^2(t)+9{\cos }^2(t)}$

This represents elliptical motion with varying speed.

Displacement and Distance Using Definite Integrals

The essential knowledge emphasizes that for a particle in planar motion over an interval of time, the definite integral of the velocity vector represents the particle's displacement (net change in position) over the interval of time, from which we might determine its position.

Displacement Vector

Displacement is the net change in position from time $t=a$ to $t=b$:
$\text{Displacement}={\int }_a^b\vec{v}(t)dt=⟨{\int }_a^b\frac{dx}{dt}dt,{\int }_a^b\frac{dy}{dt}dt⟩=⟨x(b)-x(a),y(b)-y(a)⟩$

This gives the straight-line distance and direction from the starting point to the ending point.

Total Distance Traveled

Total distance is the actual distance traveled along the curve:
$\text{Total Distance}={\int }_a^b|\vec{v}(t)|dt={\int }_a^b\sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2}dt$

This accounts for all the twists, turns, and changes in direction along the path.

Example 3: Displacement vs. Distance

A particle moves with velocity $\vec{v}(t)=⟨2\cos (t),2\sin (t)⟩$ from $t=0$ to $t=2\pi$. Find the displacement and total distance.

Displacement:

${\int }_0^{2\pi }\vec{v}(t),dt={\int }_0^{2\pi }⟨2\cos (t),2\sin (t)⟩dt$

$=⟨{\int }_0^{2\pi }2\cos (t)dt,{\int }_0^{2\pi }2\sin (t)dt⟩$

$=⟨2\sin (t){|}_0^{2\pi },-2\cos (t){|}_0^{2\pi }⟩$

$=⟨2\sin (2\pi )-2\sin (0),-2\cos (2\pi )+2\cos (0)⟩$$ $=⟨0-0,-2(1)+2(1)⟩=⟨0,0⟩$

Total Distance:

First, find the speed: $|\vec{v}(t)|=\sqrt{(2\cos (t){)}^2+(2\sin (t){)}^2}=\sqrt{4{\cos }^2(t)+4{\sin }^2(t)}=\sqrt{4}=2$

${\int }_0^{2\pi }|\vec{v}(t)|dt={\int }_0^{2\pi }2dt=2t{|}_0^{2\pi }=4\pi$

The particle returns to its starting point (displacement = $0$) but travels a total distance of $4\pi$ units around a circle.

Solving Initial Value Problems for Position

When given velocity and initial position, we can determine the position function by integration.

Example 4: Finding Position from Velocity

A particle has velocity $\vec{v}(t)=⟨3t^2,4t-1⟩$ and initial position $\vec{r}(0)=⟨2,-3⟩$. Find the position function $\vec{r}(t)$.

Step 1: Integrate the velocity to find the general position: $\vec{r}(t)=\int \vec{v}(t)dt=\int ⟨3t^2,4t-1⟩dt=⟨\int 3t^{2}dt,\int (4t-1)dt⟩=⟨t^3+C_1,2t^2-t+C_2⟩$

Step 2: Apply the initial condition $\vec{r}(0)=⟨2,-3⟩$: $\vec{r}(0)=⟨0^3+C_1,2(0{)}^2-0+C_2⟩=⟨C_1,C_2⟩=⟨2,-3⟩$

Therefore: $C_1=2$ and $C_2=-3$

Step 3: The position function is: $\vec{r}(t)=⟨t^3+2,2t^2-t-3⟩$

Multi-Step Motion Problems

Example 5: From Acceleration to Position

A particle has acceleration $\vec{a}(t)=⟨6t,4⟩$, initial velocity $\vec{v}(0)=⟨1,0⟩$, and initial position $\vec{r}(0)=⟨0,5⟩$. Find the position function and the position at $t=3$.

Step 1: Find velocity from acceleration: $\vec{v}(t)=\int \vec{a}(t)dt=\int ⟨6t,4⟩dt=⟨3t^2+C_1,4t+C_2⟩$

Using $\vec{v}(0)=⟨1,0⟩$: $⟨3(0{)}^2+C_1,4(0)+C_2⟩=⟨C_1,C_2⟩=⟨1,0⟩$

Therefore: $C_1=1$ and $C_2=0$ $\vec{v}(t)=⟨3t^2+1,4t⟩$

Step 2: Find position from velocity: $\vec{r}(t)=\int \vec{v}(t)dt=\int ⟨3t^2+1,4t⟩dt=⟨t^3+t+D_1,2t^2+D_2⟩$

Using $\vec{r}(0)=⟨0,5⟩$: $⟨0^3+0+D_1,2(0{)}^2+D_2⟩=⟨D_1,D_2⟩=⟨0,5⟩$

Therefore: $D_1=0$ and $D_2=5$ $\vec{r}(t)=⟨t^3+t,2t^2+5⟩$

Step 3: Position at $t=3$: $\vec{r}(3)=⟨3^3+3,2(3{)}^2+5⟩=⟨27+3,2(9)+5⟩=⟨30,23⟩$

Motion Along Specific Curves

Example 6: Circular Motion Analysis

A particle moves with $x(t)=5\cos (2t)$ and $y(t)=5\sin (2t)$.

The particle moves on a circle of radius 5 with velocity $\vec{v}(t)=⟨-10\sin (2t),10\cos (2t)⟩$, constant speed $|\vec{v}(t)|=10$, and acceleration $\vec{a}(t)=⟨-20\cos (2t),-20\sin (2t)⟩=-4\vec{r}(t)$ pointing toward the center with magnitude $20$.

Distance and Displacement Applications

Example 7: Path Analysis

A particle moves with velocity $\vec{v}(t)=⟨t,t^2⟩$ from $t=0$ to $t=2$.

Displacement: ${\int }_0^2⟨t,t^2⟩dt=⟨\frac{t^2}{2}{|}_0^2,\frac{t^3}{3}{|}_0^2⟩=⟨2,\frac{8}{3}⟩$

Total Distance:

$|\vec{v}(t)|=t\sqrt{1+t^2}$

${\int }_0^{2}t\sqrt{1+t^2}dt$.

Evaluate using substitution. Let $u=1+t^2$, so $du=2tdt$: When $t=0$: $u=1$. When $t=2$: $u=5$.

$\frac{1}{2}{\int }_1^5\sqrt{u}du=\frac{1}{2}{\int }_1^5{u}^{1/2}du$

$=\frac{1}{2}\ast \frac{u^{3/2}}{3/2}{|}_1^5=\frac{1}{2}\ast \frac{2}{3}{u}^{3/2}{|}_1^5=\frac{1}{3}{u}^{3/2}{|}_1^5$

$=\frac{1}{3}(5^{3/2}-1^{3/2})=\frac{1}{3}(5\sqrt{5}-1)$

Distance travelled is $\frac{1}{3}(5\sqrt{5}-1)$

Practice Section

Test your understanding with these multiple-choice questions: