Motion with Parametric Equations (College Board AP® Calculus BC): Study Guide

Written by: Mark Curtis

Reviewed by: Dan Finlay

Updated on 29 January 2025

Motion with parametric equations

How do I model a particle moving in 2D using parametric equations?

A particle moving in two-dimensions follows a path (curve) in the $x y$ -plane
- This path can be given parametrically as $x (t)$ and $y (t)$
  - $x (t)$ is the displacement in the $x$ -direction
  - $y (t)$ is the displacement in the $y$ -direction
- Displacement is measured relative to a fixed origin, $O$
- The parameter, $t$ , represents time, where $t \geq 0$
- The coordinates of the particle's position at time $t$ are $(x (t), y (t))$

The direction of motion is the direction along the curve as $t$ increases
The initial position of the particle is the point at which $t = 0$
- The coordinates of the initial position are $(x (0), y (0))$
  - This may not be at the origin
Sometimes a time interval for the motion is given, $t_{1} \leq t \leq t_{2}$

Diagram showing a curved path labelled "Path of Motion" with arrows for vertical and horizontal directions; x and y depend on time, t.

How do I find the velocity and acceleration in 2D?

Recall the ideas in the study guide on 'Motion in a Straight Line'
- Velocity, $v$ , is the rate of change of displacement, $s$ , with respect to time
  - $v = \frac{d s}{d t}$
- Acceleration, $a$ , is the rate of change of velocity, $v$ , with respect to time
  - $a = \frac{d v}{d t} = \frac{d^{2} s}{d t^{2}}$

In two dimensions, the displacement, velocity and acceleration of a particle each have two components in perpendicular directions
- Displacement splits into $x$ and $y$ components
  - Often horizontal and vertical
- Derivatives of these are as follows:

Derivative	Shorthand	Interpretation
$\frac{d x}{d t}$	$x' (t)$	The velocity of the particle in the $x$ -direction at time $t$ If the $x$ -axis is horizontal then $\frac{d x}{d t} > 0$ means moving to the right and $\frac{d x}{d t} < 0$ means moving to the left
$\frac{d y}{d t}$	$y' (t)$	The velocity of the particle in the $y$ -direction at time $t$ If the $y$ -axis is vertical then $\frac{d y}{d t} > 0$ means moving upwards and $\frac{d y}{d t} < 0$ means moving downwards
$\frac{d^{2} x}{d t^{2}}$	$x'' (t)$	The acceleration of the particle in the $x$ -direction at time $t$
$\frac{d^{2} y}{d t^{2}}$	$y'' (t)$	The acceleration of the particle in the $y$ -direction at time $t$

Examiner Tips and Tricks

If no units are given in a question, you don't need units in your answers.

How do I calculate the speed of a particle in 2D?

Diagram of a velocity triangle showing speed on the hypotenuse, dx/dt on the base, dy/dt as the height, and the direction of motion. — **A velocity triangle**

The speed of a particle at time $t$ is given by the formula:

$\sqrt{{(\frac{d x}{d t})}^{2} + {(\frac{d y}{d t})}^{2}}$

This is because speed is the magnitude (hypotenuse) of a velocity triangle formed out of:
- the velocity of the particle in the $x$ -direction, $\frac{d x}{d t}$ , at time $t$
- and the velocity of the particle in the $y$ -direction , $\frac{d y}{d t}$ , at time $t$
  - By Pythagoras' theorem, this gives $\sqrt{{(\frac{d x}{d t})}^{2} + {(\frac{d y}{d t})}^{2}}$
The velocity triangle also gives the direction of motion of the particle at that time
The particle is at rest if the speed is zero

Examiner Tips and Tricks

Do not confuse the word 'speed' with 'velocity'!

How do I calculate the slope of a line that is tangent to the path of the particle?

Graph with a parametric curve and its tangent at point P, marked at time t equals t0; formula shows derivative of y with respect to x.

The slope of the line that is tangent to the path of the particle (a tangent to the curve) at time $t$ is given by the formula:

$\frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{y' (t)}{x' (t)}$

See the study guide on 'Derivatives of Parametric Equations'
This is the same slope as the velocity triangle at that time (see above)
- Note that the direction of motion could be either way along the tangent
  - It will follow the direction in which $t$ is increasing

How do I use indefinite integration for particles moving in 2D?

Indefinite integration can be used to find
- displacement, $x (t)$ or $y (t)$ , by integrating velocity, $x' (t)$ or $y' (t)$
- velocity, $x' (t)$ or $y' (t)$ , by integrating acceleration, $x'' (t)$ or $y'' (t)$
Always remember to add a constant of integration each time you integrate
- Then use information in the question to find the constant
  - e.g. you may be told that $x = 5$ when $t = 0$
Indefinite integration is good when you need an algebraic expression in terms of time $t$
- You can then evaluate it at different times

How do I use definite integration for particles moving in 2D?

If an algebraic expression in terms of $t$ is not required but a specific value of a quantity is required (e.g. find $x$ at time $t = 2$ ) then definite integration can be used
- Below are the commonly used definite integrals in motion questions:

Integral	Shorthand	Interpretation
$\int_{t_{1}}^{t_{2}} \frac{d x}{d t} d t = x (t_{2}) - x (t_{1})$	$\int_{t_{1}}^{t_{2}} x' (t) d t = x (t_{2}) - x (t_{1})$	The change in the position of the $x$ -coordinates
$\int_{t_{1}}^{t_{2}} \frac{d y}{d t} d t = y (t_{2}) - y (t_{1})$	$\int_{t_{1}}^{t_{2}} y' (t) d t = y (t_{2}) - y (t_{1})$	The change in the position of the $y$ -coordinates
$\int_{t_{1}}^{t_{2}} \frac{d^{2} x}{d t^{2}} d t = \frac{d x}{d t} (t_{2}) - \frac{d x}{d t} (t_{1})$	$\int_{t_{1}}^{t_{2}} x'' (t) d t = x' (t_{2}) - x' (t_{1})$	The change in the velocity of the particle in the $x$ -direction
$\int_{t_{1}}^{t_{2}} \frac{d^{2} y}{d t^{2}} d t = \frac{d y}{d t} (t_{2}) - \frac{d y}{d t} (t_{1})$	$\int_{t_{1}}^{t_{2}} y'' (t) d t = y' (t_{2}) - y' (t_{1})$	The change in the velocity of the particle in the $y$ -direction

Definite integrals often need rearranging to find the specific value of a quantity at either end of the time interval
- e.g. $\int_{t_{1}}^{t_{2}} x' (t) d t = x (t_{2}) - x (t_{1})$ rearranges to either
  - $x (t_{2}) = x (t_{1}) + \int_{t_{1}}^{t_{2}} x' (t) d t$
  - or $x (t_{1}) = x (t_{2}) - \int_{t_{1}}^{t_{2}} x' (t) d t$

Examiner Tips and Tricks

Motion questions are often found in the calculator sections of the exam, where you are expected to use your calculator to:

evaluate any definite integrals,
solve any equations.

How do I calculate the distance traveled by a particle in 2D?

The distance traveled by a particle between time $t_{1}$ and time $t_{2}$ is the arc length of the path traveled, given by the definite integral:
- $\int_{t_{1}}^{t_{2}} \sqrt{{(\frac{d x}{d t})}^{2} + {(\frac{d y}{d t})}^{2}} d t$
  - See the study guide on 'Arc Lengths of Parametric Equations'
- This is always positive

Graph depicting a parametric curve from t1 to t2 with arrows showing the length of a section of the curve (a dashed line alongside), and an integral formula for distance traveled, L, above. — **L is the distance traveled between time t1 and time t2**

Examiner Tips and Tricks

Do not confuse the word 'distance' with 'displacement'!

Worked Example

A particle travels along a curve in the $x y$ -plane. The particle is at the point $(x (t), y (t))$ at time $t$ where $t \geq 0$ . The derivatives of $x (t)$ and $y (t)$ are:

$\frac{d x}{d t} = 6 t^{2} - 2 t \frac{d y}{d t} = \sqrt{4 e^{t} + 1}$

At time $t = 3$ , the particle is at the point $(50, 15)$ .

(a) Find the acceleration of the particle in the $x$ -direction at time $t = 3$ .

Differentiate the velocity of the particle in the $x$ -direction, $\frac{d x}{d t}$ , to get the acceleration in the $x$ -direction, $\frac{d^{2} x}{d t^{2}}$

$\frac{d^{2} x}{d t^{2}} = 12 t - 2$

Substitute in $t = 3$

(b) Find an expression for $x$ in terms of time, $t$ .

Integrate the velocity of the particle in the $x$ -direction, $\frac{d x}{d t}$ , to find $x$

Remember to add a constant of integration

$x = 2 t^{3} - t^{2} + C$

Use the fact that $x = 50$ when $t = 3$ (in the question) to find $c$

$\begin{array}{rcl} 50 & = & 2 {(3)}^{3} - {(3)}^{2} + C \\ 50 & = & 45 + C \\ C & = & 5 \end{array}$

Substitute this value of $C$ back into the expression for $x$

$x = 2 t^{3} - t^{2} + 5$

You know the displacement of the particle in the $y$ -direction at time $t = 3$ from the question ( $y = 15$ )

You can use a definite integral to find $y$ at time $t = 6$

$\begin{array}{rcl} \int_{3}^{6} \frac{d y}{d t} d t & = & y (6) - y (3) \\ \int_{3}^{6} \sqrt{4 e^{t} + 1} d t & = & y (6) - 15 \end{array}$

Make $y (6)$ the subject, then evaluate the definite integral on your calculator

$\begin{array}{rcl} y (6) & = & 15 + \int_{3}^{6} \sqrt{4 e^{t} + 1} d t \\ = & 15 + 62.5019 \end{array}$

77.502

(d) Find the slope of the line tangent to the path of the particle at time $t = 6$ .

The slope of the line tangent to the path of the particle has the formula $\frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}}$

$\frac{d y}{d x} = \frac{\sqrt{4 e^{t} + 1}}{6 t^{2} - 2 t}$

Substitute in $t = 6$

0.197

(e) Show that the particle is not initially at rest.

To be initially at rest, the speed $\sqrt{{(\frac{d x}{d t})}^{2} + {(\frac{d y}{d t})}^{2}}$ must equal zero when $t = 0$

Find an expression for the speed

$\sqrt{{(6 t^{2} - 2 t)}^{2} + {(\sqrt{4 e^{t} + 1})}^{2}}$

Substitute in $t = 0$ to see if the speed is zero

$\sqrt{{(6 {(0)}^{2} - 2 (0))}^{2} + {(\sqrt{4 e^{0} + 1})}^{2}} = 2.2360 . . . \neq 0$

The particle is not initially at rest, as its initial speed is 2.236

(f) Find the total distance traveled by the particle from time $t = 3$ to time $t = 6$ .

The total distance traveled is given by $\int_{t_{1}}^{t_{2}} \sqrt{{(\frac{d x}{d t})}^{2} + {(\frac{d y}{d t})}^{2}} d t$ where $t_{1} = 3$ and $t_{2} = 6$

$\int_{3}^{6} \sqrt{{(6 t^{2} - 2 t)}^{2} + {(\sqrt{4 e^{t} + 1})}^{2}} d t$

Evaluate this definite integral on your calculator

356.534

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Motion with Parametric Equations (College Board AP® Calculus BC): Study Guide

Motion with parametric equations

How do I model a particle moving in 2D using parametric equations?

How do I find the velocity and acceleration in 2D?

How do I calculate the speed of a particle in 2D?

How do I calculate the slope of a line that is tangent to the path of the particle?

How do I use indefinite integration for particles moving in 2D?

How do I use definite integration for particles moving in 2D?

How do I calculate the distance traveled by a particle in 2D?

Sign up now. It’s free!

Join the 100,000+ Students that ❤️ Save My Exams

Unit 1: Limits & Continuity

Limits

Definition of a Limit

Infinite Limits & Limits at Infinity

Properties of Limits

Evaluating Limits Analytically

Evaluating Limits Numerically & Graphically

Squeeze Theorem & Trigonometric Limits

Summary of Limits

Selecting Procedures for Determining Limits

Continuity

Continuity

Removable Discontinuities

Non-removable Discontinuities

Intermediate Value Theorem

Unit 2: Differentiation Definition & Fundamental Properties

Definition of Differentiation

Average Rate of Change

Instantaneous Rate of Change

Derivatives & Tangents

Estimating the Derivative at a Point

Differentiability & Continuity

Fundamental Properties of Differentiation

Derivative Rules

Derivatives of Exponentials and Logarithms

Derivatives of Sine and Cosine Functions

The Product Rule

The Quotient Rule

Derivatives of Tangent and Reciprocal Trig Functions

Unit 3: Differentiation of Composite, Implicit & Inverse Functions

Differentiation of Composite & Inverse Functions

The Chain Rule

The Inverse Function Theorem

Derivatives of Inverse Trigonometric Functions

Higher-Order Derivatives

Summary of Differentiation

Selecting Procedures for Calculating Derivatives

Unit 4: Contextual Applications of Differentiation

Rates of Change & Related Rates

Meaning of a Derivative in Context

Motion in a Straight Line

Related Rates

Linearization

Approximating Values of a Function

L'Hospital's Rule

L'Hospital's Rule

Unit 5: Analytical Applications of Differentiation

Graphs of Functions & Their Derivatives

Mean Value Theorem

Extreme Value Theorem

Critical Points

Increasing & Decreasing Functions

First Derivative Test for Local Extrema

Candidates Test for Global Extrema

Concavity of Functions

Second Derivative Test for Local Extrema

Graphs of f, f' & f''

Optimization Problems

Behaviors of Implicit Relations

Second Derivatives of Implicit Functions

Critical Points of Implicit Relations

Unit 6: Integration & Accumulation of Change

Integration & Antiderivatives

Indefinite Integrals

Derivatives & Antiderivatives

Indefinite Integral Rules

Constant of Integration

Riemann Sums & Definite Integrals

Accumulation of Change