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AP®MathsCollege BoardCalculus BCStudy GuidesUnit 9: Parametric Equations, Vector-Valued Functions & Polar CoordinatesVector-Valued FunctionsMotion with Vector-Valued Functions

Motion with Vector-Valued Functions (College Board AP® Calculus BC)

Study Guide

Mark Curtis

Written by: Mark Curtis

Reviewed by: Dan Finlay

Updated on

Motion with vector-valued functions

How can a displacement vector be used to model the motion of a particle in 2D?

  • A particle moving along a two-dimensional path in the x y-plane can be modelled by a displacement vector open angle brackets x open parentheses t close parentheses comma space y open parentheses t close parentheses close angle brackets that depends on time, t

    • The first component is the x-coordinate of the particle at time t

    • The second component is the y-coordinate of the particle at time t

  • These coordinates move with time and are measured relative to an origin, O

    • x open parentheses t close parentheses and y open parentheses t close parentheses are parametric equations of the path of the particle

    • Motion using vector-valued functions can be thought of as motion with parametric equations using vector notation

Examiner Tips and Tricks

A displacement vector may also be referred to as a position vector.

How do I find the velocity and acceleration vectors from a displacement vector?

  • The velocity vector, open angle brackets x apostrophe open parentheses t close parentheses comma space y apostrophe open parentheses t close parentheses close angle brackets, is the derivative of the displacement vector open angle brackets x open parentheses t close parentheses comma space y open parentheses t close parentheses close angle brackets

    • The velocity vector always points in the direction of motion of the particle

  • The acceleration vector, open angle brackets x apostrophe apostrophe open parentheses t close parentheses comma space y apostrophe apostrophe open parentheses t close parentheses close angle brackets, is the derivative of the velocity vector open angle brackets x apostrophe open parentheses t close parentheses comma space y apostrophe open parentheses t close parentheses close angle brackets

    • This makes it the second derivative of the displacement vector, open angle brackets x open parentheses t close parentheses comma space y open parentheses t close parentheses close angle brackets

Diagram showing a curved path with labelled vectors: displacement, velocity, and acceleration vectors at a point P, where the velocity vector points in the direction of motion.
The displacement, velocity and acceleration vectors of a particle, P

  • Further detail on motion in 2D can be found in the study guide on 'Motion with Parametric Equations'

    • Recall that the speed of the particle is given by square root of x apostrophe open parentheses t close parentheses squared plus y apostrophe open parentheses t close parentheses squared end root

    • Recall that, if x is horizontal and y is vertical:

      • x apostrophe open parentheses t close parentheses greater than 0, the particle moves to the right ( x apostrophe open parentheses t close parentheses less than 0 is to the left)

      • y apostrophe open parentheses t close parentheses greater than 0, the particle moves upwards ( y apostrophe open parentheses t close parentheses less than 0 is downwards)

How do I find the displacement vector from the velocity or acceleration vector?

  • To find the displacement vector from the velocity vector:

    • Use indefinite integration to integrate both components of the velocity vector

    • Add constants of integration to each component

    • Use information in the question to find these constants

      • This gives the velocity vector

  • To find the displacement vector from the acceleration vector:

    • Use indefinite integration and add integration constants to each component

    • Use information in the question to find these constants

      • This gives the velocity vector

    • Then use indefinite integration a second time and add a new set of integration constants

    • Use additional information in the question to find these new constants

      • This gives the displacement vector

Worked Example

A particle, P, moves in the x y-plane so that at time t greater or equal than 0 the position vector of the particle is open angle brackets t cubed minus 12 t plus 1 comma space 2 t to the power of 5 close angle brackets.

(a) Find the acceleration vector at time t equals 1.

Differentiate the displacement vector once to get the velocity vector

open angle brackets 3 t squared minus 12 comma space 10 t to the power of 4 close angle brackets

Then differentiate the velocity vector again to get the acceleration vector

open angle brackets 6 t comma space 40 t cubed close angle brackets

Substitute in t equals 1 (give your answer in vector notation)

open angle brackets 6 comma space 40 close angle brackets

(b) Find the time at which the particle travels parallel to the y-axis.

The question describes the direction of motion of the particle, which is the same as the direction of the velocity vector, open angle brackets 3 t squared minus 12 comma space 10 t to the power of 4 close angle brackets

For the velocity vector to be parallel to the y-axis, it must have a zero x-component, i.e. open angle brackets 0 comma space... close angle brackets

Form and solve an equation by setting 3 t squared minus 12 equal to zero

table row cell 3 t squared minus 12 end cell equals 0 row cell t squared end cell equals 4 row t equals cell plus-or-minus 2 end cell end table

Time cannot be negative, so use the positive solution only

t equals 2

(c) A second particle, Q, has the acceleration vector open angle brackets 12 t plus 2 comma space minus 4 sin space t close angle brackets. If the particle Q is initially at rest at the point open angle brackets negative 2 comma space 1 close angle brackets, find its displacement vector in terms of t.

Integrate the acceleration vector to find the velocity vector

Add different constants of integration to each component, C and D

open angle brackets 6 t squared plus 2 t plus C comma space 4 cos space t plus D close angle brackets

Initially the particle is at rest, so the velocity vector must be equal to open angle brackets 0 comma space 0 close angle brackets when t equals 0

Substitute in t equals 0 and simplify (remember cos open parentheses 0 close parentheses equals 1)

open angle brackets 6 open parentheses 0 close parentheses squared plus 2 open parentheses 0 close parentheses plus C comma space 4 cos space open parentheses 0 close parentheses plus D close angle brackets equals open angle brackets C comma space 4 plus D close angle brackets

Set open angle brackets C comma space 4 plus D close angle brackets equal to open angle brackets 0 comma space 0 close angle brackets to find c and d

C equals 0 D equals negative 4

Substitute these values back into the velocity vector

open angle brackets 6 t squared plus 2 t comma space 4 cos space t minus 4 close angle brackets

Integrate the velocity vector to find the displacement vector

Add some new constants of integration, F and G

open angle brackets 2 t cubed plus t squared plus F comma space 4 sin space t minus 4 t plus G close angle brackets

Initially, the particle is at open angle brackets negative 2 comma space 1 close angle brackets so the displacement vector must equal open angle brackets negative 2 comma space 1 close angle brackets when t equals 0

Substitute in t equals 0 and simplify

open angle brackets 2 open parentheses 0 close parentheses cubed plus open parentheses 0 close parentheses squared plus F comma space 4 sin space open parentheses 0 close parentheses minus 4 open parentheses 0 close parentheses plus G close angle brackets equals open angle brackets F comma space G close angle brackets

Set open angle brackets F comma space G close angle brackets equal to open angle brackets negative 2 comma space 1 close angle brackets to find F and G

F equals negative 2 G equals 1

Substitute these values back into the displacement vector to get the final answer

open angle brackets 2 t cubed plus t squared minus 2 comma space 4 sin space t minus 4 t plus 1 close angle brackets

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    Author
    Reviewer
    Mark Curtis

    Author: Mark Curtis

    Expertise: Maths

    Mark graduated twice from the University of Oxford: once in 2009 with a First in Mathematics, then again in 2013 with a PhD (DPhil) in Mathematics. He has had nine successful years as a secondary school teacher, specialising in A-Level Further Maths and running extension classes for Oxbridge Maths applicants. Alongside his teaching, he has written five internal textbooks, introduced new spiralling school curriculums and trained other Maths teachers through outreach programmes.

    Dan Finlay

    Author: Dan Finlay

    Expertise: Maths Lead

    Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.