MKMM 1213 Advanced Engineering Mathematicsabuhasan/content/teaching/mkmm1213/... · Faculty of...

Faculty of Mechanical EngineeringEngineering Computing Panel

MKMM 1213 Advanced Engineering Mathematics

Vector-Valued Functions and Motion in Space

Abu Hasan Abdullah

April 2015

noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vector-Valued Functions & Motion 1 / 32

Outline I

1 Curves in Space and Their Tangents

Overview

Limits and Continuity

Derivatives and Motion

Differentiation Rules

Vector Functions of Constant Length

2 Integrals of Vector Functions; Projectile Motion

Integrals of Vector Functions

3 Arc Length in Space

Arc Length Along a Space Curve

Speed on a Smooth Curve & Unit Tangent Vector

4 Curvature and Normal Vectors of a Curve

Curvature of a Plane Curve

5 Curvature and Normal Vectors of a Curve

6 Tangential and Normal Components of Acceleration

7 Velocity and Acceleration in Polar Coordinates

Motion in Polar and Cylindrical Coordinates

Planets Move in Planes

8 Bibliographynoone�dev.null MKMM 1213 Advanced Engineering Mathematics Vector-Valued Functions & Motion 2 / 32

Curves in Space and Their TangentsOverview

When a particle moves through space

during a time interval I, we think of the

particle’s coordinates as functions defined

on I:

x = f(t), y = g(t), z = h(t), t ∈ I. (1)

The points (x, y, z) = (f(t), g(t), h(t)), t ∈ I,

make up the curve in space that we call the

particle’s path. The equations and interval

in Eq. (1) parametrize the curve.

A curve in space can also be represented in

vector form:

r(t) =⇀

OP = f(t)i + g(t)j + h(t)k (2)

from the origin to the particle’s position

P(f(t), g(t), h(t)) at time t is the particle’s

position vector—see Figure 1).

Figure 1: The position vector r =

⇀OP of a

particle moving through space is afunction of time.


Curves in Space and Their TangentsOverview

The functions f , g, and h are the

component functions (components)

of the position vector. We think of

the particle’s path as the curve

traced by r during the time interval

I.

Eq. (2) defines r as a vector function

of the real variable t on the interval

I. More generally, a vector-valued

function or vector function on a

domain set D is a rule that assigns a

vector in space to each element in D.

For now, the domains will be

intervals of real numbers resulting

in a space curve.

When the domains are regions in

the plane, vector functions will then

represent surfaces in space.

Vector functions on a domain in the

plane or space also give rise to

vector fields, which are important to

the study of the flow of a fluid,

gravitational fields, and

electromagnetic phenomena.

Real-valued functions are called scalar functions to distinguish them from vector

functions. The components of r in Eq. (2) are scalar functions of t. The domain of a

vector-valued function is the common domain of its components.noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vector-Valued Functions & Motion 4 / 32

Curves in Space and Their TangentsLimits and Continuity

DEFINITION: Limits of vector-valued functions

Let r = f(t)i + g(t)j + h(t)k be a vector function with domain D, and L a vector.

We say that r has limit L as t approaches t0 and write

limt→t0

r(t) = L

if, for every number ǫ > 0, there exists a corresponding number δ > 0 such that

for all t ∈ D

|r(t) − L| < ǫ whenever 0 < |t − t0| < δ.

If L = L1i + L2j + L3k, then it can be shown that limt→t0 r(t) = L precisely when

limt→t0

f(t) = L1, limt→t0

g(t) = L2, and limt→t0

h(t) = L3

Omitting the the proof (for now), the equation

limt→t0

r(t) =

„

limt→t0

f(t)

«

i +

„

limt→t0

g(t)

«

j +

„

limt→t0

h(t)

«

k (3)

provides a practical way to calculate limits of vector functions.noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vector-Valued Functions & Motion 5 / 32

Curves in Space and Their TangentsLimits and Continuity

DEFINITION: Continuity for vector functions

A vector function r(t) is continuous at a point t = t0 in its domain if limt→t0 r(t) =r(t0). The function is continuous if it is continuous at every point in its domain.

From Eq. (3), we see that r(t) is continuous at t = t0 if and only if each component

function is continuous there.

Example 1:

Graph the vector function

r(t) = (cos t)i + (sin t)j + tk

Example 2:

If r = (cos t)i + (sin t)j + tk, find

limt→π/4 r(t).

Example 3:

The function

g(t) = (cos t)i + (sin t)j + tk

is discontinuous at every integer,

where the greatest integer function t

is discontinuous.noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vector-Valued Functions & Motion 6 / 32

Curves in Space and Their TangentsDerivatives and Motion

Suppose that r(t) = f(t)i + g(t)j + h(t)k is the position vector of a particle moving

along a curve in space and that f , g, and h are differentiable functions of t. Then

the difference between the particle’s positions at time t and time t + ∆t is

∆r = r(t + ∆t) − r(t)

Figure 2: Derivative of a vector function.

In terms of components,

∆r = r(t + ∆t) − r(t)

= [ f(t + ∆t)i + g(t + ∆t)j + h(t + ∆t)k] − [ f(t)i + g(t)j + h(t)k]noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vector-Valued Functions & Motion 7 / 32


The quotient ∆r/∆t approaches the limit

lim∆t→0

∆r

∆t=

»

f(t + ∆t)f(t)

∆t

–

i +

»

g(t + ∆t)g(t)

∆t

–

j +

»

h(t + ∆t)h(t)

∆t

–

k

=

»

df

dt

–

i +

»

dg

dt

–

j +

»

dh

dt

–

k

Figure 3: Derivative of a vector function.noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vector-Valued Functions & Motion 8 / 32


DEFINITION: Derivative of a vector function

The vector function r(t) = f(t)i + g(t)j + h(t)k has a derivative (is differentiable)

at t if f , g, and h have derivatives at t. The derivative is the vector function

r′(t) =dr

dt= lim

∆t→0

r(t + ∆t) − r(t)

∆t=

df

dti +

dg

dtj +

dh

dtk

A vector function r is differentiable if it is differentiable at every point of its

domain. The curve traced by r is smooth if dr/dt is continuous and never 0, that is,

if f , g, and h have continuous first derivatives that are not simultaneously 0.

The vector r′(t), when different from 0, is defined to be the vector tangent to the

curve at P. The tangent line to the curve at a point f(t0), g(t0), h(t0) is defined to

be the line through the point parallel to r′(t0).



We require dr/dt 6= 0 for a smooth

curve to make sure the curve has a

continuously turning tangent at

each point.

On a smooth curve, there are no

sharp corners or cusps!

A curve that is made up of a finite

number of smooth curves pieced

together in a continuous fashion is

called piecewise smooth (Figure 4).Figure 4: A piecewise smooth curve made upof five smooth curves connected end to end ina continuous fashion. The curve here is notsmooth at the points joining the five smoothcurves.



DEFINITION: Velocity & Acceleration

If r is the position vector of a particle moving along a smooth curve in space, then

v(t) =dr

dt

is the particle’s velocity vector, tangent to the curve. At any time t, the direction

of v is the direction of motion, the magnitude of v is the particle’s speed, and

the derivative a = dv/dt, when it exists, is the particle’s acceleration vector. In

summary,

1 Velocity is the derivative of position: v =dr

dt.

2 Speed is the magnitude of velocity: Speed = |v|.

3 Acceleration is the derivative of velocity: a =dv

dt=

d2r

dt2.

4 The unit vector v/|v| is the direction of motion at time t.noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vector-Valued Functions & Motion 11 / 32


Example 4:

Find the velocity, speed, and acceleration of

a particle whose motion in space is given

by the position vector

r(t) = 2 cos ti + 2 sin tj + 5 cos 2tk.

Sketch the velocity vector v(7π/4).

Figure 5: The curve and the velocityvector when t = 7π/4 for the motion givenin Example 4.


Curves in Space and Their TangentsDifferentiation Rules

Let u and v be differentiable vector functions of t, C a constant vector, c any scalar, and f

any differentiable scalar function.

Constant Function Rule:

d

dtC = 0

Scalar Multiple Rule:

d

dt[cu(t)] = cu′(t)

d

dt[ f(t)u(t)] = f

′(t)u(t) + f(t)u′(t)

Sum Rule:

d

dt[u(t) + v(t)] = u′(t) + v′(t)noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vector-Valued Functions & Motion 13 / 32

Curves in Space and Their TangentsDifferentiation Rules

Continued . . .

Difference Rule:

d

dt[u(t) − v(t)] = u′(t) − v′(t)

Dot Product Rule:

d

dt[u(t) · v(t)] = u′(t) · v(t) − u(t) · v′(t)

Cross Product Rule:

d

dt[u(t) × v(t)] = u′(t) × v(t) − u(t) × v′(t)

Chain Rule:

d

dt[u(f(t))] = f

′(t)u′(f(t)t)noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vector-Valued Functions & Motion 14 / 32

Curves in Space and Their TangentsVector Functions of Constant Length

Continued . . .

For a particle moving on a sphere,

Figure 6, the position vector has a

constant length equal to the radius of

the sphere. The velocity vector dr/dt,

tangent to the path of the motion, is

tangent to the sphere and hence

perpendicular to r.

The vector and its first derivative are

orthogonal. By direct calculation

r(t) · r(t) = c2

d

dt[r(t) · r(t)] = 0

r′(t) · r(t) + r(t) · r′(t) = 0

2r′(t) · r(t) = 0

Figure 6: Particle moving on a sphere.

The vectors r′(t) and r(t) are

orthogonal because their dot product

is 0. In summary,

r · dr

dt= 0 (4)noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vector-Valued Functions & Motion 15 / 32

Integrals of Vector Functions; Projectile MotionIntegrals of Vector Functions

A differentiable vector function R(t) is an antiderivative of a vector function r(t) on

an interval I if dR/dt = r at each point of I.

DEFINITION: Indefinite Integral

The indefinite integral of r with respect to t is the set of all antiderivatives of r,

denoted byR

r(t)dt. If R is any antiderivative of r, then

Z

r(t)dt = R(t) + C

Example 1:

To integrate a vector function, we integrate each of its components.

Z

((cos t)i + j − 2tk)dt =

„

Z

cos tdt

«

i +

„

Z

dt

«

j −„

Z

2tdt

«

k

= (sin t + C1)i + (t + C2)j − (t2 + C3)k

= (sin t)i + tj − t2k + Cnoone�dev.null MKMM 1213 Advanced Engineering Mathematics Vector-Valued Functions & Motion 16 / 32


DEFINITION: Definite Integral

If the components of r(t) = f(t)i + g(t)j + h(t)k are integrable over [a, b], then so

is r, and the definite integral of r from a to b is

Z b

a

r(t)dt =

„

Z b

a

f(t)dt

«

i +

„

Z b

a

g(t)dt

«

j +

„

Z b

a

h(t)dt

«

k

Example 2:As in Example 1, we integrate each component.

Z π

0

((cos t)i + j − 2tk)dt =

„

Z π

0

cos t dt

«

i +

„

Z π

0

dt

«

j −„

Z π

0

2t dt

«

k

= [sin t]π0 i + [t]π0 j −h

t2iπ

0k

= [0 − 0] i + [π − 0] j −h

π2 − 02i

k

= πj − π2knoone�dev.null MKMM 1213 Advanced Engineering Mathematics Vector-Valued Functions & Motion 17 / 32


Example 3:

Suppose we do not know the path of a

hang glider, but only its acceleration

vector:

a(t) = −(3 cos t)i − (3 sin t)j + 2k.

We also know that initially (at time

t = 0) the glider departed from the

point (4, 0, 0) with velocity v(0) = 3j.Find the glider’s position as a function

of t.

Figure 7: Flight path of a hang glider.


Arc Length in SpaceArc Length Along a Space Curve

Smooth space and plane curves have a

measurable length, thus allowing us

to locate points along these curves by

giving their directed distance s along

the curve from some base point, see

Figure 8. Figure 8: Arc length of a curve.

DEFINITION: Arc Length of a Curve

The length of a smooth curve r(t) = x(t)i + y(t)j + z(t)k, a ≤ t ≤ b, that is traced

exactly once as t increases from t = a to t = b, is

L =

Z b

a

s

„

dx

dt

«2

+

„

dy

dt

«2

+

„

dz

dt

«2

dt (5)

The square root in Eq. 5 is |v|, i.e. the length of a velocity vector dr/dt, thus

L =

Z b

a

|v|dt (6)noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vector-Valued Functions & Motion 19 / 32


Example 1:

A glider is soaring upward along the

helix


Figure 9. How long is the glider’s path

from t = 0 to t = 2π?

Figure 9: A helix.



If we choose a base point P(t0) on a

smooth curve C parametrized by t,

each value of t determines a point

P(t) = (x(t), y(t), z(t)) on C and a

directed distance

s(t) =

Z t

to

|v(τ )| dτ (7)

measured along C from point P(t0). Figure 10: The directed distance.

If t > t0, s(t) is the distance along the curve from P(t0) to P(t). If t < t0, s(t) is the

negative of the distance.

Each value of s determines a point on C, and this parametrizes C with respect to s.

We call s an arc length parameter for the curve.

s(t) =

Z t

to

p

[ x′(τ )]2 + [ y′(τ )]2 + [ z′(τ )]2 =

Z t

to

|v(τ )| dτ (8)noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vector-Valued Functions & Motion 21 / 32


Example 2:

This is a continuation of Example 1 for which we can actually find the arc length

parametrization of a curve. If t0 = 0, the arc length parameter along the helix


from t0 to t is

s(t) =

Z t

to

|v(τ )|dτ =

Z t

0

√2 dτ =

√2 t

Solving this equation for t gives t = s/√

2. Substituting into the position vector r

gives the following arc length parametrization for the helix:

r(t(s)) =

„

coss√2

«

i +

„

sins√2

«

j +

„

s√2

«

k

P.S.: Unlike this example, the arc length parametrization is generally difficult to find analytically for a

curve already given in terms of some other parameter t.noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vector-Valued Functions & Motion 22 / 32

Arc Length in SpaceSpeed on a Smooth Curve & Unit Tangent Vector

The arc length s is a differentiable function of t with derivative

ds

dt= |v(t)| (9)

Eq. 9 is the speed with which a particle moves along its path is the magnitude of v.

The velocity vector v = dr/dt is

tangent to the curve r(t) and that the

vector

T =v

|v|ds

dt= |v(t)| (10)

is the unit tangent vector, Figure 11.

Figure 11: Unit tangent vector.noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vector-Valued Functions & Motion 23 / 32


Example 3:

Find the unit tangent vector of the

curve

r(t) = (1 + 3 cos t)i + (3 sin t)j + t2k

representing the path of a glider.

Figure 12: Counterclockwise motion aroundthe unit circle.



The velocity vector is the change in the position vector r with respect to time t, but

how does the position vector change with respect to arc length, s? More precisely,

what is the derivative dr/ds?

Since ds/dt > 0 for the curves we are considering, s is one-to-one and The

derivative of the inverse is

dt

ds=

1

ds/dt=

1

|v|

This makes r a differentiable function of s whose derivative can be calculated with

the Chain Rule to be

dr

ds=

dr

dt

dt

ds= v

1

|v| =v

|v| = T (11)


Curvature and Normal Vectors of a CurveCurvature of a Plane Curve

As a particle moves along a smooth

curve in the plane, T = dr/ds turns as

the curve bends. Since T is a unit

vector, its length remains constant

and only its direction changes as the

particle moves along the curve. See

Figure 13.Figure 13: Curvature of a curve.

DEFINITION: Curvature

Curvature, κ, is the rate at which T turns per unit of length along the curve,

κ =

˛

˛

˛

˛

dT

ds

˛

˛

˛

˛

If dT/ds is large, T turns sharply and the curvature at P is large. If dT/ds is close to

zero, T turns more slowly and the curvature at P is smaller.noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vector-Valued Functions & Motion 26 / 32

Velocity and Acceleration in Polar CoordinatesMotion in Polar and Cylindrical Coordinates

When a particle at P(r, θ) moves along

a curve in the polar coordinate plane,

we express its position, velocity, and

acceleration in terms of the moving

unit vectors

ur = (cos θ)i + (sin θ)j (12a)

uθ = −(sin θ)i + (sin θ)j (12b)

See Figure 14.

We differentiate ur and uθ with

respect to θ to get

dur

dθ= −(sin θ)i + (cos θ)j

duθ

dθ= −(cos θ)i − (cos θ)j

Figure 14: Position vector in polar coordinates.

We differentiate ur and uθ with

respect to t to see how they change

with time

ur =dur

dθθ = θuθ (13a)

uθ =duθ

dθθ = −θur (13b)noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vector-Valued Functions & Motion 27 / 32


The velocity vector is expressed in

terms of ur and uθ as

v = r =d

dt(rur)

= rur + rur

= rur + rθuθ

See Figure 15.

The acceleration is

a = v

= (rur + rur) + (rθuθ + rθuθ + rθuθ)

Figure 15: Velocity vector in polar coordinates.

Acceleration in terms of ur and uθ is

a = (r − rθ2)ur + (rθ + 2rθ)uθ.



To extend these equations of motion to space, we add zk to the right-hand side of

the equation r = rur. Then, in cylindrical coordinates, we have

r = rur + zk (14a)

v = rur + rθuθ + zk (14b)

a = (r − rθ2)ur + (rθ + 2rθ)uθ + zk (14c)

The vectors ur, uθ, and k make a

right-handed frame in which

ur × uθ = k, uθ × k = ur, k × ur = uθ.

See Figure 16.

Figure 16: Position vector and basic unitvectors in cylindrical coordinates.noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vector-Valued Functions & Motion 29 / 32

Velocity and Acceleration in Polar CoordinatesPlanets Move in Planes

Newton’s law of gravitation says that if r is

the radius vector from the centre of a sun

of mass M to the center of a planet of mass

m, then the force F of the gravitational

attraction between the planet and sun is

F = −GmM

|r2|r

|r|

where G ≈ 6.6726× 10−11 Nm2 kg−2 is the

universal gravitational constant. See

Figure 17.

Combining the gravitation law with

Newton’s second law, F = mr, for the force

acting on the planet gives

mr = −GmM

|r2|r

|r| =⇒ r = −GM

|r2|r

|r|

Figure 17: The force of gravity is directedalong the line joining the centers of mass.

The planet is accelerated toward

the sun’s center of mass at all

times.


Velocity and Acceleration in Polar CoordinatesPlanets Move in Planes

Since r is a scalar multiple of r, we have

r × r = 0.

and from this

d

dt(r × r) = r × r + r × r = r × r = 0.

It follows that

r × r = C (15)

for some constant vector C.

Eq. (15) tells us that r and r always lie in a

plane perpendicular to C. Hence, the planet

moves in a fixed plane through the center

of its sun. See Figure 18.

Figure 18: A planet that obeys Newton’slaws of gravitation and motion travels inthe plane through the sun’s center of mass.

The planet is accelerated toward

the sun’s center of mass at all

times.


Bibliography

1 PETER V. O’NEIL (2012): Advanced Engineering Mathematics, 7ed,ISBN-13: 978-1-111-42741-2, Cengage Learning

2 DANIELA FLEISCH (2012): A Student’s Guide to Vectors and Tensors,ISBN: 978-0-521-17190-8, Cambridge University Press

3 ERWIN KREYSZIG (2011): Advanced Engineering Mathematics, 10ed,ISBN: 978-0-470-45836-5, John Wiley & Sons

4 ALAN JEFFREY (2002): Advanced Engineering Mathematics,ISBN: 0-12-382592-X, Harcourt/Academic Press

5 GLYNN JAMES ET AL. (2011): Advanced Modern Engineering Mathematics, 4ed,ISBN: 978-0-273-71923-6, Pearson Education

6 L. BRIGGS ET AL. (2013): Calculus for Scientists and Engineers: Early Transcendentals,ISBN-13: 978-0-321-78537-4, Pearson Education


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