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465Chapter 10 Vector calculus

10 10A Position, velocity and acceleration 10B Cartesian equations and

antidifferentiation of vectors 10C Applications of vector calculus 10D Projectile motion

Vector calculusPosition vector as a function of time • r(t); deriving the Cartesian equation of a path given r(t) and sketching the pathDifferentiation and anti-differentiation of a • vector function with respect to timeApplication of vector calculus to curvilinear • motion, in simple cases where the vectors

r (position),

r (velocity) and

r (acceleration) are

given as functions of timeMotion along a path in two or three dimensions; • examples could include:

i A particle moves so that its position vector r

at time t, t ≥ 0, is given in a vector parametric t ≥ 0, is given in a vector parametric tform such as:

r = 3 cos (2t)

i + 4 sin (2t)

j

or r = tan (t)

i + 4 sec2 (t)

j . Answer questions

about the equation of the path, and the magnitude and direction of the velocity and acceleration.

iir is the position vector at time t of a particle t of a particle tmoving in a plane in such a way that

r ir ir i=r i5r i5r i always, and when t = 0,

r = 0 and

r ir i jr i= =r i7 1r i7 1r i= =7 1= =r i= =r i7 1r i= =r i 0 . Find the position vector at time tand answer questions about the motion and its path.

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Vector calculus

Position, velocity and accelerationintroductionWhen any object — from a golf ball to an athlete — travels in two or more dimensions, vectors must be used if we are to fully record and study its motion. The general principles of kinematics still apply; but the position, velocity and acceleration of the object are vector quantities and these involve both magnitude and direction.

In addition, it is worth recalling the following.

1. The dot product or scalar product of two vectors b and

c where

b = bx

i + by

j and

c = cx

i + cy

j is given by

b c. =

c b. = bxcx + bycy

2. The magnitude or length of a vector, b is given by

b = b =

b b b bx y. = +2 2

3. The dot or scalar product of two vectors b and

c is related to their respective magnitudes

and the angle θ between them:

b c. =

b c cos (θ )

10a

466

4. Two non-zero vectors are perpendicular when b c. = 0. This is because

b c. =

b c cos (θ )

= 0 if and only if cos (θ ) = 0 that is, if θ = 90° and hence

b and

c are perpendicular.

PositionLet us first consider a particle which moves along the curved path ABCD as shown at right.

The point O is the origin. The position of the particle from the origin at any time, t, is given by the vector

r t( ). At a time T, the particle is

at the point B on the curve and at a later time, T + ∆ t, the particle has moved to the point C. The position vector at time T, when the particle is at B, is

r T( ); likewise, at C it is

r T t( )+ ∆ .

DisplacementThe displacement of a particle is the change of its position over a given time interval. The displacement of the particle in the diagram above during the time interval ∆ t is given by:

displacement = r T t r T( ) ( )+ −∆

which is the difference between the final position at the time T + ∆ t and the initial position at time T.

DistanceThe distance from the origin of a particle at any time, t, is given by the magnitude of

r t( ):

r t r t r t r t( ) ( ) ( ) ( )= = .

WorkeD examPle 1

Find the distance from the origin of the body described by the position vector

r t ti t j k( ) ( )= + − −2 1 52 at t = 2.

Think WriTe

1 Substitute t = 2 into r(t) to find

r(2).

r i j k( )2 4 3 5= + −

2 Evaluate r(2) =

r r( ) ( ).2 2 to find the distance from

the origin. r( )2

= + + −

=

24 3 5

50

2 2 ( )

3 Simplify the surd. = 5 2

4 State the distance from the origin when t = 2. The body is 5 2 units from the origin when t = 2.

average velocityThe average velocity

vav during the time interval ∆ t is given by the displacement vector divided

by the magnitude of the time interval. Thus:

vr T t r t

tav = + −( ) ( )∆∆

O

A

B

CD

r(T )~

r(T + ∆t )~

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WorkeD examPle 2

Find the average velocity of the body with position vector

r t t i

tj k( ) = + −

+13

52

between t = 0 and t = 3.

Think WriTe

1 Find r when t = 0 and when t = 3.

r i k( )0 5= +

r i j k( )3 2 3 5= − +

2 Simplify r r( ) ( )3 0− .

r r i j k

i j

( ) ( ) ( ) ( ) ( )3 0 2 1 3 0 5 5

3

− = − + − − + −

= −

3 Calculate the average velocity by simplifying

r r( ) ( )3 0

3 0−−

.

vi j

i j

i j

av =−−

=−

= −

3

3 03

313

4 State the average velocity. The average velocity is 13 i j− .

instantaneous velocityTo obtain the instantaneous velocity, v(T), of the particle at the point B, we apply a limit and make ∆ t → 0. Thus the instantaneous velocity of the particle at time T is:

v Tr T t r T

tt( ) lim

( ) ( )= + −→∆

∆∆0

= dr

dt

This means that the instantaneous velocity vector v t( ) is given by the time derivative of

the position vector r t( ). As ∆ t → 0 the point C approaches B and so the velocity vector has a

direction parallel to the tangent of the curved path of the particle.

The direction of motion of a particle is in the direction parallel to the velocity vector.

rules for differentiating vectorsThere are four intuitively obvious results which are used in differentiating vector quantities.Rule 1. The derivative of a unit vector equals zero:

di

dt = 0;

dj

dt = 0 .

Since both the unit vectors i and

j have constant magnitude and direction they are constants

and hence their derivative equals zero.Rule 2. The derivative of a constant vector ai

equals zero. Using the product rule:

d ai

dta

di

dtdadt

i

a i

( )

= +

= × + ×=

0 0

0

468

Rule 3. The derivative of a vector xi whose scalar coeffi cient x(t) is a function of time is given as:

d xi

dtx

di

dtdxdt

i

dxdt

( )

=

+

= +

0

=

i

dxdt

i

Rule 4. The derivative of the sum of two vectors is the sum of the derivative of each vector. In the case of the position vector

r x t i y t i= +( ) ( ) :

drdt

d xi yj

dtdxdt

idydt

j

v t

=+

= +

=

( )

( )

SpeedThe speed of a particle is a scalar quantity; that is, it has magnitude but not direction. Therefore, the speed of a particle is the magnitude of the velocity vector.

Speed = v t( ) = =.

v t v t v t( ) ( ) ( )

WorkeD examPle 3

An object has a position vector in metres r t t t i e t jt( ) ( sin ( )) ( )= − + +−3 22 π ;

t ≥ 0 seconds.a Find the velocity vector

v t( ).

b Find the velocity of the object at t = 2 s.c Find the speed of the object at t = 2 s.d Find the average velocity in the fi rst 2 seconds of the body’s motion.

Think WriTe

a 1 The velocity vector

v tdrdt

( ) = is found by

differentiating each vector component of

r t( ) with respect to time.

a

v tdrdtddt

t t i e t jt

( )

[ sin ( )] ( )

=

= − + +( −3 22 π )) = − + +− −[ cos ( )] ( )6 2t t i e jtπ π

2 State the velocity. The velocity of the body is given by

v t t t i e jt( ) [ cos ( )] ( )= − + +− −6 2π π .

b 1 The velocity at time t = 2 is found by substitution into the expression for

v t( ).

b v i e j( ) [ cos ( )] ( )2 12 2 22= − + +− −π π

2 Simplify the value of v( )2 .

= − + −

( )12 212

π i

ej

3 State the velocity. The velocity of the object at a time t =2 s is

v i

ej( ) .2 12 2

12

= −( ) + −

π

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Worked example 3



c 1 The speed of the object at time t = 2 is given by the magnitude of the vector

v( )2

— that is, the speed is equal to v v( ) ( ).2 2 .

c speed, v v v( ) ( ) ( ).2 2 2=

2 Calculate the speed using the velocity vector,

v( )2 , found in part b .

= −( ) + −

12 212

2

2

πe

3 Evaluate, correct to 2 decimal places. ≈ 9.05

4 State the approximate speed. The speed of the object is 9.05 m/s.

d 1 The average velocity in the first 2 seconds is the displacement of the object divided by the time interval (that is, 2 seconds).

d v r rav = − ÷[ ( ) ( )]2 0 2

2 The displacement is given by r r( ) ( )2 0− ,

this being the difference in the position of the object from t = 0 until t = 2.

= −( ) + −

− +( )

÷12 2

10 1 2

2π i

ej i j

3 Simplify.

= −( ) + −

÷12 1

12

2π i

ej

4 Calculate each component correct to 3 decimal places.

≈ 4.429i + 0.432

j

5 State the approximate average velocity. The average velocity in the first two seconds is

v i jav = +4 429 0 432. . m/s.

accelerationWe can, in a similar way to that shown in the previous section, obtain a vector which describes the acceleration of an object from the instantaneous velocity vector. Consider a particle travelling on the path ABCD, as before. At the point B the object has a velocity given by

v T( ). At a later time,

T + ∆ t, the body is now at the point C with a velocity v T t( )+ ∆ . This situation is shown below.

A

B

CD

v(T + ∆t )~

a∆t~ −v(T )~

v(T )~

v(T + ∆t )~

average accelerationThe average acceleration of the object,

aav, is given as the change in velocity, ∆

v, divided by the

time interval, ∆ t. Thus:

av T t v T

tav =+ −( ) ( )∆

∆.

instantaneous accelerationTo obtain the instantaneous acceleration,

a T( ), of the particle at the point B, we apply a limit and

make ∆ t → 0. Thus, the instantaneous acceleration of the particle at time T is:

a Tv T t v t

tt( ) lim

( ) ( )=

+ −→∆

∆∆0

=

dvdt

470

This means that the instantaneous acceleration vector a t( ) is given by the time derivative of

the velocity vector v t( ).

WorkeD examPle 4

Find the acceleration vector a t( ) for the object whose position vector is:

r t e i t t j t t kt

e( ) ( ) [ log ( )]= − + +2 3 .

Think WriTe

1 Find v t( ) by differentiating each vector component of

r t( ) with respect to time.

v t e i t j t kte( ) ( ) [ log ( )]= − + + +2 3 1 12 2

2 Find a t( ) by differentiating each vector component of

v t( ) with respect to time.

a t e i tjt

kt( ) = − +

4 612

WorkeD examPle 5

A body moves in such a way that its position vector in metres at an instant t seconds is given by:

r t t t i t t j t( ) ,= − +

+ −

1512

1 1213

2 3 ≥≥ 0.

Find:a the velocity vector

v t( )

b the acceleration vector a t( )

c the angle between the velocity vector and the acceleration vector of the body at a time t = 1 s, to the nearest degree

d the time when the body has an acceleration of magnitude 9.8 m/s2.

Think WriTe

a 1


Differentiate r, the position vector, with respect to t, using a CAS calculator, by completing the following steps.Defi ne the position function r(t). To do this, tap:• Action• Command• Defi neComplete the entry line as shown at right and press E.To fi nd v(t), differentiate r with respect to t. Tap:• Active• Matrix-Create• matToListComplete the line matToList(r(t),1) E.Complete the entry line as:ddt

(ans)

Then press E.


2 Write the solution using correct vector notation.

v tdrdt

( ) =

v t t i t j( ) ( ) ( )= − + −15 12 2

b 1


a tdvdt

( ) =

a t i tj( ) = −− 2

c 1

To fi nd the acceleration vector a(t), complete the following steps.Defi ne the velocity function v(t) by completing the entry line as:

Defi ne v(t) = 15

12 2

−−

t

tThen press E.Differentiate v(t) with respect to t, by completing the entry line as:matToList(v(t),1) then pressing E. Enter:ddt

(ans)

Then press E.

To fi nd the angle between the velocity and acceleration vectors, complete the following steps.Defi ne the acceleration function:a t( )a t( )a t [ ]t[ ]t= [ ]− −[ ]1 2[ ]1 2[ ][ ]− −[ ]1 2[ ]− −[ ]Calculate the velocity vector when t = 1, by fi nding v (1) and the acceleration vector when t = 1, by fi nding a(1).Calculate the magnitude of vectors v(1) and a(1) by completing the entry lines as:norm(v(1))norm(a(1))Press E after each entry.Calculate the dot product a(1) . v(1)(Refer to the previous chapter, Vectors, to review the dot product).

472

2 The angle, θ, between the two vectors

v( )1 and

a( )1 is found using the equation

a v a v( ) ( ) ( ) ( ) cos ( ).1 1 1 1= × × θ

This can be done by hand or using the calculator. Write the solution using correct vector notation.

v i j( )1 14 11= +

a i j( )1 2= −−

cos ( )( ) ( )

( ) ( )

.θ =

×

a v

a v

1 1

1 1

v( )1 317=

a( )1 5=

a v( ) ( ).1 1 36= −

cos ( )θ =− 36

317 5

θ = °155 to the nearest degree.The angle between the two vectors is approximately 155°.

d a t i tj( ) = −− 2

a t t( ) = +1 4 2

Solving 1 4 9 82+ =t . for t gives t t≈ ≥4 87 0. , .

When t = 4.87 s the magnitude of the acceleration is 9.8 m/s2.


Use the CAS calculator to solve for twhen the acceleration is equal to 9.8 m/s2. Complete the entry line as:

Solve( )( ) . , )( )1 4( ) 9 8. ,9 8. ,2( )2( )( )+ =( )( )1 4( )+ =( )1 4( )t t. ,t t. ,t t( )t t( ) 9 8t t9 8. ,9 8. ,t t. ,9 8. ,( )2( )t t( )2( )+ =t t+ =( )+ =( )t t( )+ =( )( )2( )+ =( )2( )t t( )2( )+ =( )2( )Then press E.Note: Only positive solutions for t are allowed.


Two non-zero vectors 1. a and

b are perpendicular when

a b. = 0.

The angle between two vectors 2. θ is given by cos (θ ) =

a ba b

. where

a and

b are vectors.

Position 3. r t( ), velocity

v t( ) and acceleration

a t( ) are vector quantities. In two

dimensions they can be written as:

r t xi yj( ) = +

v t v i v jx y( ) = + [Note:

v t r tdrdt

dxdt

idydt

j( ) ( )= = = + ]

a t a i a jx y( ) = + , respectively. [Note:

a t r t

d r

dt

d x

dti

d ydt

j( ) ( )= = = +2

2

2

2

2]

The average velocity during the time interval 4. ∆ t is

vr t t r t

tav = + −( ) ( )∆∆

.

The instantaneous velocity 5.

v t r tdrdt

dxdt

idydt

j( ) ( )= = = + .

The instantaneous speed is given by 6. v v v vx y. = +2 2 which is the magnitude of the

velocity vector v.

The direction of motion is in the direction parallel to the velocity vector.7.

The instantaneous acceleration 8.

a tdv

dt

dv

dti

dv

dtj

d x

dti

d y

dtx y( ) = = + = +

2

2

2

2 j .

rememBer

Position, velocity and acceleration 1 We 1 For the body described by the following position vectors, find the distance from the

origin at the times listed using distance = r r r. = .

a r t i t j( ) = +3 2 2 at t = 2

b r t ti t j k( ) ( )= − + +2 1 42 at t = 3

c r t t i t j( ) cos ( ) sin ( )= −4 4 at t = 3π

d r t t i t j t ke( ) log ( ) ( )= + + − −2 10 1 2 12 at t = 1

2 We 2 Find the average velocity of the bodies described in question 1 from t = 0 to the given time.

3 We 3a Find the velocity vector for each of the following position vectors:

a r t i tj( ) = −3 2

b r t( ) = t i t j t k2 32 1

+ − + +( )

c r t t i t j( ) sin ( ) cos ( )= + −3 4 2

d r t e i t jt( ) = −− −4 22 1

e r t t i t t j t ke( ) log ( ) ( ) ( )= − + + − −1 3 2 43 2

f r t t t i te jt( ) sin ( )= +2

g r t t i t j( ) cos ( ) sin ( )= +3 2 3 2

exerCiSe

10a

474

4 We 3b,c,d Find, by calculating v v. , the speed of the objects having the following position

vectors. In addition, calculate: i the speed at t = 2 in each case ii the average velocity in the first 2 seconds of the object’s motion.

a r t t i t j( ) ( )= − +2 2 1

b r t t t i tj t k( ) ( )= − − +3 22

c r t t i tje( ) log ( )= + −3 1 4

d r t t i t j( ) cos ( ) sin ( )= −4 4

5 We 4 Find the acceleration vector a t( ) for the following objects whose position vectors

are:a r t i t j( ) = −3 2

b r t t i t j t k( ) ( )= − + +2 2 1 12

c r t t i t j t t k( ) sin ( ) cos ( ) ( )= + − +−3 4 2 3

d r t e i t jt( ) = −− −4 22 1

e r t t i t t j

tke( ) log ( ) ( )= − + + −

1 3 213

f r t t i te jt( ) sin ( ) .= +

6 mC The position of an object is given by r t t i tj( ) = −2 6 .

a The velocity at t = 2 is equal to:

A 52 B 4 12 i j− C 4 6

i j− D 2 3

i j− E 4 10

b The average velocity for the interval [0, 2] is:

A 2 3 i j− B 2 6

i j− C 13 D 4 12

i j− E 4 6

i j−

c The acceleration at any time, t, is equal to:

A 2 6ti j

− B 2 6 i j− C 2

i D 2 E

i j− 3

7 mC The velocity of a particle at any time, t, is given by

r t t i

tj( ) sin ( )

( ).= +

+2

9

1 2π

a The acceleration at t = 2 is equal to:

A − 3j B 2 3

i j− C 2 3π

i j+ D 2 3

i j+ E 2 2

3π i j−

b The angle between the velocity vector and acceleration vector when t = 2, to the nearest degree, is:A 96° B 90° C 64° D 146° E 34°

8 An object moves with a position vector r t t t i t j( ) ( )= − + −3 12 for t > 0.

a At t = 3, determine the position vector and hence the distance of the object from the origin.

b Find the velocity vector v t( ).

c Find the speed of the object at t = 2.d At what time is the object travelling parallel to the unit vector

j?

9 An object moves with a position vector r t t i tj( ) sin ( )= +3π for t ≥ 0.

a At t = 3, determine the position vector and hence the distance of the object from the origin.

b Find the velocity vector v t( ).

c Find the speed of the object at t = 2.d What is the maximum speed of the object?



10 An object moves with a position vector r t( ), expressed in metres, at a time, t seconds:

r t

ti t j( ) =

++1

1 for t ≥ 0

a At t = 2, determine the position vector and hence the distance of the object from the origin.b Find the velocity vector

v t( ).

c Find the speed of the object at t = 2.d Show that after a long time the body travels at a speed of 1 m/s parallel to the unit

vector j .

11 We 5 The position of an object at any time, t, is given as follows: r t t i t j t k( ) = + −3 62 3 .

Find:a the velocity vector

v t( )

b the acceleration vector a t( )

c the angle between the velocity and acceleration vectors when t = 1d when the object is moving in a direction parallel to the unit vector

k.

12 A body moves with a velocity

r t t i t j( ) cos( ) sin ( )= +4 3 4 3 .

a Show that the speed is constant.b Find the magnitude of the acceleration at any time, t.

c When is the velocity first equal to − +2 2 2 2 i j?

d Find the angle between r t( ) and

r t( ) when t = π

6.

Cartesian equations and antidifferentiation of vectorsCartesian equationsWe sometimes need to know the equation for the path of a body being described by its position vector

r t( ). The equation for the path y(x) is established from the position vector

r t x t i y t j( ) ( ) ( )= + . The function y(x) will be referred to as the Cartesian equation for the motion. The functions x(t) and y(t) are referred to as parametric equations.

Generally, three steps are followed in order to obtain the Cartesian equation.1. From the parametric equation x(t), express t as a function of x.2. Substitute t(x) into the parametric equation y(t) to obtain y(x).3. Simplify.Note: The domain of the Cartesian equation is the range of x(t). The range of the Cartesian equation is the range of y(t).

WorkeD examPle 6

Find the Cartesian equation for the path of the object with position vector given by

r t ti t t j( ) ( )= + −5 42 .

Think WriTe

1 Write down the parametric equation for x(t). x(t) = 5t

2 Express t as a function of x. tx=5

3 Write the parametric equation for y(t). y(t) = t2 − 4t

10B

476

4 Substitute tx=5

into this equation to obtain the

Cartesian equation.

or yx x=

−5

45

2

5 Simplify the equation for y(x). yx x= −

2

2545

6 State the solution. The Cartesian equation is yx x= −

2

2545

; that

is, the trajectory of the object is parabolic.

WorkeD examPle 7

A body moves with a position vector r t t i t t j( ) ( ) ( )= + + −5 10 2 2 , t ∈ [0, 4].

a Find the initial position, velocity and acceleration of the body.b Find the Cartesian equation for motion and show that it is a parabola.c Plot the trajectory stating the domain and range for y(x).

Think WriTe

a 1 Write the equation. a r t t i t t j( ) ( ) ( )= + + −5 10 2 2

2 Substitute t = 0 into r t( ) to find the initial

position. r i( )0 5=

3 Find v t( ) by differentiating

r t( ).

v t i t j( ) ( )= + −10 2 2

4 Substitute t = 0 into v t( ) to find the initial

velocity. v i j( )0 10 2= +

5 Find a t( ) by differentiating

v t( ).

a t j( ) = −2

6 Substitute t = 0 into a t( ) to find the initial

acceleration. a j( )0 2= −

7 State the solution. The initial position is 5i , the initial velocity is

10 2 i j+ and the initial acceleration is −2

j .

b 1 Write the parametric equation for x(t). b x(t) = 5 + 10t

2 Express t as a function of x. t

x= − 510

3 Write the parametric equation for y(t). y(t) = 2t − t2

4 Substitute tx= − 510

into this equation to

obtain the Cartesian equation.

y xx x

( )( ) ( )= − − −2 510

5

10

2

2

5 Simplify the equation for y(x). = − − + −20 100 10 25100

2x x x

= − +− ( )x x2 30 125100

6 State the type of function of the graph of y(x).

The graph of y(x) is a parabola since it is of the form y ax bx c= + +2 , where a, b and c are real constants.



c 1 The function y xx x

( )( )= − +− 2 30 125

100

can be plotted on a graphics or CAS calculator or computer application.

c

10

−5

20 4030

(5, 0)

(45, −8)

0 x

y

2 The domain and range are found from the time interval t ∈ [0, 4].

Using t ∈ [0, 4]

3 For the domain fi nd x when t = 0 and t = 4 by direct substitution into x(t).

x(0) = 5x(4) = 45

4 The domain is [5, 45] since x(t) = 5 + 10t is an increasing linear function.

Domain is [5, 45]

5 The range is found by inspection of the graph y(x). The maximum value for y occurs when x = 15 and has the value y = 1.Alternatively, the range of y(x) is the range of y(t). Examine the range of y(t) = 2t − t2, t ∈ [0, 4].

Range is [−8, 1] (by inspection of the graph over the domain [5, 45]).

antidifferentiation of vectorsYou will recall from the work that you have done in calculus that each time you antidifferentiate a scalar function you introduce a constant called the integration constant. Additional information is required to fi nd the value of this constant. The same thing happens when you antidifferentiate a vector function of acceleration

a t( ) to obtain the velocity

v t( ), or antidifferentiate a velocity

vector function to fi nd the position vector r t( ). However, the constant is a vector and generally,

but not always, describes the initial conditions. This situation is summarised below.

Position vector: r(t)~

Velocity vector: v(t)~

Acceleration vector: a(t)~

Differentiate Antidifferentiate and determine constant

Antidifferentiate and determine constantDifferentiate

WorkeD examPle 8

For the velocity vector v t e i t j tkt( ) = − +3 3 103 2 with

r(0) =

i + 2k,

determine expressions for:a r t( ) b

a t( ) .

Think WriTe

a 1 The position vector is obtained by antidifferentiating

v t( ) with respect to

time.

a v t e i t j t kt( ) = − +3 3 103 2

r t e i t j t k dtt( ) ( )= − +∫ 3 3 103 2

2 Antidifferentiate and include the constant vector of integration,

c c i c j c kx y k= + + .

= − + + + +e i t j t k c i c j c ktx y z

3 3 25

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Worked example 8

478

3 Substitute t = 0 into r t( ) and equate to the

result that r i k( )0 2= + .

r i c i c j c k i kx y z( )0 2= + + + = +

4 Determine c by equating the components

of i , j and

k.

1 + cx = 1, 0 + cy = 0, 0 + cz = 2

5 Solve for cx, cy and cz. cx = 0, cy = 0, cz = 2

6 Substitute the values for cx, cy and cz into the expression for

r t( ).

r t e i t j t k kt( ) = − + +3 3 25 2

7 Simplify r t( ).

r t e i t j t kt( ) ( )= − + +3 3 25 2

b The acceleration vector is obtained by differentiating

v t( ) with respect to time.

b

a tdvdt

e i tj kt( ) = = − +9 6 103

WorkeD examPle 9

Given that the acceleration vector for an object is a t

ti t j( )

( )=

+−6

212

32 with

v i( )0 3

4= − and

r i j( )0 2 3= + , find:

a v t( ) b

r t( ).

Think WriTe

a 1 The velocity vector is obtained by antidifferentiating the acceleration with respect to time.

a a t

ti t j( )

( )=

+−6

212

32

v t

ti t j dt( )

( )=

+−

∫ 6

212

32

2 Antidifferentiate and include the constant vector

c.

=+

− + + +− 3

24

23

( )ti t j c i c j c kx y z

3 Evaluate v( )0 and equate to the given

result that v i( )0 3

4= −.

v i c i c j c kx y z( )0 3

4= − + + + =− 3

4 i

4 Solve for c.

− −+ =34

34cx , cy = 0, cz = 0

cx = cy = cz = 0

5 State the velocity vector v t( ).

v t

ti t j( )

( )=

+−

− 3

24

23

b 1 The position vector is obtained by antidifferentiating the velocity with respect to time.

b

r tt

i t j dt( )( )

=+

−

−

∫ 3

24

23

2 Antidifferentiate and include the constant vector

d . (Since

c was used in part a .)

=+

− + + +32

4

ti t j d i d j d kx y z

3 Evaluate r( )0 and equate to the given

result that r i j( )0 2 3= + .

r i d i d j d k i jx y z( )0 2 33

2= + + + = +

4 Solve for d . 3

22 3 0+ = = =d d dx y z, ,

d d dx y z= = =12 3 0, ,



5 State the position vector r t( ).

r t

ti t j( ) ( )=

++

+ −32

12

3 4

WorkeD examPle 10

An object is thrown off a building (t = 0 s) on a windy day. The acceleration of the object in m/s2

is given by a t i e j

t

( ) .= −−

112

109 8 . At a time, t = 1, the object has a velocity in m/s of v i j( )1 2 3= − .

The building is 50 m above the ground and hence the initial position of the object is taken to be

r i j= +0 50 .

a What is the initial acceleration of the object?b Determine the velocity vector

v t( ) for all times t ≥ 0.

c Determine the position vector r t( ) for all times t ≥ 0.

Think WriTe

a 1

2 Write the solution a i j( ) .0

12

9 8= −

b 1

2 The vector constant of integration must be found. To do this using the CAS calculator, defi ne the velocity vector

v t( )

by completing the entry line as:

Defi ne v t

tm e n

t

( ) ,= + +

−

1298 10

Then press ENTER ·.

Defi ne the acceleration function a(t) by completing the entry line as:

Defi ne a(t) =

1

12

109 8−9 8−9 8

−

9 8.9 8et

Then press E.Substitute t = 0 into a(t) to fi nd the initial acceleration.

The velocity vector is found by integrating the acceleration a(t) with respect to time.The vector constant of integration must be found. To do this using the CAS calculator, defi ne the velocity vector v(t) by completing the entry line as:

Defi ne v(t) =

tm

e nt

12

98 10e n10e n

+

+e n+e n

−

Then press E.

480

3 State the integrand and the velocity vector

v t( ) with the vector constant of

integration expressed as mi nj

+

v t a t dt c( ) ( )= +∫ v t

ti e j mi nj

t

( ) = + + +−

1298 10

4

5 State the values of m and n and the velocity vector. m = 23

12 and n = −91.674

v t

ti e

t

( ) .= +

+ −

−

122312

98 91 67410

j

c 1

2

3 State the integrand and the position vector

r t( ), with the vector constant of

integration expressed as pi qj

+ r t v t dt d( ) ( )= +∫ r t

tt p i e

t

( ) .= + +

+ −−−210

242312

980 91 674tt q j+

4


Evaluate the vector constant of integration using the fact that v(1) = 2i − 3j.Complete the entry lines as:matToList(v(1), 1)

solve e m n{ }{ }m n{ }m n e m{ }e m+ ={ }+ =m n+ =m n{ }m n+ =m nm n+ =m n{ }m n+ =m n + ={ }+ =

{ }−{ }{ }1{ }+ ={ }+ =1+ ={ }+ ={ }

12{ }m n{ }m n

12m n{ }m nm n+ =m n{ }m n+ =m n

12m n+ =m n{ }m n+ =m n{ }1{ }{ }2 9{ }m n{ }m n2 9m n{ }m n + ={ }+ =2 9+ ={ }+ ={ }8 3{ }e m{ }e m8 3e m{ }e m+ ={ }+ =8 3+ ={ }+ =e m+ =e m{ }e m+ =e m8 3e m+ =e m{ }e m+ =e me m−e m{ }e m−e m8 3e m−e m{ }e m−e m{ }10{ }8 3{ }10{ }+ ={ }+ =10+ ={ }+ =8 3+ ={ }+ =10+ ={ }+ =e m+ =e m{ }e m+ =e m10e m+ =e m{ }e m+ =e m8 3e m+ =e m{ }e m+ =e m10e m+ =e m{ }e m+ =e m, ,e m, ,e m{ }, ,{ }e m{ }e m, ,e m{ }e m+ ={ }+ =, ,+ ={ }+ ={ }2 9{ }, ,{ }2 9{ }m n{ }m n2 9m n{ }m n, ,m n{ }m n2 9m n{ }m n + ={ }+ =2 9+ ={ }+ =, ,+ ={ }+ =2 9+ ={ }+ ={ }8 3{ }, ,{ }8 3{ }e m{ }e m8 3e m{ }e m, ,e m{ }e m8 3e m{ }e m+ ={ }+ =8 3+ ={ }+ =, ,+ ={ }+ =8 3+ ={ }+ =e m+ =e m{ }e m+ =e m8 3e m+ =e m{ }e m+ =e m, ,e m+ =e m{ }e m+ =e m8 3e m+ =e m{ }e m+ =e m{ ,e m{ ,e m }

Press E after each entry.

Redefi ne the velocity vector v(t) with the correct values of m and n.

The position vector r(t) is found by integrating the vector v(t) with respect to time.

The vector constant of integration must be found. To do this using the CAS calculator, defi ne the position vector r(t) by completing the entry line as:

Defi ne r(t) =

t tp

e tt

2t t2t t

10e t10e t

46t t46t t24

980 91e t91e t674e t674e t

t t+t t +

− −e t− −e te t10e t− −e t10e t980− −980

−

e t.e t

Then press E.Use r(0) = 0i + 50j to solve vector constants p and q. To do this, complete entry lines as: r(0)matToList(r(0), 1)solve({p = 0, q − 980 = 50}, {p, q})Press E after each entry.


5 Use r i j( )0 0 50= + to solve the vector

constants p and q. To do this, complete the entry line as:solve(r (0) = [0, 50],{p, q})then press ENTER ·.

6 State the values of p and q and the position vector.

p = 0 and q = 1030

r t

tt i e t

t

( ) .= +

+ − +−−210

242312

980 91 674 11030

j

WorkeD examPle 11

A balloon is released from a point on the ground. It moves in a plane such that its position, relative to the point at any time t seconds after its release, can be described by:

r i= 5 , t ∈[0, 100] (where

r is in metres) and when

t = 0, r = 0 and

r i j= +7 10 .

a Find the position vector at any time t.b State the position of the balloon at i t = 2 and ii t = 4.c The balloon is attacked by a disgruntled magpie and bursts after 100 seconds. How far,

horizontally, is the furthest the balloon gets from its take-off point? Give its height at this point.d Determine the Cartesian equation of the path and sketch it.e What is the angle of trajectory of the balloon at take-off?

Think WriTe

a 1 Write down the acceleration. a

r i= 5

i~

j~

2 Antidifferentiate to fi nd the velocity.

r i dt= ∫ 5

= +5ti c

3 Substitute t = 0 and

r i j= +7 10 to determine the vector constant

c.

r c( )0 =

= +7 10 i j

4 Write the velocity vector.

r t i j= + +( )5 7 10

5 Antidifferentiate to fi nd the position.

r t i j dt= + +∫ [( ) ]5 7 10

= +

+ +52

7 102t t i tj d

6 Substitute t = 0 and r = 0 to

determine the vector constant d .

r d( )0 = =

0

7 Write the position vector.

r t t i tj= +

+52

7 102

eBookpluseBookplus

Tutorialint-0413

Worked example 11

482

b 1


i r i j( )2 24 20= +At t = 2 the position is (24, 20)

ii r i j( )4 68 40= +At t = 4 the position is (68, 40)

c 1

2 Write the solution r i j( )100 25 700 1000= +

At t = 100, the balloon is 25 700 metres from where it started.It has reached a height of 1000 metres above the ground.

d 1 Write the parametric equation for y. d y =10t

2 Express t as a function of y. ty=

10

3 Write the parametric equation for x. x t t= +52

72

4 Substitute t = y

10 for y. x

y y=

+

52 10

710

2

5 Simplify. xy y= +5

200710

2

= +y y2

40710

6 State the type of relation. This represents a ‘sideways’ parabola.


Use CAS to fi nd r(2) and r(4) by completing the entry lines as:

Defi ne r(t) = 52

2 7

10

t t2t t2 7t t7

t

t t+t t

r(2)r(4)Press E after each entry.

Complete the entry line as:r(100)Then press E.


7 Sketch its graph using the coordinates obtained in parts b and c .

x

1000

y

(0, 0)

(24, 20)(68, 40)

(25700, 1000)

e 1 Write the initial velocity. e

r i j( )0 7 10= +

2 The direction is given by tan( )( )

−

1 y tx t

.

θ =

−tan( )( )

1 y tx t

3 Substitute t = 0 and simplify.

=

−tan 1 107

= 55.0°4 Give the initial direction relative to

the horizontal.The angle of trajectory is 55° from the horizontal (or 55° above the horizontal and to the right).

The velocity vector 1. v t( ) can be found from the acceleration vector

a t( ) by

antidifferentiation provided the velocity is given at a particular time, t; generally this is at time t = 0 so that the vector

c can be found. Thus:

v t a t dt c( ) ( )= +∫

The position vector 2. r t( ) can be found from the velocity vector

v t( ) by

antidifferentiation provided the position is given at a particular time, t; generally this is at time t = 0 so that the vector

c can be found. Thus:

r t v t dt c( ) ( )= +∫

If the position vector is 3. r t x t i y t j( ) ( ) ( )= + then the parametric equations for the

coordinates of the path are x(t) and y(t). The Cartesian equation for the motion of the particle is y(x) which can be obtained from the parametric equations x(t) and y(t).

rememBer

Cartesian equations and antidifferentiation of vectors 1 We 6 Find the Cartesian equation for the paths of objects with position vectors given

below.

a r t ti tj( ) = +2 b

r t ti t t j( ) ( )= + −30 10 202

c r t t i t j( ) ( )= − +2 1

12 d

r t t i t je( ) log ( ) ( )= + −2 4 1

e r t e i tjt( ) = +−2 f

r t t i t j( ) sin ( ) cos ( )= +2 2

2 We 7 A particle travels in a path so that its position at any time, t, is given by

r t ti t j t( ) ( ) ; [ , ]= + − ∈2 1 4 0 5

a Find the initial position, initial velocity and initial acceleration of the particle.b Find the Cartesian equation for the motion.c Plot the trajectory stating the domain and range for y(x).

exerCiSe

10B

484

3 A body moves so that its position vector is r t e i t j t( ) ; [ , ]= + ∈

12 2 0 4 .

a Find the initial position, initial velocity and initial acceleration.b Find the Cartesian equation for the motion.c Plot the trajectory stating the domain and range for y(x).

4 mC An object moves so that its position at any time, t, is r t t i t j te( ) log ( ) ;= + − ≥3 1 3 0 .

a The initial position of the particle is:A i B − 3

j C

i j− 3

D 0 E 2 i j+

b The Cartesian equation for motion is:A y = loge (x) B y = 1 − ex C y = ex − 1D y = loge (x + 1) E y = x2 + x

c Which one of the following graphs best shows the particle’s trajectory?

A

x

y

0

B

x

y

0

C

x

y

0

D

x

y

0

E

x

y

0

5 mC The acceleration of an object which is initially at rest, in m/s2, is:

r t ti t j( ) ( )= + + −6 1 2 .

a The velocity vector of the object is:

A

r t i t j( ) ( )= − + −6 2 1 3 B

r t t i t j( ) ( )= + + −2 11 C

r t t i t j( ) ( )= − + −3 2 12 1

D

r t t i t j( ) ( )= − + −3 12 1 E

r t t i t t j( ) ( )= + + −3 12 1

b If r i( )1 = , then the distance the object is from the origin when t = 2 is nearest to:

A 8.02 m B 7.92 m C 9.1 m D 3.03 m E 3.01 m

6 We 8 For the following velocity vectors, determine expressions for i r t( ) and ii

a t( ).

a v t ti j k( ) = + +3 5 2

r i j k( )0 2 3= − +

b v t t i t j( ) sin ( ) cos ( )= +3 2

r i j( )0 5 3= +−

c v t t i tj t k( ) ( )= + − + −3 5 2 12 1

r i j( )0 2= +

d v t e i e jt t( ) = + −3 5

r i( )0 =

e v t ti j( ) = +3 5

r i j( )1 5= +

f v t

ti tj k( ) = + +1

25 4

r i j k( )2 2 3 3= − +

g v t t i t j( ) sin ( ) cos ( )= + −3 4 2

r j( )0 2=

7 We 9 For the following acceleration vectors a t( ), find i

v t( ) ii

r t( ) .

a a t i t j( ) .= +−4 9 with

v( )0 0= and

r i j( )0 = +

b a t

ti tj( )

( )=

++

− 2

16

2 with

v( )0 0= and

r i j( )0 = +

c a t e jt( ) = −4 with

v i( )0 3= and

r( )0 0=

d a t t i t j( ) cos ( ) sin ( )= − −25 5 25 5 with

v j( )0 5= and

r i( )0 5=



8 We 10 The acceleration of an aircraft is given by

r t i e j( ) . .= −

−0 1 9 8

110 . If the initial velocity

is

r i j( )0 3 2= + and the initial position is

r j( )0 20= , then find:

a the initial accelerationb the velocity vector at any time, tc the position vector at any time, t.

9 The acceleration of an object is given by

r t j( ) = −10 . If

r i j( )0 20 5= + and

r i j( )1 12 16= + ,

then:a determine the velocity vector

r t( )

b determine the position vector r t( ).

10 We 11 A balloon is released from a point on the ground. It moves in a plane such that its position, relative to the point at any time t seconds after its release, can be described by:

r i= 2 , t ∈ [0, 60]

where r is in metres, and when t = 0,

r = 0 and

r i j= +4 5 .

a Find the position vector at any time t.b State the position of the balloon at

i t = 2 and ii t = 4.c The balloon is attacked by an angry wasp and bursts after 60 seconds. How far,

horizontally, is the balloon from its take-off point? Give its height at this point.d Determine the Cartesian equation of the path and sketch it.e What is the angle of trajectory of the balloon at take-off?

11 A balloon is released from a point on the ground. It moves in a plane such that its position, relative to the point at any time t seconds after its release, can be described by:

r i= −4 , t ∈ [0, 80]


r = 0 and

r i j= +8 4 .

a Find the position vector at any time t.b State the position of the balloon at

i t = 4 and ii t = 10.c The balloon is attacked by a neurotic parakeet, and bursts after

80 seconds. What is the furthest the balloon gets, horizontally, left and right from its take-off point? Give its height at these points.

d Determine the Cartesian equation of the path and sketch it.e What is the direction of the balloon:

i when it bursts? ii halfway through its fl ight?

applications of vector calculusVector calculus may be used to solve a variety of problems. The following example illustrates some of the methods employed. In general, each problem in the exercise that follows requires you to interpret the question and use appropriate techniques.

r (0) = 20~

exam TiP A very, common mistake is to omit the constant of integration.

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Digital docWorkSHEET 10.1

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Interactivityint-0351

Applications of vector calculus

10C

486

WorkeD examPle 12

A golf ball is struck so that its position vector, in metres, at any time, t seconds, is given by

r t ti t t j e k t( ) ( ) ; [ , ]= + − + ∈30 25 5 0 52

110 .

a Find an expression for the velocity at any time, t.b Find an equation y(x) for the path of the golf ball and sketch

it if 0 ≤ t ≤ 5. (Ignore the

k component of

r t( ) for this part of the question.)

c State the velocity vectors at t = 0 and t = 2.d Find the speed and direction of the golf ball at t = 0 and t = 2.

(Give the direction as an angle from the unit vector, i .)

e Find an expression for the acceleration of the golf ball at time t, and find its initial magnitude.

Think WriTe

a Differentiate r t( ) with respect to time to obtain the

velocity vector v t( ).

a r t ti t t j e k

t

( ) ( )= + − +30 25 5 2 10

v t i t j e k

t

( ) ( ) .= + − +30 25 10 0 1 10

b 1 Write r t ti t t j( ) ( )= + −30 25 5 2 . b

r t ti t t j( ) ( )= + −30 25 5 2

2 Write the parametric equation for x(t). x(t) = 30t

3 Express t as a function of x. tx=

304 Write the parametric equation for y(t). y(t) = 25t − 5t2

5 Substitute tx=

30 into this equation to

obtain the Cartesian equation.

or yx x=

−

2530

530

2

6 Simplify the equation for y(x). yx x= −5

6 180

2

7 Since t ∈ [0, 5], the domain of the Cartesian equation needs to be determined.

8 Find x when t = 0 and t = 5. When t = 0, x = 30(0) = 0.

When t = 5, x = 30(5) = 150.

9 State the domain. x ∈ [0, 150]

10 Use a graphics calculator to obtain the sketch of the trajectory using x ∈[0, 150].

x

y

0 150

(75, 31.25)

c 1 Evaluate v( )0 . c

v i j k( ) .0 30 25 0 1= + +

2 Evaluate v( )2 .

v i j k( ) .2 30 5 0 122= + +

d 1 Evaluate v( )0 to determine the speed

when t = 0.d When t = 0,

speed = = + +v( ) .0 30 25 0 12 2 2



2 Give the speed correct to 2 decimal places.

≈ 39.05

3 Use the angle-between-two-vectors rule:

cos (θ ) =

a b

a b

. to find the angle between

v( )0 and

i .

cos ( )( )

( )

.θ =

v i

v i

0

0

4 Simplify with the appropriate values found previously.

cos ( )( . )

.

.

.θ =

+ +

×

=

30 25 0 1

39 05 1

3039 05

i j k i

5 Express θ as the subject of the equation and evaluate correct to 1 decimal place.

θ == °

−cos ( . )

.

1 0 7682

39 86 State the speed and angle required when

t = 0.Therefore, the golf ball is initially moving with a speed of 39.05 m/s at an angle of 39.8° to the unit vector

i .

7 Evaluate v( )2 to determine the speed

when t = 2.When t = 2,

speed =

= + +v( )

.

2

30 5 0 1222 2 2

8 Give the speed correct to 2 decimal places.

≈ 30.414

9 Use the angle-between-two-vectors rule to find the angle between

v( )2 and

i . cos ( )

( )

( )

.θ =

v i

v i

2

2

10 Simplify with the appropriate values found previously. cos ( )

( . )

.

.

.θ =

+ +×

=

30 5 0 122

30 414 1

3030 4

i j k i

11411 Express θ as the subject of the equation

and evaluate correct to 1 decimal place.θ =

= °

−cos ( . )

.

1 0 9864

9 5

12 State the speed and angle required when t = 2.

Therefore when t = 2 the golf ball is moving at 30.41 m/s at an angle of 9.4° to the unit vector

i .

e 1 Determine a t( ) by differentiating

v t( )

with respect to time.e a t j e k

t

( ) .= +−10 0 01 10

2 Evaluate a( )0 to determine the

acceleration vector when t = 0. a j k( ) .0 10 0 01= +−

3 Evaluate a a( ) ( )0 0= to determine the

magnitude of the acceleration when t = 0. a( )0

= +=

−( ) ( . )

.

10 0 01

10 0

2 2

4 State the solution. Therefore the initial magnitude of the acceleration is 10.0 m/s2.

CollisionsIf two objects are to collide then they must be at the same point at the same time.Using vector notation, the two objects must share the same position vector at a specific time.

488

WorkeD examPle 13

A body moves with a velocity of 2 i j− and is at a position − +

i j2 when a second body, moving with a

velocity of i j+ , is at a position 2 4

i j− at time t = 0.

a Determine the position vectors for both objects at time t.b By finding the time at which the two position vectors are equal, show that the two objects will

collide.c When and where do they collide?

Think WriTe

a 1 Let v t i j1 2( ) = − represent the velocity of

the first object.

a Let v t i j1 2( ) = −

2 Antidifferentiate v t1( ) with respect to time

to obtain r t1( ).

r t i j dt c

ti tj c i c jx y

1 2

2

( ) ( )= − +

= − + +∫

3 Use the result that r i j1 0 2( ) = +−

to evaluate the constant vector of antidifferentiation.

r c i c j

i j

x y1 0

2

( ) = +

= +−

cx = −1 and cy = 2

4 Write the position vector for the first object,

r t1( ) at any time, t.

r t t i t j1 2 1 2( ) ( ) ( )= − + −

5 Let v t i j2 ( ) = + represent the velocity of

the second object.Let v t i j2 ( ) = +

6 Antidifferentiate v t2 ( ) with respect to time

to obtain r t2 ( ).

r t i j dt d

ti t j d i d jx y

2 ( ) ( )= + +

= + + +∫

7 Use the result that r i j2 0 2 4( ) = −

to evaluate the constant vector of antidifferentiation.

r d i d j

i j

x y2 0

2 4

( ) = +

= −

dx = 2 and dy = −4

8 Write the position vector for the second object,

r t2 ( ), at any time, t.

r t t i t j2 2 4( ) ( ) ( )= + + −

b 1 Set the two position vectors r t1( ) and

r t2 ( )

equal.b If

r t r t1 2( ) ( )=

then ( ) ( ) ( ) ( )2 1 2 2 4t i t j t i t j− + − = + + −

2 Equate their i components to find the

time when the objects have the same x position.

2t − 1 = t + 2 (i component)

3 Solve for t. t = 3

4 Equate their j components to find the time

when the objects have the same y position.and 2 − t = t − 4 (

j component)

5 Solve for t. 2t = 6

6 Since both values of t are equal, the two objects occupy the same position at this time.

t = 3

7 The two objects collide when t = 3. The two objects have the same position when t = 3.



c 1 Substitute t = 3 into r t1( ) (or

r t2 ( )) to fi nd

where the two objects collide.c r i j1 3 2 3 1 2 3( ) ( ) ( )= × − + −

2 Simplify r1 3( ). = −5

i j

3 State the solution. The two objects collide when their position vectors are 5

i j− at t = 3.

WorkeD examPle 14

Two boats, A and B, start at a common point (0, 0) and move with velocity vectors

v t ti t t jA ( ) ( )= + −2 and

v t ti t t jB ( ) ( )= + +− 2 ; t ∈ [0, 4].

Distance is measured in metres and time in seconds.a How far apart are they at t = 2?b What is the angle between their respective directions at t = 2?c At what time, if any, are the acceleration vectors of the two boats perpendicular

to each other?

Think WriTe

a 1 Express r tA ( ) as the antiderivative of

v tA ( ) with respect to time.

a

r ti t t j dt cA A= + − +∫ [ ( ) ]2

2 Antidifferentiate v tA ( ) and express the

constant of antidifferentiation vector,

cA, as c i c jx yA A

+ .

r

tc i

t tc jx yA A A= +

+ − +

2 3 2

2 3 2

3 Simplify rA ( )0 and equate it to the

given result that r i j( )0 0 0= + .

r c i c j

i j

x yA A A( )0

0 0

= +

= +

4 Solve for the vector constants c xA and c yA .

cAx = 0 and cAy = 0

5 State the position vector of boat A,

r tA ( ).

rt

it t

jA =

+ −

2 3 2

2 3 2

6 Express r tB( ) as the antiderivative of

v tB( ) with respect to time.

r ti t t j dt cB B= + + +−∫ [ ( ) ]2

7 Antidifferentiate v tB( ) and express

the constant of antidifferentiation vector,

cB, as c i c jx yB B

+ .

r

tc i

t tc jx yB B B= +

+ + +

− 2 3 2

2 3 2

8 Simplify rB( )0 and equate it to the

given result that r i j( )0 0 0= + .

r c c

i j

x yB B B( )0

0 0

= +

= +

9 Solve for the constants c cx yB B+ . cBx = 0 and cBy = 0

10 State the position vector of boat B,

r tB( ).

r tt

it t

jB( ) = −

+ +

2 3 2

2 3 2

eBookpluseBookplus

Tutorialint-0414

Worked example 14

490

11 (Optional) Represent rA ( )2 and

rB( )2 on a vector diagram.

rA ( )2 = 2

23

i j+

rB( )2 = − +2

143

i j

rB

j

iO−2 2

4.66

0.66

~

rA~

~

~

rA − rB~~

12 Find the vector r rA B( ) ( )2 2−

which gives their separation at time t = 2.

r r i j i jA B( ) ( ) ( )2 2 2 22

3143− = + − − +

13 Simplify. = −4 4 i j

14 Their distance apart at t = 2 is the magnitude of the vector

r rA B( ) ( )2 2− .

r rA B− = + −

≈

4 4

5 66

2 2( )

.

15 State the solution. The two boats are separated by about 5.7 m at a time t = 2 s.

b 1 The direction in which each of the boats travels is parallel to the velocity vector for each of the boats, namely

vA

and vB respectively.

b

2

3

4


To calculate the magnitude of vA(2) and vB(2) using CAS, complete the entry line as:

Defi ne va(t) = t

t t2t t2t tt t−t t

Defi ne vb(t) = −

t

t t+t t+2t t2t tnorm(va(2))norm(vb(2))Press E after each entry.

To calculate the dot product of vA(2) and vB(2), complete the entry line as:dotP(va(2),vb(2))Then press E.

The angle between the boats at t = 2 is

given by cos (θ ) = v vA Bv vA Bv v

A Bv vA Bv v( )v v( )v vA B( )A Bv vA Bv v( )v vA Bv v, (A B, (A Bv vA Bv v, (v vA Bv v )( ) ( )

2 2v v2 2v vv vA Bv v2 2v vA Bv v( )2 2( )v v( )v v2 2v v( )v vv vA Bv v( )v vA Bv v2 2v vA Bv v( )v vA Bv v, (2 2, (v v, (v v2 2v v, (v vA B, (A B2 2A B, (A Bv vA Bv v, (v vA Bv v2 2v vA Bv v, (v vA Bv v2 2v v2 2v vv vA Bv v2 2v vA Bv v( )2 2( ) ( )2 2( )| |v v| |v vA B| |A Bv vA Bv v| |v vA Bv vA B| |A Bv vA Bv v| |v vA Bv v( )| |( )v v( )v v| |v v( )v vA B( )A B| |A B( )A Bv vA Bv v( )v vA Bv v| |v vA Bv v( )v vA Bv v2 2| |2 2v v2 2v v| |v v2 2v vv vA Bv v2 2v vA Bv v| |v vA Bv v2 2v vA Bv v( )2 2( )| |( )2 2( )v v( )v v2 2v v( )v v| |v v( )v v2 2v v( )v vv vA Bv v( )v vA Bv v2 2v vA Bv v( )v vA Bv v| |v vA Bv v( )v vA Bv v2 2v vA Bv v( )v vA Bv v| |A B| |A Bv vA Bv v| |v vA Bv v ( )| |( )2 2| |2 2v v2 2v v| |v v2 2v vA B2 2A B| |A B2 2A Bv vA Bv v2 2v vA Bv v| |v vA Bv v2 2v vA Bv v ( )2 2( )| |( )2 2( )

.

Complete the entry line as:

cos−

×

1 8

2 22 2 2 12 10Then press E.


5 Write the solution cos ( )( ) ( )

( ) ( )θ =

.

v v

v vA B

A B

2 2

2 2

vA ( )2 2 2=

vB( )2 2 10=

v vA B( ) ( )2 2. = 8

θ =×

=8

2 2 2 1063 4.

The angle between the directions in which the boats are travelling is 63.4°.

c 1

2 Write the solution. a t i t jA ( ) ( )= + −2 1

a t i t jB( ) ( )= + +− 2 1

solve A B( ( ) ( ) , ). a t a t t= 0

t = ±0 71.Only the positive solution is required.At time t = 0.71 s the boats have accelerations that are perpendicular.

WorkeD examPle 15 A particle moves so that its position vector

r at any time t, t ≥ 0, is given as

r t i t j= +2 5cos ( ) sin ( ) .

a Determine the Cartesian equation of the path and sketch it.b Find its velocity at any time t.

c Hence give the direction of the particle when t is π6

.

d Determine the minimum and maximum speeds of the particle and state the fi rst times at which they occur.

e Find the particle’s acceleration at any time t.f Verify that the maximum speed occurs when the magnitude of the acceleration is at a minimum.

Think WriTe

a 1 State the position vector. a r t i t j= +2 5cos ( ) sin ( )

The acceleration vectors will be perpendicular when the dot product of the vectors is zero.Use a CAS calculator to fi nd the acceleration vectors and then calculate the dot product of the acceleration vectors. Equate this to zero and solve for t, by completing the entry lines:matToList(va(t),1)ddt

( )( )an( )( )s( )

matToList(vb(t),1)ddt

( )( )an( )( )s( )

dotP 1

2 1

1

2 1t t2 1t t2 1 2 1t t2 12 1t t2 1−2 1t t2 1

t tt t

t tt tt tt t

−2 1+2 1

t tt t

t tt tt tt t

,

solve(ans = 0, t): Press E after each entry.

492

2 Write the parametric equation for x. x t= 2 cos ( )

3 Make cos t the subject. cos ( )tx=2

4 Write the parametric equation for y. y t= 5 sin ( )

5 Make sin t the subject. sin ( )ty=5

6 Use the identity cos ( ) sin ( )2 2 1t t+ = to obtain the Cartesian equation.

cos ( ) sin ( )2 2 1t t+ =

x y2 2

4 251+ =

7 State the type of relation. An ellipse with centre (0, 0), a = 2, b = 5

8 Sketch the graph.

0−2

−5

5

2

~

x

y x2

4y2

25+ = 1

start r (0) = (2, 0)

b Differentiate the position vector to obtain the velocity.

b

r t i t j= − +2 5sin ( ) cos ( )

c 1 Evaluate r

π6

. c

r i

π π π6

26

56

=

+

− sin cos jj

= +

−

i j

5 32

2 Use θ =

−tan 1 yx

to determine the angle

the velocity makes with the positive i

direction.

θ =

−tan 1 yx

=

−−

tan 1

5 321

(2nd quadrant)

= −

−

−180 5 3

21tan

= 180 − 77= 103° from the positive

i direction.

d 1 The speed is the magnitude of the velocity. dr t t= +[ sin ( )] [ cos ( )]2 52 2

= +4 252 2sin ( ) cos ( )t t

2 Simplify by using the identitycos ( ) sin ( )2 2 1t t+ = .

= +4 21 2cos ( )t

3 Minimum speed occurs when cos ( )t = 0. Min. speed when cos ( )t = 0

4 Evaluate the minimum speed and give the first time it occurs.

⇒ Min. speed = 4 when t = π2

= 2

5 Maximum speed occurs when cos ( )t = 1.

Max. speed when cos ( )t = 1



6 Evaluate the maximum speed and give the first time it occurs.

⇒ Max. speed = 4 21+ when t = 0

= 25 = 5

e Differentiate the velocity vector to obtain the acceleration.

e

r t i t j= − −2 5cos ( ) sin ( )

f 1 The maximum speed first occurs when t = 0.

f Max. speed when t = 0

2 Determine the magnitude of the acceleration.

r t t= +[ cos ( )] [ sin ( )]2 52 2

= +4 252 2cos ( ) sin ( )t t

3 Simplify by using the identitycos ( ) sin ( )2 2 1t t+ = .

= +4 21 2sin ( )t

4 The minimum acceleration occurs when sin ( )t = 0.

The acceleration is minimum when sin ( )t = 0 ⇒ =t 0

5 The maximum speed occurs when the acceleration is at a minimum.

∴ The maximum speed occurs when the acceleration is at a minimum.

Read each question carefully so that it is interpreted correctly and the question being 1. asked is answered.Apply vector calculus methods to analyse curvilinear motion where the vectors 2.

r, r and

r are functions of time.For inexact solutions give your answers to an appropriate number of decimal places.3. If two objects are to collide then they must be at the same point at the same time.4.

rememBer

applications of vector calculus 1 We 12 A ball is released from a cliff at time t = 0. It moves so that its position is given by

r t ti t j( ) = −10 5 2 , t ≥ 0

a Find an expression for the velocity at a time, t.b Find an equation y(x) for the path of the ball and sketch it for 0 ≤ t ≤ 2.c State the velocity vectors at t = 0, 1 and 2 and consequently find the speed and direction

of the ball at those times. (Give the direction as an angle to the nearest degree from the unit vector

i .)

d Find an expression for the acceleration of the ball at time t, and describe the type of acceleration.

2 A particle moves so that its position, r (metres) at time t (seconds) is given by:

r t t i e je

t( ) log ( )= + + −1 2 , t ≥ 0

a Find the distance of the particle from the origin at times t = 0, 1 and 2.b Find expressions for the velocity and acceleration at time t.c Find the angle, to the nearest degree, between the velocity and acceleration vectors at

times t = 0, 1 and 2.

exerCiSe

10C

494

3 An object has a position vector r t t t i t j( ) [ cos ( )] sin ( )= − + for 0 ≤ t ≤ 2π.

a Find expressions for the velocity and acceleration.b At what time is the speed a minimum?c What is the position of the object at this minimum speed?d What is the acceleration at this minimum speed?e Explain why this does not correspond to a time when the acceleration is zero.

4 The position of an object is given by r t t i t j( ) [ cos ( )] [ sin ( )]= + + +1 2 4 5 2 4 .

a Show that the velocity vector is always perpendicular to the acceleration vector.b Show that the velocity vector and acceleration vectors are both constant in magnitude.c Explain why the speed is a constant and yet the acceleration is not equal to zero.

5 The position of an object, r (metres) at time t (seconds) is given by:

r t t i t j( ) cos ( ) sin ( )= +2 3 2 3

a Show that the velocity vector is always perpendicular to the acceleration vector.b Show that the velocity vector and acceleration vectors are both constant in magnitude.c Show that the acceleration vector is parallel to the position vector at all times.d Show that the object moves in a circular path of radius 2 m.e Show that the object moves with a constant speed and find that speed.

6 An object has a position vector given by r t

ti t j t( ) ,= + >1

4 0.

a Find the instantaneous velocity vectors at t = 1 and 2.b Find the average velocity vector during the time interval t = 1 to t = 2.c Find the instantaneous acceleration vectors at t = 1 and 2.d Find the average acceleration vector during the time t = 1 to t = 2.

7 An object has a position vector given by r t t i t j t( ) ( ) ,= − − ≥2 1 3 02 .

a Find the instantaneous velocity vectors at t = 2 and 3.b Find the average velocity vector during the time interval between t = 2 and t = 3.c Find the instantaneous acceleration vectors at t = 2 and 3.d Find the average acceleration vector during the time interval between t = 2 and t = 3.e Find the angle between the acceleration vector and the velocity vector at t = 3.

8 A body moves with a velocity given by:

r t

ti t j( ) =

++1

1, 0 ≤ t ≤ 10,

r i j( )0 0 2= +

a Find the displacement of the body during the time t = 0 to 10.b From part a find the average velocity of the body for t ∈ [0, 10].c Find the time when the speed is a minimum.

(Hint: Use calculus and/or a graphics or CAS calculator to find the approximate value of t which gives the minimum of the expression for the square of the speed).

9 The velocity of an object is given by the vector

r t e i e jt t( ) = +− −3 4 2 , t ≥ 0 and the initial

position is r i j( )0 = + .

a Find the position vector for the object and hence the distance of the object from the origin at t = 1, 10 and 100.

b Show that the object will eventually stop and give the final position of the object.c Find the angle between the position vector and velocity vector at t = 1.d Find the Cartesian equation for the motion given by

r t( ).

10 A body moves with a position given by r t

ti t j( ) =

++1

1, t ≥ 0.

a Find the displacement of the body during the time interval t = 0 to t = 10.b From part a find the average velocity of the body for t ∈ [0, 10].c Show that the body will eventually travel at a constant velocity parallel to the unit vector

j.



11 We 13 A body has a constant velocity of 3 2 i j+ and is at

r j= −4 when a second body

commences moving from − 4i with a velocity of 2ti j

+ at time t = 0.

a Determine the position vectors for both objects at time t.b By finding the time at which the two position vectors are equal, show that the two objects

will collide.c When and where do they collide?d What is the acceleration of the second object at time t?e At what time is the second body travelling at twice the speed of the first body?f Find the Cartesian equations for the paths of both bodies and sketch them using either a

graphics calculator or graphing application for a computer.

12 We14 Two toy boats start from the origin and move with velocity vectors

v t i tj1

2 1 2= − +( ) and

v t i t j2

21= + +( ) ; t ∈ [0, 5]. Velocity is in m/s.

a By obtaining the position vectors for both toy boats, calculate their distance apart when t = 2.

b What is the angle between their respective directions at t = 2?c At what time, if at all, are the acceleration vectors of the two bodies perpendicular to

each other?

13 Two ferries, A and B, travelling at constant velocities, have position and velocity vectors at 10 am given by:

r i j

r i j

v i j

v i

A

B

A

B

= −

= +

= +

= +−

−6 3

2

2 3

2 jj

The distance is measured in kilometres and the time in hours.a Show that the ferries will collide if they maintain their current velocities.b Determine the time of the collision.

14 Two boats, A and B, have positions (in km) given by the coordinates (1, 3) and (2, 7) respectively. Boat A is travelling with the velocity 3 5

i j+ km/h. Boat B sets off to meet boat A

at time t = 0. Both boats travel at the same speed.a When will the boats meet?b What is the velocity vector for boat B?

15 We15 A particle moves so that its position vector r at any time t, t ≥ 0, is given as

r t i t j= +4 3cos ( ) sin ( ) .

a Determine the Cartesian equation of the path and sketch it.b Find its velocity at any time t.c Hence give the direction of the particle when t is π

6.

d Determine the minimum and maximum speeds of the particle and state the first times at which they occur.

e Find the particle’s acceleration at any time t.f Verify that the maximum speed occurs when the magnitude of the acceleration is at a

minimum.

16 A particle moves so that its position vector r at any time t, t ≥ 0, is given as

r t i t j= +3 2 4 2cos ( ) sin ( ) .

a Determine the Cartesian equation of the path and sketch it.b Find its velocity at any time t.c Hence give the direction of the particle when t is π

2.

d Determine the minimum and maximum speeds of the particle and state the first time at which they occur.

496

e Find the particle’s acceleration at any time t.f Verify that the minimum speed occurs when the magnitude of the acceleration is at a

maximum.

17 A particle moves so that its position vector r at any time t, t ≥ 0, is given as

r t i t j= − + −[sin ( ) ] [ cos ( )]3 1 3 2 3 .

a Determine the Cartesian equation of the path and sketch it.b Find the particle’s velocity at any time t.c Hence give the exact speed and the direction of the particle when t is π

9.

d Determine the minimum and maximum speeds of the particle and state the first times at which they occur.

e Find the particle’s acceleration at any time t.f When is the acceleration perpendicular to the velocity?

18 A particle moves so that its position vector r at any time t, t ∈ [0,

π2), is given as

r t i t j= +tan ( ) sec ( )4 2 .

a Determine the Cartesian equation of the path and sketch it.b Find its velocity at any time t.c Hence give the exact speed and the direction of the particle when t is π

4.

d Determine the minimum speed of the particle and state when it occurs.e Find the particle’s acceleration at any time t.f Find the magnitude and direction of the acceleration: i when the speed is minimum

ii when t = π4

.

19 A particle moves so that its position vector at any time t ∈[0, π2

) is given by

r = 2 sec (t)

i + 3 tan (t)

j.

a Determine the Cartesian equation of its path and sketch it.b State the domain and range of the relation.c Find the velocity at any time t.d Find the speed when t equals:

i 0 ii π6

20 A particle moves so that its position vector at any time t ∈[0, π2

) is given by

r = sec2 (t)

i + 4 tan2 (t)

j.

a Determine the Cartesian equation of its path and sketch it.b State the domain and range of the relation.

Projectile motionThe term projectile motion refers to the motion of a particle thrown or projected through the air. The motion can be modelled and analysed.

Examples include the motion of a golf ball or basketball through the air, a bullet from a gun aimed at a target and a car that is airborne during an action sequence in an adventure movie.

In studying projectile motion it is necessary to make some simplifying assumptions. Firstly we assume that the projectile is a point and has no spin. Secondly we assume that the force due to air resistance is negligible.

Finally we consider only projectiles moving close to the Earth’s surface. These three assumptions mean that the only force acting on the projectile is the constant force due to gravity and hence the acceleration of the projectile is constant in both magnitude and direction. This motion can be summarised as in the following diagram.

10D



The initial velocity v(0) can be considered to have two components, one in the i direction and

one in the j direction. The acceleration due to gravity close to the surface of the Earth will be

taken to be − 9 8.j m/s2.

O

θ

y j

i

xR

r(t)~

~~

~

~v(0)~

g = −9.8 j m/s2

θV sin ( )j~

θV cos ( )i~

The initial velocity v V i V j( ) cos ( ) sin ( )0 = +θ θ

where V is the initial speed and θ is the launch angle of the projectile. The vector

r t( ) gives the position of the projectile at time t. The

vector g represents the acceleration of the projectile due to the force of gravity alone. The

projectile will be in the air for a time, T, and thus the domain for the model is 0 ≤ t ≤ T. During that time the projectile will travel a horizontal distance, R, called the range.

accelerationThe acceleration

a t( ) is due solely to the weight force acting on the particle. Thus:

a t i g j( ) = −0

We can write a t( ) in component form:

a t a i a jx y( ) = +

where ax and ay are the components of acceleration in the i and

j directions respectively.

The above equation simply states that the acceleration in the i direction is zero while the

acceleration in the j direction is a constant, equal to −g.

Note: g = 9.8 m/s2 downwards near the surface of the Earth.

VelocityTo obtain the velocity vector

v t( ) we simply need to antidifferentiate the acceleration vector with

respect to time. The constant of integration will be the initial velocity of the projectile v( )0 as

shown in the previous diagram.

v t a t dt v

g j dt V i V

( ) ( ) ( )

( ) [ cos ( )

= +

= + +

∫−

0

θ ssin ( ) ]

cos ( ) [ sin ( ) ]

θ

θ θ

j

V i V gt j

∫= + −

We can write v t( ) in component form:

v t v t i v t jx y( ) ( ) ( )= + where vx(t) and vy(t) are the

components of velocity in the i and

j directions respectively.

We can write these components explicitly:vx(t) = V cos (θ ) vy(t) = V sin (θ ) − gt.

This tells us that the horizontal component of the velocity is a constant equal to V cos θ and that the vertical component is given by the expression V sin (θ ) − gt. The speed of the projectile is

given by the magnitude of the velocity vector v t v t( ) ( ). or

v t( ) or v(t).

The projectile will be at the apex (top) of its flight when the vertical component of the velocity is zero and thus we can determine the time after launch when this occurs:

vy(t) = V sin (θ ) − gt = 0

tV

g= sin ( )θθ .

498

In general, the time that the projectile will be in the air will be twice this amount, provided that the launching point and the impact point are at the same height.

PositionTo obtain the vector equation for the position of the projectile as a function of time,

r t( ), the

velocity vector needs to be further antidifferentiated with respect to time. The integration constant here is the initial position of the projectile. In general, but not always, the initial position is taken to be the origin:

r i j( )0 0 0= + .

r t v t dt r

V i V gt

( ) ( ) ( )

cos ( ) [ sin ( ) ]

= +

= + −

∫ 0

θ θ jj dt r

Vt i Vt gt j

( ) +

= + −∫

( )

cos ( ) [ sin ( ) ]

0

12

2θ θ

As we have seen, we can write r t( ) in component form:

r t x t i y t j( ) ( ) ( )= +

where x and y are the components of the position in the i and

j directions, respectively, as

functions of time.

x(t) = (V cos (θ )) ty(t) = [V sin (θ )] t − 1

2 gt2

These two equations give the parametric equations for projectile motion.

WorkeD examPle 16

A golf ball is hit so that it leaves the ground at 15° to the horizontal at a speed of 60 m/s over a horizontal fairway.a What are the horizontal and vertical components of the ball’s initial velocity?b What is the speed of the ball, to the nearest m/s, one second after being struck?c Determine the position vector,

r t( ), for the motion and consequently find the distance of the ball

from its initial position one second after being struck.

Think WriTe

1 Make the unit vector i be directed horizontally

right and the unit vector j be directed vertically

up.

i~

j~

i~

j~

15°

V = 60 m/s

2 (Optional) Draw a vector diagram of the ‘hit-off’ taking the position of hit-off as

r i j( )0 0 0= + .

Let r i j( )0 0 0= + .

a 1 The horizontal component of the velocity vector is vx(t) = V cos (θ ).

a vx = V cos (θ )

2 Evaluate vx(0) where V = 60 and θ = 15°. vx(0) = 60 cos (15°)= 57.96

3 The vertical component of the velocity is vy = V sin (θ ) − gt.

vy = V sin (θ ) − gt

4 Evaluate vy(0) where V = 60, g = 9.8, θ = 15° and t = 0.

vy(0) = 60 sin (15°) − 9.8(0)= 15.53



b 1 Use the components that were found in part a to write down the velocity vector

v t V i V gt j( ) cos ( ) ( sin ( ) )= + −θ θ .

b v t i t j( ) . ( . . )= + −57 96 15 53 9 8

2 Evaluate v( )1 .

v i j( ) . .1 57 96 5 73= +

3 Calculate v( )1 .

v( ) . .

.

1 57 96 5 73

58 24

2 2= +=

4 State the speed when t = 1. The speed of the golf ball after 1 second is 58.24 m/s.

c 1 Express r t( ) as the antiderivative, with

respect to t, of v t( ).

c r t i t j dt c( ) [ . ( . . ) ]= + − +∫ 57 96 15 53 9 8

2 Antidifferentiate and include the constant vector of integration,

c c i c jx y= + .

= + − + +57 96 15 53 4 9 2. ( . . )ti t t j c i c jx y

3 Substitute t = 0 into r t( ) and equate to the

initial position 0 0 i j+ .

r c i c j i jx y( )0 0 0= + = +

4 Determine cx + cy by equating the components of

i and

j.

cx = 0 and cy = 0

5 Simplify the position vector r t( ).

r t ti t t j( ) . ( . . )= + −57 96 15 53 4 9 2

6 Evaluate r( )1 .

r i j( ) . .1 57 96 10 63= +

7 Calculate the displacement in the fi rst second

r r( ) ( )1 0= .

r r i j i j( ) ( ) . . ( )

.

1 0 57 96 10 63 0 0

57 96

− = + − +

= i j+10 63.

8 The distance from hit-off is the magnitude of r r( ) ( )1 0− .

r r( ) ( ) . .

.

1 0 57 96 10 63

58 93

2 2− = +=

9 State the distance. The distance of the ball from its hit-off position after 1 second is 58.93 m.

WorkeD examPle 17

20°

A golf ball is hit so that it leaves the tee at a speed of 35 m/s at an angle of 20° to the horizontal.a State the initial velocity vector of the ball.b Find v t( ), the velocity of the ball as a function of time.c Determine the maximum height that the ball reaches and the location of the maximum height

from the tee.d Find the time that the ball is in the air.e Find the horizontal distance (range) covered by the golf ball.

eBookpluseBookplus

Tutorialint-0415

Worked example 17

500

Think WriTe

a 1 Make the unit vector i be directed

horizontally and the unit vector j be

directed vertically upwards.

a

θ

Note: V = |v| = speed

θ v y j = V cos ( )j~~

θ v x i = V sin ( )i~~

~

v~

2 (Optional) Draw a vector diagram of the situation at ‘tee-off’.

3 Write the general rule for the velocity of a projectile at any time, t:

v t V i V gt j( ) cos ( ) [ sin ( ) ]= + −θ θ .

v t V i V gt j( ) cos ( ) [ sin ( ) ]= + −θ θ

4 Substitute V = 35, θ = 20° and t = 0 to find the initial velocity.

v i j( ) cos ( ) sin ( )0 35 20 35 20= +

5 Evaluate v( )0 with the components correct

to 1 decimal place. v i j( ) . .0 32 9 12 0≈ + m/s

6 State the initial velocity. The initial velocity is approximately32 9 12 0. . i j+ .

b 1 Write the general rule for the velocity of a projectile.

b v t V i V gt j( ) cos ( ) [ sin ( ) ]= + −θ θ

2 Substitute V = 35, g = 9.8 and θ = 20° and evaluate.

v t i t j( ) . ( . . )= + −32 9 12 0 9 8 m/s

c 1 Since the maximum height is reached when the vertical velocity,

vy, is zero, set

vy to zero.

cv ty = −

=

12 0 9 8

0

. .

2 Solve for t to determine the time when the golf ball reaches its maximum height.

t ≈ 1.225 s.

3 Express r t( ) as the antiderivative of

v t( )

with respect to time. r t v t dt c( ) ( )= +∫

4 Antidifferentiate. = + + − +( . ) ( . . )32 9 12 0 4 9 2t c i t t c jx y 5 Simplify

r( )0 and set it equal to the given

result 0 0 i j+ .

r c i c j i jx y( )0 0 0= + = +

6 Solve for the integration constant. cx = cy = 0.7 Give the position vector of the golf ball at

any time, t. r t ti t t j( ) . ( . . )= + −32 9 12 0 4 9 2

8 Evaluate r( . )1 225 to determine the position

of the maximum height. r i j( . ) . .1 225 40 3 7 35≈ + metres

9 For r( . )1 225 the

i component gives the

horizontal displacement of the golf ball and the

j component gives the vertical

component of the maximum height position.

The coordinate for the turning point of the projectile trajectory, taking the tee-off position as the origin O (0, 0) is (40.3, 7.35). The ball reaches a maximum height of 7.35 m above the ground 40.3 m horizontally away from the point where the ball was struck.

d 1 Write the position vector of the golf ball,

r t( ).

d r t ti t t j( ) . ( . . )= + −32 9 12 0 4 9 2

2 State the vertical component of r t( ), y(t). y(t) = 12.0t − 4.9t2

Golf ball is at ground level when y(t) = 0.



3 Set y(t) = 0 to find the times when the golf ball is at ground level.

(12 − 4.9t)t = 0

4 Solve for t. t = 0 or 12 04 9

..

(≈ 2.45 s)

5 The difference between the two times, t = 0 and t = 2.45, gives the time that the golf ball is in the air.

The ball is in the air for 2.45 seconds.

Note: We could have obtained the same result simply by multiplying the time taken to reach the apex of the ball’s flight (see part c) by 2 since symmetry dictates that the time to rise is the same as the time to fall when only acceleration due to gravity is considered.e 1 Write the

i (horizontal) component of

r t( ),

x(t).e x(t) = 32.9t

2 Evaluate x(2.45) to determine the range of the golf ball.

R = x(2.45) = 32.9 × 2.45 = 80.6 m

3 State the range. The range is 80.6 metres.

Note: Alternatively, we could have achieved the answer by simply multiplying the x-ordinate for the apex, namely x = 40.3, by 2 — again because of the symmetry of the path taken by the golf ball. Any small differences in the two answers result from rounding errors in evaluating cos (20°) and sin (20°).

WorkeD examPle 18

A projectile is fired at a time t = 0 from the top of a cliff whose base is taken to be the origin, as shown at right.The position vector

r t( ) (metres) at a time, t (seconds), of the

projectile is given by r t t i t t j( ) ( . )= + + −20 50 25 4 9 2 .

a How high is the cliff top above its base?b Give an expression for the velocity vector of the projectile.c Find the initial speed and launch angle of the projectile.d Find the speed and angle at which the projectile strikes

the ground.e Find the distance between the launching point and the point where the projectile hits the ground.

Think WriTe

a 1 Write the projectile's initial position, r( )0 . a

r i j( )0 0 50= +

2 The j component of

r( )0 gives the initial

height of the projectile.The top of the cliff is 50 m directly above its base.

b 1 Write the position vector, r t( ). b

r t ti t t j( ) ( . )= + + −20 50 25 4 9 2

2 The velocity vector, v t( ), is obtained by

differentiating r t( ) with respect to time.

v t i t j( ) ( . )= + −20 25 9 8

c 1 Evaluate v t( ) when t = 0 to determine the

initial velocity of the projectile.c v i j( )0 20 25= +

2 Evaluate v( )0 to determine the initial

speed of the projectile.The speed is

v( )0

=

= +≈

v v( ) ( )

/ .

.0 0

20 25

32

2 2

m s

O

Cliff

Projectile

i~

j~

v(0)~

502

3 The launch angle (angle of elevation) is determined from the velocity vector

v t v i v jx y( ) = + using the tan rule,

θ =

−tan 1v

vy

x.

θ =

−tan 1v

vy

x

4 Substitute vx = 20 and vy = 25 to determine the launch angle of the projectile.

θ =

−tan 1 2520

5 Evaluate θ correct to 1 decimal place. θ ≈ 51.3°6 State the launch speed and angle. The projectile was launched at a speed of

32 m/s at an elevation angle of 51.3°.

d 1 Write the position vector, r t( ). d

r t ti t t j( ) ( . )= + + −20 50 25 4 9 2

2 Give the vertical position (j component),

y(t), of the position vector.y t t t( ) .= + −50 25 4 9 2

3 Set y(t) equal to zero to determine when the projectile reaches the ground.

Projectile hits the ground when y(t) = 0.50 25 4 9 02+ − =t t.

4 Solve for t using the quadratic formula. t =± − × ×

×

− −

−

25 25 4 4 9 50

2 4 9

2 ( . )

.5 Give the (first) positive value for t only. t = 6.64

6 State when the projectile strikes the ground.

The projectile strikes the ground 6.64 seconds after launch.

7 Write the velocity vector, v t( ), of the

projectile. v t i t j( ) ( . )= + −20 25 9 8

8 Evaluate v (6.64) to find the velocity of

the projectile on impact. v i j( . ) .6 64 20 40 07= −

9 Determine v( . )6 64 to find the speed of

the projectile on impact. v( . )6 64

= +=

−20 40 07

44 8

2 2( . )

.

10 The impact angle is found using

θ =

−tan 1v

vy

x where vx = 20 and

vy = −39.7.

θ =

−−

tan.1 40 07

20

11 Evaluate θ correct to 1 decimal place. θ ≈ −63.5°

12 The negative sign indicates that the angle of impact is a depression angle, below the horizontal as would be expected.

13 State the impact speed and angle of the projectile.

The projectile strikes the ground at 44.8 m/s at an angle of 63.5°.The negative sign indicates that the angle of impact is a depression angle below the horizontal, as you would expect.



e 1 Write the position vector, r t( ). e

r t ti t t j( ) ( . )= + + −20 50 25 4 9 2

2 Evaluate r( . )6 64 to find the impact

position of the projectile. r i j( . ) .6 64 132 8 0= +

3 Evaluate r( )0 to find the launch position

of the projectile (found in part a). r i j( )0 0 50= +

4 Evaluate r r( . ) ( )6 64 0− to determine the

displacement of the projectile. r r i j( . ) ( ) .6 64 0 132 8 50− = −

5 Determine r r( . ) ( )6 64 0− to find the

distance between the launch point and the impact point.

r r( . ) ( )6 64 0− = + −132 8 502 2. ( )

6 Evaluate. = 141 9.7 State the solution. The distance between the point of launching

and landing point is 141.9 metres.

The Cartesian equation for the path of a projectileIn many cases it is useful to know the path that a projectile takes. For example, we may want to determine whether a basketball (thrown from a particular position with a known initial velocity) will pass cleanly through the hoop; or whether a ball struck with a cricket bat will result in a six. It is necessary to plot the y-ordinate as a function of the x-ordinate to obtain the Cartesian equation for the projectile. In the previous section we established the parametric equations x(t) and y(t) for the motion:

x t Vt( ) cos ( )= θ y t Vt gt( ) sin ( )= −θ 12

2

The diagram at right illustrates the projectile trajectory with position vector

r t x t i y t j( ) ( ) ( )= + .

Using the first of these parametric equations, namely x = Vt cos (θ ), we can write t as the subject:

tx

V=

cos ( )θand substitute this into the second equation to obtain y in terms of x:

y t Vt gt( ) sin ( )= −θ 12

2

where tx

V=

cos ( )θ

y xVxV

gx

V

x

( )sin ( )

cos ( ) cos ( )

tan

= −

=

θθ θ

2

2 22

(( )cos ( )

θθ

− gx

V

2

2 22

y x xgx

V( ) tan( )

cos ( )= −

θθ2 2 2

The above equation tells us that the trajectory followed by a projectile, under the assumptions outlined earlier, is a parabola. The equation is a quadratic function of the form bx − ax2 = x(b − ax).

The Cartesian graph has solutions x = 0 and ba and a stationary point, which is a maximum at

x = ba2

where ag

V=

2 2 2cos ( )θ and b = tan (θ ). The non-zero solution tells us the range of the

y

x

r(t) = xi + yj~ ~ ~

ORange (R)

R

504

projectile, R, that is, the horizontal distance covered by the projectile. Thus:

Rba

Vg

Vg

=

=

=

2

2

2 2

2

cos ( ) tan ( )

cos ( ) )

θ θ

θ θsin (

and since sin (2θ ) = 2 cos (θ ) sin (θ )R

Vg

=2 2sin ( )θθ

The range, R, can therefore be found directly from the launching conditions of the projectile, namely the initial speed, V, and launch angle, θ. The horizontal ordinate for the apex of the fl ight occurs at precisely half the range. As an example of the above equation consider a golf ball struck from a tee with a speed V = 70 m/s and launch angle θ = 15°. The range of the golf ball would be calculated to be:

RV

g=

= =

2

2

2

70 309 8

250

sin ( )

sin ( ).

θ

Thus the ball would land 250 m away from the tee having a maximum height at a horizontal distance from the tee of 125 m. The assumption here is that the tee and point where the ball returns and hits the ground are at the same altitude.

Further, golf balls have signifi cant air resistance and spin forces acting on them such that the paths they take through the air differ from purely parabolic trajectories.

WorkeD examPle 19

An object is launched at the origin at a speed of V m/s and angle of θ ° to the horizontal. It passes through the point (60, 80); that is, 60 m horizontally from the launch point and 80 m vertically up at time t = 3 s.a Find the initial velocity vector,

v( )0 m/s, for the object.

b Find the Cartesian equation, y(x), for the trajectory of the object.c Find the range, R, of the object.d Find the maximum height above the launching point reached by the object.

Think WriTe

a 1 Write the general rule for the position of a projectile,

r t( ).

a r t Vt i Vt gt j( ) cos ( ) ( sin ( ) )= + −θ θ 1

22

2 Evaluate r( )3 and equate to

the given position when t = 3,

r i j( )3 60 80= + .

r V i Vg

j( ) cos ( ) sin ( )3 3 392

60

= + −

=

θ θ

ii j+ 80

3 Equate the i components. 3V cos (θ ) = 60

4 Solve for V cos (θ ). V cos (θ ) = 20

5 Equate the j components. 3

92

80Vg

sin ( )θ − =

6 Solve for V sin (θ ). V sin (θ ) ≈ 41.4.7 Write down the general rule for

v( )0 of a projectile.

v V i V j( ) cos ( ) sin ( )0 = +θ θ

8 Substitute V cos (θ ) = 20 and V sin (θ ) = 41.4 into the rule.

= +20 41 4 i j.

eBookpluseBookplus

Tutorialint-0416

Worked example 19



b 1 Find the initial speed v( )0 of the

projectile.bv( )0 = + ≈20 41 4 462 2. m/s

2 Calculate the projection angle

using θ =

−tan 1

v

vy

x

.

The projection angle is tan.−

≈ °1 41 4

2064 .

3 Write the general rule for y(x). y x xgx

V( ) tan ( )

cos ( )= −θ

θ

2

2 22 in general.

4 Substitute V = 46, θ = 64 and g = 9.8 into the equation.

= −x xtan ( ).

( ) cos ( )64

9 8

2 46 642 22

5 Simplify. y x x x( ) . .= −2 05 0 012 2

c 1 Write the general rule for the range of a projectile.

c RV

g=

2 2sin ( )θ

2 Substitute θ = 64°, V = 46 and g = 9.8 and evaluate.

=

≈

46 1289 8

170

2 sin ( ).

3 State the range. The range is 170 m.

d 1 Write the Cartesian equation for this projectile.

d y(x) = 2.05x − 0.012x2

2 The maximum height occurs when x is half the value of the range, that

is, x = =1702

85.

Maximum height occurs when

R2

1702

85

=

=3 Evaluate y(85). y(85) = 87.6

4 State the maximum height. The maximum height is 87.6 metres.

Projectiles launched close to the surface of the Earth (assuming that air resistance is 1. negligible) have an acceleration given by

a t j( ) .= −9 8

where the unit vector j is in the vertically upwards direction.

Projectiles launched with a speed of 2. V, at an elevation angle of θ, have an initial velocity vector of

v V i V j( ) cos ( ) sin ( )0 = +θ θ

The instantaneous velocity, 3. v t( )v t( )v t , at time t, is

v t V i V gt j( ) cos ( ) [ sin ( ) ]= + −θ θ where g = 9.8 m/s2

The position 4. r t( )r t( )r t , at any time t, is

r t Vt gt y j( )r t( )r t [ cVt[ cVt os ( ) ] [

] [

n ( ) ]) ]gt) ]gt y j) ]y j= +[ c= +[ cVt[ cVt= +Vt[ cVt os= +os ( )= +( ) + −) ]+ −) ]) ]+) ]θ θx iθ θx i Vtθ θVt( )θ θ( ) ] [θ θ] [x i] [x iθ θx i] [x iθ θsiθ θsin (θ θn ( ) ]θ θ) ]= +θ θ= +( )= +( )θ θ( )= +( ) + −θ θ+ −Vt+ −Vtθ θVt+ −Vt] [+ −] [θ θ] [+ −] [ si+ −siθ θsi+ −sin (+ −n (θ θn (+ −n (0θ θ0θ θx iθ θx i0x iθ θx i 1) ]1) ]2) ]2) ]2) ]2) ]0y j0y jy j) ]y j0y j) ]y j

where (x0, y0) are coordinates of the initial position of the projectile. In most instances this is the origin (0, 0).

rememBer

506

Projectile motionUse g = 9 8. m/s2 downwards for questions in this exercise (unless otherwise indicated).

1 We 16 A golf ball is hit with a two-iron so that it leaves the tee at 10° to the horizontal with a speed of 80 m/s over a horizontal fairway.a What are the horizontal and vertical components of the ball’s initial velocity?b What is the speed of the ball one second after being struck?c Determine the position vector,

r t( ), for the motion and consequently fi nd the distance of

the ball from the tee one second after being struck.

2 A bullet is fired into the air from a gun at ground level at a speed of 150 m/s at an angle of 60° to the horizontal.a Give an expression for the velocity vector,

v t( ).

b Calculate the speed of the bullet after 2.0 seconds.c Give an expression for the position vector,

r t( ).

d Calculate the distance of the bullet from the gun at t = 2.0 s.

e Find the time when the bullet is at the apex of its motion.

3 We 17 A clay disc is fired at an elevation angle of 12° with a speed of 45 m/s. The disc is used as a moving target for rifle practice.a State the initial velocity vector of the clay disc.b Find

v t( ), the velocity of the ball as a function of

time.c Determine the maximum height the clay disc

reaches and the location of the maximum height from the fi ring point.

d Find the time during which the clay disc is in the air.e Find the horizontal distance (range) of the clay disc.

4 A football is kicked into the air at an angle of 30° to the horizontal so that it reaches a maximum height in 0.50 s.a Find the maximum height reached by the ball.b Find the speed at which the ball was kicked.c Find the horizontal distance that the ball travels before striking the ground.d Find the speed at which the ball strikes the ground.e Find the angle at which the ball strikes the ground.

5 A baseball player hits a ball to the outfield 50 m away. The ball was in the air for 1.8 seconds.a At what angle to the horizontal

was the ball hit?b At what speed was it hit?

6 An object is thrown such that its position,

r t( ), is given by the

vector

r t ti t t j( ) ( . )= + −30 50 4 9 2

where distances are measured in metres and time in seconds.a Find the position of the object at t = 3.b Find the velocity vector,

v t( ), and hence the speed of the object at t = 3.

exerCiSe

10D

u = 150 m/s

60°

50 m



c Find the acceleration vector, a t( ).

d Find the time when the velocity vector is perpendicular to the acceleration vector and explain where this occurs in the object’s motion.

7 We 18 A projectile is thrown across a horizontal field such that its position, r t( ), is given by

the vector

r t ti t t j( ) . ( . )= + + −4 9 1 22 4 9 2

where distances are measured in metres and time in seconds.a From what height is the projectile thrown?b Give an expression for the velocity vector of the projectile.c Find the initial speed and launch angle of the projectile.d Find the speed and angle at which the projectile strikes the ground.e Determine how far from the launching site the projectile lands.

8 We 19 A projectile is launched at the origin at a speed V m/s and angle θ ° to the horizontal. It passes through the point (50, 60.4) after 2 seconds.a Find the initial velocity vector,

v( )0 , for the projectile.

b Find the Cartesian equation, y(x), for the trajectory of the projectile.c Find the range, R, of the projectile.d Find the maximum height above the launching point made by the projectile.

9 A rock is thrown with a velocity vector v i= 10 off

a cliff as shown at right. The rock is in the air for 3 seconds before it lands in the sea at a point, S. Take the initial position of the rock to be the origin.a How high is the cliff above sea level?b How far out from the cliff does the rock strike

the sea?c At what speed and angle does the rock hit the sea?d Find the position vector,

r t( ), for the

rock, t ∈ [0, 3.0].e Find the Cartesian equation for the motion of the

rock.

10 An archer shoots an arrow at a horizontal speed of 160 m/s directly at a target positioned 30 m away. The arrow was aimed directly at the target centre.a How far below the centre will the arrow hit the target?b If the arrow is to hit the target centre, at what angle to the horizontal should the arrow be

aimed?

11 A student skateboarding at a constant speed of 5.0 m/s throws a ball vertically into the air at 4.0 m/s. She catches the ball as it returns to the ground. Let the ball be released from her hand at time t = 0, the moment she passes a marker, as shown at right.a What is the initial velocity of the ball?b Show that the ball will follow a parabolic

path and give the position vector, r t( ).

c How long is the ball in the air?d How far past the marker is the student when she catches the ball? (Assume she catches

the ball at the same height above the ground that she threw the ball.)

12 Show for a projectile with a range R and a time of flight T that:

tan ( )θ = gTR

2

2 where θ is the initial angle of projection.

OS

v = 10i~ ~

g = −9.8j

j~

i~

~~

Ball released at(0, 0) at t = 0

y

x

g = −9.8j~ ~r(t)~

508

13

A cannonball is fired out to sea from a gun located on a cliff 50 m above the ocean. The cannon has the position vector

r i j= +0 50 . The initial velocity of the cannonball is given by

the velocity vector v i j= +120 50 .

a Give the Cartesian equation for the trajectory of the cannonball.b Find the maximum altitude of the cannonball above the surface of the water.c How far out to sea is the cannonball when it hits the water?d For how long is the cannonball in the air?e At what angle does the cannonball strike the water?

14 A golf ball passes horizontally through its maximum point 20 m above a horizontal fairway 98 metres from where it was struck. Find the Cartesian equation that describes the trajectory and the total time that the ball is in the air.

15 An object is thrown horizontally with a speed of V m/s from the top of a cliff-face H metres high. The object strikes the ground at a distance R metres from the base of the cliff after a time of T seconds. Show that:

a THg

= 2

b R = VT

c the speed at which it hits the ground is V Hg2 2+

d the angle at which it strikes the ground is tan−

1 2HR

e the ball travels in a parabolic path y Hx

R= −

12

2.

16 A baseball is initially hit from a distance of 1.5 metres above the ground at an angle of 35°. The ball reaches a maximum height of 8 metres above the ground. Find:a the initial speed of projectionb the distance of the outfielder from the hitter if the outfielder just catches the ball at ground

levelc the speed and angle at which the baseball strikes the fielder’s hands.

17 A footballer is 30 metres from goal, and kicks the ball from an initial height of one metre above the ground. The ball takes half a second to travel through the goalposts at a height of 2 metres. Find:a the initial kicking speedb the angle at which the ball is kicked.

18 A particle is projected under gravity with an initial speed of V, from an origin O, to pass through a point a distance d from the point of projection and height h above it. Show that the angle of projection a above the horizontal satisfies the equation:

tan ( ) tan ( )22 2

2

21

20α α− + +

=Vgd

V h

gd. Deduce that the two possible angles of projection,

a1 and a2, satisfy the equation tan ( ) .α α1 2+ =− d

h

50 m

j

i~

~



19 Three points — A, B and C — are in a straight line on a horizontal ground with d (AB) = d (BC) = a. Vertical poles of heights 3b and 2b are erected at B and C.A ring is fi xed to the top of the pole at B and a target is attached to the top of the pole at C.A basketball is thrown from A so as to pass through the ring and hit the target.

a Show that the time required to hit the target is 4bg

.

b Show that it is projected at an angle of tan−

1 5ba

.

c Show that the speed of projection is given by g a b

b

2 225

2

+( ).

20 A golf ball is struck from level ground, giving it an initial velocity of 30i + 20

k m/s where

i is a unit vector

directed horizontally and k is a unit vector directed vertically upwards. The acceleration of the ball due to

wind resistance and gravity is −2i − 10

k m/s2. Find expressions for the velocity vector and position vector of

the ball at any time t seconds. Hence find:a when and where the ball reaches its highest pointb when and where the ball hits the ground. eBookpluseBookplus

Digital docWorkSHEET 10.2

510

Summary

Position, velocity and acceleration

Position, •r t( ), velocity,

v t( ) and acceleration,

a t( ), are vector quantities. In two dimensions they can be

written as:

r t xi yj( ) = +

v t v i v jx y( ) = + [Note:

v t r t

drdt

dxdt

idydt

j( ) ( )= = = + ]

a t a i a jx y( ) = + , respectively. [Note:

a t r t

d r

dt

d x

dti

d ydt

j( ) ( )= = = +2

2

2

2

2

]

The average velocity during the time interval • ∆ t is

vr t t r t

tav = + −( ) ( )∆∆

.

The instantaneous velocity •

v t r tdrdt

dxdt

idydt

j( ) ( )= = = + .

The instantaneous speed is given by • v v v vx y. = +2 2 which is the magnitude of the velocity vector

v.

The direction of motion is in the direction parallel to the velocity vector.•

The instantaneous acceleration •

a tdv

dt

dv

dti

dv

dtj

d x

dti

d y

dtx y( ) = = + = +

2

2

2

2 j .

Two non-zero vectors, •a and

b, are perpendicular when

a b. = 0.

The angle, • θ, between two vectors is given by cos (θ ) =

a b

a b

. where

a and

b are vectors.

Cartesian equations and antidifferentiation of vector functions

If the position vector is • r t x t i y t j( ) ( ) ( )= + then the parametric equations for the coordinates of the path are

x(t) and y(t). The Cartesian equation for the motion of the particle is y(x) which can be obtained from the parametric equations x(t) and y(t).The velocity vector, •

v t( ), can be found from the acceleration vector,

a t( ), by antidifferentiation provided that

the velocity is given at a particular time, t. Generally this is at time t = 0 so that the vector c can be found. Thus:

v t a t dt c( ) ( )= +∫

The position vector, •r t( ), can be found from the velocity vector,

v t( ), by antidifferentiation provided that the

position is given at a particular time, t. Generally this is at time t = 0 so that the vector c can be found. Thus:

r t v t dt c( ) ( )= +∫

Applications of vector calculus

Read each question carefully so that it is interpreted correctly and the question being asked is answered.•Apply vector calculus methods to analyse curvilinear motion where the vectors •

r, r and

r are functions of

time.For inexact solutions give your answers to an appropriate number of decimal places.•If two objects are to collide then they must be at the same point at the same time.•

Projectile motion

Projectiles launched close to the surface of the Earth (assuming that air resistance is negligible) have an •acceleration given by

a t j( ) .= − 9 8 where the unit vector,

j, is in the vertically upwards direction.

Projectiles launched with a speed • V, at an elevation angle θ, have an initial velocity vector:

v V i V j( ) cos ( ) sin ( )0 = +θ θ



The instantaneous velocity is •v t( ):

v t V i V gt j( ) cos ( ) [ sin ( ) ]= + −θ θ where g = 9.8 m/s2

The position of the projectile, •r t( ), is:

r t V t x i V t gt y( ) [ cos ( ) ] [ sin ( ) ]= + + − +θ θ0

12

20j

where (x0, y0) are the coordinates of the initial position of the projectile. In most instances this is the origin, (0, 0).

512

ChaPTer reVieW

ShorT anSWer

1 The velocity of a projectile in m/s is given by

v t t t i t j( ) ( ) cos ( )= − −4 6 8 72 π . Find the magnitude of the acceleration of the projectile at time t = 3 s.

2 A body moves according to

x t j( ) .= − 9 8 for all

times t ≥ 0. It starts at the point (0, 20) with an initial velocity

x i j( )0 20 49= + . Find the position

vector, x t( ), for all times t ≥ 0.

3 The velocity vector of a particle at time t seconds is given by

r t t i t j( ) cos ( ) sin ( )= +5 3 12 3 (m/s).

Find:a the position vector,

r t( ), given that

r j= − 4

at t = 0b the acceleration vector,

a t( )

c the times when the position vector is perpendicular to the velocity vector

d the Cartesian equation for the pathe the maximum speed of the particle. [Hint: Find

an expression for the square of the speed and find its maximum value.]

4 Two bodies, A and B, move with position vectors

x t i t jA = +3 43 2 and

x t i tjB = −3 42 respectively.

At time t = 2 determine the distance between the two bodies.

5 An object moves with a velocity in m/s given by

v t t i t j( ) cos ( ) sin ( )= +8 2 8 2π π .

a Show that the speed is a constant and state its value.

b Show that the acceleration and velocity vectors are perpendicular at all times.

6 An object moves with the following position vector:

r t e i e jt t( ) ( ) ( )= + + −− −3 1 2 12 4

a Find the initial position, velocity and acceleration vectors describing the motion of the object.

b Find the angle between the initial velocity and the initial acceleration of the object.

c Show that the object will eventually stop moving.

d By finding the Cartesian equation of the object, sketch the trajectory labelling the initial and final position of the object.

7 Particles A and B move with position vectors:

r ti tjA = −4 2

r t i t jB = +4 4sin ( ) cos ( )α α

where a is a positive constant.Distance is measured in metres and time in seconds.a Find the speed of B at any time, t.b Find the Cartesian equations of the paths of A

and B.c On the same set of axes, sketch the paths of A

and B indicating the directions of travel.d Find the coordinates for the points where the

paths of A and B intersect.

8 A particle moves so that its position vector r at any

time t, t ≥ 0, is given as

r t i t j= +3 2cos ( ) sin ( )

a Determine the Cartesian equation of the path and sketch it.

b Find its velocity at any time t.c Hence give the direction of the particle when

t is π3

.

d Find the particle’s acceleration at any time t.e Verify that the maximum speed occurs when

the magnitude of the acceleration is at a minimum.

9 A particle moves in the Cartesian plane with position vector

r xi yj= + where x and y are

functions of t.If its velocity vector is

v yi xj= +− , find the

acceleration vector of the particle in terms of the position vector

r.

exam TiP Answers to this question were poor, and many students appeared to have difficulty in understanding what was being asked. Some students used a ‘pattern’ approach to obtain the correct expression for a, but without any indication of differentiation or understanding of the situation. Where incorrect assumptions were used that still occasionally allowed the student to reach the correct answer, credit could not be given.

©VCAA Assessment report 2007

[©VCAA 2007]

10 The coordinates of three points are A (1, 0, 5), B (−1, 2, 4) and C (3, 5, 2).

Express the vector →AB in the form:



a xi + yj + z k.

b Find the coordinates of the point D such that ABCD is a parallelogram.

c Prove that ABCD is a rectangle.

mulTiPle ChoiCe

1 The position vector of an object is given by

r t ti t j( ) = +2 2 . The average velocity in the time

interval [1, 2] is:A 3 3 i j+ B 3 2

i j+

C 3 3 i j− D 2 3

i j+

E 3 2 i j−

2 The position vector of an object is given by

r t t t i t t j( ) ( ) ( )= − + −3 2 . The magnitude of the velocity at t = 2 is:

A 112 B 40

C 130 D 32

E 153

3 If r t ti t t j( ) ( cos )= − +3 2 π then

r( )2 is equal to:

A (π 2 − 6)j B (6 + π 2)

j

C (6 − π 2)j D (−6 − π 2)

j

E −(6 − π 2) j

4 Two objects, M and N, have position vectors given by:

r t i t jM = + −( )4 6 1

22

r t i tjN = − +( )2 1 3The nearest whole angle between their respective directions at t = 2 is:A 37° B 53°C 127° D 143°E 83°

5 The acceleration of a ball at any time, t, is given by

a t i t j( ) ( )= + −5 4 1 with the initial velocity of the

ball v j= 2 . At time t = 6 the velocity vector is:

A 4j B 30 12

i j+

C 5 72 i j+ D 30 72

i j+

E 30 68 i j+

6 The velocity vector of a moving object is given by the expression:

v t t i t j( ) sin ( )= +25 4 2

When the component of the velocity in the i

direction first equals 25 m/s, the value of the instantaneous acceleration is:

A π4

j B 25

2 4

2π π i j+

C 25

2 2π π i j+ D

π 2

4 j

E π 2

2 j

7 A tennis ball is launched with a speed of 20 3 m/s at an angle of 30° to the horizontal. If

i

is the unit vector in the horizontal and j is the unit

vector vertically upwards in the plane of motion of the ball, the initial velocity (in m/s) is:

A 30 15 i j+ B 15 3 45

i j+

C 45 10 3 i j+ D 15 45 3

i j+

E 30 10 3 i j+

8 A particle starts at position − + i j and moves so that

its velocity at time t is 2ti j

− . The position of the particle at any time t,

r t( ), is given by:

A ( ) ( )t i t j2 1 1− + +

B ( ) ( )t i t j2 1 1− + −

C ( ) ( )t i t j2 1 1+ + −

D ( ) ( )t i t j2 1 1+ + − +

E ( ) ( )t i t j2 1 1− − + 9 The position vector of a projectile at time t is

r t ti t t j( ) ( . . )= + −10 19 6 4 9 2 (metres) where

i

is horizontal and j is vertically upwards. The

maximum height of the projectile is:A 4.9 m B 9.8 m C 19.6 mD 29.4 m E 39.2 m

10 An object is launched with a velocity

v t i j( ) = +40 15 m/s. The horizontal range of the object, to the nearest ten metres, is:A 110 m B 120 m C 130 mD 140 m E 150 m

11 A particle initially at the origin starts from rest at t = 0. The particle moves in a straight line in such a way that its acceleration at time t is given by e i t jt− +0 1 6. ( )

.

The velocity of the particle at time t is given by:

A − − +( . ) ( ).0 1 60 1e i t jt

B − − +( ) ( ).10 30 1 2e i t jt

C 10 1 30 1 2( ) ( ).− +−e i t jt

D 0 1 1 30 1 2. ( ) ( ).− +−e i t jt

E 10 10 10 30 1 2( ) ( ).− − +−t e i t jt

exam TiP Don’t forget the constant of integration.

[©VCAA 2005]

514

12 P, Q and R are three collinear points with position vectors p, q and r respectively, where Q lies between P and R.

If →QR= 1

2 →PQ, then r is equal to:

A 32

12

q p− B 32

12

p q−

C 32

32

q p− D 12

32

p q−

E 32

32

p q− [©VCAA 2008]

13 The velocity v m s–1 of a body which is moving in a straight line, when it is x m from the origin, is given by v x= −sin ( ).1 The acceleration of the body in m s–2 is given by:

A −cos−1 (x) B − cos ( )

sin ( )

x

x2

C −cot (x) cosec2 (x) D 1

1 2− x xsin ( )

E sin ( )−

−

1

21

x

x [©VCAA 2008]

exTenDeD reSPonSe

1 Two particles, R and S, move so that their position vectors, r and

s, are given by

r t i t j= − + +[ sin ( )] [ cos ( )]2 2 1

s t i t j= +cos ( ) sin ( )2

respectively, where t seconds (t > 0) is the time elapsed since the start of their motions.a Find

r s. and determine the exact time that the position vectors of the two particles are first at right

angles.b Show that the particles are always moving perpendicular to each other.c i Find the Cartesian equation of the path of S. ii Find the Cartesian equation of the path of R.d Sketch the Cartesian equations of the paths of S and R on the same set of axes, indicating the directions

of motion.e Use the graph to verify that the motion of the particles is perpendicular when t = 0 and t =

π2

.

2 A balloon is released from a point on the ground. It moves in a plane such that its position, in relation to the point at any time t seconds after its release, can be described by:

r i t= ∈6 0 100, [ , ]


r = 0 and

r = 6i + 4j.

a Find the position vector at any time t.b State the position of the balloon at i t = 4 and ii t = 10.c The balloon is attacked by a rogue canary, and bursts after 100 seconds. How far, horizontally, is the

furthest the balloon gets from its take-off point? Give its height at this point.d Determine the Cartesian equation of the path and sketch it.e What is the angle of trajectory of the balloon when it reaches half its maximum height?

3 An object is launched from the origin (x = 0, y = 0) and passes through the coordinates (60, 80) and (120, 80). Distances are measured in metres and time in seconds. The acceleration of the object is given by

a t j( ) = −10

at all times.a Find the initial velocity vector for the trajectory.b Find the range and the time of flight for the trajectory.A second object is located at (20, 0) and can be launched at the same speed as the first one. It is planned that the second object collide with the first at the coordinates (120, 80).c At what angle to the horizontal should the second object be launched to pass through the point

(120, 80)?d What time delay in the launching of the second object is required so that the two objects collide at the

point (120, 80)?



4 Two projectiles, R and S, are launched in the same plane, at the sametime, a horizontal distance of 40 metres apart as shown.Projectile R is launched with an initial speed of 20 m/s at an elevation of 45° from a height of h metres.Projectile S is launched with an initial speed of 30 m/s at an elevation of 30°.a Find the position vectors,

r and

s, for the projectiles, R and S respectively,

at any time, t.b Find the time at which the projectiles collide and the value of h, to the

nearest centimetre, for a collision to occur.

5 A particle moves so that its position vector r at any time t, t ∈ [0, π

4), has its velocity vector given as:

r t i t j= +

− +

sec2 2

4 4π π

cosec

The initial position of the particle is 4 i j+ .

a i Find the derivative of cot x. ii Hence state an antiderivative of cosec2 (x).b Find the particle’s position at any time t.c Determine the Cartesian equation of the path and sketch it.d Give the exact speed and the direction of the particle when t is

π12

.

e Find the particle’s acceleration at any time t.f Find the exact magnitude and the direction of the acceleration when t is π

12.

g Determine the angle between the velocity vector and the acceleration vector when t is π12

.

6 A town is situated on a straight shoreline (see O on diagram below). Let i and

j be unit vectors in the east

and north directions respectively. Displacements are measured in kilometres. At 12.00 midday, a cargo ship is at C (2, −6) and a sailing ship is at S (8, −4).

Town

C

Ocean

North

(2, −6)

Land

Shoreline

S (8, −4)

O (0, 0)

a Find →CS in terms of

i and

j and hence determine the distance between the cargo ship and the sailing ship

at 12.00 midday. Give your answer correct to the nearest tenth of a kilometre.b Let P (6m, −2m) denote a point on the shoreline, where m > 0.

i Express the scalar product →OP •

→PS in terms of m.

ii Hence, or otherwise, find the coordinates of the closest point on the shoreline to the sailing ship at 12.00 midday.

iii To the nearest tenth of a kilometre, how close is the sailing ship to the shoreline at 12.00 midday?c Let t be the time in hours after 12.00 midday. The velocities of the cargo ship and the sailing ship are

respectively given by v i jc = −15 5 and

v i t js = + −12 3 8( sin ) .

i Find the position vector at time t of the cargo ship. ii Find the position vector at time t of the sailing ship.d Show that after two hours (at exactly 2.00 pm), the cargo ship is directly south of the sailing ship.

7 An aircraft approaching an airport with velocity v i j k= − −30 40 4 is observed on the control tower radar

screen at time t = 0 seconds. Ten seconds later it passes over a navigation beacon with position vector − +500 2500

i j relative to the base of the control tower, at an altitude of 200 metres. Let

i and

j be horizontal

x

y

O

R

S

30 m/s

20 m/s

45°

4030°

j~

i~

~ ~g = –10j m/s2

h

516

orthogonal unit vectors and let k be a unit vector in the vertical direction. Displacement components are

measured in metres.a Show that the position vector of the aircraft relative to the base of the control tower at time t is given by

r t t i t j t k( ) ( ) ( ) ( )= − + − + −30 800 2900 40 240 4

b When does the aircraft land and how far (correct to the nearest metre) from the base of the control tower is the point of landing?

c At what angle from the runway, correct to the nearest tenth of a degree, does the aircraft land?d At what time, correct to the nearest second, is the aircraft closest to the base of the control tower? e What distance does the aircraft travel from the time it is observed on the radar screen to the time it lands? Give your answer correct to the nearest metre.

exam TiP Remember to use tildes to indicate vectors. When calculating the direction in which the aircraft lands do not use position vectors. Remember to use your CAS calculator for complex calculations to avoid errors.

©VCAA Assessment report 2007

[©VCAA 2007] 8 The position vector

r(t) of the front of a toy train at time t seconds on a closed track is given by

r t

ti

tj t( ) sin sin ,=

+ +

≥3

12

23

0

where displacement components are measured in metres.a i If the front of the train is at the point P(x, y) at time t, show that

yt t2 2 2

3 3=

sin cos .

ii Hence fi nd the Cartesian equation of the path of the train.b Sketch the path of the train on the axes below.

−0.2

x

0.2

0.4

−0.4

−0.6

−0.2−0.4−0.6−0.8−1−1.2 0 0.2 0.4 0.6 0.8 1 1.2

0.6y

c Find the exact time, in seconds, that it takes the train to complete one circuit of the track.d Find the exact speed, in m/s, of the train as the front of the train passes through the origin.e The distance travelled by the train between times t = t0 and t = t1 is given by

v t dtt

t

n

( )1∫

where v t( ) is the speed of the train at time t. i Write down a defi nite integral, involving only scalar quantities, which gives the distance travelled by

the train when it completes exactly one circuit of the track. ii Find the distance in metres, for one circuit of the track, correct to one decimal place.

[©VCAA 2008]

eBookpluseBookplus

Digital docTest Yourself

Chapter 10



eBookpluseBookplus aCTiViTieS

Chapter openerDigital doc

10 Quick Questions: Warm up with ten quick •questions on vector calculus. (page 465)

10A Position, velocity and accelerationTutorial

We3 • int-0411: Watch a tutorial on how to fi nd the velocity vector, the speed at a given time, and the average speed over a period of time, given the position vector. (page 468)

10B Cartesian equations and antidifferentiationof vectors

Tutorial

We8 • int-0412: Watch how to fi nd the position vector and acceleration vector given a velocity vector. (page 477) We11 • int-0413: Watch a tutorial on how to model the movement of a balloon. (page 481)

Digital doc

WorkSHEET 10.1: Calculate speed, velocity, •acceleration, times and distances of particles with motion defi ned as vector functions, and the respective Cartesian equations. (page 485)

10 C Applications of vector calculusInteractivity

Applications of vector calculus • int-0351: Consolidate your understanding of vector calculus by applying your skills. (page 485)

Tutorial

We 14 • int-0414: Watch how to fi nd the distance between two boats at a particular time, their angular difference and if the acceleration vectors are perpendicular. (page 489)

10D Projectile motionTutorial

We 17 • int-0415: Watch a tutorial on how to make calculations about the movement of a golf ball. (page 499) We 19 • int-0416: Watch a tutorial on how to fi nd the initial velocity vector of an object, the Cartesian equation of its path and the range and the maximum height it reaches. (page 504)WorkSHEET 10.2: More calculations on speed, •velocity and acceleration of bodies. (page 509)

Chapter reviewDigital doc

Test Yourself: Take the end-of-chapter test to test •your progress. (page 516)

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