applications of exponential growth and decay · web viewme-v1 introduction to vectors this document...

ME-V1 Introduction to vectors

This document contains sample questions and solutions from the sample unit ME-V1 Introduction to vectors.

Identifying scalars and vectorsA scalar is a quantity that has magnitude only. A vector is a quantity that has both magnitude and direction.

Identify each quantity as a scalar or vector: length, displacement, area, volume, velocity, acceleration, speed, time, mass, weight, density, force, temperature

Scalar Vector

Length Displacement

Area Velocity

Volume Acceleration

Speed Weight

Time Force

Mass

Density

Temperature

Scalar Vector

Length Displacement

Area Velocity

Volume Acceleration

Speed Weight

Time Force

Mass

Density

Temperature

© NSW Department of Education, 2019

https://education.nsw.gov.au/about-us/copyright

Vector notation 1. Write vector OA

→as a column vector, an ordered pair and in component form.

Column vector: OA→

=[42 ]Ordered pair: OA

→

=(4 ,2)

Component form: OA→

=4 i+2 j

2. On a Cartesian plane:a. Plot two points A and B

b. Plot two additional points C and D and draw vectors AC→

and DB→

such that AC→

=3 i+ j

and DB→

=−5 i+3 j

2 ME-V1 Introduction to vectors

Multiplying a vector by a scalar 1. If u=5 i−2 j calculate 4 u.

4 u=4 (5 i−2 j )4 u=20 i−8 jGeometrically, this can be shown as:

2. On a blank page draw a vector v. Use this to draw vectors 3v ,−2vand 12v .

3. If u=3i+7 j, find 3u ,−2uand 12u.

3u=3 (3 i+7 j )¿9 i+21 j

−2u=−2(3 i+7 j )¿−6 i−14 j

12u=12 (3 i+7 j )¿

32i+72j



Adding and subtracting vectors 1. Given OA=¿ 8i−3 jand OB=3i+2 j:

a. graph vectors OA and OB on a Cartesian plane

b. Find the vector AB .AB=−5i+5 j

c. Explain the relationship between AB, OAand OB . AB=?

AB=−(OA )+OB=OB−OAd. Confirm the relationship numerically.

OB−OA=3 i+2 j−(8 i−3 j )¿3 i+2 j−8 i+3 j¿−5 i+5 j¿ AB

2. If v=3 i+7 j and u=−5 i−3 j, calculate v+u

v+u=3 i+7 j+(−5 i−3 j )¿3 i+7 j−5 i−3 j¿−2i+4 j

3. Consider the vectors u=3i−2 j and v=i+4 j. Find:

a. u+v

u+ v=3 i−2 j+(i+4 j )¿4 i+2 j

b. u−v

u−v=3i−2 j−(i+4 j)¿3 i−2 j−i−4 j¿2 i−6 j

c. v−u

v−u=i+4 j−(3i−2 j)¿ i+4 j−3 i+2 j¿−2i+6 j

d. Confirm the results for a. to c. by geometrically representing these on a Cartesian plane.u+v


u−v

v−u



4. If v=5 i+4 j, u=−3 i−7 j and w=5 i−4 j ,match each statement to its component form.

Statement Component form expressions

u+v 25 i−20 j

v−w 2 i−3 j

v+w−u 13 i+7 j

5w 0 i+8 j

−v 15 i+4 j

2v+w −5 i−4 j

Statement Component form expressions – solutions

u+v 2 i−3 j

v−w 0 i+8 j

v+w−u 13 i+7 j

5w 25 i−20 j

−v −5 i−4 j

2v+w 15 i+4 j

Magnitude of a vector1. If u=8 i−6 j, find the:

a. magnitude of the vector u

|u|=√82+(−6 )2¿√64+36¿√100¿10

b. unit vector ´u


´u=u

|u|¿8 i−6 j

10¿ 810

i− 610

j¿ 45i−35j

2. If v=i+7 j, find the:

a. magnitude vector v.

|u|=√1+72¿√1+49¿√50¿5√2

b. unit vector ´v

´v=v

|v|¿i+7 j

5√2¿ 15√2

i+ 75√2

j¿ √210

i−7√210

j

3. Given OA=¿ 4i−3 jand OB=−2i+3 j, find the magnitude of AB

AB=OB−OA¿−2i+3 j−(4 i−3 j )¿−2i+3 j−4 i+3 j¿−6 i+6 j



Direction of a vector1. A particle is projected from the origin with an initial velocity of 5 i+3 j ms−1. What is the angle

of projection?

tanθ=35θ=tan−1( 35 )¿30.96…°

2. A particle is projected from the origin with an initial velocity of −2 i+3 jms−1. What is the

angle of projection?

tan α=32α=tan−1( 32 )θ=180−tan−1( 32 )¿123.69…°


Scalar (dot) product1. u=−3 i−7 j, If v=5 i+4 j and w=5 i−4 j :

a. Calculate u ⋅v

u ⋅v=x1 x2+ y1 y2

u ⋅v=−3 ⋅5+ (−7 ) ⋅4

u ⋅v=−15−28

u ⋅v=−43

b. Find the angle between v and u

u.v=¿u||v|cosθ

cosθ=u . v

¿u∨¿ v∨¿¿

|u|=√ (−3 )2+(−7)2¿√9+49¿√58

|v|=√52+42¿√25+16¿√41

cosθ= −43√58 ⋅√41

θ=cos−1( −43√58 ⋅√41 )¿151.85…°

c. Show the distributive property holds, i.e. v ⋅ (u+w )=v ⋅u+v ⋅w

u+w=−3i−7 j+5 i−4 j¿2 i−11 j

v ⋅ (u+w )=5 ⋅2+4 ⋅ (−11)¿10−44¿−34

v ⋅u+v ⋅w=5⋅ (−3 )+4 ⋅ (−7 )+5 ⋅5+4 ⋅ (−4 )¿−15−28+25−16¿−34

2. A, B and C are points defined by the position vectors a=i+3 j, b=2 i+ j and c=i−2 j respectively. Find the size of ∠ ABC.



∠ ABC is the angle between vectors BA and BCu.v=¿u||v|cosθ

cosθ=u . v

¿u∨¿ v∨¿¿ , or in our case, letting ∠ ABC=θ, cosθ=BA . BC

¿ BA∨¿ BC∨¿¿

BA=a−b¿ i+3 j−(2 i+ j )¿−i+2 j

BC=c−b¿ i−2 j−(2i+ j )¿−i−3 j

BA . BC=−1×−1+2×−3¿−5

¿ BA∨¿ √(−1 )2+22¿√5

¿ BC∨¿ √(−1 )2+(−3)2¿√10

cosθ= BA . BC¿ BA∨¿ BC∨¿¿

cosθ= −5√5×√10

cosθ= −5√5×√5×√2

cosθ= −55×√2

cosθ=−1√2

θ=cos−1(−1√2 )


¿180 °−cos−1( 1√2 )¿180 °−45°¿135 °

∴∠ ABC=135 °

3. If v=i+7 j and u=8 i−6 j, find u ⋅v and hence find the angle between uandv (to the nearest degree). u ⋅v=x1 x2+ y1 y2

u ⋅v=1×8+7×−6u ⋅v=−34

u ⋅v=|u||v|cosθ−34=√50×√100 cosθ

−34=50√2cosθ

−1725√2

=cosθ

θ=119o



Parallel and perpendicular vectors 1. If v=6 i+4 j, u=2i−3 j, w=12 i−18 j

a. Show uand vare perpendicular

u . v=x1 x2+ y1 y2¿2×6+(−3 )×4¿12−12¿0

∴ uand v are perpendicular

b. Show uand ware parallel

Show w=λu or if show x1x2

=y1y2

x1x2

=122

=6

y1y2

=−18−3

=6

∴uand w are parallel.

Alternate method: Show u ⋅w=¿u ||w | or u ⋅w=−¿u ||w |

2. If a=x i+2 j is perpendicular to vector b=2 i− j, find x.

a ⋅b=0 as they are perpendicular.

x×2+2×−1=02 x−2=02 x=2x=1

3. If a=−6 i+ y j is parallel to vector b=2 i−4 j, find y.

Two vectors are parallel if they are scalar multiples of one another.y

−4=−62

y−4

=−3

∴ y=12


Vector projection 1. Let a=i+3 j and b=i− j,

a. find the scalar projection of a onto b.

Scalar projection of a onto b=¿a∨cosθ

Note: The scalar projection will be negative as a opposes b

cosθ=a .b

¿a∨¿b∨¿¿

|a|cosθ=a .b

¿b∨¿¿

|a|cosθ=1×1+3×(−1)

√12+ (−1 )2

|a|cosθ=−2√2

|a|cosθ=−√2

b. find the vector projection of a in the direction of b.

projba=scalar projection of aontob× theunit vector of b¿−√2×i− j

√2¿−i+ j



2. Let a=i+3 j and b=i− j, find the vector projection of a in the direction b.

a ∙b=( i+3 j) (i− j)a ∙b=1×1+3× (−1 )

a ∙b=−2

b ∙b=(i− j )(i− j )b ∙b=1×1+(−1)× (−1 )

b ∙b=2

projba=a ∙b

b ∙bb

projba=−22 ( i− j )

projba=−i+ j


Proving geometric results1. The diagonals of a parallelogram meet at right angles if and only if it is a rhombus

(ACMSM039)

( u+ v ) ∙ ( u− v )=u . u− v . v¿|u|2−|v|2

If the diagonals meet at a right angle then ( u+ v ) ∙ ( u− v )=0

∴|u|2−|v|2=0

|u|2=|v|2

|u|=|v|∴ The parallelogram is a rhombus

2. The midpoints of the sides of a quadrilateral join to form a parallelogram (ACMSM040)Part 1: First show the line joining the midpoints of two sides of a triangle is parallel to the third side and is half its length.

LM=LJ+ JM¿ 12IJ+ 1

2JK¿ 1

2( IJ+ JK )¿ 1

2IK

Part 2: Consider quadrilateral ABCD with midpoints F, G, E and H.

FE=12BD and GH=1

2BD

FG=12AC and EH=1

2AC

∴ FGHE is a parallelogram as opposite sides are equal in length.

3. The sum of the squares of the lengths of the diagonals of a parallelogram is equal to the sum of the squares of the lengths of the sides (ACMSM041)



Proof using vectors:

|u+ v|2+|u−v|2=( u+v ) ∙ (u+ v )+( u−v ) ∙ (u− v )¿ u ∙u+ u∙ v+ v ∙ u+ v ∙ v+u ∙ u−u ∙ v−v ∙u+ v ∙ v¿2 u∙ u+2 v ∙ v

|u+ v|2+|u−v|2=2|u|2+2|v|2 i.e. the sum of the squares of the length of the sides

∴ The sum of the squares of the lengths of the diagonals of a parallelogram is equal to the sum of the squares of the lengths of the sides

Proof without vectors:

Consider the cosine rule for angle C in triangle DBC:

cosC=s22+s3

2−d12

2 s2 s3

Consider the cosine rule for angle D in triangle DAC:

cosD=s32+s4

2−d22

2 s3 s4cosD=cos (π−C )=−cosC (Cointerior angles are supplementary when BC||AB)

−cosC=s32+s4

2−d22

2 s3 s4

cosC=−s3

2−s42+d2

2

2 s3 s4


s22+s3

2−d12

2 s2 s3=

−s32−s4

2+d22

2 s3 s4

¿ s2=s4(Opposite sides of a parallelogram are equal)

s22+s3

2−d12

2 s4 s3=

−s32−s4

2+d22

2 s3 s4

Multiply both sides by2 s3 s4

s22+s3

2−d12=−s3

2−s42+d2

2

s22+s3

2+s32+s4

2=d12+d2

2

¿ s1=s3(Opposite sides of a parallelogram are equal)

s22+s1

2+s32+s4

2=d12+d2

2

s12+s2

2+s32+s4

2=d12+d2

2

∴ The sum of the squares of the lengths of the diagonals of a parallelogram is equal to the sum of the squares of the lengths of the sides.



Modelling motion1. In still water, Jacques can swim at 3 m/s. Jacques is at point A on the edge of a canal, and

considers point B directly opposite. A current is flowing from the left at a constant speed of 1 m/s. If Jacques dives in straight towards B, and swims without allowing for thecurrent, what will his actual speed and direction be? (answers correct to 1 decimal place)

Direction is the deviation of the desired path.

tanθ=13

θ=tan−1( 13 )¿18.4349…°≈18.4 ° (correct¿1decimal place )

Let S=speed of travel .Using Pythagoras’ Theorem,

S2=32+12

S=√10¿3.1622…≈3.2ms

( correct¿1decimal place )


2. Jacques wants to swim directly across the canal to point B. At what angle should Jacques aim to swim in order that the current will correct his direction? And what will Jacques’ actual speed be? (answers correct to 1 decimal place)

Direction Jacques should aim to swim:

sin θ=13

θ=sin−1( 13 )θ=19.4712…°θ≈19.4 ° (correct ¿1decimal place)

Let S=speed of travel . Using Pythagoras’ Theorem,

32=S2+12

9=S2+18=S2

S=√8¿2.824…≈2.8m /s (correct ¿1decimal place)



Projectile motion1. A projectile is launched off a cliff 15m above the ground at an angle of 30 ° to the horizontal

with an initial velocity of 12m /s. Find:

a. Initial velocities

sin 30=¿uy

12¿

uy=12sin 30¿6m /s

cos30=ux

12ux=12cos30¿12× √3

2 ¿6√3m /s


b. Horizontal componentsx=0

x=∫0 ⋅dt¿c1Whent=0 , x=6√3

6√3=c1∴ x=6√3x=∫6√3 ⋅dt¿6√3 t+c2Whent=0 , x=0

0=6√3×0+c2c2=0

∴ x=6√3 t

c. Vertical componentsy=−9.8

y=∫−9.8 ⋅dt¿−9.8 t+c3

Whent=0 , y=6

6=−9.8×0+c3c3=6

∴ y=−9.8 t+6

y=∫ (−9.8 t+6 )⋅ dt y=−4.9 t2+6 t+c4

Whent=0 , y=15

15=−4.9×02+6×0+c4c4=15

∴ y=−4.9 t 2+6 t+15

Note: The projectile motion Geogebra applet can be used to check solutions for a, b and c.

d. The time of the flight

The time of the flight is the time taken for the projectile to hit the ground, i.e. the vertical displacement is zero.



https://www.geogebra.org/m/BXBMnZPS

Solve y=−4.9 t2+6 t+15for twheny=0

0=−4.9 t 2+6 t+15Use the quadratic formula with a=−4.9 ,b=6 , c=15

t=−6±√62−4× (−4.9 )×152(−4.9)

¿ −6±√330−9.8

t=−6−√330−9.8

,−6+√330−9.8

t=−1.2414… ,2.4659…∴the time of the flight is 2.47 seconds (correct to 2 decimal places)

The solution can be checked by graphing y=−4.9 t2+6 t+15 using graphing software. (use x in place of t in the graphing software)

e. The range of the projectile

The range of the projectile is the horizontal displacement when the projectile strikes the ground.

Solve x=6√3 t for xwhent=−6−√330−9.8

x=6√3×−6−√330−9.8

x=25.6264…

∴ the range of the projectile is 25.63m (correct to 2 decimal places)

f. The final velocity of the projectile


To find the final velocity we need to obtain the x and y velocity components when

t=−6−√330−9.8

, the time when the projectile hits the ground.

The x component of the velocity is a constant: x=6√3.

To find the y component of the final velocity, solve y=−9.8 t+6 for y when t=−6−√330−9.8

.

y=−9.8×−6−√330−9.8

+6 y=−√330m /s or −18.1659m /s

Note: The negative signifies the object is travelling downwards.

g. Final velocity

v final2 =(6√3 )2+(√330 )2¿438

v final=√438¿20.9284…

tanθ=√3306√3

θ=tan−1( √3306√3 )¿60.2271…

∴ the final velocity is 20.93m /sat an angle of 60.23 ° to the horizontal. (correct to 2 decimal places)

h. The maximum height the projectile reaches

The maximum height of the projectile is the vertical displacement when y=0

Step 1: Solve y=−9.8 t+6 for t when y=0.

0=−9.8 t+6−6=−9.8 t −6−9.8

=t t=3049

∨0.6122…

Step 2: Solve y=−4.9 t2+6 t+15 for y when t=3049

y=−4.9( 3049 )2

+6( 3049 )+15¿16.8376…≈16.84(correct¿ thenearest centimetre)



Derive the Cartesian equation of the trajectory ( y as a function of x)x=6√3 t

t= x6√3

Substitute t=x6√3 into y=−4.9 t2+6 t+15

y=−4.9( x6√3 )

2

+6( x6 √3 )+15

y=−4.9x2

108+ √3 x3

+15


2. A projectile is launched from the ground at an angle of θ ° to the horizontal with an initial velocity of um /s. Assuming no air resistance, show the maximum range is achieved when

θ=45 °∨ π4 .

Initial velocities

sin θ=¿u y

u¿uy=u sinθ

cosθ=ux

uux=ucosθ

Horizontal componentsx=0

x=∫0 ⋅dt¿c1Whent=0 , x=u cosθ

ucosθ=c1∴ x=ucos θ

x=∫ucosθ ⋅dt x=(ucosθ )t+c2

Whent=0 , x=0

0=ucosθ×0+c2c2=0

∴ x=ut cosθ

t= xucosθ

Vertical componentsy=−9.8

y=∫−9.8 ⋅dt¿−9.8 t+c3

Whent=0 , y=u sin θ

u sinθ=−9.8×0+c3c3=u sin θ

∴ y=−9.8 t+u sin θ

y=∫ (−9.8 t+u sin θ ) ⋅dt y=−4.9 t2+(u sin θ ) t+c4



Whent=0 , y=0

0=−4.9×02+6×0+c4c4=0

∴ y=ut sin θ−4.9 t2

Derive the Cartesian equation of the trajectory by substituting t=x

ucosθ

y=u ( xucos θ )sin θ−4.9( x

ucosθ )2

¿ x tan θ− 4.9 x2

u2cos2θ

The range of the projectile is the horizontal displacement, x, when the vertical displacement, y, is 0.

0=x tanθ− 4.9 x2

u2cos2θ

0=x (tan θ− 4.9 xu2cos2θ )

Solve tanθ−4.9 x

u2cos2θ=0

u2 cos2θ tan θ−4.9 x=0u2 sin θ cosθ−4.9 x=0u2 sin θ cosθ=4.9 xx=u2 sinθ cos θ4.9

x=u22sin θcosθ9.8

¿ u2sin 2θ9.8

The maximum value of the functions is when sin 2θ=¿1 as −1≤sin 2θ≤1sin 2θ=12θ=90°

θ=45 °∨ π4

3. 2011 Mathematics Extension 1 HSC, question 6, part b

The diagram shows the trajectory of a ball thrown horizontally, at speed v ms–1, from the top of a tower h metres above ground level.


The ball strikes the ground at an angle of 45 °, d metres from the base of the tower, as shown in the diagram. The equations describing the trajectory of the ball are x=vt and

y=h−12g t2, (Do NOT prove this.) where g is the acceleration due to gravity, and t is time in

seconds.

a. Prove that the ball strikes the ground at time t=√ 2hg seconds.

The ball will strike the ground when y=0

Solve y=h−12g t2 for twhen y=0

0=h−12g t 212g t2=hg t2=2ht

2=2hg t=±√ 2hg ∴The ball strikes the ground at time,t=√ 2hg (

t must be positive)

b. Hence, or otherwise, show that d=2h.

d is the horizontal displacement when ,t=√ 2hg , x=vt

The ball strikes the ground at 45 ° so the horizontal and vertical compoants of velocity are equal in magnitude.

tan135= yx−1= y

x x= y

Horizontal component of velocity

x= ddt

( x )¿ ddt

(vt )¿ v

Vertical component of velocity

y= ddt

( y )¿ ddt (h−12 g t2)¿−¿



x=− y , ∴ v=¿

Substitute v=¿ into (1)

x=g t 2

Substitute t=√ 2hg to find the horizontal displacement when the ball strikes the ground, d.

x=g(√ 2hg )2

¿ g× 2hg ¿2h,

∴d=2h


applications of exponential growth and decay · web viewme-v1 introduction to vectors this document...

Documents