So lu tions

Ques tion 1 E

The bottom line of the fraction providedthe vertical asymptote. Here, it will bex = 0.Removing the fraction (x ® ¥), the graphtends towards f x x( ) = 2 . Thus, there aretwo asymptotes. A quick use of CAS (oradding ordinates if your batteries are flat)gives this:

The graph has a local maximum. Don’tbother calculating (- -1 3, ). Just scan thealternatives and look at the sketch.

Ques tion 2 D

Once again, the best method here is tosketch two graphs. Since they are notfunctions, most CAS devices will havetrouble sketching them. Find thex-intercepts first.

1. ( )x y-

+ =1

9 41

2 2

\( )x -

=1

91

2

\ x - = ±1 3\ x = -2 or 4.

2. x y2 2 4- = \ x2 4=\ x = ± 2

Using your ‘graphing by hand’ skills withthis information gives:

Thus, the graph of ( )x y-

+ =1

9 41

2 2

and the

graph of x y2 2 4- = have three points incommon.

Ques tion 3 D

Using domain information:x b- = -1 is one extreme andx b- = 1 is the other.

Thus, 2 1- = -b making b = 3. (Note thatthis is true for x b- = 1 as well.)

Using range information:The range of y x= -cos ( )1 is [0, p]. Thus a = 6 would make the range of f x a x b( ) cos ( )= --1 become [0, 6p].Thus, b = 3 and a = 6.

Ques tion 4 A

The curve given byx = -1 + 2 sec ( )t 1.y = 1 + 3 tan ( )t 2.We will use the fact that tan ( ) sec ( )2 21x x+ = which is derived from the equation sin ( ) cos ( )2 2 1x x+ = .

From equation 1, sec ( )( )2

21

4x

x=

+

From equation 2, tan ( )( )2

21

9x

y=

-

Thus, ( ) ( )y x-

+ =+1

91

1

4

2 2

1

0

y

x

y = 2x

y

x(–2, 0)

Solutions for Specialist Mathematics Exam 2 2009

which is rearranged to give ( ) ( )x y+

--

=1

4

1

91

2 2

Ques tion 5 C

The equation x ax y by2 22 2 4 16 0+ + + + =is for ellipse with centre ( , )3 2- ? Thatmeans that we should find the equation ofthe ellipse in the form ( ... ) ( ... )x y k+ + + =2 2

x ax y by2 22 2 2 16 0+ + + + =( )\ x ax a a2 2 22+ + - +

2 2 2 16 02 2 2( )y by b b+ + - + =\ ( ) ( )x a a y b b+ - + + - + =2 2 2 22 2 16 0\ ( ) ( )x a y b k+ + + =2 22Where k is a constant. Since the centre is ( , )3 2- , we have that a is -3 and b = 2.

Ques tion 6 A

The distance between z and - z in thecomplex plane is

z z- -( )= x iy x iy+ - - -( ( ))= x iy x iy+ + -( )= 2x= 2´ Re( )z

Ques tion 7 C

Two factors of P z( ) must be ( )z - 2 and ( ( ))z i- - +2 . But multiplying these twofactors together would produce a quadraticwith complex number coefficients (and weare told they are real coefficients.). So a third factor, using the conjugate of z i= - +2 will be ( ( ))z i- - -2 , making P z( ) a cubic.

Ques tion 8 E

( )1+ =i ain

\ ( ) ( ) ( )1 1 1+ + = +i i ai in

\ ( ) ( )1 11+ = ++i ai in

\ (( ) ) ( ( ))1 11 2 2+ = ++i ai in

,\ ( ) ( )1 12 2 2 2 2+ = ++i a i in

= - + +a i i2 21 2( )= -a i2 2( )= -2 2a i

Ques tion 9 B

The main idea of a direction field is that ifyou knew the equation of one of the sets of lines joined together, and differentiatedthat equation then you would get the

required expression for dy

dx.

So here is a typical graph chosen from theavailable sets:

It reasonable to assume it is an ellipse with centre (6, 3). That is,

( ) ( )x

a

y

b

-+

-=

6 31

2

2

2

2

Now differentiate using implicitdifferentiation:

2 6 2 30

2 2

( ) ( )x

a

y

b

dy

dx

-+

-=

\ dy

dx

x

a

b

y=

- -´

-

2 6

2 32

2( )

( )

\ dy

dx

x

a

b

y=

- -´

-

( )

( )

6

32

2

\ dy

dx

b x

a y=

- -

-

2

2

6

3

( )

( )

or

\ dy

dx

b x

a y=

-

-

2

2

6

3

( )

( )

Answer B with suitable a and b values fits.

2

y

x1 2 3 4 5 6 7 8 9

1

2

3

4

5

O

Solutions for Specialist Mathematics Exam 2 2009

Ques tion 10 B

Let f R:[ , ]- ®p p2 , where f x x( ) sin ( )= 3 .Since we are being asked for an area, agraph would be good. From my CAS,

Note th three equal sections. By findingarea B and tripling it, the total area isfound.

That is, A x dx= ò3 3

0

sin ( )p

Use the substitution u x= cos ( ),

noting that du

dxx= -sin ( )

\ A x x dxx

x

= - - -=

ò3 1 2

0

sin( )( sin ( ))=p

= - -ò3 1 2

0

du

dxx dx( sin ( ))

p

= 3 1 2

1

1

( sin ( ))--

ò x dx

= 3 1 2

1

1

( )--

ò u dx

Ques tion 11 C

It is a good idea here to sketch somepossible graph sections.1. ¢ >f x( ) 0 Positive slope

2. ¢¢ <f x( ) 0 The slope becomes lesspositive.

In order for our graph to have bothcharacteristics, the gradient must bepositive always and must become flatter.That is, the graph increases, havingdecreasing gradient with increasing x.

Ques tion 12 E

The velocity equation has v on the left anda function of x (not t) on the right. Thatrules out a being the ‘f dashed’ notation.

That leaves us with the famous pair:

a vdv

dx=

and

a =

d1

2v

dx

2æ

èç

ö

ø÷

Neither of which look to be immediatelypresent in the alternatives. However have a close look at E. In this question, thevelocity is f x( ), so E becomes v f x¢( ).Now since t is not present, that ¢f x( )

means dv

dx, which would leave us with

v f x¢( ).

Ques tion 13 A

A diagram:

3

x

y

-0.75p -0.5p -0.25p 0 0.25p 0.5p 0.75p p 1.25p 1.5p 1.75p

-1.5

-1

-0.5

0.5

1

1.5

–p p 2pB

A C

y = sin (x)3

x

y

x

y

10 litres/min

10 litres/min

fresh water

100 litres initially

When t = 0, x = 4x grams attime t

Solutions for Specialist Mathematics Exam 2 2009

Here, dx

dt= inflow - outflow

= 010

100- ´ x

Or, to match one of their alternatives,dx

dt

x= -

10

\ 10 0dx

dtx+ =

Ques tion 14 B

The vectors u i j k~ ~ ~ ~

= + +m , v i j k~ ~ ~ ~

= + +mand w i j k

~ ~ ~ ~= + + m where m is a real

To be linearly dependent,u v w~ ~ ~

= +a b where a, b are real.

\ m a m bi j k i j k i j mk~ ~ ~ ~ ~ ~ ~ ~ ~

+ + = + + + + +( ) ( )

Compare coefficients of i j k~ ~ ~

, , :

i~

m a b= + ... 1.j~

1 1= +am ... 2.k~

1= +a bm ... 3.

From equation 2., either a = 0 or m = 0.If m = 0, then from equation 3., a = 1 and fromequation 1., b = -1. That would lead to

0 0i j k i j k i j 0k~ ~ ~ ~ ~ ~ ~ ~ ~

+ + = + + - + +( ) ( )

\ j k i k i j~ ~ ~ ~ ~ ~

+ = + - -

\ 2 j 0~ ~

= which not necessarily true.

So, from 1 1= +am , take a = 0.. From 1, that means b m= and from 3., m2 1= .The nearest alternative answer is m= 1.

Note: A faster method would be to testeach alternative and look for a possibleanswer. But this way seems so muchmore... pleasing.

Ques tion 15 C

A diagram of the two vectors, noting the i~

and j~

directions, will help here:

I will choose an algebraic approach here.Some may wish to use geometry morefully.The first vector, has an easterly component x.

cos ( )454

° =x

\ x = ° =4 45 2 2cos ( ) .

Using symmetry, the vector can be written

as 2 2 2 2i j~ ~

+ .

The second vector can be written as 3 i~

The sum of the two vector is

2 2 2 2 3i j i~ ~ ~

+ +

= ( )2 2 3 2 2+ +i j~ ~

The magnitude of this vector is

( ) ( )2 2 3 2 22 2+ +

= 8 12 2 9 8+ + +

= 25 12 2+

Ques tion 16 D

Start by finding the dot product betweenpairs of these vectors.

a b~ ~. = - - - = -6 12 4 22

a c~ ~. = - + =26 30 4 0

b c~ ~

. = - + - =39 40 1 0

Thus, c~ is perpendicular to both a

~ and b

~.

4

east45

4 N

3 Neast

j

i

~

~

Solutions for Specialist Mathematics Exam 2 2009

Ques tion 17 D

The vectors a~ , b

~ and c

~ form a triangle

for which the cosine rule applies. Here isthe triangle (without vectors):

The cosine rule:

c a b a b~ ~ ~ ~ ~

2 2 2

2 120= + - °cos ( )

= c a b a b~ ~ ~ ~ ~

2 2 2

21

2= + - ´ -

æ

èç

ö

ø÷

= c a b a b~ ~ ~ ~ ~

2 2 2

= + +

Ques tion 18 B

A diagram will help:

Resolving in the j~

direction:

N g= °2 30cos ( )

Resolve in the i~ direction, noting that

friction will act up the plane:

2 30 01g N masin ( ) .° - =\ g g a- ° =01 2 30 2. ( cos ( ))Keep an eye on the answers at this stage...

\ g g a- =023

22. )

\ g a1 013

12-

æ

èçç

ö

ø÷÷ =.

\ g a13

102-

æ

èçç

ö

ø÷÷ =

Ques tion 19 A

You could use the projectile motionequation, but I will leave that for thephysics people. Start with a diagram:

Resolving the velocity in the vertical j~direction,

v = °20 45cos ( )

= 10 2

The acceleration acting on the ball in the j~direction is -g

The time taken for the velocity to reachzero (the top of the ball's journey) is foundusing

v u at= +

\ 0 10 2= - gt

\ tg

=10 2

Thus, the time to return to the ground willbe twice that value,

tg

=20 2

Ques tion 20 E

We wish to find an equation with x and ttogether.

v x=

\dx

dtx=

\dt

dx x=

1

5

a

b

c~

~

~

120

N

2g

30

ij

30

m = 0.1~~

i

j

~

~Initial movement

45

v = 20 m/s

Solutions for Specialist Mathematics Exam 2 2009

\ t x ce= +log ( )When x = 1, t = 3.\ 3 1= +log ( )e c\ c = 3\ t xe= +log ( ) 3Look at the alternatives. They all have theform x = ...\ t xe- =3 log ( )\ e xt- =3

\ x e t= -3

Ques tion 21 D

Momentum is the product of mass andvelocity. Thus we need to know thevelocity when the displacement is 21metres. We will need an equation with bothv and x.

a = 2

\d

dxv

1

222æ

èç

ö

ø÷ =

\1

222v x c= +

Let ‘initial’ mean that when t = 0, x = 0 and v = 4.

\1

24 2 02 = +( ) c

\ c = 8

\1

22 82v x= +

When the displacement is 21 m,

\1

22 21 82v = +( )

\1

2502v =

\ v = ±10 m/sThe momentum is mv = ±50.Thus, the magnitude of the momentum is50 kg ms–1.

Ques tion 22 B

The distance of the body form the startingpoint is the sum of A and C subtract B,where A, B and C are the areas shown inthe diagram:

The area of triangle A is 1

2´ ´base height

=1

22 10 10´ ´ =

The area of triangle C is 1

2´ ´base height

=1

24 10 20´ ´ =

Shape B may be considered as twotriangles and a rectangle, the rectanglebeing between 3 and 4. The area of shapeB is

1

21 5 1 5

1

22 5´ ´ + + ´ ´( )( )

=5

25 5 12

1

2+ + =

Thus, the distance travelled is

10 20 121

217

1

2+ - =

Ques tion 23

Start by sketching the graph of v t= 3 2/ onyour CAS. Don't forget brackets around the 3/2.. The exam paper graph scales give you the settings for your CAS window. Thesedays, the examiners require you to sketchprecisely, so the end (9, 27) and at leastone point in the middle should beaccurately drawn in. I chose (5, 11.1). Next draw the line v = 27 between t = 9 to t = 39,

and finally, v t= -æ

èç

ö

ø÷27

2439cos ( )

p noting

6

12

3

4

567

8

910

1 2 3 4 5 6 7 8 9 10O

–6

–5

–4–3–2

–1

v

t

A

B

C

Solutions for Specialist Mathematics Exam 2 2009

the comments above, from 39 to 51. Notealso the cos graph has a turning point att = 39. A suitable ‘middle point’ is(45, 19).

a.

b. The distance travelled by the carduring the first nine seconds of its motionis

t dt3 2

0

9/

ò

=t 5 2

0

9

5 2

/

/

é

ëê

ù

ûú

=9

5 2

0

5 2

2 3

50

5 2 5 2 5/ /

/ /- =

´-

=486

5 metres

» 97.2 metres. I chose to change to decimals after doingthe next part.

c. This refers to the third graph section,and the decimal value is a hint to use CAS.

2724

3939

51

cos ( )p

t dt-æ

èç

ö

ø÷ò

I obtained the value of 206.26 metres» 206.3 metres.

d. The average speed

=distance travelled

time taken

=486 5 30 27 20626

51

/ ( )( ) .+ +

= 21.8 m/s .

e. The question is asking us to findwhen the graph has a height (v-value)

greater than 200

9 ms -1 . The first and third

sections will have points where v =200

9.

1. t 3 2 200

9

/ =

\ t =æ

èç

ö

ø÷

200

9

2

3

» 7.9042

2. 2724

39200

9cos ( )

pt -

æ

èç

ö

ø÷ =

For this, I used my CAS and sketched the

Cos graph and the graph of v =200

9 and

looked for the intersections between 39 and 51. There is only one, at t = 43.614

Given that t t t1 2< < , then t 1 79= . and t 2 436= . seconds, correct to 1 decimalplace.

f. To overtake, it must be true that thetwo cars have travelled the same distance.

Consider the first 9 seconds. The car willhave travelled 97.2 metres (found earlier).The faster motorcycle will have travelled 9 20 180´ = metres. So at some time after 9second (when the car is faster) they willpass. We want the area under the middlesection of the graph to be 180 972 828- =. .plus the additional distance travelled by the motorcycle.Let t 3 the time after t = 9 seconds.

\ 27 8283´ =t . ++

ò 209

9 3

dtt

\ 27 8283´ =t . + 180 + 20 1803t -\ 7 8283´ =t .\ t 3 1183= .They will pass at time 9 + 11.83 = 20.83.Solution: they pass at 20.8 seconds, whenthe distance is 20 2083 4166 417´ = ». . m.(use pre rounded-down numbers).

7

10 20 30 40 50

5

10

15

20

25

30

v

t

O

Solutions for Specialist Mathematics Exam 2 2009

Ques tion 2

In the complex plane, L is the line with

equation z z i- = - -11

2

3

2

a. Verify that the point (0, 0) lies on L.

z z i- = - -11

2

3

2

\ x iy x iy i+ - = + - -11

2

3

2

LHS = x iy+ -1= 0 0 1 1+ - =i

RHS = x iy i+ - -1

2

3

2

= 0 01

2

3

2+ - -i i

=1

2

3

21

2 2

æ

èç

ö

ø÷ +

æ

èçç

ö

ø÷÷ =

LHS = RHS, so it is true that the point (0, 0) lies on L

b. There is a geometric verification tothis question, but an algebraic proof is besthere.Given:

x iy x iy i+ - = + - -11

2

3

2

\ ( )x y x y- + = -æ

èç

ö

ø÷ + -

æ

èçç

ö

ø÷÷1

1

2

3

2

2 2

2 2

\ x x y2 22 1- + + =

x x y y2 21

43

3

4- + + - +

\ - + = - + -2 1 1 3x x y

\ - = -x y3

\ y x=1

3

As required.

c. The line y x=1

3 has gradient

1

3..

Note that tanp

6

1

3

æ

èç

ö

ø÷ = and that

tan7

6

1

3

pæ

èç

ö

ø÷ = .

In the third quadrant, the angle 7

6

p should

be expressed as -5

6

p. That is a.

d. All points lying n the graph of z = 2

will have a distance of 2 units from theorigin. That is, x y2 2 4+ = . We must

intersect this with the graph of . y x=1

3.

\ x x2

21

34+

æ

èçç

ö

ø÷÷ =

\ x x2 21

34+ =

\4

342x =

\ x = ± 3 and so y = ± = ±1

33 1.

One such point is ( , )3 1 and the other is (

- -3 1, ).

e. The two graphs look like this:

f. The area in the first quadrant that isenclosed by L and the graphs of

z = 2, z = 1 and Arg( )z =p

3 is

shown shaded in the following diagram:

8

Re (z)

Im (z)

O

–1

1

1–1–2 2

–2

3–3

2L|z| = 2

Solutions for Specialist Mathematics Exam 2 2009

Note that the area is 1

12 of the full

‘doughnut shape*’ area enclosed betweenthe two circles. That area is

p p p2 1 32 2- =Thus the required area is

1

123´ p

=p

4 square units.

(*Mmm doughnuts...)

Ques tion 3

The position of the centre of the mat can be assumed to be at the top of the slide at time t = 0. Since it is only the k

~ vector that

determines the height,

24 58

2

. -æ

èçç

ö

ø÷÷

tk~ becomes 24 5. k

~ when t = 0.

That is, the top of the slide is 24.5 metreshigh.

b. Here we want the height aboveground to be zero.

\ 24 58

02

. - =t

\t 2

824 5= .

\ t = 14 seconds as require.(the negative value is rejected since t ³ 0.)

c. Both the cos and the sin functions

here have period 2

612

p

p /= seconds. That is

the time for one loop.

d. Differentiating the three vectorcomponents,

&( ) cos sin~ ~ ~ ~r t t i t j

tk=

æ

èç

ö

ø÷ -

æ

èç

ö

ø÷ -

5

6 6

5

6 6 4

p p p p

e. The child is on the ground at time t = 14. The velocity vector at that time is

&( ) cos sin~ ~ ~

r t i t j k=æ

èç

ö

ø÷ -

æ

èç

ö

ø÷ -

5

6 6

5

6 6

14

4

p 14p p 14p~

&( ) . . .~ ~ ~ ~r t i j k= - -1309 2267 35

The speed is 4.37067 » 4.4 m/s.

f. Differentiating the three velocityvector components,

&&( ) sin cos~ ~ ~

r t t i t j=æ

èç

ö

ø÷ -

æ

èç

ö

ø÷ -

5

36 6

5

36 6

1

4

2 2p p p pk~

The square of the magnitude of this is

5

36 6

5

36 6

22

2p p p psin cost t

æ

èç

ö

ø÷

æ

èçç

ö

ø÷÷ +

æ

èç

ö

ø÷

æ

èçç

ö

ø÷÷ +

1

16

= 5

36 6 6

1

16

22 2p p p

sin cost tæ

èç

ö

ø÷ +

æ

èç

ö

ø÷

æ

èçç

ö

ø÷÷ +

= 5

36 6 6

1

16

22 2p p p

sin cost tæ

èç

ö

ø÷ +

æ

èç

ö

ø÷

æ

èçç

ö

ø÷÷ +

= 5

36

1

16

2p+

Which is a constant (no t), making themagnitude a constant as required.

g. i. Return to the velocity vector:

&( ) cos sin~ ~ ~ ~r t t i t j

tk=

æ

èç

ö

ø÷ -

æ

èç

ö

ø÷ -

5

6 6

5

6 6 4

p p p p

Using the techniques of the previous

question we require &( )~r t . The square of

this magnitude is,

=25

36 6 6 16

22 2

2p 5p pcos sint t

tæ

èç

ö

ø÷ +

æ

èç

ö

ø÷

æ

èçç

ö

ø÷÷ +

=25

36 16

2 2p+

t

9

Re (z)

Im (z)

O

–1

1

1–1–2 2

–2

3–3

2

Solutions for Specialist Mathematics Exam 2 2009

Thus, &( )~r t

t= +

25

36 16

2 2p

which is in the required format.

ii. To find the distance travelled by thechild, we require

25

36 16

2 2

0

14p

+òt

dt

This has the look of a task for CAS. Usingthe integration function, I obtained ananswer of 48.6845= 48.7 metres, correct to the nearesttenth of a metre.

Question 4

a. Despite the fact that part b. sayssketch the graph, I prefer to have a graph in front of me for any domain or rangequestion. Just by looking at the equation,

we know that the graph of yx

x=

-4

2

1 will

have x-intercepts at x = ±1 and a verticalasymptote at x = 0. Using the CAS SOLVEcommand, I found that when y = 10, x = ±31779. and that when y = -10, x = ±03147. . These values are marked onthe rough CAS sketch below. Note that thegraph kept inside the domain is shown inbold.

From that sketch, the domain can bewritten down. Note that the question asksfor one decimal place figures.

x Î - - È[ . , . ] [ . , . ]32 03 03 32 . That wouldmean that a = -32. and b = -03. .

b. A repeat of that graph with a littlemore formality is asked for here. Be sure to transfer the graph from the CASaccurately, even to the extent of finding the coordinates of some dots in the middlesomewhere, to scale it properly.

c. We are told that rule relating x to ymay be rearranged to give x yx4 2 1 0- - = .Treat this as a quadratic with w x= 2

x yx4 2 1 0- - =\ w yw2 1 0- - =

\ wb b ac

a=

- ± -2 4

2

=- - ± - -( ) ( )y y2 4 1

2

=y y± +2 4

2Now, since w x= 2 , w ³ 0. Note also that

y2 4+ is greater than y, so we reject the

solution of y y- +2 4

2 because the top

line would be negative, making w negative.

Thus, xy y2

2 4

2=

+ + .

d. i. The required integral (to give thevolume of the contents of the glass) is

V x dy=-

òp 2

10

10

10

10

5

0

–5

–10

–3 –2 –1 1 2 3 x

y

3.17–3.17

0.32–0.32

10

5

0

–5

–3 –2 –1 1 2 3 x

y

3.17–3.17

(0.32, –10)(–0.32, –10)

(3.17, 10)(–3.17, 10)

(0.32, 0)(–0.32, 0)

Solutions for Specialist Mathematics Exam 2 2009

but we found before that xy y2

2 4

2=

+ +

\ Vy y

dy=+ +æ

è

çç

ö

ø

÷÷

-

òp2

10

10 4

2

= V y y dy= + +-

òp

242

10

10

( )

ii. Evaluating this definite integral on the CAS calculator gives 174.72» 174.7 cubic cm.(The number of brackets you need incalculating this is amazing...)

Liquid is poured into the glass at a rate of1.5 cm3 per second.

e. A related rates question. We wish to

find dy

dt, the rate at which the surface is

rising. We are told that dV

dt= 15. .

dy

dt

dy

dV

dV

dt= ´

=1

152px

´ .

Note, that when the level is 6 cm ‘from thetop of the glass’, the y-value is 4. But wewill need to know the x-value for thatexpression.

xy y2

2 4

2=

+ +

=4 4 4

24 236

2+ += .

\dy

dt x= ´

115

2p.

=1

4 23615

p( . ).´

= 0.1127» 0.11 cms–1

Question 5

a. The a diagram shows the forcesacting on thedevice Takingupwards aspositive for the vectors, theresultant forcewill be 4 2v g-newtonsupwards.

F ma=

\ aF

m=

=4 2

2

v g-

= 2v g- upwards.That is, the acceleration downward is2v g- . Hmm, looks like they wantdownwards positive. Ok, so be it.The acceleration down is g v- 2 ms–2.

b. a g v= - 2

\dv

dtg v= - 2

\dt

dv g v=

-

1

2

\ tg v

dv=-ò1

2

= --

-ò1

2

2

2g vdv

= - - +1

22log ( )e g v c

We are told that when the time is zero, thevelocity is also 0 (‘released from rest’).

\ 01

2= - +log ( )e g c

\ t g v ge e= - - +1

22

1

2log ( ) log ( )

\ tg

g ve=

-

æ

èçç

ö

ø÷÷

1

2 2log

as required.

11

2v

4v

+v

Solutions for Specialist Mathematics Exam 2 2009

c. As time becomes very large (that is,the left side of the equation

tg

g ve=

-

æ

èçç

ö

ø÷÷

1

2 2log becomes large), so the

right side, the value of g v- 2 mustapproach zero. Hence the velocity

approaches g

2 ms–1.

d. tg

g ve=

-

æ

èçç

ö

ø÷÷

1

2 2log

The velocity is g

4,

\ tg

g ge=

-

æ

èçç

ö

ø÷÷

1

2 2 4log

( / )

\ tg

g ge=

-

æ

èçç

ö

ø÷÷

1

2

4

4 2log

\ t e= log ( )2 seconds.

The relation given in part b. may be

rearranged to vg

e t= - -

21 2( ).

e. The relation given in part b. may be

rearranged to vg

e t= - -

21 2( )

\dx

dt

ge t= - -

21 2( )

\ xg

e dtt= - -

ò 21 2

0

180

( )

Using my CAS, I obtained an answer of 879.55.» The ocean is 880 m deep, correct tothe nearest metre.

If the device takes 180 seconds to hit thesea bed, how deep is the ocean at that location? Give your answer correct to thenearest metre.

f. We require x when the velocity is g

3ms–1. The relation given in part b. may berearranged to

vg

e t= - -

21 2( )

\dx

dt

ge t= - -

21 2( ) ...1

Using this, we will needs the two times. Atthe start, the time is 0. When the velocity is g

3, using the techniques of part d.

tg

g ve=

-

æ

èçç

ö

ø÷÷

1

2 2log

\ tg

g ge=

-

æ

èçç

ö

ø÷÷

1

2 2 3log

( / )

\ t e= log ( )3 seconds.

from ...1,

xg

e dtte

= - -

ò 21 2

0

3

( )log ( )

Using my CAS, I obtained an answer of 1.05827» 1.1 metres below sea level, correct tothe nearest 0.1 metre.

g. Once again use vg

e t= - -

21 2( )

\dx

dt

ge t= - -

21 2( )

\ xg

e dtt= - -

ò 21 2( )

=g

t e ct

2

1

2

2( )+ +-

When t = 0, it s at the surface (x = 0)

\ 02

1

2

0= +g

e c( )

\ cg

= -4

\ xg

t egt= + --

2

1

2 4

2( )

When x is 1200,

\ 12002

1

2 4

2= + --gt e

gt( )

From the CAS, t = 245.398 seconds. Some CAS calculators will only give thenegative answer for t, but t ³ 0. It evenwarns you. You can prompt it to give thepositive answer by putting | t > 0 or

12

Solutions for Specialist Mathematics Exam 2 2009

similar at the end of the SOLVE command, or you could use a graphical method by

sketching both yg

x egx= + --

2

1

2 4

2( ) and

y = 1200 and looking for the intersectionover the correct domain and range.

Now, the boat is drifting. The velocity is 2m/s so that it has drifted 2 245398´ . = 490.796 metres.

To complete thequestion, weneed to use good oldPythagoras. Sideways, thesituation lookslike this,

Tha means

x = +1200 4907962 2.

= 1296.49= 1296 m to the nearest metre.

13

1200 m depth

490.796

x

Solutions for Specialist Mathematics Exam 2 2009