Maths Assignment- With Roller Coaster Question

8/9/2019 Maths Assignment- With Roller Coaster Question

1/12Mark Riley 3107631608 Pure Mathematics Assessment Item #2 1

ASSESSMENT

MATHEMATICS (PURE)1

ASSESSMENT ITEM #2

UESTION 1

a. i) The Remainder Theorem

In algebra, polynomial long division is an algorithm for dividing a polynomial by another polynomial of the same o

lower degree, a generalised version of the familiar arithmetic technique called long division. It can be done easily

hand, because it separates an otherwise complex division problem into smaller ones. The polynomial remainder

theorem is an application ofpolynomial long division. It states that the remainder of a polynomial P(x) divided by

linear divisor (x-a) is equal to Q(x)

Pxx a = Qx +

R

x a

ii) The Factor TheoremThe factor theorem is a theorem for finding out the factors of a polynomial. It is a special case of the polynomial

remainder theorem. It states that a polynomial P(x) has a factor (x a) if a nd only if P(a) = 0.

P(x) = x aQ(x)

b. i)

+ + 33 + 162 + 4 7

152 + 0 012 + 4 7

5 01 7

5 2 remainder

ii)

+

+

83 + 02 + 0 142 + 0 042 + 0 1

2 02 1

10
http://en.wikipedia.org/wiki/Algebrahttp://en.wikipedia.org/wiki/Algorithmhttp://en.wikipedia.org/wiki/Polynomialhttp://en.wikipedia.org/wiki/Degree_of_a_polynomialhttp://en.wikipedia.org/wiki/Long_divisionhttp://en.wikipedia.org/wiki/Polynomial_long_divisionhttp://en.wikipedia.org/wiki/Remainderhttp://en.wikipedia.org/wiki/Polynomialhttp://en.wikipedia.org/wiki/Linearhttp://en.wikipedia.org/wiki/Divisorhttp://en.wikipedia.org/wiki/Divisorhttp://en.wikipedia.org/wiki/Linearhttp://en.wikipedia.org/wiki/Polynomialhttp://en.wikipedia.org/wiki/Remainderhttp://en.wikipedia.org/wiki/Polynomial_long_divisionhttp://en.wikipedia.org/wiki/Long_divisionhttp://en.wikipedia.org/wiki/Degree_of_a_polynomialhttp://en.wikipedia.org/wiki/Polynomialhttp://en.wikipedia.org/wiki/Algorithmhttp://en.wikipedia.org/wiki/Algebra


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Mark Riley 3107631608 Pure Mathematics Assessment Item #2 2

QUESTION 1

c. i) 0 = 3

+ 32

4 = 1 + 1 = 0 &

+ + 3 + 32 + 0 4

2 + 0 042 + 0 4

4 04 4

40

+ 2 + 2( 1)

ii) 0 = 33 + 142 + 7 4 = 1 + 1 = 0 &

+ + 33 + 142 + 7 4

32 + 0 0112 + 7 4

11 04 4

40

+ 4 + 1(3 1)


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QUESTION 1

d. 0 = 23 112 + 18 9 = 3 3 = 0 &

+ 23 112 + 18 9

62 + 0 + 052 + 18 9

15 + 03 9

90

1 32 3 = 0

= 1 = 3 = 32


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QUESTION 2

Top curve (T)

Turning point5,4 T(x) = a(x 5)2 + 4 Passes thru origin 0,0 0 = a (5)2 + 4 a = 425

T

x

=425

(x

5)2 + 4 EXPAND

= +

Bottom curve (B)

a = 2m d = 0 c = 0 T = 20 B =220

=

10

() = ()

1.25km = 1250m 1250m 39.3989m2 = 49248.625m3

Cost is $120.00 per m3 estimated cost cost of excavation = 49248.625 120 = $5909835Estimated cost of excavation to nearest 1000 = $5,910,000.00


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QUESTION 3

It is the year 2059. A thriving colony on Mars has been set up and the population is growing rapidly. Unfortunately there is a problem getting

supplies from earth. Destruction of the launch facilities by Techno Terrorists means that it is likely that no supply ship will be able to arrive within

three years. To meet the needs of the growing colony a hydroponics agriculture plant is to be established.

The colony has 180 tonnes of food in storage at present.

Rate of consumption of food in the colony:

xexA

06.01.2)( tonnes per month

Rate of growth of food production:

xexB

03.07.2)( tonnes per month

The total food consumed within the 3 year period until the next supply ship arrives must not exceed the amount of food in storage a

present + the amount of food able to be produced in the 3 year period.

3 + (3 )Food available

B

x

=

2.7e0.03x dx

Bx = 2.70.03

e0.03x + c

Bx = 90e0.03x + cwhere c is the amount of food already in storage

F(t) = 90e0.03t + 180

t = time in months

F = amount of food in tonnes

() = (.) +

= .

Food consumed of this amount

Ax = 2.1e0.06x dx

Ax = 2.10.06

e0.06x + c

= 350.06 + = 0 = 0 c = 0

FC

t

= 35e0.06t

t = time in months

FC = Food consumed in tonnes

FC36 = 35e(0.0636) = .

. < . *THIS ONLY PROVES THAT OVERALL THERE WILL BE MORE FOOD AVAILABLE THAN FOOD CONSUMED- BUT IS THERE ANY POINT

WHERE THE TWO LINES INTERESECT OVER THE 36 MONTH PERIOD? See next page


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QUESTION 3

= . + = .

The graph shows that the amount of food consumed will never exceed the amount of food available during the 36

month period.


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QUESTION 4

= + . () | =. .

( + .) ()

| = 7 5 = 4.4 104 0.2256

(1+ 690.2256)2 = 10.63565594 & = 26.89988406

10.6 = 28001 + 6 9(0.2256052 10.6) = 382.85 26.9 =

2800

1 + 6 9(0.2256052 26.9) = 2414.49

75

382 & 2415

=

| = 125 = 4.4 104 0.2256

(1 + 690.2256)2 = 14.39730438 & = 23.13823562

14.4 = 28001 + 6 9(0.2256052 14.4) = 761.11 23.1 =

2800

1 + 6 9(0.2256052 23.1) = 2034.44

125 762 & 2034 =

The farmer should remove 2000 (or close to) amount of fish when the population reaches 2400 and then again approximately

every 16 months when the population reaches 2400 in order to maintain a constant growth rate of more than 75 fish per month

She may like or need to remove fish more frequently (or even less possibly) for business purposes. To ensure a growth rate of a

least 75 fish per month, she shouldnt remove so many fish that the population falls below 400 or so little that is rises above 240

I would definitely suggest to the farmer that she should keep the population between 762 and 2034 fish to maintain a high grow

rate of 125 fish per month, removing 1272 fish when the population first reaches 2034 then again approximately every 9 month

when the population again reaches 2034.


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QUESTION 5

2 simple rules to solve this problem

1. All segments (AB, BC and CD) are parabolic and will have the general formu

= 2 + + & | = 2 + 2. The gradient at the start of segments BC & DC will be equal to the gradient

the end of the previous segment.

SECTION AB B(-1.5, 10) A(-9.5, 30) where m=-tan10

10 = 1.52 + 1.5+ . + . =

30 = 9.52 + 9.5+

. + . = . + . = .+ .

1.5 9.5 = 30 10 90.25 + 2.258 = 20 88

=

| = 2 + 110 = 2 9.5 + = () +

= () +

88 20 = 8(10) + 152

= (

) +

= . = + . = .

= ..+ .. = .

= . .+ .


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QUESTION 5

SECTION BC B(-1.5, 10) C(1.5,10)

= 0.2904591272 5.695050402 + 2.110957397| = 0.580918254(1.5) 5.695050402 = 2(1.5)+

0.871377381 5.695050402+ 3 =

.+ =

10 = 1.52 + 1.5+

. + . =

10 = 1.52 + 1.5+

.. =

.. = . + .1.5 = 1.5

3 = 0 =

.+ = = .

= . . + . = .

= . +.


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QUESTION 5

SECTION CD C(1.5, 10) D(12.5,y) where m=0

= 1.6078910072 + 6.382245234

| = 3.215782014 (1.5) = 2

(1.5) +

.=

10 = 1.52 + 1.5+ .. =

..(.) = 10 2.25 7.235509532 + 4.5 =

.+ . =

| = 0 = 212.5+

=

= 12.52 + 12.5 + .. =

..() = 156.25 + 312.5 =

+ . =

+ . = .+ . = .

.=

= . = .

= .. = . = .= .

= .+ .. = .

= . + .+ .


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QUESTION 5

y=-0.290459127x^2-5.695050402x+2.110957397{-9.5,-1.5} white liney=1.607891007x^2+6.382245234{-1.5,1.5} red liney=-0.219257864x^2+5.4814466x+2.271160275{1.5,12.5} green line

NOW THAT TO ME- LOOKS LIKE A DARN SMOOTH ROLLER COASTER (very pleased!)

The height of the rollercoaster is .metres


12/12

M k Ril 3107631608 P M th ti A t It #2 12

QUESTION 5

Finding the length of a curve

f(x) = -0.290459127x^2-5.695050402x+2.110957397

Increment= 0.1

Start = -9.5

Finish = -1.5

x f(x) Section Length Total Length

-1.5 9.999999964 0.490 22.107


f(x) = 1.607891007x^2+6.382245234

Increment= 0.1

Start = -1.5

Finish = 1.5


1.5 10 0.477 8.095


f(x) = -0.219257864x^2+5.4814466x+2.271160275

Increment= 0.1

Start = 1.5

Finish = 12.5


12.5 36.53020153 0.100 29.691

=22.107+8.095+29.691=59.893m = . $. = $,.

Maths Assignment- With Roller Coaster Question

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