mathematical methods 2019 v1 · 3. communicate using mathematical, statistical and everyday...
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Mathematical Methods 2019 v1.2 IA2 sample marking scheme November 2019
Examination (15%) This sample has been compiled by the QCAA to model one possible approach to allocating marks in an examination. It matches the examination mark allocations as specified in the syllabus (~ 60% simple familiar, ~ 20% complex familiar and ~ 20% complex unfamiliar) and ensures that all the objectives are assessed.
Assessment objectives This assessment instrument is used to determine student achievement in the following objectives:
1. select, recall and use facts, rules, definitions and procedures drawn from all Unit 3 topics 2. comprehend mathematical concepts and techniques drawn from all Unit 3 topics 3. communicate using mathematical, statistical and everyday language and conventions
4. evaluate the reasonableness of solutions 5. justify procedures and decisions by explaining mathematical reasoning 6. solve problems by applying mathematical concepts and techniques drawn from all Unit 3
topics.
Mathematical Methods 2019 v1.2 IA2 sample marking scheme
Queensland Curriculum & Assessment Authority November 2019
Page 2 of 14
Instrument-specific marking guide (ISMG) Criterion: Foundational knowledge and problem-solving
Assessment objectives 1. select, recall and use facts, rules, definitions and procedures drawn from all Unit 3 topics
2. comprehend mathematical concepts and techniques drawn from all Unit 3 topics 3. communicate using mathematical, statistical and everyday language and conventions 4. evaluate the reasonableness of solutions 5. justify procedures and decisions by explaining mathematical reasoning
6. solve problems by applying mathematical concepts and techniques drawn from all Unit 3 topics
The student work has the following characteristics: Cut-off Marks
• consistently correct selection, recall and use of facts, rules, definitions and procedures; authoritative and accurate command of mathematical concepts and techniques; astute evaluation of the reasonableness of solutions and use of mathematical reasoning to correctly justify procedures and decisions; and fluent application of mathematical concepts and techniques to solve problems in a comprehensive range of simple familiar, complex familiar and complex unfamiliar situations.
> 93% 15
> 87% 14
• correct selection, recall and use of facts, rules, definitions and procedures; comprehension and clear communication of mathematical concepts and techniques; considered evaluation of the reasonableness of solutions and use of mathematical reasoning to justify procedures and decisions; and proficient application of mathematical concepts and techniques to solve problems in simple familiar, complex familiar and complex unfamiliar situations.
> 80% 13
> 73% 12
• thorough selection, recall and use of facts, rules, definitions and procedures; comprehension and communication of mathematical concepts and techniques; evaluation of the reasonableness of solutions and use of mathematical reasoning to justify procedures and decisions; and application of mathematical concepts and techniques to solve problems in simple familiar and complex familiar situations.
> 67% 11
> 60% 10
• selection, recall and use of facts, rules, definitions and procedures; comprehension and communication of mathematical concepts and techniques; evaluation of the reasonableness of some solutions using mathematical reasoning; and application of mathematical concepts and techniques to solve problems in simple familiar situations.
> 53% 9
> 47% 8
• some selection, recall and use of facts, rules, definitions and procedures; basic comprehension and communication of mathematical concepts and techniques; inconsistent evaluation of the reasonableness of solutions using mathematical reasoning; and inconsistent application of mathematical concepts and techniques.
> 40% 7
> 33% 6
• infrequent selection, recall and use of facts, rules, definitions and procedures; basic comprehension and communication of some mathematical concepts and techniques; some description of the reasonableness of solutions; and infrequent application of mathematical concepts and techniques.
> 27% 5
> 20% 4
Mathematical Methods 2019 v1.2 IA2 sample marking scheme
Queensland Curriculum & Assessment Authority November 2019
Page 3 of 14
The student work has the following characteristics: Cut-off Marks
• isolated selection, recall and use of facts, rules, definitions and procedures; partial comprehension and communication of rudimentary mathematical concepts and techniques; superficial description of the reasonableness of solutions; and disjointed application of mathematical concepts and techniques.
> 13% 3
> 7% 2
• isolated and inaccurate selection, recall and use of facts, rules, definitions and procedures; disjointed and unclear communication of mathematical concepts and techniques; and illogical description of the reasonableness of solutions.
> 0% 1
• does not satisfy any of the descriptors above. 0
Task See the sample assessment instrument for IA2 (available on the School Portal).
Mathematical Methods 2019 v1.2 IA2 sample marking scheme
Queensland Curriculum & Assessment Authority November 2019
Page 4 of 14
Sample marking scheme Criterion Allocated marks Marks awarded
Foundational knowledge and problem-solving Assessment objectives 1, 2, 3, 4, 5 and 6 15 –
Total 15 –
The annotations are written descriptions of the expected response for each question and are related to the assessment objectives.
Note: = 12 mark
1a.
select and use:
• rules to change from log form to index form
• index facts 36 = 62 • facts to solve
1c.
comprehend the information is in factorised form
select and use:
• null factor procedure • rules to change from
index form to log form to c solutions for 𝑥𝑥
2a.
select and use:
• derivative of 𝑒𝑒𝑥𝑥 • procedure for chain
rule (recognise inner and outer function)
• derivative of a trigonometric function
Marking scheme — Paper 1 (technology-free) Question 1 (SF 7 CF 4 marks)
a. 𝑥𝑥 = log6 36 36 = 6𝑥𝑥 𝑥𝑥 = 2
b. log3(4𝑥𝑥 − 7) = 2
4𝑥𝑥 − 7 = 32 4𝑥𝑥 = 16 → 𝑥𝑥 = 4
c. (𝑒𝑒𝑥𝑥 − 2)(𝑒𝑒𝑥𝑥 − 3) = 0
Use null factor theorem: 𝑒𝑒𝑥𝑥 = 2 and 𝑒𝑒𝑥𝑥 = 3 𝑥𝑥 = ln 2 and 𝑥𝑥 = ln 3
d. ln𝑥𝑥 + ln(2− 𝑥𝑥) = 0
ln�𝑥𝑥(2 − 𝑥𝑥)� = 0 𝑥𝑥(2− 𝑥𝑥) = 𝑒𝑒0 𝑥𝑥2 − 2𝑥𝑥 + 1 = 0 (𝑥𝑥 − 1)2 = 0 𝑥𝑥 = 1
Question 2 (SF 16 marks) a. 𝑓𝑓(𝑥𝑥) = 𝑒𝑒𝑥𝑥 + sin(2𝑥𝑥)
𝑓𝑓′(𝑥𝑥) = 𝑒𝑒𝑥𝑥 + 2 cos(2𝑥𝑥)
b. 𝑓𝑓(𝑥𝑥) = 𝑒𝑒sin(𝑥𝑥) 𝑓𝑓′(𝑥𝑥) = 𝑒𝑒sin(𝑥𝑥) × cos(𝑥𝑥)
1b.
select and use:
• rules to change from log to index form
• analytic procedure to solve for 𝑥𝑥
1d.
select and use:
• logarithm of a product rule
• rules to change from log form to index form
comprehend (make connection) to quadratic form of equation
recall and use analytic procedure to determine solution for 𝑥𝑥
2b.
select and use:
• procedure for chain rule to differentiate an exponential function and trigonometric function (recognise inner and outer function)
Mathematical Methods 2019 v1.2 IA2 sample marking scheme
Queensland Curriculum & Assessment Authority November 2019
Page 5 of 14
2c.
select and use:
• procedure for chain rule to differentiate a polynomial and a trigonometric function
2e.
select and use:
• procedure for differentiating a quotient
• rule to differentiate polynomials
• rules to simplify
response
3a.
select and use:
• rules for integrating a polynomial function
• procedure for calculating the value of a definite integral
• facts to determine the value of the definite integral
c. 𝑓𝑓(𝑥𝑥) = cos3(𝑥𝑥) 𝑓𝑓′(𝑥𝑥) = 3(cos(𝑥𝑥))2 × − sin(𝑥𝑥)
or −3 cos2(𝑥𝑥) sin(𝑥𝑥)
d. 𝑓𝑓(𝑥𝑥) = 𝑥𝑥 + 𝑥𝑥 ln(𝑥𝑥)
𝑓𝑓′(𝑥𝑥) = 1 + 𝑥𝑥 ×1𝑥𝑥
+ ln(𝑥𝑥) × 1
𝑓𝑓′(𝑥𝑥) = 2 + ln(𝑥𝑥)
e. 𝑓𝑓(𝑥𝑥) = 2𝑥𝑥+3𝑥𝑥2+3𝑥𝑥
Use quotient rule:
𝑓𝑓′(𝑥𝑥) =(𝑥𝑥2 + 3𝑥𝑥) × 2 − �(2𝑥𝑥 + 3) × (2𝑥𝑥 + 3)�
(𝑥𝑥2 + 3𝑥𝑥)2
𝑓𝑓′(𝑥𝑥) = 2𝑥𝑥2+6𝑥𝑥−4𝑥𝑥2−6𝑥𝑥−6𝑥𝑥−9(𝑥𝑥2+3𝑥𝑥)2
𝑓𝑓′(𝑥𝑥) = −2𝑥𝑥2−6𝑥𝑥−9(𝑥𝑥2+3𝑥𝑥)2
Question 3 (SF 11 marks) a. ∫ 4𝑥𝑥2𝑑𝑑𝑥𝑥3
1 = 4
3𝑥𝑥3|31
= 4
3(33 − 13)
= 104
3
b. ∫ 6𝑒𝑒2𝑡𝑡 + 𝑡𝑡 𝑑𝑑𝑡𝑡20
= 6𝑒𝑒2𝑡𝑡
2+ 𝑡𝑡2
2|20
= �3𝑒𝑒4 + 22
2� − �6𝑒𝑒
0
2+ 0
2�
= 3𝑒𝑒4 + 2 − 3 = 3𝑒𝑒4 − 1
2d.
select and use:
• rule for derivative of a polynomial
• rule for differentiating a product
• rule for derivative of natural logarithm
• facts to simplify response
3b.
select and use:
• rules for integrating polynomial function
• rules for integrating an exponential function
• procedure for calculating the value of a definite integral
recall rule 𝑎𝑎0 = 1
Mathematical Methods 2019 v1.2 IA2 sample marking scheme
Queensland Curriculum & Assessment Authority November 2019
Page 6 of 14
4a.
select and use:
• substitution to generate equation
• rule log𝑎𝑎 1 = 0
determine 𝐻𝐻
communicate solution in metres
5.
identify 𝑥𝑥′(𝑡𝑡) = 𝑣𝑣(𝑡𝑡)
select and use procedure for differentiating a product to determine 𝑣𝑣(𝑡𝑡)
translate information into mathematically workable format (determine 𝑣𝑣(𝑡𝑡) = 0)
select and use procedures (factorising, null factor theorem)
recall log laws/exponential function asymptote to identify solution that is ‘not possible’
rearrange to generate trigonometric equation
recall common ratios to determine solution to trigonometric equation
c. Using the given information
𝑓𝑓′(𝑥𝑥) = 2𝑥𝑥𝑥𝑥2+4
∴ ∫ 4𝑥𝑥𝑥𝑥2+4
20 𝑑𝑑𝑥𝑥
= 2 ln(𝑥𝑥2 + 4)| 20 = 2 (ln 8− ln 4) = 2 ln(2) 𝑜𝑜𝑜𝑜 ln(4 )
Question 4 (SF 7 marks) a. Substitute 𝑡𝑡 = 1 into the model:
𝐻𝐻 = 6 + 6 log31 log3 1 = 0 ∴ 𝐻𝐻 = 6 metres
b. 18 = 6 + 6 log3 𝑡𝑡 126
= log3 𝑡𝑡
𝑡𝑡 = 32
𝑡𝑡 = 9 years
Question 5 (CF 6 marks) Given 𝑥𝑥(𝑡𝑡) = 𝑒𝑒𝑡𝑡 sin(𝑡𝑡) 𝑥𝑥′(𝑡𝑡) = 𝑣𝑣(𝑡𝑡)
𝑥𝑥′(𝑡𝑡) = 𝑒𝑒𝑡𝑡 cos 𝑡𝑡 + sin 𝑡𝑡 × 𝑒𝑒𝑡𝑡
Particle is at rest when 𝑥𝑥′(𝑡𝑡) = 0
0 = 𝑒𝑒𝑡𝑡 cos 𝑡𝑡 + sin 𝑡𝑡 × 𝑒𝑒𝑡𝑡
0 = 𝑒𝑒𝑡𝑡(cos(𝑡𝑡) + sin(𝑡𝑡))
∴ 𝑒𝑒𝑡𝑡 = 0 not possible
and cos(𝑡𝑡) + sin(𝑡𝑡) = 0
cos(𝑡𝑡) = − sin(𝑡𝑡)
tan(𝑡𝑡) = −1
Recall common ratios and CAST rule
𝑡𝑡 = 3𝜋𝜋4
and 7𝜋𝜋4
determine value of the definite integral
3c.
select and use rules/procedures
• for differentiating functions of the form ln𝑓𝑓(𝑥𝑥)
• to determine the indefinite integral using information about the derivative of a given function
• to determine the value of a definite integral
use log laws to determine value of the definite integral in simplest form
4b.
select and use:
• substitution to generate equation
• procedure to rearrange equation
• rules to change equation from log form to index form
determine solution for 𝑡𝑡
communicate solution in years
Mathematical Methods 2019 v1.2 IA2 sample marking scheme
Queensland Curriculum & Assessment Authority November 2019
Page 7 of 14
6.
translate information into mathematically workable format (maximum point occurs when 𝑓𝑓′(𝑥𝑥) =0)
select and use:
• procedure for differentiating a quotient, differentiating ln(𝑥𝑥) and polynomials to determine 𝑓𝑓′(𝑥𝑥)
• procedures for rearranging the equation
• rules for changing from log form to index form
• substitution to determine 𝑦𝑦-coordinate of maximum point
• rules for changing from log form to index form
communicate the coordinates of the maximum point using appropriate terminology
7.
identify critical elements:
• 𝑃𝑃(3)=50 • 𝑃𝑃′(3) is required
use algebraic skills to determine 𝑏𝑏
use rules to:
• determine an expression for 𝑃𝑃′(𝑡𝑡)
select and use procedure to:
• generate an equation using substitution
• solve equation
communicate findings
Question 6 (CF 6 marks) Given 𝑓𝑓(𝑥𝑥) = ln(2𝑥𝑥)
𝑥𝑥, 𝑥𝑥 > 0
Maximum occurs when 𝑓𝑓′(𝑥𝑥) = 0
Use quotient rule:
𝑓𝑓′(𝑥𝑥) =𝑥𝑥× 2
2𝑥𝑥−(ln(2𝑥𝑥)×1)
𝑥𝑥2
Maximum point 𝑓𝑓′(𝑥𝑥) = 0
0 = 1−ln(2𝑥𝑥) 𝑥𝑥2
0 = 1 − ln(2𝑥𝑥)
ln(2𝑥𝑥) = 1
2𝑥𝑥 = 𝑒𝑒
𝑥𝑥 = 𝑒𝑒2
𝑓𝑓 �𝑒𝑒2� =
ln�2×𝑒𝑒2�𝑒𝑒2
=2ln𝑒𝑒𝑒𝑒
= 2𝑒𝑒
Maximum point of the function �𝑒𝑒2
, 2𝑒𝑒�
Question 7 (CF 5 marks) Given at time 𝑡𝑡 = 3 population is 50:
𝑃𝑃(3) = 50 = 1001+𝑒𝑒𝑏𝑏−𝑡𝑡
50(1 + 𝑒𝑒𝑏𝑏−2) = 100
1 + 𝑒𝑒𝑏𝑏−2 = 2
𝑒𝑒𝑏𝑏−3 = 1
∴ 𝑏𝑏 = 3
Using chain rule: 𝑃𝑃′(𝑡𝑡) = −100(1 + 𝑒𝑒3−𝑡𝑡)−2 × −𝑒𝑒3−𝑡𝑡
𝑃𝑃′(3) =−100 × 𝑒𝑒0
(1 + 𝑒𝑒0)2
𝑃𝑃′(3) =100
4= 25
The flu is spreading at a rate of 25 students/day on day 3.
Mathematical Methods 2019 v1.2 IA2 sample marking scheme
Queensland Curriculum & Assessment Authority November 2019
Page 8 of 14
8.
recall velocity is rate of change of displacement w.r.t. time
comprehend information requires use of reciprocal function
recall how to determine the indefinite integral of a power function
solve for the function 𝑡𝑡(𝑠𝑠) using given information
use substitution process and appropriate analytic processes to solve for 𝑠𝑠
Question 8 (CU 7 marks) Given 𝑣𝑣 = √4 + 4𝑠𝑠
𝑣𝑣 = 𝑑𝑑𝑑𝑑𝑑𝑑𝑡𝑡
= √4 + 4𝑠𝑠
∴ 𝑑𝑑𝑡𝑡𝑑𝑑𝑑𝑑
= 1√4+4𝑑𝑑
= (4 + 4𝑠𝑠)−12
𝑡𝑡 = 2(4+4𝑑𝑑)12
4+ 𝑐𝑐 or 𝑡𝑡 = √𝑠𝑠 + 1 + 𝑐𝑐
Given 𝑠𝑠 = 0 when 𝑡𝑡 = 0
∴ 0 = 1×412
2+ 𝑐𝑐
𝑐𝑐 = −1
So 𝑡𝑡 = √4+4𝑑𝑑2
− 1
Determine 𝑠𝑠 when 𝑡𝑡 = 2
2 =√4 + 4𝑠𝑠
2− 1
6 = √4 + 4𝑠𝑠
36 = 4 + 4𝑠𝑠
𝑠𝑠 = 8 metres
Mathematical Methods 2019 v1.2 IA2 sample marking scheme
Queensland Curriculum & Assessment Authority November 2019
Page 9 of 14
1.
use technology to produce sketch
label graphical display appropriately
1a.
select and use definition for domain
use appropriate notation (symbols) for domain
1b.
select and use definition
1c.
select and use procedure for describing the shape of a graph
use appropriate notation (symbols) for increasing/decreasing intervals
2a.
translate information into a mathematically workable format (substitute)
Marking scheme — Paper 2 (technology-active) Question 1 (SF 11 marks) Sketch 𝑓𝑓(𝑥𝑥) = ln(𝑥𝑥)
𝑥𝑥
a. 𝑥𝑥 > 0
b. No 𝑦𝑦-intercept 𝑥𝑥-intercept (1,0)
c. Graph is increasing for 0 < 𝑥𝑥 < 2.72
Graph is decreasing for 𝑥𝑥 > 2.72
d. Maximum (2.72, 0.368)
e. Horizontal asymptote 𝑦𝑦 = 0 or 𝑥𝑥-axis Vertical asymptote 𝑥𝑥 = 0 or 𝑦𝑦-axis
Question 2 (SF 8 marks) a. Given 𝑁𝑁 = 4200 when 𝑡𝑡 = 8,𝑛𝑛(𝑡𝑡) = 𝐴𝐴𝑒𝑒𝑏𝑏𝑡𝑡
4200 = 𝐴𝐴𝑒𝑒0.55×8
𝐴𝐴 =4200𝑒𝑒0.55×8
1d.
use technology to determine the maximum point
1e.
select and use definition
Mathematical Methods 2019 v1.2 IA2 sample marking scheme
Queensland Curriculum & Assessment Authority November 2019
Page 10 of 14
determine 𝐴𝐴
2b.
use substitution to generate equation for estimate
comprehend ‘12 years from now’ requires 𝑡𝑡 value of 20
determine 𝑛𝑛(12)
interpret calculator notation
3.
use technology to determine both points of intersection
use technology to solve for bounded region
communicate method used
𝐴𝐴 = 51.5648
b. Determine 𝑛𝑛(12)
𝑛𝑛(12) = 51.5648𝑒𝑒0.55×20 𝑛𝑛(12) = 3.0874 × 106 𝑛𝑛(12) = 3087400
c. 𝑛𝑛(𝑡𝑡) = 51.5648𝑒𝑒0.55𝑡𝑡
𝑛𝑛′(𝑡𝑡) models the rate of change of the population 𝑛𝑛′(𝑡𝑡) = 28.3606𝑒𝑒0.55𝑡𝑡 Determine when 𝑛𝑛′(𝑡𝑡) = 250000
250000 = 28.3606𝑒𝑒0.55𝑡𝑡 8815.05 = 𝑒𝑒0.55𝑡𝑡
𝑡𝑡 = ln(8815.05)0.55
𝑡𝑡 = 16.5168 years
Question 3 (SF 4 marks)
Points of intersection are (0.662,1.89) and (1.55,4.76) Bounded area =
∫ (−3𝑒𝑒−𝑥𝑥 + 𝑥𝑥2 + 31.55.662 ) − (2𝑒𝑒𝑥𝑥 − 3𝑥𝑥)𝑑𝑑𝑥𝑥
Bounded area =
Bounded area = 0.298649
2c.
comprehend information requires use of derivative function
generate equation to solve using given information
use appropriate method to determine time (include units in years)
Mathematical Methods 2019 v1.2 IA2 sample marking scheme
Queensland Curriculum & Assessment Authority November 2019
Page 11 of 14
4a.
use technology (or otherwise) determine 𝑣𝑣(2)
4b.
recall 𝑣𝑣′(𝑡𝑡) = 𝑎𝑎(𝑡𝑡) and use rules to determine derivative of exponential, polynomial and a constant
use substitution to determine acceleration at time 𝑡𝑡 = 2
communicate using appropriate units
4c.
comprehend a suitable method to identify times when acceleration is positive (e.g. graphical method)
clearly communicate technique used
determine interval for positive acceleration by first calculating x-intercept and then identifying where function is positive
communicate using correct notation
5.
comprehend information to identify critical elements:
• translate information into a mathematically workable format
• determine derivatives
Question 4 (SF 9 marks) a. 𝑣𝑣(2) = 𝑒𝑒4
2− 5 × 2− 1
2
𝑣𝑣(2) =
b. 𝑣𝑣′(𝑡𝑡) = 𝑎𝑎(𝑡𝑡) 𝑎𝑎(𝑡𝑡) = 𝑒𝑒2𝑡𝑡 − 5 Determine acceleration after 2 seconds: 𝑎𝑎(2) = 𝑒𝑒4 − 5 = 49.5982 metres/second/second
c. Sketch 𝑎𝑎(𝑡𝑡)
Acceleration function is zero at .805
∴
𝑡𝑡 > .805 seconds
Question 5 (CF 5 marks) The gradients will be equal when:
𝑓𝑓′(𝑥𝑥) = 𝑔𝑔′(𝑥𝑥)
𝑓𝑓′(𝑥𝑥) = 33𝑥𝑥−2
𝑔𝑔′(𝑥𝑥) = 2 sin(0.5𝑥𝑥)
Mathematical Methods 2019 v1.2 IA2 sample marking scheme
Queensland Curriculum & Assessment Authority November 2019
Page 12 of 14
use an appropriate method to determine the values of 𝑥𝑥
6.
translate information into mathematically workable format by
• sketching • identifying half the
area equates to half the area under the parabola
use procedure for determining definite integrals to determine the area under the parabola and therefore determine half the area
use procedure to determine 𝑚𝑚 in terms of intersection point 𝑎𝑎
use procedure to determine 𝑎𝑎 using the area between the curves
Solve by graphing the derivatives and identifying points of intersection:
Gradients are the same at 𝑥𝑥 = 1.43 and 𝑥𝑥 = 6.1
Question 6 (CU 8 marks) Sketch functions to identify area:
Line contains (0, ,0) and 𝑥𝑥 intercepts of parabola at points (0,0) and (2,0) Total area = ∫ 2𝑥𝑥(2− 𝑥𝑥)𝑑𝑑𝑥𝑥2
0
= 4𝑥𝑥2 − 2𝑥𝑥3
3|20 = 8 − 16
3= 8
3
∴ half of the area = 4
3 (technology may be used here)
Let 𝑦𝑦 = 𝑚𝑚𝑥𝑥 and 𝑦𝑦 = 2𝑥𝑥(2 − 𝑥𝑥) intersect at 𝑥𝑥 = 𝑎𝑎 Then 𝑚𝑚𝑎𝑎 = 2𝑎𝑎(2− 𝑎𝑎) 𝑚𝑚 = 4 − 2𝑎𝑎 Hence 𝑦𝑦 = (4− 2𝑎𝑎)𝑥𝑥
� (4𝑥𝑥 − 2𝑥𝑥2)𝑎𝑎
0− (4− 2𝑎𝑎)𝑥𝑥 𝑑𝑑𝑥𝑥 =
43
∫ −2𝑥𝑥2𝑎𝑎0 + 2𝑎𝑎𝑥𝑥 𝑑𝑑𝑥𝑥 = 4
3
→−2𝑎𝑎3
3+
(2𝑎𝑎)𝑎𝑎2
2=
43
→−4𝑎𝑎3 + 6𝑎𝑎3
6=
43
2𝑎𝑎3 = 8
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solve for 𝑎𝑎
solve for equation of the line
evaluate the reasonableness of the solution
7.
select and use procedure for producing regression models using technology
identify incorrect value
use regression model to generate equation in 𝑉𝑉
solve for 𝑉𝑉
8.
comprehend concept to decide on method of solution:
• graph function • translate information
into a mathematical representation
generate translated curve and/or identify area between the two curves as the cross-sectional area
𝑎𝑎 = √43 (Note: both the approximate solution and use of technology to determine a solution are acceptable.) e.g.
The equation of the line is 𝑦𝑦 = �4− √43 �𝑥𝑥 Use technology to determine the area to verify the line divides the area in half.
Question 7 (CU 4 marks) Scatterplot used to display information
(variables log10 𝑉𝑉and log10 ℎ)
The value for ℎ = 70 produces the greatest residual value for the linear regression model
Determine model using non-outliers:
ln(𝑉𝑉) = 0.499648 ln(ℎ) + 2.0805
Substitute ℎ = 70
𝑉𝑉 = 66.90
Question 8 (CU 7 marks)
Given the run terminates at a place 1 kilometre above sea level:
∴ determine 𝑥𝑥 value when 𝑦𝑦 = 1
𝑥𝑥 = 1.1499
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recall rule for determining cross-sectional area
solve for the volume of snow on the run
communicate using mathematical symbols and conventions (e.g. units)
represent ideas in a way that makes sense — relate parts in an orderly, consistent way (justifying procedures)
Equation for curve vertically 2 metres (.002 kilometres)
𝐻𝐻2 = 1.8𝑒𝑒−𝑥𝑥 + 0.43 + 0.002
= 1.8𝑒𝑒−𝑥𝑥 + .432
Cross-sectional area using integration
Area = ∫ (𝐻𝐻2 −1.14990 𝐻𝐻)𝑑𝑑𝑥𝑥
= ∫ 0.002𝑑𝑑𝑥𝑥1.14990
= 0.0023
Volume = area × width
= 0.0023 × .3
= 0.00069 km3
(Communication of response )