Course: HNC in Construction and the Built Environment

Assignment - Assisting with Construction Mathematical problems.

Learning Outcome 1: Use analytical and computational methods to solve construction related problems

Learning Outcome 2: Investigate applications of statistical techniques to interpret, organise and present data by using appropriate computer software packages

Learning Outcome 3: Illustrate the wide-ranging uses of calculus within different construction disciplines by solving problems of differential and integral calculus

Learning Outcome 4: Use mathematical methods to solve vector analysis, arithmetic progression and dimensional analysis examples.

Unit 8: Maths for Construction

You are asked to provide a pamphlet with a range of common conversions found generally in construction that can be given as a useful aide memoire to the many tradesmen (plumbers, electricians, plasterers etc, i.e. non-technical audience) employed by your company. You decide to provide a handy conversion table and twenty or so worked examples. Complete the table below and provide six worked examples for each of the following three factors: length, weight and volume.

Conversion Table

 METRIC TO IMPERIAL IMPERIAL TO METRIC From To Multiply by From To Multiply by meters yards 1.094 yards metres 0.914 metres feet 3.281 feet metres 0.305 centimetres Inches 0.394 inches centimetres 2.540 kilometres miles 0.621 miles kilometres 1.609 grams ounces 0.035 ounces grains 437.5 kilograms pounds 2.205 pounds kilograms 0.454 litres quarts 1.057 quarts litres 0.950 litres gallons 0.264 gallons litres 3.785

(b) A building services engineer is to design a water tank for a project. The tank has a rectangular area of 26.5m2. With the design specifics of the width being 3.2m shorter than the length, calculate the length and width to 3 significant figures for resource requirements.

Given, Area of tank = 26.5 m2 and Width = Length - 3.2m
Length * Width = 26.5 m2
Length * (Length -3.2m) = 26.5 m2
Lengtb - 3.2* Length -26.5 = 0
Length = 3.2 ± sqrt((-3.2)2 - 4*1*(-26.5))/(2*1)
Length = (3.2 ± sqrt(10.24 +106))/2
Length = (3.2 ± sqrt(116.24))/2
Length = (3.2 ± 10.781)/2
Length = 13.981/2 or -7.581/2
Length = 6.9905 or -3.7905

Since Length is always positive , Length = 6.991 m and Width = 6.991 - 3.2 = 3.791m

(c) You have received a letter regarding the project your company is working on. It has a penalty clause that states one of the sub-contractors working on the building for you will forfeit a certain sum of money each day for late completion. (i.e. the contractor gets paid the value of the original contract less any sum forfeit). If she is 5 days late, she receives £4250 and if she is 12 days late, she receives £2120. Calculate the daily forfeit and determine the original contract.

Given 5 days late, the contractor receives 4250
Therefore, the equation is a - 5b = 4250 where a = original compensation and b = daily forfeit

Similarly for 12 days late, the contractor receives 2120
Equation is a - 12b = 2120

a - 5b = 4250
a - 12b = 2120

Subtracting equation 1 and 2 we get

7b = 2130 and b = 304.286
Substituting b in first equation we get a = 5*304.286 + 4250 = 5771.43

Therefore, daily forfeit is £304.286 and original contract is £5771.43

Comprehensive Support for Unit 21: Mastering Site Supervision and Operations in Your HND in Construction and the Built Environment

(d) You have been asked to investigate the following arithmetic sequences
An arithmetic sequence is given by b, 2b/3, b/3 , 0.......

Common difference, d = 2b/3 - b = -b/3
First term, a' = b
Nth term, an = a' + (n-1)d

Determine the sixth term
6th term, a6= a' + (6-1)*d = b + 5*(-b/3) = b -5b/3 = -2b/3

State the kth term
Kth term, ak = a + (k-1)d = b + (k-1)*(-b/3) = b -kb/3 + b/3 = 4b/3 - kb/3

If the 20th term has value of 15 find the value of b and the sum of the first 20 terms
20th term = 15
4b/3 - 20b/3 = 15
-16b/3 = 15
b = -2.813

Sum of first n terms = n/2[2a + (n-1) * d]
Sum of first 20 terms = 20/2[2b + (20-1)* (-b/3)]
= 10[2(-2.8130) + 19*(-2.813/3)]
= 10[-5.626 -17.816] = -234.42

(e) For the following geometric progression 1, 1/2, 1/4 ........ determine
An = nth term
First term, a1 = 1
Common ratio, r = an+1/a1 = a2/a1 = (1/2)/1 =1/2
Nth term = a * rn -1

The 20th term of the progression
Nth term = a * rn -1
20th term = 1 * (1/2)20 -1 = (1/2)19 = 1/524288 = 1.907 * 10-6

The value of the sum when the number of terms in the sequence tends to infinity and explain why the sequence tends to this value Sn = ∑n=0n→∞ arn

S = a/1-r = 1/1 - ½ = 1/1/2 =2

(f) Solve the following Equations for x
2Log (3x) + Log (18x) = 27
log(3x)2 + log(18x) = 27 => alog(3) = log(3)a
log(162x3) = 27
162x3 = 1027
X3 = 1027/162
X = 1027/3/1621/3
X = 109/5.451 = 183440402.7

2LOGe(3x) + LOGe(18x) = 9
Loge(3x)2 + Loge(18x) = 9 => alog(3) = log(3)a
Loge(162x3) = 9
162x3 = e9
X3 = e9/162
X = e9/3/1621/3
X = e3/5.451 = 3.685

(g)
Given the flowing 2 by 2 matrices A and B

A= [(4&1@1&3)] B= [(-1&4@1&2)]

Calculate A + B, A - B, AB, A-1, B-1, B-1A-1and (AB)-1, AT, BT, (AB)T, BTAT
Comment on the matrices B-1A-1and (AB)-1and on the value of (AB)T and BTAT

A + B = [(4&1@1&3)]+ [(-1&4@1&2)] = [(4-1&1+4@1+1&3+2)] = [(3&5@2&5)]

A - B = [(4&1@1&3)]- [(-1&4@1&2)] = [(4-1&1-4@1-1&3-2)] = [(5&-3@0&1)]

AB = [(4&1@1&3)]* [(-1&4@1&2)] = [((4*-1)+(1*1)&(4*4)+(1*2)@(1* -1)+(3*1)&(1*4)+(3*2) )]
= [(-4+1&16+2@-1+3&4+6)] = [(-3&18@2&10)]

A-1= 1/((4*3-1*1)) [(4&1@1&3)]=1/(12-1) [(3&-1@-1&4)]=1/11 [(3&-1@-1&4)]= [((3/11&-1/11@-1/11&4/11)@)]

B-1= 1/((-1*2-4*1) ) [(2&-4@-1&-1)]=1/(-2-4) [(2&-4@-1&-1)]=1/(-6) [(2&-4@-1&-1)]= [((-2)/6&4/6@1/6&1/6)]
= [((-1/3&2/3@1/6&1/6)@)]

B-1A-1 = `[((-1/3&2/3@1/6&1/6)@)] * [((3/11&-1/11@-1/11&4/11)@)] = [((-1)/3*3/11+ 2/3*(-1)/11&(-1)/3*(-1)/11+ 2/3*4/11@1/6*3/11+ 1/6*(-1)/11&1/6*(-1)/11+ 1/6*4/11)]

= [(-1/11-2/33&1/33+8/33@1/22-1/66&-1/66+2/33)] = [(-5/33&3/11@1/33&1/22)]

(AB)-1= [(-3&18@2&10)]^(-1)= 1/(-30-36) [(10&-18@-2&-3)] = [(-10/66&(-18)/(-66)@(-2)/(-66)&(-3)/(-66))] = [(-5/33&3/11@1/33&1/22)]

AT= [(4&1@1&3)]^T = [(4&1@1&3)]

BT= [(-1&4@1&2)]^T = [(-1&1@4&2)]

(AB)T = [(-3&18@2&10)]^T = [(-3&2@18&10)]

BTAT = [(-1&1@4&2)]* [(4&1@1&3)] = [(-1*4+1*1&-1*1+1*3@4*4+2*1&4*1+2*3)]

= [(-4+1&-1+3@16+2&4+6)]

= [(-3&2@18&10)]

Therefore, we can conclude that matrices B-1A-1= (AB)-1and the matrices (AB)T = BTAT

Task 1 covers the following Learning Outcomes and Assessment Criteria LO1; P2, P7, P8, M1,D1

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In your capacity as maths consultant to the company, you work across several departments and each department have a range of mathematical queries that they wish you to investigate for them. Using appropriate computer software packages, interpret, organise and present relevant data in answer to the following queries.

From the finance department. The finance director comes to you to ask for some assistance in presenting some revenue figures. You are asked to investigate the following data for the company.

 Revenue Number of customers January June Less than 5 27 22 5 and less than 10 38 39 10 and less than 15 40 69 15 and less than 20 22 41 20 and less than 30 13 20 30 and less than 40 4 4

(a) Produce a histogram for each of the distributions scaled such that the area of each rectangle represents frequency density and find the mode.Number of customers - January

The maximum class frequency is 40 and the class interval corresponding to this frequency is 10 - 14. Thus, the modal class is 10 - 14.
Lower limit of the modal class (l) = 10
Size of the class interval (h) = 4
Frequency of the modal class (f1) = 40
Frequency of the class preceding the modal class (f0) = 38
Frequency of the class succeeding the modal class (f2)= 22
Substituting these values in the formula we get;
Mode = l + ((f_1-f_0 ))/(2f_1-f_0-f_2 ) = 10 + (40-38)/(80-38-22) = 10 +2/20 = 10+0.1 =10.1

Number of customers - June

The maximum class frequency is 69 and the class interval corresponding to this frequency is 10 - 14. Thus, the modal class is 10 - 14.
Lower limit of the modal class (l) = 10
Size of the class interval (h) = 4
Frequency of the modal class (f1) = 69
Frequency of the class preceding the modal class (f0) = 39
Frequency of the class succeeding the modal class (f2)= 41
Substituting these values in the formula we get;
Mode = l + ((f_1-f_0 ))/(2f_1-f_0-f_2 ) = 10 + (69-39)/(138-39-41) = 10 +30/58 = 10+0.517 =10.517

Produce a cumulative frequency curve for each of the distributions and find the median, and interquartile range.

 Revenue Number of customers January Cumulative Frequency - January Upper Limit Less than 5 27 27 4 5 and less than 10 38 65 9 10 and less than 15 40 105 14 15 and less than 20 22 127 19 20 and less than 30 13 140 29 30 and less than 40 4 144 39

N = sum of cf = 144, N/2 = 144/2 = 72
Cumulative frequency greater than and nearer to 72 is 105 which belongs to the class interval 10-14
Thus, median class = 10 - 14
l = Lower limit of the median class = 10
N = Sum of frequencies = 144
cf = Cumulative frequency of the class preceding the median class = 65
f = Frequency of median class = 40
h = Class height = 4
Q3 = l + [(3N/4 - cf)/f] × h
= 10 + [(144/2 - 65)/40] × 4
= 10 + [(7/40)] × 4
=10 + 0.7
= 10.7
Therefore, the median of the given distribution is 10.7.

Q1

= sum of cf = 144, N/4= 144/4 = 36
Cumulative frequency greater than and nearer to 36 is 65 which belongs to the class interval 5-9
Thus, first quartile class = 5 - 9
l = Lower limit of the median class = 5
N = Sum of frequencies = 144
cf = Cumulative frequency of the class preceding the median class = 27
f = Frequency of median class = 38
h = Class height = 4
Q1 = l + [(N/4 - cf)/f] × h
= 5 + [(144/4 - 27)/38] × 4
= 5 + [(9 / 38)] × 4
=5 + 0.947
= 5.947

Q3= sum of cf = 144, 3N/4= 3*(144/4)= 3*36 =108

Cumulative frequency greater than and nearer to 108 is 127 which belongs to the class interval 15-19
Thus, third quartile class = 15 - 19
l = Lower limit of the median class = 15
N = Sum of frequencies = 144
cf = Cumulative frequency of the class preceding the median class = 105
f = Frequency of median class = 22
h = Class height = 4
Q3 = l + [(N/4 - cf)/f] × h
= 15 + [(3*144/4 - 105)/22] × 4
= 15 + 12/22
=15 + 0.545
= 15.545
Interquartile range, Q3 - Q1 = 15.545 - 5.947= 9.598

Revenue Number of customers
June Cumulative Frequency - June Upper Limit
Less than 5 22 22 4
5 and less than 10 39 61 9
10 and less than 15 69 130 14
15 and less than 20 41 171 19
20 and less than 30 20 191 29
30 and less than 40 4 195 39

N = sum of cf = 195, N/2 = 195/2 = 97.5
Cumulative frequency greater than and nearer to 97.5 is 130 which belongs to the class interval 10-14
Thus, median class = 10 - 14
l = Lower limit of the median class = 10
N = Sum of frequencies = 195
cf = Cumulative frequency of the class preceding the median class = 61
f = Frequency of median class = 69
h = Class height = 4
Median = l + [(N/2 - cf)/f] × h
= 10 + [(195/2 - 61)/69] × 4
= 10 + 2.116
= 12.116
Therefore, the median of the given distribution is 12.116.

Q1 = sum of cf = 195, 195/4= 48.75
Cumulative frequency greater than and nearer to 48.75 is 61 which belongs to the class interval 5-9
Thus, first quartile class = 5 - 9
l = Lower limit of the median class = 5
N = Sum of frequencies = 195
cf = Cumulative frequency of the class preceding the median class = 22
f = Frequency of median class = 39
h = Class height = 4
Q1 = l + [(N/4 - cf)/f] × h
= 5 + [(195/4 - 22)/39] × 4
=5 + 2.744
=7.744

Q3= sum of cf = 195, 3N/4= 3*(195/4)= 3*48.75 =146.25

Cumulative frequency greater than and nearer to 147 is 171 which belongs to the class interval 15-19
Thus, third quartile class = 15 - 19
l = Lower limit of the median class = 15
N = Sum of frequencies = 195
cf = Cumulative frequency of the class preceding the median class = 130
f = Frequency of median class = 41
h = Class height = 4
Q3 = l + [(3*N/4 - cf)/f] × h
= 15 + [(3*195/4 - 130)/41] × 4
=15 + 1.585
=16.585
Interquartile range, Q3 - Q1 = 16.585- 7.744 = 8.842

For each distribution find the:
• the mean
• the range
• the standard deviation

 Lower limit Upper Limit Frequency Midpoint Frequency * Midpoint Mean Midpoint - Mean (Midpoint - Mean)^2 Frequency * (Midpoint - Mean)^2 0 5 27 2.5 67.5 11.82292 -9.322916667 86.91677517 2346.75293 5 10 38 7.5 285 11.82292 -4.322916667 18.68760851 710.1291233 10 15 40 12.5 500 11.82292 0.677083333 0.45844184 18.33767361 15 20 22 17.5 385 11.82292 5.677083333 32.22927517 709.0440538 20 30 13 25 325 11.82292 13.17708333 173.6355252 2257.261827 30 40 4 35 140 11.82292 23.17708333 537.1771918 2148.708767 Sum 144 1702.5 8190.234375

Mean = ∑(Frequency * Midpoint)/∑Frequency = 1702.5/144 = 11.823
Standard deviation = √(∑(Frequency * (Midpoint - Mean)^2@/( Frequency -1))) = 7.568
Range = 40-0 = 40

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(b)

 Lower limit Upper Limit Frequency Midpoint Frequency * Midpoint Mean Midpoint - Mean (Midpoint - Mean)^2 Frequency * (Midpoint - Mean)^2 0 5 22 2.5 55 13.16667 -10.66666667 113.7777778 2503.111111 5 10 39 7.5 292.5 13.16667 -5.666666667 32.11111111 1252.333333 10 15 69 12.5 862.5 13.16667 -0.666666667 0.444444444 30.66666667 15 20 41 17.5 717.5 13.16667 4.333333333 18.77777778 769.8888889 20 30 20 25 500 13.16667 11.83333333 140.0277778 2800.555556 30 40 4 35 140 13.16667 21.83333333 476.6944444 1906.777778 Sum 195 2567.5 9263.333333

Mean = ∑(Frequency * Midpoint)/∑Frequency = 2567.5/195 = 13.167
Standard deviation = √(∑(Frequency * (Midpoint - Mean)^2@/( Frequency -1))) = 6.91

Range = 40-0 = 40

From procurement, they have a query about light bulbs:

At present the firm is using type A light bulbs which have a lifetime which is normally distributed with mean 360 days and standard deviation of 60 days. At present there are 5000 bulbs in a large building and they are all changed at 282 days
How could the assumption that the bulb life is normally distributed be tested?
Using the Standard normal Z - distribution and centre limit theorem
What is the probability that a single light bulb fails before 282 days?
Mean = 360 and SD = 60 days , n = 5000 and
P(X<282) = P(z < 282 - 360/60) = P(z < -1.3)
P(Z<-1.3) = 96.8% =0.0968

What practical considerations might dictate such a replacement policy?
Faulty bulbs in place

(iv) A single room has 10 light bulbs. What is the probability that more than one bulb has failed after 282 days? (Hint use binomial distribution)

More than one bulb failed = 1-P(more than no one bulb failed)10 - P(only one bulb failed) = 1 - (P(y=0) - P(y=1)) = 1-(1-0.0968)10 - (10C1 * (0.0968) * (1 -0.0968)9) =1 - 0.3613 - 0.3872=0.2515

(v) What is the approximate probability that more than 10% of the 5000 total light bulbs have failed after 282 days, when they are all changed? (Hint use normal approximation of binomial distribution)

Approximate probability , P is calculated using the formula N(np, np(1-p))
Approximate P = N(5000*0.0968, 5000*0.0968*(1-0.0968))
Approximate P = N(484, 437.149)
P(X>500) =P( Z> (500.5 - 484)/ sqrt(437.149))
= P(Z > 16.5/ 20.9081) = P(Z > 0.789) = 1- 0.789 = 0.211

(vi) Type B bulbs which have a lifetime which is normally distributed at 432 days with a standard deviation of 45 days.
At what time would you change these bulbs such that the probability of failure is the same at the probability of failure of type A bulbs at 282 days?

P =0.0932
432 -X/45 = 1.3
432 -X =58.5
X = 432 - 58.5 =373.5 days

(b) If bulb B costs 25% more than bulb A, which is more economic to use all other factors being equal?

Bulb B as it burns more time i.e., 373.5 days compared to 282 days hence value for money.

373.5 - 282 = 91.5 days

Hence we can use additional 91.5 days which would be value for money hence go with type B bulb

(c)
From the human resource section of the company:
They have been looking at the average age of people employed by the company. A simple random sample of 10 people from the employment pool has a mean age of 27 years. Can we conclude that the mean age of all the people employed by the company is not 30 years? The variance of the ages of those employed is known to be 20. Test your chosen hypothesis at a 5% level of significance using both a two tailed test and a one tailed test and explain your conclusions.
Two tailed test
The null hypothesis is that μ = 30. We begin with computing the test statistic.
Sample mean = 27 years
Hypothesis value = 30 years
Standard deviation = sqrt(20)
Sample size =10
Z = (27-30)/(sqrt(20)/sqrt(10)) = 3/sqrt(2) = 2.12

The critical value at 5% test of significance is -1.96 to 1.96

As the value doesn't lie in between we reject the null hypothesis, therefore we can conclude that mean age of all people employed by company is not 30 years

One tailed test

The null hypothesis is that μ ≥ 30. We begin with computing the test statistic.
Sample mean = 27 years
Hypothesis value = 30 years
Standard deviation = sqrt (20)
Sample size =10
Z = (27-30)/(sqrt (20)/sqrt (10)) = 3/sqrt(2) = 2.12

The critical value at 5% test level of significance is -1.64. As test static doesn't lie in the level of significance, we reject the null hypothesis.
Therefore we can conclude that mean age of all people employed by company is not 30 years

Task 2 covers the following Learning Outcomes and Assessment Criteria LO2; P3, P4, M2 and D1

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You are approached by a team from the company who are working on a project to develop a new school and have the following queries.
(a)
The bending moment, M of a beam is given by
M = 3000 - 550x -20 x2
Plot the bending moment and determine where the bending moment is zero

 x M = 3000 -550x-20x^2 -5 5250 -4 4880 -3 4470 -2 4020 -1 3530 0 3000 1 2430 2 1820 3 1170 4 480 5 -250

M = 0
3000 - 550x -20 x2 = 0
20x2 + 550x -3000 = 0
X = ((-550±√([550]^2-4*20*(-3000) )))/(2*20) = ((-550±√([550]^2-4*20*(-3000) )))/(2*20) = ((-550±736.546))/(2*20) = (186.546 )/40 or (-1286.55 )/40
= 4.664 or -32.164

Investigate and state the range of values where the above Bending Moment Function ls maximum or Minimum, decreasing or increasing or neither.
Maximum and minimum at dM/dx = 0
-550 -40x = 0
40x + 550 = 0
X = -550/4
Maximum and minimum at -550/4 and 0.

Let's evaluate f′at each interval to see if it is increasing or decreasing

 Interval x - value F1(x) Observation X < -550/4 -554/4 4990 increasing -550/4 0 1 -590 decreasing

M is increasing when x < -550/4 and decreasing when x>0 and -550/4<x<0

(b)
The temperature θ°C at time t(mins) of a body is given by
θ = 300 +100e-0.1t
Evaluate θ for t = 0, 1, 2 and 5
Θt=0 = 300 + 100e-0.1*0 = 300 +100e0 = 300 + 100*1 = 400
Θt=1 = 300 + 100e-0.1*1 = 300+100e-0.1= 300 + 100*(0.905) = 390.484
Θt=2 = 300 + 100e-0.1*2 = 300+100e-0.2= 300 + 100*(0.819) = 381.873
Θt=5 = 300 + 100e-0.1*5 = 300+100e-0.5= 300 + 100*(0.607) = 360.653

Determine the range of the temperature for positive t
-10e-0.1t = 0
0 to infinity
Determine the rate of change of the temperature at time 2. (Hint find dθ/dt)
dθ/dt = 0 + 100*-0.1*e-0.1*t
At time 2 , dθ/dt = 0 + 100*-0.1*e-0.1*2 = -8.187

(c) The cost of manufacture for a particular component, £C, is related to the production time (t)minutes, by the following formula
C=16t-2+2t
Investigate the variation of cost over a range of production times from 1 minute to 8 minutes
Variation of cost =( 16/8^2 +16 ) - (16/1 + 2) = 16.25 -18 = -1.75
Dc/dt = -32t + 2 = (-32(8)+2)-(-32(1)+2) = -256 +2 +32 -2 = -224
Plot the cost function over the given range

Explain how calculus may be used to find an analytical solution to this problem of optimisation.
Dc/dt = -32t + 2
Use calculus to find the production time at which the cost is at a turning point.
T = 1/16
C = 4096.125
Show that the turning point is a mathematical minimum.
-32(1/16) + 2 = 0

Discuss whether there would still be a minimum cost of production.
No there would not be minimum cost of production

(d) Differentiate the function f(x) = 3sin(2x) + 5loge(x)
Df(x)/dx = 3*cos(2x)*2 + 5/x = 6 cos2x + 5/x
Differentiate the function g(x) = ecos(x) (Hint use function of a function differentiation)
We differentiate it using chain rule

d/dx [f(g(x) ) ] is f^' (g(x) )g'(x) where f(x) = ex and g(x) = cos(x)

For applying the chain rule, set u as cos(x)
d/du[eu] d/dx [cos(x)]

Based on the exponential rule we state that d/du [au] is au ln(a) where a = e

(e^u d)/dx[cos?(x)]

Substitute cos(x) in u

(e^(cos?(x)) d)/dx[cos?(x)]

The derivative of cos(x) with respect to x is -sin(x).

e cos(x) * (-sin(x)) = - e cos(x) sin(x)

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(e) A table top is to be made in the shape of the area between the x axis, the line x = 1.5 and the quadratic y = x2.Calculate the area of the table top. (Hint use polynomial integration)

Shaded area = ∫_0^(1.5)[ydx = ∫_0^(1.5)[x^2 dx=⌈[x^3/3]_0^(1.5) ⌉=[1.5]^3/3-0^3/3= (3.375)/3]] = 1.125

(f) Integrate the following

(i) ∫3/x dx

= 3∫1/x dx = 3 ∫x^(-1) dx = -3 ln x + c = 3 ln |x| + c

(ii) ∫[Sin3θ dθ]

Let u = 3θ

Du = 3dθ

Du/3 = dθ

=∫[Sinu du/3]

= 1/3 ∫[Sinu du]

= 1/3 * (-cos(u))+ c

= 1/3 * (-cos(3θ)) + c

(g) An asymmetric heat conductor within a building is such that the differential of the rate of heat transfer with respect to temperature difference T is e0.03T. T is the difference between the outside and inside temperatures in degrees Celsius That is
dP/dT=e^0.03T
Given that P = 0 when T = 0, determine P as a function of T and plot the function P(T) for values of T ranging from- 20 degrees to + 20 degrees (Hint use integration)
Applying integral on both sides

∫[dP/dT=∫e^(0.03T) ]

P = 1/0.03 e0.03T = 33.33 e0.03T

 T P -20 18.29372 -15 21.25427 -10 24.69394 -5 28.69027 0 33.33333 5 38.72781 10 44.99529 15 52.27707 20 60.73729

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