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Unit 13
1- Suppose that it takes three hours to fill in 2m in a swimming pool of total depth =
14m. Assume that the entire swimming pool will be filled with the same rate.
a. Setup a mathematical model to find out the time it takes the water to reach any
height in the swimming pool by answering the following
1. Choose appropriate variables
2. Make an assumption that leads to a proportionality relationship.
3. Find the constant of proportionality
4. Write down this relationship using the constant of proportionality above.
5. Graph the proportionality relationship.
b. Use the model to estimate how long it will take to fill in
1. 5m of the swimming pool
2. Half of the swimming pool
3. The entire swimming pool
c. What will be the water depth after 6 hours.
Solution:
a. Mathematical model:
1. Let h be the water height in meters and let t be the elapsed time in hours.
2. Assume a constant filling rate of 2m per 3 hours
3. 2/3 m/h
4. h = 2/3 * t
5.
h
3
2
t
(Notice that the true proportional relationship should start from the point (0,0).
That is the y-intercept is zero)
b. Mathematical model:
a. t = 3/2 * h  t = 3/2 * 5  t = 15/2 = 7.5 hrs
b. t = 3/2 * 7  t = 21/2 = 10.5 hrs
c. t = 3/2 * 14  t = 21hrs
d. h = 2/3 * t  h = 2/3 * 6 = 4m
2- If a cake recipe is scaled from a 12-inch diameter round cake tin to a 10-inch
diameter round cake tin, and you want to add almond and icing paste on top of the
cake,
a. find the factor by which the almond and icing paste should scale?
b. Express the proportional relationship between the area A and diameter d?
Describe this relationship? What is the constant of proportionality in this
relationship ?
c. Draw the proportional relationship between the diameter d and the area A.
Solution :
Mu120 – Shatha Habra
a. The amount of icing and almond paste is given by the area of the circle on top of
the cake which is A1 = r2 = (d/2)2 = /4 *d2 . When the diameter is scaled from
12 to 10. That is the scale factor is 12/10, then the area will be A2 = /4*(12/10d)2
= /4*(12/10)2d2 = (12/10)2 * A1, which means the area will be scaled by the
square of this factor, that is (12/10)2.
b. A  d2. This is a quadratic relationship. The constant of proportionality = /4
c.
A
d
3- A cake recipe is prescribed for a cuboid tin with square base of side length = 4 inch
and height = 0.65 of the base side length. If you have instead a tin with base side
length = 3inch and height = 0.65 of the base side length.
a. Describe the relationship between volume and the 4 inch cuboid side
dimension X. What is the constant of proportionality?
b. How would the ingredient of the cake be scaled to fit into the tin you have?
c. Draw the proportional relationship between the tin dimension and its
volume.
Solution :
a.
V = X * X * 0.65X = 0.65X3.  V  X. The constant of proportionality is 0.65
b.
For the first tin, V1 = 0.65X13. Since the cuboid side length is scaled by a factor of
3/4  V2 = 0.65 (3/4X1)3 = (¾)3 * 0.65 * X13 = (¾)3 * V1. Which means the
ingredients of the ingredient of the cake should be scaled by a the cube of the scale
factor of the cuboid side length.
c.
V
X
4- A cake recipe requires a total of 10 hours to be prepared. You would like to prepare
the cake in a less time frame by getting assistant from your friends.
a. Describe the relationship between the time each participant would take in
preparing the cake and the number of participants. What is the constant of
proportionality?
Mu120 – Shatha Habra
b. If the number of participants has doubled, how would the participant time
be affected?
c. Draw the proportional relationship between participant time and the
number of participants
Solution :
a.
Let T be the participant time and let N be the number of participant  T = 10/N. This
is inversely proportional relationship with proportionality factor = 10.
b. Clearly, the time takes will be halved. Because T1 = 10/N1  T2 = 10/2N1 = ½ *
10/N1 = ½ T1
c.
T
N
5- If a square-base tin of side length = 40 cm is used to cook ingredient of a cake
recipe instead of the recommended square-base tin with side length = 35 cm.
a. How would the depth of the cake in the 40 cm be affected?
b. Draw the proportional relationship between cake height and the side length
of the square base.
Solution :
a.
v = hx2
Since the ingredients have not changed, so will be the volume v  h  1/x2 with v
being the constant of proportionality. Now h is scaled from 35 to 40, which means a
scale factor of 40/35 = 4/3.5
Hence if the side of the square base is scaled by a factor of 4/3, the height is scaled
will be scaled by a factor of 1/(4/3.5)2 = (3.5/4)2 which is 0.766 to 3 d.p.
b.
h
x
(Please review table 2 which describes the proportional relationships, power law and
their graph.)
6 - Given the following set of data :
Mu120 – Shatha Habra
X
Y
75
87
70
78
65
69
60
65
55
59
a. Use power regression to fit the above data and find the value of the power to the
nearest decimal point
b. Is the regression accurate? Explain.
Solution :
a.
y = axb a = 0.418
power = b = 1.2 to nearest d.p.
b.
r = 0.9898 which is close to 1. therefore the regression is an accurate one.
7- Given that f(x) = g’’(x) and g(x) = h’’’(x) where f, g, h are all polynomials
(‫)حدوديات‬. If the degree of h is 7,
a. find the degree of polynomial f(x)
b. How many times does f(x) cross the x-axis
c. How many turning points (local minimum and maximum) does polynomial f(x)
has
Solution :
a.
Since g is the third derivative of h and h is of degree 7, then g should be of degree 7-3
= 4. Now f is the second derivative of g, so its degree should be 4-2 = 2.
b.
f(x) crosses the x-axes 2 times
c.
The number of turning points (vertices) = 2-1 = 1
8- Consider the following exponential functions :
a. y = 5 – 3exp(-0.3x)
b. y = 5 + 3exp(-0.3x)
c. y = 3exp(-0.3x)
d. y = 3exp(0.3x)
e. y = 5 – 3exp(0.3x)
f. y = 5 + 3exp(0.3x)
Classify the above functions as :
a. Growth to a limit
b. Decay to nothing
c. Decay to a limit
d. Infinite growth
Solution :
Mu120 – Shatha Habra
You can answer this question by simply drawing the functions
a. y = 5 – 3exp(-0.3x)
5
2
This is growth to a limit
b. y = 5 + 3exp(-0.3x)
8
5
This is decay to a limit
c. y = 3exp(-0.3x)
3
This is decay to nothing (0)
d. y = 3exp(0.3x)
3
This is infinite growth
Mu120 – Shatha Habra
e. y = 5 – 3exp(0.3x)
5
2
This is decay to a limit
f. y = 5 + 3exp(0.3x)
8
5
This is decay to a limit
9- Consider the following proportional relationships :
W  r3
A  r2
a. Find the proportional relationship between W and A
b. From your answer to a, find the scale factor by which A should change if
W is doubled
Solution :
a.
Since W  r3  W1/3  r  W2/3  r2 . Now since A  r2  W2/3  A
b.
W12/3  A1, If W2 = 2W1, then A2 = (2W1)2/3 = 22/3 (W1)2/3 = 22/3 A1. This
means A will be scaled by a factor of 3 4
10- Save following data in L1 and L2 in your calculator:
Mu120 – Shatha Habra
X
Y
2
8
3
25
4
50
5
84
6
130
a. Use calculator to plot scatter plot and sketch the same in box 1 given below.
b. Use calculator to find best fit linear regression function and plot it. Sketch in
box2
c. Use calculator to find best fit exponential regression function and plot it.
Sketch in box3
d. Use calculator to find best fit power regression function and plot it. Sketch in
box4
e. Now observe your regression graphs and determine the most appropriate
regression for this data
f. Use calculator to find coefficient of regression ( r ) in the three cases and
determine, on the basis of the value of r
Solution :
a, b, c, d, e.
Y
Power Regression
*
*
Line Regression
Exponential Regression
*
*
*
X
It clear from the figure that the most appropriate regression is the power regression,
because almost all the points in the scatter plot are laying on the power regression
graph.
(consult the calculator book for the steps to draw the scatter plot)
f.
The coefficient of variation r is as follows :
a. Linear Regression
r = 0.9828
b. Exponential Regression r = 0.9797
c. Power regression r = 0.9990
Clearly, since r is closest to 1 in case of power regression, this means power
regression is the best fit for the given data
Mu120 – Shatha Habra
Unit 15
Summary :
1. A sinusoidal wave of the form y = Asin(2/T)t is characterized by the following
two parameters :
a. Amplitude A : The maximum height above or below the central line. So if yu
know the minimum and maximum data you have
b. Period T : The time in seconds to complete one cycle
2. The frequency f if the number of completed cycles in time unit and is measured in
hertz  f = 1/T. That is T and F are inversely proportional to each other.
3. The angular frequency  (speed or velocity) is given by  = 2/T = 2f.
Therefore the sinusoidal wave can be put as y = Asint
4. The values of y in this case will be in the range [-A,A]
5. Before you can draw the sin curve, you should change the mode of your calculator
to radian
6. A typical sinusoidal wave starting from value 0 will reach its amplitude for the
first time at the quarter of the period, that is at T/4.
7. The general form of the sinusoidal wave is y = M + Asin((2/T)t + ), where M is
a vertical shift and  is a horizontal shift which is called phase shift.
8. The sine and cosine functions are shifted version of each other. The following
identities hold:
i. cos = sin(+/2)
sine is shifted left
ii. cos = -sin(-/2) OR sin(-/2) = -cos
sine is shifted right
iii. sin = cos(-/2)
cosine is shifted right
iv. sin = -cos(+/2) OR cos(-/2) = -sin
cosine is shift left
9. The above identities can be obtained either from applying right or left phase shifts
or by applying the following identities:
i. Sin() = sincos  cossin
ii. cos() = coscos  sinsin
For example, if you want to investigate what cos(-/2) is equivalent to, then you
should apply the following :
cos(-/2) = coscos/2 + sinsin/2
= cos * 0 + sin * 1
= sin
10. The following identities are also true:
i. sin = sin(  n) where n is an even number
ii. cos = cos(  n) where n is an even number
(This means if you add or subtract full cycles (multiple of 2) then the sine
and cosine do not change)
iii. sin = -sin(  n) where n is an odd number
iv. cos = -cos(  n) where n is an odd number
(This means if you add or subtract half cycles (multiple of ) then the sine
and cosine get reversed)
Mu120 – Shatha Habra
11. If you add two sinusoidal waves of the form y1 = Asinat, y2 = Bsinbt the result
will also be a sinusoidal wave whose amplitude has a maximum height of A+B
and whose period = 2/(greatest common factor)
For example consider the following set of sinusoidal waves:
1- y1 = 2sin3t, y2 = 4sin6t
If y = y1 + y2 then the maximum amplitude of y = 6 and its period
= 2/3
2- y1 = 2sin3t, y2 = 4sin2t
If y = y1 + y2 then the maximum amplitude of y = 6 and its period
= 2/1 = 2 (If there is no common factor then the period is simply
2)
ab
ab
)sin(
)
2
2
sin (Y+X) + sin (Y-X) = 2cos(X)sin(Y)
Fourier series is a summation of infinite number of sinusoidal terms and is used
to approximate any random wave. The frequency in the first term in the series is
called the fundamental frequency, which is the same as the original frequency. It
is also known as the first harmonic. The amplitude of the first harmonic is the
same as that of the original wave.
The second term in the Fourier series is called the third harmonic because the
frequency of the wave will be 3 times the fundamental frequency and its
amplitude will by 1/3 that of the fundamental frequency
Therefore Fourier series is given by :
y = sin(2f t) + 1/3sin(2*3f t) + 1/5sin(2*5f t) + ……Or equivalently
y = sin(t) + 1/3sin(3t) + 1/5sin(5t) + ……
12 sin a + sin b = 2 cos(
13
14
15
16
Questions
1
Two pure tones of the same amplitude but with frequencies 16Hz and 18Hz
respectively are sounded together to produce beats. Each tone is modeled as a sine
wave. Which two options model the resulting waveform?
a. sin 16t + sin18t
b. sin 4t + sin 36t
c. sin 2t + sin 34t
d. sin 32t + sin 36t
e. 2cos(16t)sin(18t)
f. 2cos(4t)sin(32t)
g. 2cos(2t)sin(34t)
h. 2cos(32t)sin(36t)
Solution :
Since the sum of the two sinusoidal waves is also a sin wave then the sum will be of
the form:
sin 2f1t + sin 2f2t
where f1 and f2 are the frequencies of the individual sine
waves. By substituting we get
sin 2*16t + sin 2*18t = sin 32t + sin 36t
Hence answer d is correct
ab
ab
Now since sin a + sin b = 2 cos(
)sin(
)  sin 32t + sin 36t
2
2
32  36
32  36
= 2 cos(
)sin(
)
2
2
Mu120 – Shatha Habra
= 2cos2 sin34 Hence answer g is also correct
( Note : cos x = cos (–x) )
2 Which of the following identities are correct :
a. sin x = sin (x + 5)
b. cos (x + /2) = -cos x
c. sin x = -sin (-x)
d. cos x = -sin (x-/2)
e. sin (x + ) = cos (x+3/2)
f. cos (x + 4) = cos (x+2)
g. sin (-x) = sin x
h. cos(-x) = -cos x
Solution :
You can easily find exclude many of the above assertions by using substitution, but
be careful that you might select a value of x which accidentally satisfy some of the
above assertions. now
a. sin x = sin (x + 5)
is incorrect, check x = /2. Remember also that
when you add multiple of half periods the sign of the sine function will be
reversed.
b. cos (x + /2) = -cos x
is incorrect, check x = 0. Remember also that the
sign of the cosine function will be reversed when adding multiples of half
periods not quarter periods.
c. sin x = -sin (-x)
correct since sin (-x) = - sin x  sin x = -sin (-x)
d. cos x = -sin (x-/2) correct, check the identities in the summary above
e. sin (x + ) = cos (x+3/2) incorrect, check x = /2
f. cos (x + 4) = cos (x+2) correct because the cosine remains the same if
you add any multiple of full periods that is cos (x + 4) = cos (x+2) =
cos x
g. sin (-x) = sin x
incorrect, check x = /2
h. cos(-x) = -cos x
incorrect. cos x = cos (-x) : The cosine is positive in
the first and forth quadrant.
3 Write down the equation of the sin wave that has an amplitude of 20 and a period
of 0.5 and has a phase shift of 60
Solution :
y = 20sin(2/T t + )
 = 60 * /180 = /3
y = 20sin(2/0.5 t +/3)
y = 20sin(4t +/3)
4 Given the following set of data representing weeks and sunrise times:
Week
0
4
8
12
Sunrise
08.46
08.09
07.05
05.52
Mu120 – Shatha Habra
16
20
24
28
32
36
40
44
48
52
04.42
03.48
03.31
04.01
04.53
05.47
06.42
07.41
08.33
08.46
Use sine regression to find the best fit curve for the above data.
Solution:
By using the calculator sinreg feature the wave is characterized by:
a = 2.56
b = 0.115
c =1.895
d =6.22
Approximating to 2 d.p 
y = 2.6sin(0.12x + 1.9) + 6.2
5 Using the above sinusoidal wave, re represent the wave using cosine instead of
sine by adding a left shift of /2
Solution :
We need to use an identity where the cosine is shifted to the left by an amount of /2.
The suitable one is
sin = -cos(+/2)
(note the left hand side is the sine function because we want
to find its equivalent in terms of cosine)
y = 2.6sin(0.1n2x + 1.9) + 6.2
y = -2.6cos(0.12x + 1.9 +/2) + 6.2
y = -2.6cos(0.12x + 1.9 +/2) + 6.2
y = -2.6cos(0.12x + 3.47) + 6.2
6- Given a square wave of frequency = 2 Hz,
a. What are the frequencies of the fundamental and the third, fifth and
seventh harmonic components of the wave
b. Write down the Fourier series in terms of 
Solution :
a. The fundamental frequency = 2Hz
The frequency of the third harmonic component = 3 * 2 = 6hz
The frequency of the fifth harmonic component = 5 * 2 = 10hz
The frequency of the fifth harmonic component = 7* 2 = 14hz
b. y = sin(t) + 1/3sin(3t) + 1/5sin(5t) + ……, where  = 2f = 4
Mu120 – Shatha Habra