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course Mth 173
9/30 15:00
Text_06PROBLEM 1-11
Given function P = P0e^(0.2t) which can also be written as P = P0*(e^0.2)^t
Comparing it with the standard function of P = P0(a^t)
We get a = e^0.2 = 1.2214
Thus the function can be written as P = P0*( 1.2214 ^ t )
Since the base a = 1.2214 is greater than 1 thus it represents exponential growth.
PROBLEM 1-15
a. We know that initial population P0 = 1000 people
Increase in population every year = 50 people
Let P be the population at any time t, where t is the number of years for which the population is to be increased. Thus we have P = P0 + 50*t
Which is P = 1000 + 50*t , where t is the number of years
b. Initial population Po = 1000 people
Rate at which the population increases = 5%
Thus we have growth rate = 5 / 100 = 0.05 and a growth factor of = 1 + 0.05 = 1.05
Let P be the population at any time t after the initial year of population
Thus the function obtained is P = P0*( growth factor ^ t )
Thus the function is P = 1000 * ( 1.05 ^ t )
PROBLEM 1-18
a) The average temperature remains the same over day 1 and 2 interval and day 9 and 10 interval
Increases in the day 5 and 6 interval, day 6 and 7 interval, day 7 and 8 interval, day 8 and 9 interval or we can say that the average temperature was increasing over the day 5 to 9 interval
Decreases in the day 2 and 3 interval, day 3 and 4 interval, day 4 and 5 interval or we can say that the average temperature was decreasing over the day 2 to 5 interval.
b) the pair of consecutive intervals over which the average temperature increases at a decreasing rate is the interval of day 5 and 6 and the interval of day 6 and 7.
The pair of consecutive intervals over which the average temperature increases at an increasing rate is the interval of day 7 and 8 and the interval of day 8 and 9.
PROBLEM 1 - 19
(a)
The values of the function f(x) are increasing by 5, 3, 1 and -2. Since the increase are not the same for consecutive values of x, thus it does not represent a linear function.
@&
Good.
Additional note:
The increases, however, are changing at a constant rate so the rate of increase is a linear function. This implies that the function itself is an integral of a linear function, and is therefore quadratic.
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The values of the function g(x) are increasing by 8, 12, 18 and 27. Since the increase are not the same for consecutive values of x, thus it also does not represent a linear function.
The values of the function h(x) are decreasing by 3, 3, 3 and 3. Since the decrease are same for consecutive values of x, thus it represents a linear function.
Now for h(x) function, the formula for the function can be found out by using 2 points which are (-2, 37) and (-1, 34). The run for the straight line joining the 2 points is -1 - (-2) = 1
The rise for the straight line = 34 - 37 = -3
The slope of the straight line = rise / run = -3 / 1 = -3
The formula of the line thus will be (y - 37) / (x + 2) = -3
y - 37 = -3x - 6
Thus the formula is y = -3x + 31
(b)
We have seen that h(x) is a straight line thus it cannot be exponential and f(x) increases for some values of x and decreases for some thus it too cannot be an exponential function.
For the g(x) function the ratio of consecutive values of y are
g(-1) / g(-2) = 24 / 16 = 1.5
g(0) / g(-1) = 36 / 24 = 1.5
g(1) / g(0) = 54 / 36 = 1.5
g(2) / g(1) = 81 / 54 = 1.5
Since the ratios of consecutive y value is the same thus the function g(x) is an exponential function.
We see that the value of the function at x = 0 is 36 , that is the initial quantity.
Each quantity increases by a factor of 1.5, thus the growth factor of the exponential function is 1.5.
Thus the function become g(x) = g(0)*( growth factor ^ x )
g(x) = 36*(1.5^x)0
PROBLEM 1-22
a. The world’s population in 2010 = 6.91 billions
Rate = 1.1%
Growth rate = 0.011
Growth factor = 1 + Growth Rate = 1.011
Let t be the number of years since 2010
Let P be the population at t number of years since 2010
Thus P = initial population * ( growth factor ^ t )
P = 6.91 * ( 1.011^t ) billions
b. P for the year 2020 will be calculated as
t = number of year since 2010 = 2020 - 2010 = 10 years
thus P for t = 10 years will be P = 6.91 * ( 1.011 ^ 10 ) = 6.91 * 1.1156 = 7.7087 billion
c. the population as a function of years graph is plotted with an interval of 10 years on the x axis and an interval of 1 billion on the y axis. The points plotted are ( 0 , 6.91 ) , ( 10 , 7.71 ) , ( 20 , 8.6 ) , ( 30 , 9.6) , ( 40 , 10.70) , ( 50 , 11.94 ) , ( 60 , 13.32 ) and ( 70 , 14.86 ) where 0 , 10 , 20 ….70 are years since 2010 and 6.91 , 7.71 , 8.6 ……….. 14.86 is the population in billion corresponding to the respective number of years since 2010. The initial population is 6.91 billion. Thus the doubling population will be 6.91 * 2 = 13.82 billion. In order to calculate the number of years required to reach this population we first need to locate the point 13.82 on the y axis and then draw a line parallel to x axis to meet the curve. From the point on the curve at which the line meets the graph draw another line parallel to y axis to meet the x axis at a point. The point of intersection of the line and the x axis will give us the number of years since 2010 when the initial population would double itself. We can see that, the point of intersection is between ( 60 , 0 ) and ( 70 , 0 ) but more closer to ( 60 , 0 ) thus it would be between ( 60 , 0 ) and ( 65 , 0 ). According to my scale the estimated point of intersection would be ( 63.5 , 0 ) thus the estimated doubling time would be 63.5 years, which means that it would take 63.5 years since 2010 to reach a population of 13.81 billion from 6.91 billion.
PROBLEM 1-24
Initial population = P(0)_= 10,000,000
Rate = 2%
Growth rate = 0.02
Growth factor = 1 + 0.02 = 1.02
The double population will thus be P(doubling time) = P(0) * 2 = 20,000,000
Let doubling time be t2
Thus P(t2) = P(0) * ( growth factor ^ t2 )
20,000,000 = 10,000,000 * ( 1.02 ^ t2 )
2 = 1.02 ^ t
t = ln(2) / ln(1.02) = 35.00278 years
This is done with the pre knowledge of logarithms. Now instead using the guessing and checking rule we do it as
1.02 ^ 1 = 1.02
1.02 ^ 5 = 1.104
1.02^10 = 1.2189
1.02^15 = 1.3458
1.02^25 = 1.6406
1.02^30 = 1.811
1.02^35 = 1.9998
This value is too close to 2 thus
1.02^ 36 = 2.03988
Thus the value of t is less than 36 and greater than 35. Since the value 2 is much close to 1.9998 , thus the value of t must too be too close to 35 , so approximately we can say that t must be something like 35.0
PROBLEM 1-28
The graph (a) must be a straight line, thus the function must have values increasing by the same value for consecutive values of x. the function g(t) has values increasing by 1, 2 , 3 , 4 , 5 thus not a straight line. For function h(t) has values increasing by 10 , 9 , 8 , 7 , 6 thus not a straight line. For function k(t) has values increasing by 0.3 , 0.3 , 0.3 , 0.3 , 0.3 thus it represents the straight line.
Graph (b) is an increasing graph with decreasing rate. We can see that the difference between consecutive values of y in the function h(t) decreases constantly and the values increase. Thus (b) mus be h(t)
Graph (c) is an increasing graph with increasing rate. We can see that the difference between consecutive values of g(t) is increasing and the values itself are increasing. Thus (c) must be g(t).
Thus summarizing we obtain
(a) - k(t)
(b) - h(t)
(C) - g(t)
PROBLEM 1-31
The graph looks like an exponential graph. Passing through the points are ( 0 , 3 ) and ( 2 , 12 )
Since for x = 0 it passes through 3, thus initial quantity( value ) is 3.
Thus for a probable exponential function would be y = 3 * ( growth rate ^ x )
In order to find the growth rate we will use the other point which is ( 2 , 12 )
Substituting the point we get
12 = 3 * ( growth rate ^ 2 )
4 = growth rate ^ 2
Growth rate = 2
Thus the possible formula of the function would be y = 3 * ( 2 ^ x )
PROBLEM 1-32
The function possibly will be a decreasing exponential function.
Thus the function may be y = a * ( b ^ x ) where y is the function of x , a is the value at x = 0, and b is the base of the exponential function or the growth factor by which the value of y changes. Since the graph is decreasing we know that b is less than 1 and greater than 0.
Next the points ( -1 , 8 ) and ( 1 , 2 ) satisfy the function. Thus the 2 equations obtained on substituting the points in the equation is
8 = a * ( b ^ -1 ) = a / b and 2 = a * ( b ^ 1 ) = a * b
Dividing both the equation we eliminate a and obtain an expression for b, which is
4 = 1 / (b^2)
b^2 = 1 / 4
thus b = 1 / 2 = 0.5
substituting the value of b in any equation we obtain the value of a
thus we have 8 = a * ( 0.5 ^ -1 ) = a * 2
thus a = 4
thus the function obtained it y = 4 * ( 0.5 ^ x )
this is a possible estimated result, assuming that the graph is exponential. The actual function may vary.
PROBLEM 1-37
(a)
From the graph we clearly see that for a graph of population vs. time(years) is an exponential graph.
And the point ( 6 , 40,000 ) is clearly possible which means that after 6 years the total population is 40,000.
Now we also know that doubling time required for a quantity is independent of the initial value. We can approximate the point (3.75, 20,000) and the point (8.25, 80,000).
Doubling time is defined as the time required for the quantity to double its initial value. Let us assume the initial value is 20,000. The population reaches the population of 20,000 at time t = 3.75 years. The doubling value of this population is 40,000 and the population reaches this value at t = 6 years( as inferred from the graph ). Thus the doubling time = time required to reach 40,000 - time required to reach 20,000 = 6 years - 3.75 years = 2.25 years.
Next let us assume that 40,000 is the initial population. thus the doubling population is 80,000 and the time required to reach this population is 6 years and 8.25 years respectively. Thus the doubling time = 8.25 years - 6 years = 2.25 years.
Thus proved that the doubling time of a quantity is independent of the initial value chosen.
(b)
The equation for the graph is P = P0 * (a ^ t )
Let P be the population at any time t
Let the population double itself at time t + d, where d is the doubling time let P1 be the double population
Thus P1 = P0 * ( a ^ ( t + d )) = P0 * ( (a^t ) * (a^d) )
And P1 = 2 * P
Thus P0 * ( ( a^t )* (a^d) ) = 2 * P0 * ( a ^ t ) , further simplifying we get
a ^ d = 2
d = ln(2) / ln(a)
thus we see that the doubling time is independent of t which is the initial time or basically independent of the initial value, and is only dependent on a ( the growth factor )
PROBLEM 1-38
(a) initial deposit = P0
Doubling time = 50 years
Thus after 50 years the deposit would become P0 * 2 = 2P0
After another 50 years that is a total of 100 years the total would become double of the value after 50 years = 2 * 2P0 = 4 * P0
And lastly after 150 years the total deposit = 2 * deposit at 100 years = 2 * 4 * P0 = 8 * P0
(b) The number of times the quantity doubles itself in t years = t years / doubling time. This is because after every span of doubling years the quantity will double itself. Thus for this function the number of times the quantity double itself in t years = t years / 50 years
P is the amount of deposit in the bank account. Since is doubles itself after every 50 years and the number of time it double is t / 50, thus the factor by which the quantity would change would be 2 ^ ( t / 50 ) . thus the population at the end of t years would be P = P0 * 2 ^ ( t / 50 )
PROBLEM 1-40
(a)
The initial quantity of radium-226 = Q0
Half life = 1620 years
The quantity of radium at any time t = Q
The number of times the quantity reduces to half its value = t / half life = t / 1620
Thus the factor by which the quantity reduces in t years = ( ˝ ) ^ ( t/1620 )
Thus the quantity of radium at any time Q would be
Q = Q0 * [ ( ˝ ) ^ ( t/1620 ) ]
(b)
The quantity of radium after 500 years that is t = 500
Q = Q0 * [(˝) ^ (500/1620)]
Q = Q0 * [0.5 ^ 0.3086]
Q = Q0 * 0.8074
The quantity left after 500 years thus would be 0.8074*Q0
The percentage of original amount of radium left after 500 years = [(0.8074*Q0)/Q0] * 100 = 80.74 %
Thus the percentage of the original amount of radium left after 500 years would be 80.74 %
PROBLEM 1-43
(a) The quantity of biodiesel in 2005 = 91 million
Thus for quantity of biodiesel in 2006 we notice that from 2005 to 2006 the percentage change in quantity of biodiesel = 186.6 %
Thus the change of biodiesel from the previous year = (186.6 / 100) * 91 million = 169.806 million
Thus the quantity of biodiesel in 2006 = quantity of biodiesel in 2005 + change in biodiesel = 91 million + 169.806 million = 260.806 million
Now for quantity of biodiesel in 2007 we notice that from 2006 to 2007 the percentage change in biodiesel = 37.2%
Thus the change of biodiesel from the previous year = (37.2 / 100) * 260.806 million = 97.0198 million
Thus the quantity of biodiesel in 2007 = quantity of biodiesel in 2006 + change in biodiesel = 260.806 million + 97.0198 million = 357.825 million
Thus the quantity of biodiesel in 2006 = 260.806 million
And quantity of biodiesel in 2007 = 357.825 million
(b)
The graph of the function would be an uneven graph. The graph would have a x- axis of years and y axis of quantity of biodiesel in million. The points plotted are (2005 , 91) , (2006 , 260.806) , (2007 , 357.825 ) , ( 2008 , 315.959) and (2009 , 339.024). The graph would be a straight line joining the points 2005 and 2006 with a positive slope of 169.806, a straight line between 2006 and 2007 with a positive slope of 97.019 , a straight line between 2007 and 2008 with a negative slope of -41.866 and a straight line between 2008 and 2009 with a positive slope of 23.065.
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Self-critique (if necessary):
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Self-critique rating:
@&
Very well done. I did insert one note.
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