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Unit 9
1- Find the angle in radian corresponding to angle  = 80
Solution :
r =  * /180 
r = 80 * /180 = (4/9) = 1.4 (to nearest 1 d.p.)
2- Find the angle in degree corresponding to angle  r = 3/5
Solution :
 =  r * 180/ 
 = 3/5 * 180/ = = 108
3- Find the length of the circle arc corresponding to angle 90 if the radius of the
circle = 20m
Solution :
When the size of the angle enclosed by the arc = 1 radian  the arc length = r.
When the size of the angle enclosed by the arc = X radian  the arc length = Xr.
Now X = 90 * /180 = /2 = 1.57 (to 2 d.p.)
 Length of the arc = 1.57 * 20 = 31.4m
4- Find the radius of the circle if the length of the arc corresponding to angle 120 =
40m
Solution :
When the size of the angle enclosed by the arc = 1 radian  the arc length = r.
When the size of the angle enclosed by the arc = X radian  the arc length = Xr.
Now X = 120 * /180 = 2/3 = 2.1 (to 1 d.p.)
 Length of the arc = 2.1 * r = 40m  r = 40/2.1 = 19.05m
Unit 10
1. Find the equation of the line passing through the points (-3,5) , (2,-1) , find the
gradient & intercept.
Solution :
m=
Y2 – Y1
X2 – X1
=
-1–5
2+3
=
-6
X+b
5
Substitute for (2,-1) 
-6
7
-1=
X2+bb=
5
5
-6
7
X+
 Y=
5
5
-6
Gradient = m =
Intercept = b =
5
-6
5
Y=ax+b=
7
5
2. Find the equation of the line with gradient = 3 & passing through the point (1,2)
Mu120 – Shatha Habra
Solution :
Y = 3X + b
Substituting for (1,2)  2 = 3X + b  b = -1

Y = 3X -1
3. Find the intersection point of the following 2 lines :
2Y = - X + 1 …….. 1
Y = 3 X + 4 …….. 2
Solution :
Substituting 2 in 1 
2(3X + 4) = - X +1  6 X + 8 = - X + 1
 7X = -7  X = -1
Substituting for x = -1 in equation 2 
Y = (3X -1)+4 = 1
 Intersection point = (-1,1)
Alternatively, you could multiply equation 2 by -2 and add the two equations.
 2Y = - X + 1 ....….. 1
-2Y = -6 X - 8 …….. 2
 0 = -7X – 7  X = -1
 substitute for X=-1 in equation 1  y = 1  intersection point = (-1,1)
4. Find the equation of the line parallel to the line 2y = 3X +10 and has the same yintercept as the line 2y = X - 2
Solution :
Since the line is parallel to the line 2y = 3X +10, it has the same gradient. Now the by
rearranging the equation to keep y alone in the left hand side  y = (3/2)X +5 
gradient = 3/2.
Now we need a point on the line. Since our line has the same y-intercept as the line 2y
= X – 2 and y-intercept is simply the value of y when X=0, so substituting for x = 0 in
the equation 2y = X – 2 gives y = -1.  y-intercept of our line = -1  the equation of
the line is y = (3/2)X – 1 or equivalently 2y = 3X - 2
5. Find the equation of the line parallel to the line 2y = 3X +10 and has the same xintercept as the line 3y = 1 – 2X
Solution :
Since the line is parallel to the line 2y = 3X +10, it has the same gradient. Now the by
rearranging the equation to keep y alone in the left hand side  y = (3/2)X +5 
gradient = 3/2.
Now we need a point on the line. Since our line has the same x-intercept as the line 3y
= 1 – 2X, and the x-intercept is simply the value of x when y = , so substituting for y
= 0 in the equation 3y = 1 – 2X, gives X = 1/2  (1/2,0)  to our line. Now we
should find the y-intercept of our line. Substitute for (1/2,0) in the line y = (3/2)X + c
 c = -3/4  The desired equation is : y = (3/2)X – 3/4
6. Draw the curve represented by the line passing through point (1,2) &
has gradient = 3 ( as in question 2 )
Mu120 – Shatha Habra
Solution :
2
0
1
1-
7- A student studying in a 4-year college has improved his GPA from 3.0 in the first
year to 3.2 , 3.4 , 3.6 in the successive 3 years.
a. Write the algebraic equation that describes the GPA improvement of this
student over the 4 years, taking into account student GPA in the first two
years.
b. State the limitation of your modeling equation.
c. Use line regression to find the equation and evaluate on the fitness of your
regression by commenting on the parameter r (regression correlation
coefficient)
Solution :
a.
Year (n)
GPA
1
3.0
2
3.2
3
3.4
4
3.55
Clearly the GPA improved by a rate ( gradient ) of
(3.2 – 3)/(2-1) = 0.2
 GPA = 0.2 n + GPA o
Where GPA o is the initial GPA, clearly GPA o should be 2.8 you can substitute
for any of the given points above to find it 
GPA = 0.2 n + 2.8
b. Limitation of the model : 1<=n<=4
c. By using line regression LineReg feature :
a = 0.185
b = 2.825
 GPA = 0.185 n + 2.825
The equation is close but not identical to the algebraic equation. Since the parameter r
 0.997 which is very close to 1, the regression is a perfect fit.
Note that the parameter r can be activated by the following step.
1. 2nd 0 which is the catalog.
2. scroll down until you come at the DiagnosticOn line and press enter
3. The DiagnosticOn feature will be pasted on the home screen. Press enter to
confirm its selection.
Mu120 – Shatha Habra
4. Now whenever you do any type of regression, whether line, quadratic,
exponential, power, etc, two additional parameters will appear, r and r2.
5. the value of r reflects the correlation of the given data with the best fit curve.
6. The closer r to 1 or -1 the better the fit is. whenever r gets far from 1 or -1, it
means the fit is a poor one.
7. The following describes the height of water in a tank as it is filled over successive
minites :
Time
15
20
25
30
Water height
100
150
200
250
a- Use your calculator to fit the above data into a line regression equation.
Solution :
Use the following steps :
1- Stat  EDIT
2- Populate L1 with the values : 15,20,25,30
3- Populate L2 with the values : 100,150,200,250
4- Stat  CALC  press 4 ( linReg (aX + b ))
5- Type linReg(aX+b) L1,L2,Y1
L1,L2,Y1 are parameters of the function
6- To obtain Y1 ( for the purpose of drawing ), do the following :
7- VARS  Y-VARS  1  select Y1
Hence the equation of the form y = aX + b
= Y = 10X – 50
b- Find the intercept & the rate at which water level is changing over time
Solution :
Intercept = b = -50
The rate of change = gradient = slope of the line = a = 10
c- Use your calculator to draw the curve represented by the regression equation .
Solution :
Steps :
1- Make sure all the in-memory curves are cleared ( review unit7 in the
calculator book for such methods )
2- Use appropriate window size to specify Xmin, Xmax, Ymin, Ymax
3- Press GRAPH & copy the graph to your sheet
4- As an alternative steps to 3 & 4, you can use ZOOM, 6:ZSTANDARD
8. The following table illustrates the variations of the national demand and supply for
soft fruit:
Price( p / kg) 15 20 25 30 35 40
Demand Quantity ( 109 kg) 30 26 23 19 16 12
Supply Quantity 10 15 20 25 30 35
Use your calculator to find the following:
(a) The regression line for the demand for soft fruit .
Mu120 – Shatha Habra
Solution :
Populate L1 with price data and populate L2 with demand data, you will get
the following regression result:
LinReg
y=ax+b
a=-0.7085714286 (price)
b = 40.48571429 (demand)
 y = -0.7085714286X + 40.48571429
(b) Use your answer in (a) to predict the demand for soft fruit if the price is 45.
Solution :
Demand = -0.7085714286 * 45 + 40.485714298.6 = 8.6* 109 kg
(c) Find the regression line for the supply of soft fruit. )c(
Solution :
Populate L2 with supply data, you don’t need to populate the price data again,
since they are readily available in L1 from the previous question. You will get
the following regression result:
LinReg
y=ax+b
a=1
(price)
b = -5 (supply)
y=X-5
(d) Find the point of equilibrium ( i.e. the price where supply = demand ).
Solution :
There is more than one method to find this point. The following illustrates three
methods:
1- When using line regression, you can paste the output of first equation in Y2
and paste the output of second equation in Y3. Review point a in question 7
above for the calculator steps to do this.
Now using the table facility, search for the value of x when Y2 (corresponding
to demand) = Y3 (corresponding to supply) as the following table shows:
Hence, the equilibrium point is approximately (26.62, 21.62)
2- You can graph the two lines and using the trace feature of the calculator, find
the intersection point.
Mu120 – Shatha Habra
3- You can solve a system of 2 algebraic linear equations by taking the regression
equations above.
9. A furniture company produces two types of sofas. A small sofa whose size is
around 100cm and a big sofa whose size is around 250cm. The company can afford to
produce 30 sofas per month. Moreover produced sofas are likely to be stocked in a
store that has a total capacity of about 50m2.
Given that the profit of the small sofa is £45 and the profit of the large sofa is £70,
what is the optimal number of small and big sofas that the company should produce
monthly.
Solution :
Let X = number of small sofas
Let Y = number of big sofas
The following constraints (equalities) holds
1- X >= 0
2- Y >=0
3- X + 2.5Y <= 50
4- X + Y <=30
By drawing the above equalities in the plane, we get the following shaded feasibility
region:
The optimal solution exists at one of the vertices of the this feasibility region, namely
at the points (0,0) or (0,20) or (30,0) or at the intersection point of the two lines above.
You can find the intersection point by any method you like, foe example you can
graph the lines and use the trace facility, or you can solve the following simultaneous
equations :
Y = -0.4X+20 .....1
Y = -x + 30 ......2
 intersection point = (16.67,13.33)
Now you substitute these points in the following profit function :
P = 20X + 55Y
And the one which will give maximum value for P, will be the desired point.
Point
(0,0)
(0,20)
(30,0)
(16.67,13.33)
P
0
1100
600
1066.55
Mu120 – Shatha Habra
Since P has the largest value at the point (0,20), then the company shall better produce
0 small sofas and 20 big ones.
10. Select the proper answer for the following questions:
1- Which of the following lines intersect with Y = 3X + 2 at the point (0,2)
aY = 5X + 10
b2Y = 6X + 4
c4Y = 2X + 8
dY=X–2
Note : The easiest way to check which other line intersects with Y = 3X + 2 is to
substitute the point (0,2) in each of the given alternative. The one which satisfies the
equation will be the answer. In our case 4 * 2 = 2 * 0 + 8, hence c is the correct
answer.
2- Which of the following lines is parallel to y = 2x -1
a2Y = 4X – 2
b4Y = 8X – 3
cY = 4X – 2
dY = ½X – 1
Note : The line with same gradient as y = 2x -1 is parallel to it. You have to make sure
that Y should be alone in the left hand side of the equation. Choice b is the correct
answer. If you divide both sides of the equation on 4  Y = 2X – ¾.. Note that in
choice a, the line also has gradient 2, but it is just another representation of the
original line y = 2x -1 by multiplying the two sides of the equation by 2.
3- The lines : Y = 2X + 1 , Y = X + 3 intersect in the point
a(2,5)
b(3,5)
c(2,3)
d(3,2)
Note : Here again the easiest way to check which point the two lines intersects in is to
substitute the points in the given choices in the two lines. Clearly, point (2,5) satisfies
both equations. You don’t have to solve simultaneous equations.
4- The equation of the line passing through the point (3,4) and parallel to x-axis is
ay=3
by=4
cx=3
dx=4
Note : Clearly the line parallel to x-axis has the same gradient as the x-axis which is 0.
Therefore the general line equation y = mX + c, will give y = c. Now since point (3,4)
 line  the equation should be y = 4.
5- The equation of the line passing through the point (-3,3) and parallel to y-axis is
ay=3
by = -3
cx = -3
dx=3
Note : Clearly the line parallel to y-axis has infinity gradient. Therefore the general
line equation y = mX + c, will give x = b, where b is the x-intercept. Now since point
(-3,3)  line  the equation should be x = -3.
Unit 11
1- Find the equation of the parabola passing through the point (2,3) & has its vertex
at (-1,-15).
Mu120 – Shatha Habra
Solution :
Y = a ( X – k )2 + L
Y = a ( X + 1 )2 – 15
To find a, substitute for (2,3)
 3 = a (3)2 – 15  a = (3+15)/a = 18/a = 2
 Equation : Y = 2( X + 1)2 – 15
2- Find the equation of the parabola passing through the points (0,5), (1,6.5), (2,12)
a- Without using the calculator
Solution :
Construct the following set of equations that represent the parabola
Y = aX2 + bX + c
5 = c ….. 1 ( from point (0,5)
6.5 = a + b + 5  b = 1.5 – a ……….. from point (1,6.5)
12 = 4a + 2b + 5  4a + 2b = 7 …….. from point (2,12)
 4a + 2 [1.5 – a ] = 7  2a = 4  a = 2
6.5 = 2 + b + 5  b = -0.5
 Equation Y = 2X2 – 0.5X + 5
b- By using your calculator :
Solution :
Follow the steps below :
1- Populate L1 with 0, 1, 2
2- Populate L2 with 5, 6.5, 12
3- STAT  CALC  5(QuadReg)
4- QuadReg L1, L2
5- Write down the equation by filling the values of a, b, c
3. Find the vertex for the following parabolas :
a. Y – 2 = 12 ( X + 5 )2
Solution :
Y = 12 ( X + 5 )2 + 2
 vertex is at ( -5,2 )
b. -102 + 20X – X2
Solution :
-b
b2
,c)
2a
4a
Where a = -1 , b = 20 , c = -102
Vertex = (-20/(2 * -1), -102 – (400/4 * -1))
= ( 10,-2 )
Vertex is at (
4. Solve the following equations using appropriate methods :
Mu120 – Shatha Habra
a. 6X2 – 12 = 0
Solution :
0 = 6X2 – 12  6X2 = 12  X2 = 2  X ± √2
b. 8X2 + 5 = 77
Solution :
8X2 = 77 – 5  8X2 = 72  X = ± √9 = ± 3
c. 5(X+3)2 – 6 = 14
Solution :
(X+3)2 = (14+6)/5 = 4
 X+3 = ± √4  X = -3 ± 2
 X = -5 or X = -1
d. 3X2 – 4X = 7
Solution :
3X2 – 4X -7 = 0  ( 3X – 7 )( X + 1 ) = 0
 either 3X – 7 = 0  X = 7/3
Or X + 1 = 0  X = -1
e. 2X2 + 11X – 21 = 0
Solution :
Determinant = √b2-4ac where a=2 , b=11 , c=-21
= √121+168 = 17
-b ± √b2-4ac
-11 ± 17
=
X=
2a
4
 X = -7 or X = 3/2
5. The position-time graph of a thrown ball is described by :
P(t) = -5t2 + 20t
a. Find the time when the ball will be at the following heights :
i. 20m
ii. 30m
iii. 1m
solution :
you should find the intersection of p(t) with the above lines :
i. y = 20 , y = -5X2 + 20X
 -5X2 + 20X = 20  -5X2 + 20X – 20 = 0
 +X2 – 4X + 4 = 0  ( X – 2 )2 = 0
 X = t = 2 sec
Notice that since you got a single roat (solution ) to this given height, it means
that the point (2,20) is just the vertex of your parabola ( check that please )
ii. y =30 , y = -5X2 + 20
-5X2 + 20X – 30 = 0  X2 – 4X + 6 = 0
Determinant = √b2-4ac = √16-4X6 = √-24
Mu120 – Shatha Habra
Since the determinant is –ve, there is no real solution & hence this means the
ball can't reach this height. Indeed as you found in i , the maximum height
reached is 20
iii. -5X2 + 20X = 10  -5X2 + 20X – 10 = 0
 X2 – 4X + 2 = 0
Determinant = √b2-4ac = √16-4*1*2 = √8
-b ± √8
4 ± √8
√8
=
=2±
X=
2*1
2
2
So either the ball will be at 10m when it is the upward direction at time
√8
2or it will be at that height when it returns downwards at time t =2 +
2
b. Find the maximum height the ball can reach .
Solution :
This is nothing that finding the vertex of the parabola Y = -5t2 + 20t this can
be done in two ways .
Method1:
In Y = -5t2 + 20t, a=-5 , b=20 , c=0
-b
b2
-20
, c)=(
,0 vertex = (
2a
4a
2*-5
= ( 2 , 20 )  maximum height = 20m
202
4*-5
)
Method2:
By using the nderiv function of your calculator you can find when the gradient
( rate of change ) of this function = 0, this represents the point when the
velocity (derivative) = 0 & this is also the point when the ball changed its
direction, hence follow these steps :
1- Press Y= to draw the derivative
2- MATH  press 8
3- Type the function -5t2 + 20t and supply the following parameters
Nderiv ( -5t2 + 20t , t, t, )
4- By observing the drawn curve, which supposed to be a straight line, find an
appropriate TblStart value. You need to find the point of intersection of the
line with the X-axis i.e. you want to find when this gradient is 0. a suitable
Tblstart value could be 1. also use appropriate ∆Tbl value 0.5 is good
enough.
5- Press 2nd Tblset , & provide the values of TblStart & ∆Tbl
6- Find the value of X when Y=0
7- Substitute for this value in the p(t) function .
8- you can use the trace facility instead to find the same result.
c. Draw the p(t) function along with the v(t) & a(t) where v & a are the velocity &
acceleration functions :
Solution :
Mu120 – Shatha Habra
√8
2
P(t)
20
v(t)
2
4
a(t)
2
4
2
4
20
t
t
t
-20
d. Find the gradient of the function at t = 3
Solution :
Nderiv ( -5t2 + 20t , t , 3 ) = -10
e. Find the equation of the tangent ‫ ( مماس‬gradient ) line passing through the point
(3,15)
Solution :
Y = aX + b , a = -10
To find b, substitute for ( 3,15 )
15 = 3X – 10 + b  b = 45  Y = -10X + 45
5- A rocket is propelled vertically up, so that its height in meters after t seconds is
given by : H = -t2 + 7t + 54
a. find the maximum height reached by the rocket .
solution :
the maximum height = y-coordinate of the vertex
the general form of a parabola is aX2 + bX + c & the vertex coordinate is
( - b , c – b2 ) where a = -1 , b = 7 , c = 54
2a
4a
= ( - 7 , 54 – 49 ) = ( 7 , 66.25 )
2*-1
4*-1
2
 maximum height = 66.25m ( notice that the parabola opens downwards )
Another approach is by finding the intersection point of the derivative of H
with the X-axis, as illustrated in previous examples.
Mu120 – Shatha Habra
b. find the height of the rocket at time t = 6 second
solution :
simply substitute t = 6 in the equation of H  - 62 + 7 * 6 + 54 = 60m
c. time in which the rocket reaches to height 36m
solution :
this is simply found by equating the line y=36 with the parabola
 -t2 + 7t + 54 = 36
 -t2 + 7t + 18 = 0  ( -t – 2 ) ( t – 9 ) = 0
 Either t – 9 = 0  t = 9
Or -t – 2 = 0  t = -2 which is not a valid time
d. Velocity of the rocket :
i) at t = 3
ii) when the rocket hit the ground
solution :
i)
nderiv (-t2 + 7t + 54 , t , 3 ) = 1 m/s
ii)
when the rocket hit the ground. Here we need to find the value of t .
the specific thing when the rocket hits the ground is that it its equation H
intersects with the line y=0, so we should solve -t2 + 7t + 54 = 0
by using the determinert √b2 – 4ac = √72 – 4*-1*54 = √265  16.278
 t = -7 ± 16.278
-2
Either t = -7 ± 16.278 = - 4.63 which is not valid
-2
Or
t = -7 ± 16.278 = 11.639
-2
V(t=11.630) = nderiv (-t2 + 7t + 54 , t , 11.639 ) = - 16.28
e. Equation of the velocity ( use part d to answer )
solution :
From part d , we have the velocity at 2 different points ( 3,1 ),( 11.639,-16.28 )
The general form of the equation is y = aX + b
A = m = Y2-Y1 = -16.28 – 1 = -17.28  -2
X2-X1 11.639 – 3
8.639
To find b, substitute (3,1)
 1 = -2 * 3 + b b = 1+ 6 = 7
 Y = -2t + 7
Remark : any polynomial of the form y = aX2 + bX + c can be derived as follows
dy = aX2 x + b & generally Xn is derived as nXn-1 so applying this rule to the
dx
function H  P(t) = -t2 + 7t + 54
V(t) = d(p(t)) = -2 * t + 7 = -2t + 7
dt
f. use e to find the equation of the acceleration a(t) .
solution :
Mu120 – Shatha Habra
clearly the acceleration is the derivative ( gradient / rate of change ) of velocity
V(t), hence it is just the slope of the velocity line y = -2t + 7 & hence the
equation is given by a = -2
5- A rock is dropped into a well that is 200 meters in depth. The position of the rock (
relative to the top of the well ) after t seconds is given by
S = 4.8 t 2
(a) Find the position of the rock after 3 seconds. ( to 2 decimal place )
Solution
After the 3 seconds, the rock will be at 4.8 x 9 = 43.20 meters from the top of the
building  its position from the ground will be
H – S(3) = 200 – 43.20 = 156.80 meters from the ground
(b) find the velocity of the rock after 5 seconds. ( to 2 decimal place )
Solution:
You can take the derivative of S which is simply V(t) = 2 * 4.8*t  V(t) = 9.8*t 
V(5) = 9.8*5 = 49.00 m/s
Or you can simply use the nderiv(4.8 t 2 , t, 5)
(c) Find the time it will take the rock to reach the ground. ( to 2 decimal place )
Solution:
200 = 4.8t2
t=
200
≈ 6.45 seconds
4.8
(d) How fast was the rock traveling when it hits the ground.
Solution:
Substituting into the velocity equation in point b above for t = 6.45, which is the time
at which the rock will hit the ground  V(6.45) = 9.8 * 6.45 = 63.21m/t
7- Select the best answer to the following questions :
1- Which of the following curve is likely to represent the equation -2X2 + 2X + 5
5.5
5.5
a.
b.
½
½
c.
d.
½
5.5
-½
Mu120 – Shatha Habra
Note: Clearly choices a and d should be excluded right away because our parabola
should open downwards. This is clear since the parameter a is negative. Now choices
b and c are both candidates. Now by checking the vertex of the parabola, which is
(1/2, 5.5) we therefore should select choice a. Alternatively if you wish, you could
graph the above graph and use the trace facility to check the coordinates of the point
at the vertex
2- Which of the following curves is likely to represent Y = (X+2)2 + 1
a.
b.
1
1
2-
-2
c.
2
d.
1
1
-2
2
3- Which of the following curves is likely to represent the velocity time graph of
p(t) = t2 + 5t
a.
b.
5
5
5
2
c.
d.
5
-5
2
-5
2
Note : You can graph the nderv( t2 + 5t, t, t) function and find the answer easily. Also
you can easily find the derivative of p(t) which is 2t+5. Since the gradient of this
velocity graph is positive, you should expect that the proper answer is c.
Mu120 – Shatha Habra
4- Given the following velocity-time graph
1.5
-3
which of the following position-time graph is likely to have the above curve as its rate
of change :
a. –X2 – 3X + 8
c. -2X – 3X + 4
b. 2X2 – 3X + 8
d. X2 – 3X + 4
Note : first of all, you should exclude choices a and c, since in both cases the gradient
of the velocity graph will be –ve. Now either b or d will be the answers. When you
find the derivatives of 2X2 – 3X + 8 and X2 – 3X + 4, you will find it 4X-3 and 2X+4
respectively. The proper choice should satisfy the two points (1.5,0) and (0,-3).
Choice d is the only one that satisfy the two points and hence is the proper answer.
5- The curve p(t) = X + 4X + 25 intersects with the line Y = 15 in the points
a. (2,15)
b. They do not intersect
c. (1,15)
d. (2,15) , (-2,15)
Note : b is the proper answer since the determinate (b2-4ac) is < 0
6- The acceleration of an object moving according to p(t) = 2t + 5 is
a. constant
b. – ve
c. + ve
d. no acceleration
Note : This is true since the p(t) in this case is linear.
7- Which of the following three options is true:
a. If an object has zero speed, then it must be stationary.
b. If an object has constant acceleration, then it must have quadratic position-time
relationship
c. If an object is speeding up, then it has positive velocity
d. If an object is speeding up, then it has positive acceleration
e. Velocity is the rate of change of acceleration.
f. Acceleration is the rate of change of velocity
Unit 12
Mu120 – Shatha Habra
1- Write the exponential function describing the following cases :
a. A population of insects that start with 10 and doubles every day .
Solution :
Since the population is doubling  exponential factor = 2, Since we initially
have 10 insects  initial population = 10  p(n) = 10 (2)n
b. The population of students registering in the MU120 course which decreases by 1
tenth every year. Assume the initial # of student registered is 400
Solution :
Since the population is reducing by one tenth  exponential factor = 1/10,
Since we initially have 400 students  initial population = 400  p(n) = 400
(1/10)n
c. The balance of a person who has just inherited £5000 & spends 5% of the
remaining money every month
Solution :
The initial population = £5000
The exponential factor representing the decay of money in the account = ( 1 0.5) = 0.95  p(n) = 5000 (0.95)n
2- For the above examples, find the sum of all populations up to the 10th generation.
Solution :
a. S(n) = a [
bn – 1
b–1
] , b = 2 , a = 10
 S(10) = 10 [
210 – 1
] = 10230
2–1
b. S(10) = 400 [
0.110 – 1
] = 400 * [
0.1 – 1
c. S(10) = 5000 [
(0.95)10 – 1
0.95 – 0.95
1 – 0.110
]  177778
1 – 0.1
]  £ 40126
3- Write down the equation of the following exponential models :
a. initial population 75, doubling time = 5 minutes
solution :
we should put the equation in the form Y=abx a is known, b is missing, to
find it, it is given that the doubling of the population occurs every 5 minutes
 150 = 75 * b5
 b5 = 2  b = 5√2  1.148
 Y = 75 (1.148) x
b. initial population 1000, half life = 30 minutes .
solution :
Mu120 – Shatha Habra
the population halves every 30 minutes
 1000 b30 = 500  b30 = ½
 b = 30√½  0.977
 Y = 1000 (0.977)x
4- Given that the half-life of a radioactive isotope = 20.5 minutes .
a. find the quarter-life time of the isotope
solution :
There are two approaches to solve this problem, the first can be reviewed in the
resource book for block c page 33, question # 7(b). in that approach, they have found
the value of b first & solve for X in N(b)x = N
A simpler approach is as follows :
20.5
= (1/2)N
suppose the initial # of atoms of the radioactive Isotope = N  Nb
 Nb20.5 = ½, For some decay factor b.
for quarter-life time we have
(Nb20.5 )2 =(½)2,
 Nb41 =(1/4),
 quarter-life time = 41 minutes.
b. Find the third-life time of the isotope
solution :
In the second approach, we shall not bother our self finding b, we can solve it as in the
previous point since (b)20.5 = 1/2 , we want to find an exponent that makes right-hand
side = 1/3
(½)x = 1/3  2x = 3  Xlog2 = log3
 X = log3/log2  1.585
Hence [ (b)20.5 ]1.585 = 1/3
 third-life time = 20.5 * 1.585  32.5 minutes
5- given that Ln(10)  2.302 , find log10e
solution :
 Logcu =
logb4
logbc

log10e =
logee
loge10
=
Lne
=
Ln10
1
2.302
6- given the following bird statistics over 10 years
Year
0
1
2
3
4
5
Populatin 515
481
441
390
339
307
 0.434
6
275
7
227
8
195
9
164
a. Use exponential regression facility of your calculator to fit the above data & write
down the equation in the form y = abx
Solution :
follow these steps
Mu120 – Shatha Habra
1- Populate L1 with year # & L2 with population sizes
2- STAT  CALC  press 0 { for ExpReg }
3- a  555 , b  0.88
 y = 555 (0.88)X
b. Use line regression to fit the above data
Solution :
by using the lineReg facility
Y = - 40 X + 513.61
c. Find out when the birds species is going to extinct based on the exponential &
linear models
Solution :
based on exponential, let’s estimate when the population will include just a
single bird
555* (0.88)x = 1  0.88x = 1/555  0.0018
 Xlog100.88 = log100.0018
log100.0018
X=
 49 years
Log100.88
Hence after 49 years ( year 50 ) , it is expected there will be no such birds
Based on the linear model, we shall solve
Y = 1  -40X + 513.61 = 1
 X ≈ 12 years
So the birds will extinct according to the linear model in the 13th year.
7- If the doubling time of a certain population = 5 years, find the growth factor:
Solution:
bx = 2  b5 = 2  5log10b = log102
log102
 log102 =
= 0.06
5
 b = 100.06 = 1.148
8- Find the APR for an interest that is compound 3 times in a year, with a rate of 3%
Solution:
APR = (1.03)3 X 100 – 100 ≈ 9.27%
a. Find the interest rate for an APR = 8% if the interest is to be compound 2 times in
a year
Solution :
APR = 100(1+r)2 – 100
8+100 = 100 (1+r) 2
(1+r) 2 = 1.08
1+r = √1.08
r = 0.039
Interest = 3.9%
9- Suppose you want to invest your money for 5-years in a bank that offers simple and
compound interests. Your initial deposit is £1000 and you shall withdraw your
balance at the end of the 5th year.
Mu120 – Shatha Habra
a. If the simple and compound interest are 5% and 3.5% respectively, which interest
option you should go for
solution :
The balance at the end of the 5th year =
simple interest case:
P (n) = 1000 + n * 0.05 * 1000
P (5) = 1000 + 5 * 0.05 * 1000
= £ 1250
Compound interest case :
P(n) = 1000*(1.035)n
P(s) = 1000*(1.035)5  1187.68
Since the balance is greater in the simple interest case, you'd rather go for it .
b. Suppose you opt to take the compound interest offer, How many years your initial
deposit has to stay in your account, such that it will give you a higher balance than the
simple interest approach
Solution:
we want to find minimum n, that satisfy 1000 (1.035)n > 1000 + 50 n
You can do it by try and error, try n = 6,7, ,…. You will find that n = 21. of course
this way of guessing is tedious. A better approach is to use your calculator to sketch
Y1= 1000(1.033)x. Y2 = 1000 + 50 n. By using the trace facility to find the
intersection point or preferably by using the table facility with appropriate settings,
you can compare between Y1 and Y2 for various values of X and the minimum X that
just give you Y1 > Y2 will be the solution.
9- The bacteria in a laboratory culture grew exponentially from an initial number of
500 to 1500 in 3 hours.
How long will it take for the population to reach 10000?
Mu120 – Shatha Habra
Solution:
a)
the exponentia l model is expressed by an equation of the form
y  a  bx
When x = 0 in the above equation t hen y  a . the initial population in this case is y  500,
So a  500 . Substituti ng this into the above equation gives :
y  500  b x
The population after 3 hours is 1500, Substituti ng these values into the equation gives :
1500  500  b3
1500
3
500
and b  3 3  1.44
So b3 
And the exponentia l model equation is :
y  500 1.44 x
 y  10000
10000  500  1.44 x
10000
 20
500
log 10 (1.44 x )  log 10 (20)
1.44 x 
x log 10 (1.44)  log 10 ( 20)
x 
log 10 ( 20)
 8.2
log 10 (1.44)
 after
8.2 hours the population will reach 10000.
10 - If you deposited an amount of money and it was doubled in 12 years, what is the
annual compound interest rate?
Solution:
Mu120 – Shatha Habra
M  P (1  r ) t
since doubling time is 12 years
 M  2P When x  12
2P  P (1  r )12
2  (1  r )12
log 10 ( 2)  log 10 (1  r )12
log 10 ( 2)  12 log 10 (1  r )
log 10 ( 2)
12
log 10 (1  r )  0.025
log 10 (1  r ) 
10 log1 0 (1 r )  10 0.025
(1  r )  1.06
 the annual compund interest rate  6%
11 - If an initial of BD 10000 earns 8% interest compounded annually.
a) How much will be in the account after 10 years?
b) When the amount of money will double?
c) When the amount of money will triple?
d) If the interest compounded monthly, how much money will be in the account
after 10 years?
Solution:
a)
M  P (1  r ) t
M  10000(1  .08) t
M  10000(1  .08)10
M  10000(1.08)10
M  21589.25
b)
When the amount of money doubling M  2P
2P  P(1  .08) t
2  (1.08) t
log 10 (2)  log 10 (1.08) t
log 10 (2)  t log 10 (1.08)
t
log 10 (2)
 9.01 years.
log 10 (1.08)
Mu120 – Shatha Habra
c)
When the amount of money trip le M  3P
3P  P(1  .08) t
3  (1.08) t
log 10 (3)  log 10 (1.08) t
log 10 (3)  t log 10 (1.08)
t
log 10 (3)
 14.27 years.
log 10 (1.08)
d)
M  P(1  r ) tn
M  10000(1  .08)1012
M  10000(1  .08)120
M  10000(1.08)120
M  102529929.4
12-
Solve the following equations :
(a) log 10 (x) = 2.245
(d) 4 (10) 2x = 28
(b) log 10 (2x + 1) = 0.5
(e) 10 -x/4 = 0.5
(c) log(3x + 1) = log(x + 7)
(f)
4x = 5
Re mark ;
10 log1 0 ( x )  x
log 10 (10 x )  x
Solution:
a)
log 10 ( x)  2.245
10 log1 0 ( x )  10 2.245
x  10 2.245
x  175.79
b)
Mu120 – Shatha Habra
log 10 (2 x  1)  0.5
10log1 0 ( 2 x 1)  100.5
2 x  1  100.5
2 x  100.5  1
100.5  1
x
 1.08
2
c)
log 10 (3x  1)  log 10 ( x  7)
10log1 0 (3 x 1)  10log1 0 ( x  7 )
3x  1  x  7
3x  x  7  1
2x  6
x3
d)
4(10) 2 x  28
28
4
7
(10) 2 x 
(10) 2 x
log 10 (10) 2 x  log 10 (7)
2 x  log 10 (7)
x
log 10 (7)
 0.42
2
e)
10
x
4
 0.5
x
log 10 (10 4 )  log 10 (0.5)
x
)  log 10 (0.5)
4
x  log 10 (0.5)  (4)
(
x  1.20
f)
Mu120 – Shatha Habra
4x  5
log 10 ( 4 x )  log 10 (5)
x log 10 ( 4)  log 10 (5)
x
log 10 (5)
log 10 ( 4)
x  1.16
13 –
Solve the following equations :
(a)
ln(x) = – 0.5
(d)
e 6x = 17.5
(b)
ln (2x + 5) = 3
(e)
e - x /2 = 9
(c)
6 ln(2x) – 12 = 0
(f)
3e (3x + 4) – 21 = 0
Solution:
Re mark;
e ln(x )  x
a)
AND
ln( e x )  x
ln( x)  0.5
eln(x )  e  0.5
x  e  0.5
x  0.61
b)
ln( 2 x  5)  3
e ln(2 x  5)  e3
2 x  5  e3
2 x  e3  5
e3  5
x
 7.54
2
c)
Mu120 – Shatha Habra
6 ln( 2 x )  12  0
6 ln( 2 x )  12
ln( 2 x )  12 6
ln( 2 x )  2
e ln(2 x )  e 2
2 x  e2
x  e 2 2  3.69
d)
e 6 x  17.5
ln( e 6 x )  ln( 17.5)
6 x  ln( 17.5)
x  ln( 17.5) 6
x  0.477
e)
e x 2  9
e x  9  2
e  x  18
ln( e  x )  ln( 18)
 x  ln( 18)
x   ln( 18)  2.89
f)
3e( 3 x  4 )  21  0
3e( 3 x  4 )  21
e ( 3 x  4 )  21 3
e(3 x  4)  7
ln( e ( 3 x  4 ) )  ln( 7)
3 x  4  ln( 7)
3 x  ln( 7)  4
x  (ln( 7)  4) 3
x  0.68
Mu120 – Shatha Habra
14- Select the best answer to the following set of questions:
1. Which of the following growth/decay can be classified as exponential :
a. The population of locust attacking green farm which is found to be 10%
more than its previous generation.
b. A student's GPA which happens to improve from 3.00 to 3.02, 3.04, 3.06
in the successive 4 study years.
c. The height of an object rotating in a circular field with respect to the circle
axis.
d. The position of a car moving along the motorways in the forward direction
with a constant acceleration.
2. If the population of cockroaches is described by p(t) = 2500 e -0.5t, where t is the
time in years, then the population after 10 years will be approximately
a.
15
b.
1
c.
10
d.
8
3. Which of the following options is not equivalent to 2 3/4
4
a.
( 4 √ 2)3
b.
√ 23
1/4 3
c.
(2 )
d.
(1/2) 4/3
e.
(1/2) -3/4
f.
(1/2)4/3
4. If the # of students registered in the IT program is increasing by a rate = 24%
every year, then the total # of students registered in that program after 5 years =
( to the nearest whole number). (Assume an initial # of 200 students in the
program in the 1st year )
a.
381
b.
220
c.
473
d.
586
5. In the previous example, suppose that students withdrawing from this program =
20 students every year, then the total # of students at the end of the 5th year =
a.
400
b.
359
c.
390
d.
312
6. Given that log aX=C and logaY=d, then the option that is equivalent log b (X.Y) is:
a.
C +d
b.
Cxd
c.
d/c
d.
c/d
7. Given that log aX = C and log bY = d, then the option that is equivalent to loga
(X.Y) is :
a.
C +d
b.
C x d log ab
c.
C + d/log ba
d.
C + d log ab
8. Which of the following options is a solution to the exponential equation 10X = 5
a.
0.699
b.
1.43
c.
1
d.
1/2
9. The curve 4bx intersects with the Y-axis in the point
a.
(0,4)
b.
(4,b)
c.
(4,0)
d.
(0,1)
11. Which of the following curves represents the function P(t) = 3.2 t +1
a.
b.
Mu120 – Shatha Habra
3.
4.
C.
d.
4.
4
.
1
1
12. Which of the following options represents the function P(t) = 3.2 –t – 2
a.
b.
. .3
3
-2
.
2
C.
d.
.1
2
.
1
-2
13. Which of the following options represents the following function P(t) = 2 t-2
a.
b.
1
1/4
.
.
.
-2
2
C.
d.
4
.
2
.
Mu120 – Shatha Habra
14. Which of the following options represents P(t) = -2e x
a.
b.
2
.
-2
C.
.
d.
2
.
.
2
Mu120 – Shatha Habra