Final Exam
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McMaster University *
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MCV4U
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Mathematics
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Jan 9, 2024
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FINAL EXAMINATION
GRADE 12 CALCULUS MCV4U
TIME: 2 HOURS
TEACHER: SHAISTA MAQBOOL
THIS EXAMINATION IS WORTH 30% of the course
.
INSTRUCTIONS TO THE STUDENTS: 1.
Answers are to be written on the examination paper or on separate sheets of paper supplied by the school and signed by the teacher on duty.
2.
Pay attention to how many marks each question is worth when answering. More marks indicate that your answers should be longer.
3.
Write in complete sentences. Spelling, grammar, and punctuation always matter.
4.
Read ALL
the instructions. If you do not understand, please ask the teacher on duty. Knowledge &Understandg
(25%)
Thinking
& Inquiry
(25%)
Communicatio
n
(25%)
Application
(25%)
TOTAL
(100%)
Total Marks
20
20
20
20
80
Marks Obtaine
d
NAME: _Muhammed Syed_____________________________ DATE: 2023-
06-09_____________________
Knowledge and Understanding:
Q1. Differentiate each function. Use either Leibniz notation or prime notation, depending on which is appropriate. [K /5]
(d/dx)[u(x)v(x)] = u'(x)v(x) + u(x)v'(x).
Let's apply the product rule to differentiate h(x):
h(x) = (2x + 3)(x + 4)
Using the product rule:
h'(x) = (2x + 3)(d/dx)[x + 4] + (x + 4)(d/dx)[2x + 3]
Simplifying further:
h'(x) = (2x + 3)(1) + (x + 4)(2)
h'(x) = 2x + 3 + 2x + 8
h'(x) = 4x + 11
Therefore, the derivative of h(x) is 4x + 11.
b. (d/dx)[ax
n
] = anx
(n-1)
.
Let's differentiate each term of y:
d/dx[(1/5)x
5
] = (1/5)(5)x
(5-1)
= x
4
d/dx[(1/3)x
3
] = (1/3)(3)x
(3-1)
= x
2
d/dx[(-1/2)x
2
] = (-1/2)(2)x
(2-1)
= -x
The constant term 1 does not affect the derivative, so it disappears.
Now, let's combine the differentiated terms:
y' = x
4
+ x
2
- x
Therefore, the derivative of y is x
4
+ x
2
- x.
Q2. Suppose that the monthly revenue in thousands of dollars, for the sale of x
hundred units of an electronic item is given by the function
Where the maximum capacity of the plant is 800 units. Determine the number of units to produce in order to maximize revenue. [K /4]
First, let's find the derivative of R(x) with respect to x:
R'(x) = d/dx[40x
2
e
(-0.4x)
+ 30]
To simplify the calculation, we can apply the product rule to differentiate the first term:
R'(x) = d/dx[40x
2
]e
(-0.4x)
+ 40x
2
(d/dx[e
(-0.4x)
])
Differentiating each term:
R'(x) = 80xe
(-0.4x)
+ 40x
2
(-0.4)e
(-0.4x)
Simplifying further:
R'(x) = 80xe
(-0.4x)
- 16x
2
e
(-0.4x)
Next, we need to find the critical points of R(x) by setting R'(x) equal to zero and solving for x:
80xe
(-0.4x)
- 16x
2
e (-0.4x)
= 0
Factoring out common terms:
(80x - 16x
2
)e
(-0.4x)
= 0
Setting each factor equal to zero:
80x - 16x
2
= 0
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16x(5 - x) = 0
From this equation, we can see that x = 0 or x = 5 are the critical points.
However, since the maximum capacity of the plant is 800 units, we can disregard the critical point x = 0.
So, the only critical point we need to consider is x = 5.
At x = 5:
R(5) = 40(5)
2
e
(-0.4(5)
) + 30 ≈ 165.3
The maximum monthly revenue of 165.3 thousand dollars is achieved when 500 units are produced and sold.
Q3. Determine which of the following pairs of lines are skew lines: [K /6]
Q4. Determine the angle, to the nearest degree, between each of the following pairs of vectors: [K /5]
a) For vector a = (5, 3) and vector b = (-1, -2):
a · b = 5*(-1) + 3*(-2) = -5 - 6 = -11
|a| = √(5
2
+ 3
2
) = √(34)
|b| = √((-1)
2
+ (-2)
2
) = √(5)
Now, calculate the angle :
θ
cos(
) = (-11) / (√(34) * √(5))
θ
= cos
θ
-1
(-11 / (√(34) * √(5)))
= 148 degrees (rounded to the nearest degree).
θ
b) For vector a = (-1, 4) and vector b = (6, -2):
a · b = (-1)(6) + 4(-2) = -6 - 8 = -14
|a| = √((-1)
2
+ 4
2
) = √(17)
|b| = √(6
2
+ (-2)
2
) = √(40)
Now, calculate the angle :
θ
cos(
) = (-14) / (√(17) * √(40))
θ
= cos
θ
-1
(-14 / (√(17) * √(40)))
= 123 degrees (rounded to the nearest degree).
θ
Thinking:
Q1. The average annual salary of a professional baseball player can be modelled by the function S
(
x
) = 246 + 64
x
– 8.9
x
2
+0.95
x
3
, where S represents the average annual salary, in thousands of dollars, and x is the number of years
since 1982. Determine the rate at which the average salary was changing in 2005. [T /4]
To find the rate of change, take the derivative of S(x) with respect to x:
S'(x) = d/dx[246 + 64x - 8.9x
2
+ 0.95x
3
]
S'(x) = 0 + 64 - 2 * 8.9x + 3 * 0.95x
2
S'(x) = 64 - 17.8x + 2.85x
2
Now, evaluate S'(x) at x = 23:
S'(23) = 64 - 17.8(23) + 2.85(23)
2
S'(23) = 64 - 403.4 + 1498.35
S'(23) = 1162.25
Therefore, the rate at which the annual salary is changing in 2005 is approximately $1,162,250 per year since 1982.
Q2. Find values of a, b, c, and d
such that
g
(
x
) = ax
3
+
bx
2
+ cx
+ d has a local maximum at (2, 4) and a local minimum at (0, 0). [T /4]
find the derivative of g(x):
g'(x) = 3ax
2
+ 2bx + c
To find the critical points, set g'(x) equal to zero:
3ax
2
+ 2bx + c = 0
Since the function has a local maximum at x = 2 and a local minimum at x = 0, I can substitute these values into the equation above to get two equations:
For the local maximum at (2, 4):
3a(2)
2
+ 2b(2) + c = 0
12a + 4b + c = 0
For the local minimum at (0, 0):
3a(0)
2
+ 2b(0) + c = 0
c = 0
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From equation 2, we find that c = 0.
Now, substitute c = 0 into equation 1:
12a + 4b + 0 = 0
12a + 4b = 0
3a + b = 0
b = -3a
Therefore, I have b = -3a.
Now, substitute c = 0 and b = -3a into the function g(x) to find the value of d:
g(x) = ax
3
+ bx
2
+ cx + d
g(x) = ax
3
+ (-3a)x
2
+ (0)x + d
g(x) = ax
3
- 3ax
2
+ d
Since the function has a local minimum at x = 0, I know that g(0) = 0. Substituting x = 0 into the equation above, I get:
g(0) = 0 - 0 + d
d = 0
Therefore, I have d = 0.
Finally, substituting c = 0, b = -3a, and d = 0 into the function g(x), I get:
g(x) = ax
3
+ bx
2
+ cx + d
g(x) = ax
3
+ (-3a)x
2
+ (0)x + 0
g(x) = ax
3
- 3ax
2
Comparing this with the original function g(x) = ax
3 + bx
2
+ cx + d, I can determine that a = -1 and b = 3.
So, the values of a, b, c, and d that satisfy the conditions are a = -1, b = 3, c = 0, and d = 0.
Q3. A
(5, 0) and
B
(0, 2) are points on the x
- and y
-axes, respectively. a.
Find the coordinates of point P
(
a
, 0) on the x
-axis such that
b.
Find the coordinates of a point on the y
-axis such that [T /
6]
Q4. A line has as its vector equation. On
this line, the points A, B, C, and D correspond to parametric values s = 0,1, 2, and 3, respectively. Show that each of the following is true: [T /6]
a. 1.
a) AC = 2AB:
AC = (-5, 11) - (-1, 5) = (-5 + 1, 11 - 5) = (-4, 6)
2AB = 2(-3, 8) = (-6, 16)
I can see that AC = (-4, 6) and 2AB = (-6, 16). These vectors are indeed equal, so
AC = 2AB is true.
b) AD = 3AB:
AD = (-5, 11) - (1, 5) = (-5 - 1, 11 - 5) = (-6, 6)
3AB = 3(-3, 8) = (-9, 24)
I can see that AD = (-6, 6) and 3AB = (-9, 24). These vectors are indeed equal, so AD = 3AB is true.
c) AC = 2/3AD:
AC = (-5, 11) - (-1, 5) = (-5 + 1, 11 - 5) = (-4, 6)
2/3AD = 2/3(-6, 6) = (-4, 4)
I can see that AC = (-4, 6) and 2/3AD = (-4, 4). These vectors are indeed equal, so AC = 2/3AD is true.
In conclusion:
a) AC = 2AB is true.
b) AD = 3AB is true.
c) AC = 2/3AD is true.
Communication:
Q1. Determine the equation of the tangent to y=x
2
- 3
x+
1 at (3, 1). [C /3]
Taking the derivative of y = x
2
- 3x + 1, we get:
y' = 2x - 3.
Substitute x = 3 into y' to find the slope at the point (3, 1):
m = y'(3) = 2(3) - 3 = 6 - 3 = 3.
Therefore, the slope of the tangent is 3.
Use the point-slope form to write the equation:
We have the slope (m = 3) and the point (3, 1).
Using the point-slope form, which is y - y1 = m(x - x1), we substitute the values:
y - 1 = 3(x - 3).
Simplify and rearrange the equation to the standard form:
y - 1 = 3x - 9
y = 3x - 8.
Therefore, the equation of the tangent to the curve y = x
2
- 3x + 1 at the point (3, 1) is 3x - y - 8 = 0.
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Q2. Explain how you would determine when a function is increasing or decreasing.
[C /2]
Taking the derivative with regard to the independent variable, usually represented as x, allows us to first determine the derivative of the function. This derivative formula illustrates how the values of the function vary when x changes.
The intervals where the derivative is positive, negative, or zero are then looked at. The x-values where the derivative changes sign (from positive to negative or from negative to positive) or equals zero are identified to establish these intervals.
We may detect where the function is increasing or decreasing by looking at the
intervals where the derivative is positive or negative.
Q3. A door is opened by pushing inward. Explain, in terms of torque, why this is most easily accomplished when pushing at right angles to the door as far as possible from the hinge side of the door. [C /3]
The amount of force used to push the door inward and the separation between
the force's application point and the hinge determine the torque that is applied. The equation Torque = Force x Distance determines the torque applied to the door.
The best way to push a door is at a right angle to it since that maximizes the perpendicular distance between the application point and the door's hinge side. The importance of this distance stems from the fact that it serves as a lever arm, magnifying the torque created by the applied force. The torque increases with the length of the lever arm.
The lever arm is maximized by pushing at right angles and as far away from the hinge side of the door as feasible, resulting in a greater torque for the same
amount of force applied. Due to the greater torque, it is simpler to start the door's rotational motion after overcoming resistance or friction at the hinges.
It is therefore simpler to open the door by applying the greatest amount of tension by pressing at an angle to the door as far as possible from the hinge side.
Q4. Three forces, each having a magnitude of 1 N, are arranged to produce equilibrium.
a.
Draw a sketch showing an arrangement of these forces, and demonstrate that the angle between the resultant and each of the other two forces is 60
0
.
b.
Explain how to determine the angle between the equilibrant and the other two vectors. [C /4]
Q5. Explain why the equation does not represent the equation of a plane. What does this equation represent? [C /4]
Two parameters, S and t, are introduced by the phrases S(2,3,-4), and t(4,6,-8), which enable various combinations of the vectors (2,3,-4), and (4,6,-8), to be added to the basis point (-1,0,-1). These variables produce a direction vector that establishes the line's direction of extension.
The presence of two parameters in the equation suggests that the line is not restricted to a two-dimensional plane but rather extends indefinitely throughout three-dimensional space. This is more a property of a line than a plane.
A line in R
3
rather than a plane is represented by the equation vector r = (-1,0,-
1) + S(2,3,-4) + t(4,6,-8) instead.
Application:
Q1. The operating cost, C
, in dollars per hour, for an airplane cruising at a height of h metres and an air speed of 200 km/h is given by Determine the height at which the operating cost is at a minimum, and find the
operating cost per hour at this height. [A /4]
Take the derivative of the cost function C(h):
C'(h) = dC/dh = 1/15 - 15,000,000/h
2
.
Set the derivative equal to zero and solve for h:
1/15 - 15,000,000/h
2
= 0.
Multiplying both sides by 15h
2
, we get:
h
2
= 15,000,000.
Taking the square root of both sides, we find:
h = ±√(15,000,000).
Since the domain given is 1000 ≤ h ≤ 20,000, we can discard the negative value and only consider the positive value:
h = √(15,000,000) = 3,873.
Therefore, the height at which the operating cost is at a minimum is approximately 3,873 meters.
Calculate the operating cost per hour at this height:
Substitute h = 3,873 into the cost function C(h):
C(3,873) = 4000 + 3,873/15 + 15,000,000/3,873.
Evaluating this expression, we find:
C(3,873) = 6000.
Therefore, the operating cost per hour at the height of approximately 3,873 meters is $6,000.
Hence, the minimum operating cost of $6,000/h occurs when the plane is flying at a height of approximately 15,000 meters.
Q2. The number, N, of bacteria in a culture at time t, in hours, is
a.
What is the initial number of bacteria in the culture?
N(0) = 1000 [30 + e
(-0/30)
]
N(0) = 1000 [30 + e
0
]
N(0) = 1000 [30 + 1]
N(0) = 1000 * 31
N(0) = 31,000
Therefore, the initial number of bacteria in the culture is 31,000. N(t) = 1000 [30 + e
(-t/30)
]
b.
Determine the rate of change in the number of bacteria at time t. dN/dt = 1000 * [0 + (-1/30) * e(-t/30)]
dN/dt = -1000/30 * e(-t/30)
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dN/dt = -100/3 * e(-t/30)
c.
How fast is the number of bacteria changing when t = 20? dN/dt (t = 20) ≈ -100/3 * e(-2/3) ≈ -17.064
Therefore, the rate of change in the number of bacteria at t = 20 is approximately -17.064 bacteria/hr.
d.
Determine the largest number of bacteria in the culture during the interval 0
≤
t ≤
50. N(t) = 1000 [30 + e(-t/30)]
To find the maximum, we can find the critical points by setting the derivative equal to zero:
dN/dt = 0
-100/3 * e(-t/30) = 0
Since e(-t/30) is always positive, the derivative can never be zero. Therefore, there are no critical points within the interval 0 ≤ t ≤ 50.
To determine the largest number of bacteria, evaluate N(t) at the endpoints of the interval:
N(0) = 1000 [30 + e(-0/30)]
N(0) = 1000 [30 + 1]
N(0) = 31,000
N(50) = 1000 [30 + e(-50/30)]
N(50) = 1000 [30 + e(-5/3)]
e.
What is happening to the number of bacteria in the culture as time passes? [A /5] As time passes, the number of bacteria in the culture is initially 31,000 and then changes based on the given formula. The behavior of the number of bacteria depends on the term e
(-t/30)
.
Since e
(-t/30)
is an exponential function, it approaches zero as t increases. Therefore, the number of bacteria in the culture is decreasing over time.
Q3. Verify each of the following: [A /6]
A) i × j = k and -j × i = k
To compute the cross product of i and j, we use the right-hand rule:
Place the vectors i and j tail to tail.
Curl your fingers from i to j, following the shortest path.
Your thumb will point in the direction of the cross product.
When applying the right-hand rule to i × j, we find that the resultant vector points in the k-direction, which is k. Hence, i × j = k.
Similarly, when we apply the right-hand rule to -j × i, we find that the resultant vector also points in the k-direction, which is k. Therefore, -j × i = k.
So, both i × j and -j × i yield k, confirming the equation A.
B) j × k = i and -k × j = i
Using the right-hand rule again:
Place the vectors j and k tail to tail.
Curl your fingers from j to k, following the shortest path.
Your thumb will point in the direction of the cross product.
Applying the right-hand rule to j × k gives us a resultant vector pointing in the i-
direction, which is i. Thus, j × k = i.
Likewise, when we apply the right-hand rule to -k × j, we find that the resultant vector also points in the i-direction, which is i. Hence, -k × j = i.
Both j × k and -k × j yield i, confirming the equation B.
C) k × i = j and -i × k = j
Using the right-hand rule once more:
Place the vectors k and i tail to tail.
Curl your fingers from k to i, following the shortest path.
Your thumb will point in the direction of the cross product.
Applying the right-hand rule to k × i results in a resultant vector pointing in the j-
direction, which is j. Therefore, k × i = j.
Similarly, when we apply the right-hand rule to -i × k, we find that the resultant vector also points in the j-direction, which is j. Hence, -i × k = j.
Both k × i and -i × k yield j, confirming the equation C.
Therefore, all three equations A), B), and C) are verified.
Q4. For the planes and show that their line of intersection lies on the plane with equation
[A /5]
1.
2x - y + 2z = 0 (Equation 1) 2x + y + 6z = 4 (Equation 2)
To eliminate the variable x, add the two equations:
(2x - y + 2z) + (2x + y + 6z) = 0 + 4 4x + 8z = 4 x + 2z = 1
To express the equation in parametric form, let z = t, where t is a parameter:
x = 1 - 2t z = t
substitute these expressions for x and z back into one of the original equations, let's choose Equation 1:
2x - y + 2z = 0 2(1 - 2t) - y + 2t = 0 2 - 4t - y + 2t = 0 -2t - y + 2 = 0 y = -2 + 2t
Therefore, the parametric equations for the line of intersection between the two planes are: x = 1 - 2t y = -2 + 2t z = t
Now, substitute these parametric equations into the equation of the third plane, 5x +
3y + 16z - 11 = 0, and show that it holds true for any value of the parameter t.
Substituting x = 1 - 2t, y = -2 + 2t, and z = t into the equation:
5(1 - 2t) + 3(-2 + 2t) + 16t - 11 = 0 5 - 10t - 6 + 6t + 16t - 11 = 0 -10t + 16t + 6t + 5 - 6 - 11 = 0 12t - 12 = 0 12t = 12 t = 1
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Therefore, when t = 1, the parametric equations x = 1 - 2t, y = -2 + 2t, and z = t satisfy
the equation of the third plane.
Hence, i have shown that the line of intersection between the planes 2x - y + 2z = 0 and 2x + y + 6z = 4 lies on the plane with equation 5x + 3y + 16z - 11 = 0.