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# mcv4u assi#1

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MCV4U UNIT 1 Rates of Change Assessment Opportunity
Calculus And Vector (Best notes for high school - EN-CA)
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MCV4U UNIT 1: Rates of Change Assessment Opportunity
Instructions:
There are 10 questions in this Assessment Opportunity for a total of 60 marks. Show all work for each
question.
3
2
1. Determine the average rate of change of y in the function π¦ = 2π₯ + 7π₯ + 2π₯ − 3 over the
interval [3,5]. (5 marks)
Given:
3
2
π¦ = 2π₯ + 7π₯ + 2π₯ − 3
x=3, x=5
3
2
ππ π₯ = 3, π¦ = 2(3) + 7(3) + 2(3) − 3
π¦ = 2(27) + 7(9) + 6 − 3
π¦ = 54 + 63 + 6 − 3
π¦ = 120
Point (3, 120)
3
2
ππ π₯ = 5, π¦ = 2(5) + 7(5) + 2(5) − 3
π¦ = 2(125) + 7(25) + 10 − 3
π¦ = 250 + 175 + 7
π¦ = 432
Point (5, 432)
Average Rate of Change:
432−120
5−3
=
312
2
π(π)−π(π)
π−π
= 156
3
2
∴ The average rate of change of y in the function π¦ = 2π₯ + 7π₯ + 2π₯ − 3 over the interval
[3,5] is 156.
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2
2. Given the function π(π₯) = 2π₯ + 3π₯ + 1,
a. Use the secant method to ο¬nd the instantaneous rate of change when x = 1. (4 marks)
Given:
2
π(π₯) = 2π₯ + 3π₯ + 1
x= 1
Selected series of points close to x=1: 1.1, 1.01, 1.001
2
ππ π₯ = 1, π(1) = 2(1) + 3(1) + 1
π(1) = 2 + 3 + 1
π(1) = 6
Point: (1, 6)
2
ππ π₯ = 1. 1, π(1. 1) = 2(1. 1) + 3(1. 1) + 1
π(1. 1) = 2. 42 + 3. 3 + 1
π(1. 1) = 6. 72
Point: (1.1, 6.72)
2
ππ π₯ = 1. 01, π(1. 01) = 2(1. 01) + 3(1. 01) + 1
π(1. 01) = 2. 0402 + 3. 03 + 1
π(1. 01) = 6. 0702
Point: (1.01, 6.0702)
2
ππ π₯ = 1. 001, π(1. 001) = 2(1. 001) + 3(1. 001) + 1
π(1. 001) = 2. 004002 + 3. 003 + 1
π(1. 001) = 6. 007002
Point: (1.001, 6.007002)
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Find the average rates of change for each set of points
Average Rate of Change:
π(π)−π(π)
π−π
(π₯1, π¦1) = (1, 6)
(π₯2, π¦2) = (1. 1, 6. 72)
(π₯1, π¦1) = (1, 6)
(π₯2, π¦2) = (1. 01, 6. 0702)
6.72−6
1.1−1
0.72
0.1
=
6.0702−6
1.01−1
= 7. 2
0.0702
0.01
=
(π₯1, π¦1) = (1, 6)
6.007002−6
1.001−1
=
= 7. 02
(π₯2, π¦2) = (1. 001, 6. 007002)
0.007002
0.001
= 7. 002
As the point gets closer to the tangent point, the value of the average rate of change is
getting closer to 7.
2
∴ The instantaneous rate of change when x=1 in the function π(π₯) = 2π₯ + 3π₯ + 1
is approximately 7.
b. Use First Principles to ο¬nd the value of the derivative when x = 1. (4 marks)
Derivative: π'(π₯) = lim
β→0
π(π₯+β)−π(π₯)
β
Given:
2
π(π₯) = 2π₯ + 3π₯ + 1
x= 1
π'(1) = lim
β→0
π'(1) = lim
β→0
2
2
(2(1+β) +3(1)+1)−(2(1) +3(1)+1)
β
2
(2(1+2β+β )+3+1)−(2+3+1)
β
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π'(1) = lim
β→0
π'(1) = lim
β→0
2
2+4β+2β +3+1−2−3−1
β
2
4β+2β
β
π'(1) = lim 4 + 2β
β→0
π'(1) = lim 4 + 2(0)
β→0
π'(1) = 7
∴The value of the derivative when x = 1 using the First Principles is 7.
c. What did you notice?
Ans. I noticed that the value of the derivative and the instantaneous rate of change are
similar in value (7) when x = 1, which means that the derivative of the function of the curve
and the slope of the tangent line at a point on the curve are the same.
3. Explain the di erence between a secant line and a tangent line. How do they relate to the rate of
change of a function? Include a sketch of each type of line in your solution. (6 marks)
Ans. A secant line on a curve is a straight line that intersects the curve at two points and is also
known as the line that joins the two points; while a tangent line on a curve is a straight line that
touches the curve at one point, the point it touches is called the tangent point. In relations to the
rate of change of a function, the slope of the secant line is equivalent to the average rate of change
between two points of the curve; the tangent line is equivalent to the instantaneous rate of change
at one point of the curve.
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RED LINE= Tangent Line
BLUE LINE= Secant Line
4. The path of a baseball relative to the ground can be modelled by the function
π(π‘) =− π‘ + 8π‘ + 1, where d(t) represents the height of the ball in metres, and t represents
2
time in seconds.
a. Find the average rate of change of the ball between 1 and 3 seconds. (4 marks)
Given:
2
π(π‘) =− π‘ + 8π‘ + 1
t= 1, t= 3
2
ππ π‘ = 1, π(1) =− 1 + 8(1) + 1 = 8
Point: (1, 8)
2
ππ π‘ = 3, π(3) =− 3 + 8(3) + 1 = 16
Point: (3, 16)
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Average Rate of Change:
=
16−8
3−1
8
2
=
π(π)−π(π)
π−π
= 4
∴ The average rate of change of the ball between 1 and 3 seconds is 4m/s.
b. Using the secant method, ο¬nd the instantaneous rate of change at 2 seconds. (8 marks)
Given:
2
π(π‘) =− π‘ + 8π‘ + 1
t= 2
Selected series of points close to t=2: 2.1, 2.01, 2.001
2
ππ π‘ = 2, π(2) =− 2 + 8(2) + 1 = 13
Point: (2, 13)
2
ππ π‘ = 2. 1, π(2. 1) =− 2. 1 + 8(2. 1) + 1 = 13. 39
Point: (2.1, 13.39)
2
ππ π‘ = 2. 01, π(2. 01) =− 2. 01 + 8(2. 01) + 1 = 13. 0399
Point: (2.01, 13.0399)
2
ππ π‘ = 2. 001, π(2. 001) =− 2. 001 + 8(2. 001) + 1 = 13. 003999
Point: (2.001, 13.003999)
Find the average rates of change for each set of points
Average Rate of Change:
π(π)−π(π)
π−π
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(π₯1, π¦1) = (2, 13)
(π₯2, π¦2) = (2. 1, 13. 39)
(π₯1, π¦1) = (2, 13)
(π₯2, π¦2) = (2. 01, 13. 0399)
13.39−13
2.1−2
0.39
0.1
=
13.0399−13
2.01−2
=
= 3. 9
0.0399
0.01
= 3. 99
(π₯1, π¦1) = (2, 13)
13.003999−13
2.001−2
=
(π₯2, π¦2) = (2. 001, 13. 003999)
0.003999
0.001
= 3. 999
As the point gets closer to the tangent point, the value of the average rate of change is
getting closer to 4.
2
∴ The instantaneous rate of change when x=1 in the function π(π‘) =− π‘ + 8π‘ + 1
is approximately 4 m/s.
5.
3
a. Evaluate lim (3π₯ + 7π₯ − 16).
π₯→2
3
ππ π₯ = 2, lim (3(2) + 7(2) − 16)
π₯→2
= lim (24 + 14 − 16) = 22
π₯→2
3
∴ lim (3π₯ + 7π₯ − 16) = 22
π₯→2
b. What is the meaning of this value?
Ans. This value means that as x approaches 2 from the right or left, y approaches 22.
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6. Evaluate the following limits.
a.
lim
π₯→4
2
π₯ −16
π₯−4
Substitute
ππ π₯ = 4, lim
π₯→4
= lim
π₯→4
16−16
4−4
2
4 −16
4−4
= 0 ← Inconclusive
Factoring
lim
π₯→4
2
π₯ −16
π₯−4
= lim
π₯→4
(π₯ −4)(π₯+4)
π₯−4
= lim (π₯ + 4)
π₯→4
Substitute
ππ π₯ = 4, lim (4 + 4) = 8
π₯→4
b.
∴ lim
π₯→4
lim
π₯→∞
2
π₯ −16
π₯−4
3
= 8
8π₯ −5π₯+17
3
2
6π₯ +2π₯ −4π₯
Substitute
lim
π₯→∞
3
8(∞) −5(∞)+17
3
2
6(∞) +2(∞) −4(∞)
= No Conclusion
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Dividing Everything
lim
π₯→∞
8π₯
π₯
3
−
3
6π₯
π₯
3
+
3
5π₯
π₯
2
3
2π₯
π₯
2
3
+
−
5
17
π₯
4π₯
π₯
8− π₯ +
= lim
π₯→∞
6+
2
π₯
3
3
17
−
3
π₯
4
π₯
2
Substitute
ππ π₯ = ∞, lim
π₯→∞
5
17
8− ∞ +
6+
2
π₯∞
3
∞
4
−
2
∞
A constant value divided by a really large number results in a number really small.
8−0+0
6+0−0
= lim
π₯→∞
c.
3
8π₯ −5π₯+17
∴ lim
π₯→∞
lim
β→0
8
6
=
3
=
2
6π₯ +2π₯ −4π₯
=
4
3
4
3
49+β−7
β
Substitute
ππ β = 0, lim
β→0
= lim
β→0
7−7
0
=
49+0−7
0
0
0
← Inconclusive
Rationalizing (multiply both numerator and denominator by the conjugate–
( 49 + β + 7) )
lim
β→0
( 49+β−7)( 49+β+7)
= lim
β→0
= lim
β→0
= lim
β→0
β( 49+β+7)
(49+β)+(7( 49+β))+(−7( 49+β))−49
(49+β)−49
β( 49+β+7)
β( 49+β+7)
β
β 49+β+7
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1
= lim
β→0
49+β+7
Substitute
ππ β = 0, lim
β→0
1
7+7
= lim
β→0
d.
49+β−7
β
∴ lim
β→0
lim
π¦ → −2
=
1
49+0+7
1
14
1
14
=
3
π¦ −8
2
2π¦ −7π¦+12
Substitute
ππ π¦ = 2, lim
π¦ → −2
0
6
= lim
π¦ → −2
∴ lim
π¦ → −2
e.
lim
β→0
= 0
2(2) −7(2)+12
3
π¦ −8
2
= 0
2π¦ −7π¦+12
4
2+β
3
2 −8
2
−2
β
Substitute
ππ β = 0, lim
β→0
4
2+0
−2
0
= 0← Inconclusive
Simplifying
lim
β→0
4
2+β
−2
β
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4−2(2+β)
2+β
= lim
β→0
β
4−4−2β
2+β
= lim
β→0
β
−2β
2+β
= lim
β→0
= lim −
β→0
β
2β
β(2+β)
2
2+β
= lim −
β→0
Substitute
πΌπ β = 0, lim −
β→0
∴ lim
β→0
4
2+β
−2
β
2
2+0
=− 1
=− 1
2
7. Find the slope of the tangent line at point (-2,2) on the curve π(π₯) = 2π₯ + 3π₯ using First
Principles. (4 marks)
Given:
2
π(π₯) = 2π₯ + 3π₯
Point: (-2,2)
First Principles: π = lim
β→0
ππ π₯ =− 2, π = lim
β→0
= lim
β→0
= lim
β→0
2
π(π₯+β)−π(π₯)
β
2
2
(2(−2+β) +3(−2+β))−(2(−2) +3(−2))
β
(2(4−4β+β )−6+3β)−(8−6)
β
2
8−8β+2β −6+3β−8+6
β
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= lim
β→0
= lim
β→0
2
−8β+2β +3β
β
2
2β −5β
β
= lim 2β − 5
β→0
Substitute
ππ β = 0, π = lim 2(0) − 5 =− 5
β→0
2
∴The slope of the tangent line at point (-2,2) on the curve π(π₯) = 2π₯ + 3π₯ using First
Principles is -5.
2
8. Find the derivative of the function π(π₯) = π₯ − 10π₯ + 3 using First Principles. (4 marks)
Given:
2
π(π₯) = π₯ − 10π₯ + 3
First Principle: π¦' = lim
β→0
2
π(π₯+β)−π(π₯)
β
2
lim
β→0
((π₯+β) −10(π₯+β)+3)−(π₯ −10π₯+3)
β
lim
β→0
2π₯β+β −10β
β
lim
β→0
2
2
2
π₯ +2π₯β+β −10π₯−10β+3−π₯ +10π₯−3
β
2
lim 2π₯ + β − 10
β→0
Substitute
ππ β = 0, lim 2π₯ + (0) − 10 = 2π₯ − 10
β→0
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2
∴ The derivative of the function π(π₯) = π₯ − 10π₯ + 3 using First Principles is 2x-10.
9. Using ο¬rst principles, determine the equation of the tangent line at point (2,2) on the curve.
2
π(π₯) = π₯ − 7π₯ + 12
Given:
2
π(π₯) = π₯ − 7π₯ + 12
Point (2,2)
First Principles: π = lim
β→0
π(π₯+β)−π(π₯)
β
equation: π¦ = π(π₯ − π₯1) + π¦1
Calculuate the slope
ππ π₯ = 2, π = lim
β→0
= lim
β→0
= lim
β→0
2
2
2
((2+β) −7(2+β)+12)−(2 −7(2)+12)
β
4+4β+β −14−7β+12−4+14−12
β
2
4β+β −7β
β
= lim 4 + β − 7
β→0
= lim β − 3
β→0
Substitute
ππ β = 0, π = lim (0) − 3
β→0
π =− 3
The slope of the function is -3, with the point (2, 2) we can substitute the values in the formula.
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π¦ = π(π₯ − π₯1) + π¦1
π¦ =− 3(π₯ − 2) + 2
π¦ =− 3π₯ + 6 + 2
π¦ =− 3π₯ + 8
2
∴ The equation of the tangent line at point (2,2) on the curve π(π₯) = π₯ − 7π₯ + 12 is
y=-3x+8.
2
10. At what point on the parabola π¦ = 3π₯ + 2π₯ is the tangent line parallel to the line
π¦ = 10π₯ − 2? Use ο¬rst principles in your solution.
Given:
2
Parabola: π¦ = 3π₯ + 2π₯
Line: π¦ = 10π₯ − 2
First Principle: π¦' = lim
β→0
π(π₯+β)−π(π₯)
β
Find the derivative of the parabola
π¦' = lim
β→0
= lim
β→0
= lim
β→0
2
2
2
(3(π₯+β) +2(π₯+β))−(3π₯ +2π₯)
β
2
2
3π₯ +6π₯β+3β +2π₯+2β−3π₯ −2π₯
β
2
6π₯β+3β +2β
β
= lim 6π₯ + 3β + 2
β→0
ππ β = 0, π¦' = lim 6π₯ + 3(0) + 2
β→0
= 6π₯ + 2
The slope of the line π¦ = 10π₯ − 2 is its derivative, meaning that the parallel lines will have the
same slopes.
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Finding x coordinate:
π¦' = 6π₯ + 2
π = 6π₯ + 2
10 = 6π₯ + 2
8 = 6π₯
π₯ =
4
3
Use this value to ο¬nd the y coordinate.
ππ π₯ =
4
4
3
= 6( 3 ) + 2
π¦ = 8
4
, π¦ = 6( 3 ) + 2
2
∴ At point ( 3 ,8) on the parabola π¦ = 3π₯ + 2π₯ is the tangent line parallel to the line
4
π¦ = 10π₯ − 2.
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