lOMoARcPSD|28760100
MCV4U UNIT 1 Rates of Change Assessment Opportunity
Calculus And Vector (Best notes for high school - EN-CA)
Studocu is not sponsored or endorsed by any college or university
Downloaded by Inderjit Sidhu (indersidhu098@gmail.com)
lOMoARcPSD|28760100
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 𝑦 = 2𝑥 + 7𝑥 + 2𝑥 − 3 over the
interval [3,5]. (5 marks)
Given:
3
2
𝑦 = 2𝑥 + 7𝑥 + 2𝑥 − 3
x=3, x=5
3
2
𝑖𝑓 𝑥 = 3, 𝑦 = 2(3) + 7(3) + 2(3) − 3
𝑦 = 2(27) + 7(9) + 6 − 3
𝑦 = 54 + 63 + 6 − 3
𝑦 = 120
Point (3, 120)
3
2
𝑖𝑓 𝑥 = 5, 𝑦 = 2(5) + 7(5) + 2(5) − 3
𝑦 = 2(125) + 7(25) + 10 − 3
𝑦 = 250 + 175 + 7
𝑦 = 432
Point (5, 432)
Average Rate of Change:
432−120
5−3
=
312
2
𝑓(𝑏)−𝑓(𝑎)
𝑏−𝑎
= 156
3
2
∴ The average rate of change of y in the function 𝑦 = 2𝑥 + 7𝑥 + 2𝑥 − 3 over the interval
[3,5] is 156.
Downloaded by Inderjit Sidhu (indersidhu098@gmail.com)
lOMoARcPSD|28760100
2
2. Given the function 𝑓(𝑥) = 2𝑥 + 3𝑥 + 1,
a. Use the secant method to find the instantaneous rate of change when x = 1. (4 marks)
Given:
2
𝑓(𝑥) = 2𝑥 + 3𝑥 + 1
x= 1
Selected series of points close to x=1: 1.1, 1.01, 1.001
2
𝑖𝑓 𝑥 = 1, 𝑓(1) = 2(1) + 3(1) + 1
𝑓(1) = 2 + 3 + 1
𝑓(1) = 6
Point: (1, 6)
2
𝑖𝑓 𝑥 = 1. 1, 𝑓(1. 1) = 2(1. 1) + 3(1. 1) + 1
𝑓(1. 1) = 2. 42 + 3. 3 + 1
𝑓(1. 1) = 6. 72
Point: (1.1, 6.72)
2
𝑖𝑓 𝑥 = 1. 01, 𝑓(1. 01) = 2(1. 01) + 3(1. 01) + 1
𝑓(1. 01) = 2. 0402 + 3. 03 + 1
𝑓(1. 01) = 6. 0702
Point: (1.01, 6.0702)
2
𝑖𝑓 𝑥 = 1. 001, 𝑓(1. 001) = 2(1. 001) + 3(1. 001) + 1
𝑓(1. 001) = 2. 004002 + 3. 003 + 1
𝑓(1. 001) = 6. 007002
Point: (1.001, 6.007002)
Downloaded by Inderjit Sidhu (indersidhu098@gmail.com)
lOMoARcPSD|28760100
Find the average rates of change for each set of points
Average Rate of Change:
𝑓(𝑏)−𝑓(𝑎)
𝑏−𝑎
(𝑥1, 𝑦1) = (1, 6)
(𝑥2, 𝑦2) = (1. 1, 6. 72)
(𝑥1, 𝑦1) = (1, 6)
(𝑥2, 𝑦2) = (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
=
(𝑥1, 𝑦1) = (1, 6)
6.007002−6
1.001−1
=
= 7. 02
(𝑥2, 𝑦2) = (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 𝑓(𝑥) = 2𝑥 + 3𝑥 + 1
is approximately 7.
b. Use First Principles to find the value of the derivative when x = 1. (4 marks)
Derivative: 𝑓'(𝑥) = lim
ℎ→0
𝑓(𝑥+ℎ)−𝑓(𝑥)
ℎ
Given:
2
𝑓(𝑥) = 2𝑥 + 3𝑥 + 1
x= 1
𝑓'(1) = lim
ℎ→0
𝑓'(1) = lim
ℎ→0
2
2
(2(1+ℎ) +3(1)+1)−(2(1) +3(1)+1)
ℎ
2
(2(1+2ℎ+ℎ )+3+1)−(2+3+1)
ℎ
Downloaded by Inderjit Sidhu (indersidhu098@gmail.com)
lOMoARcPSD|28760100
𝑓'(1) = lim
ℎ→0
𝑓'(1) = lim
ℎ→0
2
2+4ℎ+2ℎ +3+1−2−3−1
ℎ
2
4ℎ+2ℎ
ℎ
𝑓'(1) = lim 4 + 2ℎ
ℎ→0
𝑓'(1) = lim 4 + 2(0)
ℎ→0
𝑓'(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.
Downloaded by Inderjit Sidhu (indersidhu098@gmail.com)
lOMoARcPSD|28760100
RED LINE= Tangent Line
BLUE LINE= Secant Line
4. The path of a baseball relative to the ground can be modelled by the function
𝑑(𝑡) =− 𝑡 + 8𝑡 + 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
𝑑(𝑡) =− 𝑡 + 8𝑡 + 1
t= 1, t= 3
2
𝑖𝑓 𝑡 = 1, 𝑑(1) =− 1 + 8(1) + 1 = 8
Point: (1, 8)
2
𝑖𝑓 𝑡 = 3, 𝑑(3) =− 3 + 8(3) + 1 = 16
Point: (3, 16)
Downloaded by Inderjit Sidhu (indersidhu098@gmail.com)
lOMoARcPSD|28760100
Average Rate of Change:
=
16−8
3−1
8
2
=
𝑓(𝑏)−𝑓(𝑎)
𝑏−𝑎
= 4
∴ The average rate of change of the ball between 1 and 3 seconds is 4m/s.
b. Using the secant method, find the instantaneous rate of change at 2 seconds. (8 marks)
Given:
2
𝑑(𝑡) =− 𝑡 + 8𝑡 + 1
t= 2
Selected series of points close to t=2: 2.1, 2.01, 2.001
2
𝑖𝑓 𝑡 = 2, 𝑑(2) =− 2 + 8(2) + 1 = 13
Point: (2, 13)
2
𝑖𝑓 𝑡 = 2. 1, 𝑑(2. 1) =− 2. 1 + 8(2. 1) + 1 = 13. 39
Point: (2.1, 13.39)
2
𝑖𝑓 𝑡 = 2. 01, 𝑑(2. 01) =− 2. 01 + 8(2. 01) + 1 = 13. 0399
Point: (2.01, 13.0399)
2
𝑖𝑓 𝑡 = 2. 001, 𝑑(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:
𝑓(𝑏)−𝑓(𝑎)
𝑏−𝑎
Downloaded by Inderjit Sidhu (indersidhu098@gmail.com)
lOMoARcPSD|28760100
(𝑥1, 𝑦1) = (2, 13)
(𝑥2, 𝑦2) = (2. 1, 13. 39)
(𝑥1, 𝑦1) = (2, 13)
(𝑥2, 𝑦2) = (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
(𝑥1, 𝑦1) = (2, 13)
13.003999−13
2.001−2
=
(𝑥2, 𝑦2) = (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 𝑑(𝑡) =− 𝑡 + 8𝑡 + 1
is approximately 4 m/s.
5.
3
a. Evaluate lim (3𝑥 + 7𝑥 − 16).
𝑥→2
3
𝑖𝑓 𝑥 = 2, lim (3(2) + 7(2) − 16)
𝑥→2
= lim (24 + 14 − 16) = 22
𝑥→2
3
∴ lim (3𝑥 + 7𝑥 − 16) = 22
𝑥→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.
Downloaded by Inderjit Sidhu (indersidhu098@gmail.com)
lOMoARcPSD|28760100
6. Evaluate the following limits.
a.
lim
𝑥→4
2
𝑥 −16
𝑥−4
Substitute
𝑖𝑓 𝑥 = 4, lim
𝑥→4
= lim
𝑥→4
16−16
4−4
2
4 −16
4−4
= 0 ← Inconclusive
Factoring
lim
𝑥→4
2
𝑥 −16
𝑥−4
= lim
𝑥→4
(𝑥 −4)(𝑥+4)
𝑥−4
= lim (𝑥 + 4)
𝑥→4
Substitute
𝑖𝑓 𝑥 = 4, lim (4 + 4) = 8
𝑥→4
b.
∴ lim
𝑥→4
lim
𝑥→∞
2
𝑥 −16
𝑥−4
3
= 8
8𝑥 −5𝑥+17
3
2
6𝑥 +2𝑥 −4𝑥
Substitute
lim
𝑥→∞
3
8(∞) −5(∞)+17
3
2
6(∞) +2(∞) −4(∞)
= No Conclusion
Downloaded by Inderjit Sidhu (indersidhu098@gmail.com)
lOMoARcPSD|28760100
Dividing Everything
lim
𝑥→∞
8𝑥
𝑥
3
−
3
6𝑥
𝑥
3
+
3
5𝑥
𝑥
2
3
2𝑥
𝑥
2
3
+
−
5
17
𝑥
4𝑥
𝑥
8− 𝑥 +
= lim
𝑥→∞
6+
2
𝑥
3
3
17
−
3
𝑥
4
𝑥
2
Substitute
𝑖𝑓 𝑥 = ∞, lim
𝑥→∞
5
17
8− ∞ +
6+
2
𝑥∞
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
𝑥→∞
c.
3
8𝑥 −5𝑥+17
∴ lim
𝑥→∞
lim
ℎ→0
8
6
=
3
=
2
6𝑥 +2𝑥 −4𝑥
=
4
3
4
3
49+ℎ−7
ℎ
Substitute
𝑖𝑓 ℎ = 0, lim
ℎ→0
= lim
ℎ→0
7−7
0
=
49+0−7
0
0
0
← Inconclusive
Rationalizing (multiply both numerator and denominator by the conjugate–
( 49 + ℎ + 7) )
lim
ℎ→0
( 49+ℎ−7)( 49+ℎ+7)
= lim
ℎ→0
= lim
ℎ→0
= lim
ℎ→0
ℎ( 49+ℎ+7)
(49+ℎ)+(7( 49+ℎ))+(−7( 49+ℎ))−49
(49+ℎ)−49
ℎ( 49+ℎ+7)
ℎ( 49+ℎ+7)
ℎ
ℎ 49+ℎ+7
Downloaded by Inderjit Sidhu (indersidhu098@gmail.com)
lOMoARcPSD|28760100
1
= lim
ℎ→0
49+ℎ+7
Substitute
𝑖𝑓 ℎ = 0, lim
ℎ→0
1
7+7
= lim
ℎ→0
d.
49+ℎ−7
ℎ
∴ lim
ℎ→0
lim
𝑦 → −2
=
1
49+0+7
1
14
1
14
=
3
𝑦 −8
2
2𝑦 −7𝑦+12
Substitute
𝑖𝑓 𝑦 = 2, lim
𝑦 → −2
0
6
= lim
𝑦 → −2
∴ lim
𝑦 → −2
e.
lim
ℎ→0
= 0
2(2) −7(2)+12
3
𝑦 −8
2
= 0
2𝑦 −7𝑦+12
4
2+ℎ
3
2 −8
2
−2
ℎ
Substitute
𝑖𝑓 ℎ = 0, lim
ℎ→0
4
2+0
−2
0
= 0← Inconclusive
Simplifying
lim
ℎ→0
4
2+ℎ
−2
ℎ
Downloaded by Inderjit Sidhu (indersidhu098@gmail.com)
lOMoARcPSD|28760100
4−2(2+ℎ)
2+ℎ
= lim
ℎ→0
ℎ
4−4−2ℎ
2+ℎ
= lim
ℎ→0
ℎ
−2ℎ
2+ℎ
= lim
ℎ→0
= lim −
ℎ→0
ℎ
2ℎ
ℎ(2+ℎ)
2
2+ℎ
= lim −
ℎ→0
Substitute
𝐼𝑓 ℎ = 0, lim −
ℎ→0
∴ lim
ℎ→0
4
2+ℎ
−2
ℎ
2
2+0
=− 1
=− 1
2
7. Find the slope of the tangent line at point (-2,2) on the curve 𝑓(𝑥) = 2𝑥 + 3𝑥 using First
Principles. (4 marks)
Given:
2
𝑓(𝑥) = 2𝑥 + 3𝑥
Point: (-2,2)
First Principles: 𝑚 = lim
ℎ→0
𝑖𝑓 𝑥 =− 2, 𝑚 = lim
ℎ→0
= lim
ℎ→0
= lim
ℎ→0
2
𝑓(𝑥+ℎ)−𝑓(𝑥)
ℎ
2
2
(2(−2+ℎ) +3(−2+ℎ))−(2(−2) +3(−2))
ℎ
(2(4−4ℎ+ℎ )−6+3ℎ)−(8−6)
ℎ
2
8−8ℎ+2ℎ −6+3ℎ−8+6
ℎ
Downloaded by Inderjit Sidhu (indersidhu098@gmail.com)
lOMoARcPSD|28760100
= lim
ℎ→0
= lim
ℎ→0
2
−8ℎ+2ℎ +3ℎ
ℎ
2
2ℎ −5ℎ
ℎ
= lim 2ℎ − 5
ℎ→0
Substitute
𝑖𝑓 ℎ = 0, 𝑚 = lim 2(0) − 5 =− 5
ℎ→0
2
∴The slope of the tangent line at point (-2,2) on the curve 𝑓(𝑥) = 2𝑥 + 3𝑥 using First
Principles is -5.
2
8. Find the derivative of the function 𝑓(𝑥) = 𝑥 − 10𝑥 + 3 using First Principles. (4 marks)
Given:
2
𝑓(𝑥) = 𝑥 − 10𝑥 + 3
First Principle: 𝑦' = lim
ℎ→0
2
𝑓(𝑥+ℎ)−𝑓(𝑥)
ℎ
2
lim
ℎ→0
((𝑥+ℎ) −10(𝑥+ℎ)+3)−(𝑥 −10𝑥+3)
ℎ
lim
ℎ→0
2𝑥ℎ+ℎ −10ℎ
ℎ
lim
ℎ→0
2
2
2
𝑥 +2𝑥ℎ+ℎ −10𝑥−10ℎ+3−𝑥 +10𝑥−3
ℎ
2
lim 2𝑥 + ℎ − 10
ℎ→0
Substitute
𝑖𝑓 ℎ = 0, lim 2𝑥 + (0) − 10 = 2𝑥 − 10
ℎ→0
Downloaded by Inderjit Sidhu (indersidhu098@gmail.com)
lOMoARcPSD|28760100
2
∴ The derivative of the function 𝑓(𝑥) = 𝑥 − 10𝑥 + 3 using First Principles is 2x-10.
9. Using first principles, determine the equation of the tangent line at point (2,2) on the curve.
2
𝑓(𝑥) = 𝑥 − 7𝑥 + 12
Given:
2
𝑓(𝑥) = 𝑥 − 7𝑥 + 12
Point (2,2)
First Principles: 𝑚 = lim
ℎ→0
𝑓(𝑥+ℎ)−𝑓(𝑥)
ℎ
equation: 𝑦 = 𝑚(𝑥 − 𝑥1) + 𝑦1
Calculuate the slope
𝑖𝑓 𝑥 = 2, 𝑚 = lim
ℎ→0
= lim
ℎ→0
= lim
ℎ→0
2
2
2
((2+ℎ) −7(2+ℎ)+12)−(2 −7(2)+12)
ℎ
4+4ℎ+ℎ −14−7ℎ+12−4+14−12
ℎ
2
4ℎ+ℎ −7ℎ
ℎ
= lim 4 + ℎ − 7
ℎ→0
= lim ℎ − 3
ℎ→0
Substitute
𝑖𝑓 ℎ = 0, 𝑚 = lim (0) − 3
ℎ→0
𝑚 =− 3
The slope of the function is -3, with the point (2, 2) we can substitute the values in the formula.
Downloaded by Inderjit Sidhu (indersidhu098@gmail.com)
lOMoARcPSD|28760100
𝑦 = 𝑚(𝑥 − 𝑥1) + 𝑦1
𝑦 =− 3(𝑥 − 2) + 2
𝑦 =− 3𝑥 + 6 + 2
𝑦 =− 3𝑥 + 8
2
∴ The equation of the tangent line at point (2,2) on the curve 𝑓(𝑥) = 𝑥 − 7𝑥 + 12 is
y=-3x+8.
2
10. At what point on the parabola 𝑦 = 3𝑥 + 2𝑥 is the tangent line parallel to the line
𝑦 = 10𝑥 − 2? Use first principles in your solution.
Given:
2
Parabola: 𝑦 = 3𝑥 + 2𝑥
Line: 𝑦 = 10𝑥 − 2
First Principle: 𝑦' = lim
ℎ→0
𝑓(𝑥+ℎ)−𝑓(𝑥)
ℎ
Find the derivative of the parabola
𝑦' = lim
ℎ→0
= lim
ℎ→0
= lim
ℎ→0
2
2
2
(3(𝑥+ℎ) +2(𝑥+ℎ))−(3𝑥 +2𝑥)
ℎ
2
2
3𝑥 +6𝑥ℎ+3ℎ +2𝑥+2ℎ−3𝑥 −2𝑥
ℎ
2
6𝑥ℎ+3ℎ +2ℎ
ℎ
= lim 6𝑥 + 3ℎ + 2
ℎ→0
𝑖𝑓 ℎ = 0, 𝑦' = lim 6𝑥 + 3(0) + 2
ℎ→0
= 6𝑥 + 2
The slope of the line 𝑦 = 10𝑥 − 2 is its derivative, meaning that the parallel lines will have the
same slopes.
Downloaded by Inderjit Sidhu (indersidhu098@gmail.com)
lOMoARcPSD|28760100
Finding x coordinate:
𝑦' = 6𝑥 + 2
𝑚 = 6𝑥 + 2
10 = 6𝑥 + 2
8 = 6𝑥
𝑥 =
4
3
Use this value to find the y coordinate.
𝑖𝑓 𝑥 =
4
4
3
= 6( 3 ) + 2
𝑦 = 8
4
, 𝑦 = 6( 3 ) + 2
2
∴ At point ( 3 ,8) on the parabola 𝑦 = 3𝑥 + 2𝑥 is the tangent line parallel to the line
4
𝑦 = 10𝑥 − 2.
Downloaded by Inderjit Sidhu (indersidhu098@gmail.com)