MHR • Calculus and Vectors 12 Solutions 1 Chapter 1 Rates of Change Chapter 1 Prerequisite Skills Chapter 1 Prerequisite Skills Question 1 Page 2 a) x y First Differences –4 9 –4 –3 5 –2 –2 3 0 –1 3 2 0 5 4 1 9 6 2 15 Answers may vary. For example: The first differences are not equal, but they progress by an equal amount. b) Answers may vary. For example: Yes. Since the first differences are not equal, the function is not linear. Also, since the first differences increase by the same amount each time, the curve is quadratic. Chapter 1 Prerequisite Skills Question 2 Page 2 a) slope = 3 ! 1 !2 ! 4 = ! 2 6 or ! 1 3 b) slope = !7 ! ( !1) 3 ! 0 = ! 6 3 or ! 2 c) slope = 1 ! 0 5 ! 0 = 1 5 d) slope = 4 ! 4 0 ! ( !9) = 0
102
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The first differences are not equal, but they progress by an equal amount.
b) Answers may vary. For example:
Yes. Since the first differences are not equal, the function is not linear. Also, since the first differences
increase by the same amount each time, the curve is quadratic.
Chapter 1 Prerequisite Skills Question 2 Page 2
a) slope =
3!1
!2 ! 4
=
!2
6 or !
1
3
b) slope =
!7 ! (!1)
3! 0
=
!6
3 or ! 2
c) slope =
1! 0
5! 0
=
1
5
d) slope =
4 ! 4
0 ! (!9)
= 0
MHR • Calculus and Vectors 12 Solutions 2
Chapter 1 Prerequisite Skills Question 3 Page 2
a) y =
1
2x ! 7
4; slope:
1
2; y-intercept:
!7
4
b) y = !
5
3x +
1
3; slope:
!5
3; y-intercept:
1
3
c) y = !2x ! 10
9; slope: −2; y-intercept:
!10
9
d) y =
7
5x +
2
5; slope:
7
5; y-intercept:
2
5
Chapter 1 Prerequisite Skills Question 4 Page 2 a) y = 5x + 3
b)
slope =3!1
!5!1
= !2
6 or !
1
3
Substitute the point (1, 1) into y =
!1
3x + b to find b.
1= !1
3(1) + b
b =4
3
The equation is
y = !
1
3x +
4
3.
c) Substitute the point (4, 7) into y = –2x + b to find b.
7 = !2(4) + bb = 15
The equation is y = –2x + 15.
d)
slope =0 ! (!1)
3! 2
= 1
Substitute the point (3, 0) into y = x + b to find b.
0 = (3) + bb = !3
The equation is y = x – 3.
MHR • Calculus and Vectors 12 Solutions 3
Chapter 1 Prerequisite Skills Question 5 Page 2 a)
(a + b)2
= a2+ 2ab + b2
b) (a + b)
3= a3
+ 3a2b + 3ab2+ b3
c) (a ! b)
3= a3
! 3a2b + 3ab2! b3
d)
(a + b)4
= (a2+ 2ab + b2
)2
= a4+ 4a3b + 6a2b2
+ 4ab3+ b4
e)
(a ! b)5
= (a ! b)2(a ! b)
3
= a5! 5a4b +10a3b2
!10a2b3+ 5ab4
! b5
f)
(a + b)5
= (a + b)2(a + b)
3
= a5+ 5a4b +10a3b2
+10a2b3+ 5ab4
+ b5
Chapter 1 Prerequisite Skills Question 6 Page 2 a)
2x2! x !1= (2x +1)(x !1)
b)
6x2+17x + 5 = 6x2
+15x + 2x + 5
= 3x(2x + 5) + (2x + 5)
= (2x + 5)(3x +1)
c) x
3!1= (x !1)(x2
+ x +1)
d)
2x4+ 7x3
+ 3x2= x2
(2x2+ 7x + 3)
= x2[2x(x + 3) + (x + 3)]
= x2(2x +1)(x + 3)
e)
x4! x3 – x +1= x3
(x –1) –1(x –1)
= (x –1)(x3 –1)
= (x –1)(x –1)(x2+ x +1)
=(x –1)2(x2
+ x +1)
f)
t3+ 2t2
! 3t = t(t2+ 2t ! 3)
= t(t !1)(t + 3)
Chapter 1 Prerequisite Skills Question 7 Page 2 b)
(a ! b)(a + b)
MHR • Calculus and Vectors 12 Solutions 4
c) a3! b3
d) (a ! b)(a3
+ a2b + ab2+ b3
)
e) (a ! b)(a4
+ a3b + a2b2+ ab3
+ b4)
f)
(x + h ! x)((x + h)n!1
+ x(x + h)n!2
+ x2(x + h)
n!3+ " " " + xn!3
(x + h)2
+ xn!2(x + h) + xn!1
)
= h((x + h)n!1
+ x(x + h)n!2
+ x2(x + h)
n!3+ " " " + xn!3
(x + h)2
+ xn!2(x + h) + xn!1
)
Chapter 1 Prerequisite Skills Question 8 Page 2
a)
x ! 2( ) x + 2( ) = x( )2
+ 2x ! 2x ! 2( )2
= x ! 2
b)
x +1! x( ) x +1 + x( ) = x +1( )2
+ x(x +1) ! x(x +1) ! x( )2
= x +1! x= 1
c)
x +1! x !1( ) x +1 + x !1( ) = x +1( )2
+ (x +1)(x !1) ! (x +1)(x !1) ! x !1( )2
= x +1! (x !1)
= 2
d)
3(x + h) ! 3x( ) 3(x + h) + 3x( ) = 3(x + h)( )2
+ 9x(x + h) ! 9x(x + h) ! 3x( )2
= 3(x + h)! 3x= 3h
Chapter 1 Prerequisite Skills Question 9 Page 2
a)
1
2 + h!
1
2=
2 ! (2 + h)
2(2 + h)
=!h
2(2 + h)
b)
1
x + h!
1
x=
x ! (x + h)
x(x + h)
=!h
x(x + h)
MHR • Calculus and Vectors 12 Solutions 5
c)
1
(x + h)2!
1
x2=
x2! (x + h)
2
x2(x + h)
2
=!2xh ! h2
x2(x + h)
2
= !h(2x + h)
x2(x + h)
2
d)
1
x + h!
1
xh
=
x ! (x + h)
x(x + h)
h
=!h
xh(x + h)
=!1
x(x + h)
Chapter 1 Prerequisite Skills Question 10 Page 3 a)
f (!2) = 3(!2) +12
= 6
f (3) = 3(3) +12
= 21
The points are (–2, 6) and (3, 21).
b)
f (!2) = !5(!2)2
+ 2(!2) +1
= !23
f (3) = !5(3)2
+ 2(3) +1
= !38
The points are (–2, –23) and (3, –38).
c)
f (!2) = 2(!2)3! 7(!2)
2+ 3
= !41
f (3) = 2(3)3! 7(3)
2+ 3
= !6
The points are (–2, –41) and (3, –6).
Chapter 1 Prerequisite Skills Question 11 Page 3 a)
f (3+ h) = 6(3+ h)! 2
= 16 + 6h
b)
f (3+ h) = 3(3+ h)2
+ 5(3+ h)
= 3(h2+ 6h + 9) +15+ 5h
= 3h2+ 23h + 42
c)
f (3+ h) = 2(3+ h)3! 7(3+ h)
2
= 2(h3+ 3h2x + 3hx2
+ x3)! 7(h2
+ 6h + 9)
= 2h3+11h2
+12h ! 9
MHR • Calculus and Vectors 12 Solutions 6
Chapter 1 Prerequisite Skills Question 12 Page 3
a)
f (2 + h)! f (2)
h=
6(2 + h)! 6(2)
h= 6
b)
f (2 + h)! f (2)
h=
2(2 + h)3! 2(2)
3
h
=2(2
3+ 3(2
2)h + 3(2)h2
+ h3)!16
h= 2h2
+12h + 24
c)
f (2 + h)! f (2)
h=
1
(2 + h)!
1
2
h
=
!h2(2 + h)
h
= !1
2(2 + h)
d)
f (2 + h)! f (2)
h=
!4
(2 + h)! !
4
2
"#$
%&'
h
=
!8 + 4(2 + h)
2(2 + h)
h
=4
2(2 + h)
Chapter 1 Prerequisite Skills Question 13 Page 3 a)
{x | x !!}
b) {x | x ! 8, x "!}
c) {x | x !!}
d) {x | x ! 0; x "!}
e) {x | x ! "3 and x ! 2; x #!}
f) {x | 0 ! x ! 9, x "!}
MHR • Calculus and Vectors 12 Solutions 7
Chapter 1 Prerequisite Skills Question 14 Page 3
Interval Notation Inequality Number Line
(–3, 5) !3 < x < 5 [–3, 5] !3" x " 5 [–3, 5) !3" x < 5 (–3, 5] !3 < x " 5
(–3, ∞) x > –3
[–3, ∞) x ! "3
(–∞, 5) x < 5 (–∞, 5] x ! 5 (–∞, ∞) !
Chapter 1 Prerequisite Skills Question 15 Page 3 a) domain:
{x | x !R} ; range: {y | y !!}
b) domain:
{x | x ! 0, x "R} ; range: {y | y ! 0, y "!}
c) domain: {x | x ! "2, x #R} ; range:
{y | y ! "4, y #!}
MHR • Calculus and Vectors 12 Solutions 8
d) domain: {x | x !R} ; range:
{y | y !1, y "!}
MHR • Calculus and Vectors 12 Solutions 9
Chapter 1 Section 1 Rates of Change and the Slope of a Curve
Chapter 1 Section 1 Question 1 Page 9
a)
average rate of change =6 ! (!1)
2 ! (!4)
=7
6
b)
average rate of change =17 ! (!6.7)
!5! 3.2
=23.7
!8.2
= !237
82
c)
average rate of change =
3
4! !
4
5
"#$
%&'
!11
2!
2
3
"#$
%&'
=
31
20
!13
6
= !93
130
Chapter 1 Section 1 Question 2 Page 9
a) i)
average rate of change =!3! 5
1! (!3)
= !2
ii)
average rate of change =5! 5
3! (!3)
= 0
iii)
average rate of change =45! (!3)
7 !1
= 8
MHR • Calculus and Vectors 12 Solutions 10
iv)
average rate of change =6 ! (!5)
5! (!1)
=11
6
b) Answers may vary. For example: Sketch the graph (scatter plot) to estimate the instantaneous rates by
choosing small intervals and using the formula for the average rate of change as in a).
i) –1
ii) 5
iii) 0.5
iv) 10
Chapter 1 Section 1 Question 3 Page 9
a) Choosing the points (2, 9) and (1, 7):
7 ! 9
1! 2= 2
b) Choosing the points (–2, 2) and (10, 14):
14 ! 2
10 ! (!2)= 1
c) Choosing the points (1, 0) and (5, 7):
7 ! 0
5!1=
7
4
Chapter 1 Section 1 Question 4 Page 9 Answers may vary. For example:
a) i) They are all zero. B and F are local minima and D is a local maximum.
ii) They have the same magnitude but are opposite in sign. The instantaneous rate of change is
negative at A and positive at G.
iii) They are both positive since the function is increasing at both points.
iv) They are both negative since the function is decreasing at both points.
b) i) The instantaneous rate of change is negative at B and positive at C.
ii) They are all negative since the function is decreasing at all three points.
iii) They are both positive since the function is increasing at both points.
iv) The instantaneous rate of change is negative at A and positive at C.
MHR • Calculus and Vectors 12 Solutions 11
Chapter 1 Section 1 Question 5 Page 10 a) The dependent variable is surface area in square centimetres and the independent variable is time in
seconds.
The rate of change of surface area over time is expressed in square centimetres per second.
b) i)
average rate of change =324.0 !10.0
10 ! 0
= 31.4
The average rate of change during the first 10 s is 31.4 cm2/s.
ii)
average rate of change =2836.0 !1266.0
30 ! 20
= 157
The average rate of change between 20 s and 30 s is 157 cm2/s.
iii)
average rate of change = 2836.0 !1818.6
30 ! 24
!= 169.57
The average rate of change during the last 6 s is about 169.57 cm2/s.
c) i)
instantaneous rate of change =60.24 !10.0
4 ! 0
!= 13
The instantaneous rate of change at t = 2 s is an estimated 13 cm2/s.
ii)
instantaneous rate of change =813.8! 462.16
16 !12
!= 88
The instantaneous rate of change at t = 14 s is an estimated 88 cm2/s.
iii)
instantaneous rate of change =2836.0 ! 2132.6
30 ! 26
!= 176
The instantaneous rate of change at t = 28 s is an estimated 176 cm2/s.
d) Answers may vary. For example:
i) 38 cm2/s
ii) 100 cm2/s
iii) 163 cm2/s
MHR • Calculus and Vectors 12 Solutions 12
e) Answers may vary. For example:
The instantaneous rate of change is increasing rapidly as the time is increasing. The values I found in
part d) agree with this statement.
Chapter 1 Section 1 Question 6 Page 10 C, since it is the smallest interval.
Chapter 1 Section 1 Question 7 Page 10
a) i)
ii)
Parts b)–d): Answers may vary. For example:
b) The average rates of change are equal to the first differences in part i), and they are half the value of
the first differences in part ii).
c) The values of the calculated first differences and the average rates of change of y are the not the same
in part ii) because the difference between successive x-values is two. In this case, the first differences
must be divided by two to calculate the average rate of change.
x y First Differences
–3 –50
Average Rate of Change
38 –2 –12 38
14 –1 2 14
2 0 4 2
2 1 6 2
14 2 20 14
x y First Differences
–6 –26
Average Rate of Change
52 –4 26 26
–4 –2 22 –2
–12 0 10 –6
28 2 38 14
116 4 154 58
MHR • Calculus and Vectors 12 Solutions 13
d) The first differences and average rates of change for a function will be equal if the difference
between successive x-values is equal to one.
Chapter 1 Section 1 Question 8 Page 10 Explanations may vary. For example:
a) Instantaneous rate of change: The rate of change occurs at the specific instant when the radius is 4 cm.
b) Average rate of change: The rate of change refers to a distance over an interval of 5 h.
c) Instantaneous rate of change: The rate of change occurs at the specific instant when the time is 1 P.M.
d) Average rate of change: The rate of change refers to the stock price over an interval of time of one
week.
e) Average rate of change: The rate of change refers to the water level of a lake over an interval from the
beginning of March to the end of May.
Chapter 1 Section 1 Question 9 Page 10 Answers may vary. For example:
a) The initial temperature of the water was 10° C; After 3 min the water reached its boiling point.
b) The graph shows that the rise in temperature of water is rapid during the first 40 s or so, slowing
further until it reaches its boiling point at t = 180 s. After 180 s, the curve is flat, and the instantaneous
rate of change is zero after this point.
At t = 60 s, the instantaneous rate of change is about
87 ! 57
90 ! 30 or 0.5°C/s.
At t = 120 s, the instantaneous rate of change is about
97 ! 87
90 ! 30 or 0.17°C/s.
At t = 120 s, the instantaneous rate of change is 0°C/s.
Chapter 1 Section 1 Question 10 Page 11
a) i) The average rate of change is
32 299 496 ! 23 143 192
2005!1975 or approximately 305 210 people per year.
ii) The average rate of change is
27 697 530 ! 24 516 071
1990 !1980 or approximately 318 146 people per year.
iii) From 1975 to 1985:
25 842 736 ! 23 143 192
1985!1975 or approximately 269 954 people per year
From 1985 to 1995:
29 302 091! 25 842 736
1995!1985 or approximately 345 936 people per year
From 1995 to 2005:
32 299 496 ! 29 302 091
2005!1995 or approximately 299 741 people per year
MHR • Calculus and Vectors 12 Solutions 14
Parts b)–e): Answers may vary. For example:
b) While the population has steadily increased, the rate at which it has increased varies.
The estimated instantaneous rate of change in 1983 is:
25 607 651! 25 117 442
1984 !1982 or approximately 245 105 people per year
The estimated instantaneous rate of change in 1993 is:
28 999 006 ! 28 366 737
1994 !1992 or approximately 316 135 people per year
The estimated instantaneous rate of change in 2003 is:
31 989 454 ! 31 372 587
2004 ! 2002 or approximately 308 434 people per year
c)
The rate of change of Canada’s population has increased steadily since 1975.
d) Canada’s population is increasing with respect to time.
e) Q: What do you predict will be Canada’s population in the year 2015? Explain.
A: The average rate of change of Canada’s population between 1975 and 2005 was
305 210 people per year. Therefore, the estimated population in 2015 is
32 299 496 + (10)(305 210) = 35 351 596.
Q: What do you predict will be the instantaneous rate of change of Canada’s population in the year
2015? Explain.
A: The prediction assumes a change equal to the average rate of change over the previous 30 years,
which is 305 210 people per year. Therefore, the estimated instantaneous rate of change will be the
same figure.
Chapter 1 Section 1 Question 11 Page 11 Solutions to the Achievement Checks are shown in the Teacher’s Resource.
MHR • Calculus and Vectors 12 Solutions 15
Chapter 1 Section 1 Question 12 Page 11 Answers may vary. For example:
a) The resistance increases as the voltage increases since the slope of the graph increases.
b) Use the points (0.8, 40) and (1.2, 85).
R = instantaneous rate of change of V
=85! 40
1.2 ! 0.8
!= 113
The estimated instantaneous rate of change of V when V = 60 V is 113 V/A or Ω.
Chapter 1 Section 1 Question 13 Page 12 a)
Time (min) Radius (m) Area (m2) 0 0 0
2 4 50.3
4 8 201.1
6 12 452.4
8 16 804.2
10 20 1256.6
12 24 1809.6
14 28 2463.0
16 32 3217.0
18 36 4071.5
20 40 5026.5
22 44 6082.1
24 48 7238.2
26 52 8494.9
28 56 9852.0
30 60 11 309.7
MHR • Calculus and Vectors 12 Solutions 16
b) i) The average rate of change during the first 4 min is
201.1! 0
4 ! 0 or approximately 50.3 m
2/min.
ii) The average rate of change during the next 10 min is
2463.0 ! 201.1
14 ! 4 or approximately
226.2 m2/min.
iii) The average rate of change during the entire 30 min is
11 309.7 ! 0
30 ! 0or approximately
377.0 m2/min.
c) Answers may vary. For example:
The instantaneous rate of change at t = 5 min is
452.4 ! 201.1
6 ! 4 or approximately 125.7 m
2/min.
The instantaneous rate of change at t = 25 min is
8494.9 ! 7238.2
26 ! 24 or approximately 628.4 m
2/min.
d) Answers may vary. For example: The information might be useful to determine when the oil spill will
reach shore in order to protect the birds, animals and the environment.
Chapter 1 Section 1 Question 14 Page 12 a) Since x represents t in this case: at t = 0, y = 2; at t = 5 y = 12; at t = 10, y = 22; at t = 15, y = 12;
at t = 20, y = 2
From a hand sketch of these points, it is clear that the function can be expressed using either a sine or
cosine function.
Use a cosine function of the form y = acos(b(x + c)) + d. Since the function appears upside down
compared to cos x, use a negative sign in front of the cosine. The diameter of the windmill is 10 m, so
the amplitude of the function is half the diameter, which is 5.
Thus, a = –5. It takes 20 s to complete one full revolution, so the period is 20.
Thus, b is equal to
2!
20=!
10. There does not need to be a horizontal shift so c = 0. So far, the
function is
y = !5cos
"10
t#
$%&
'(+ d . At time t = 0 s, the height of the ladybug is 2 m, so substitute the
point (0, 2) into the function to find d.
2 = !5cos"10
(0)#
$%&
'(+ d
d = 7
Graph
y = !5cos
"10
t#
$%&
'(+ 7 using radians or
y = !5cos(18t) + 7 using degrees.
MHR • Calculus and Vectors 12 Solutions 17
b) Answers may vary. For example:
No. The rate of change of the ladybug’s height will not be constant because the rate of change of the
height is affected by the position of the blade.
c) Yes. The rate of change of the height of the blade is constantly changing since the slope of the graph is
constantly changing.
Chapter 1 Section 1 Question 15 Page 12 Answers will vary. For example: a) If the wind speed increased the blades would turn faster and the period of the function would
decrease. The rate of change of the height of the ladybug would increase since the slope of the graph
would be steeper.
If the wind speed decreased the blades would turn more slowly and the period of the function would
increase. The rate of change of the height of the ladybug would decrease since the slope of the graph
would be less steep.
b) The amplitude of the function representing the motion of the ladybug would be reduced from 10 m to
8 m. Therefore the rate of change of the height would decrease since the graph is less steep.
Chapter 1 Section 1 Question 16 Page 12
a) i) first 3 s:
3.58! 0
3! 0 or approximately 1.19 cm/s; last 3 s:
5.35! 4.75
10 ! 7 or 0.20 cm/s
ii) at 3 s:
3.94 ! 3.13
4 ! 2 or approximately 0.41 cm/s; at 9 s:
5.35! 4.96
10 ! 8 or approximately 0.2 cm/s
b) i) Use the slopes of the secants from (0,0) to (3, 3.58) and from (7, 4.75) to (10, 5.35).
ii) Use the slopes of the secants from (2, 3.13) to (4, 3.94) and from (8, 4.96) to (10, 5.35).
c)
If the cup was a cylinder, the graph would be a straight line.
MHR • Calculus and Vectors 12 Solutions 18
d)
t (s) H (cm) r (cm) V (cm3) 0 0 0 0 1 2.48 1.24 4 2 3.13 1.565 8 3 3.58 1.79 12 4 3.94 1.97 16 5 4.24 2.12 20 6 4.51 2.255 24 7 4.75 2.375 28 8 4.96 2.48 32 9 5.16 2.58 36
10 5.35 2.675 40 The water is being poured at a constant rate.
Chapter 1 Section 1 Question 17 Page 12
1
x+
1
y=
x + yxy
=8
12
=2
3
Chapter 1 Section 1 Question 18 Page 12
log93
g= log
95
g log93 = log
95
g 1
2
!
"#$
%&= log
95
g = 2 log95
g = log95
2
log9
g = log9(log
95
2)
log9
g1
2 = log9(2 log
95)
1
2log
9g = log
9(2 log
95)
log9
g = 2 log9(2 log
95)
MHR • Calculus and Vectors 12 Solutions 19
Chapter 1 Section 2 Rates of Change Using Equations
Chapter 1 Section 2 Question 1 Page 20
a)
average rate of change: 4 !1
4 !1= 1
b)
average rate of change: 4
2!1
2
4 !1= 5
c)
average rate of change: 4
3!1
3
4 !1= 21
d)
average rate of change: 7 ! 7
4 !1= 0
Chapter 1 Section 2 Question 2 Page 20
a)
f (2 + h)! f (2)
h=
(2 + h)! 2
h= 1
The instantaneous rate of change is 1.
b)
f (2 + h)! f (2)
h=
(2 + h)2! 2
2
h
=h2
+ 4h + 4 ! 4
h= h + 4
The instantaneous rate of change is 4.
c)
f (2 + h)! f (2)
h=
(2 + h)3! 2
3
h
=h3
+ 3(2)h2+ 3(2)
2 h + (2)3! 8
h= h2
+ 6h +12
The instantaneous rate of change is 12.
d)
f (2 + h)! f (2)
h=
7 ! 7
h= 0
The instantaneous rate of change is 0.
MHR • Calculus and Vectors 12 Solutions 20
Chapter 1 Section 2 Question 3 Page 20
f (4 + h)! f (4)
h
Chapter 1 Section 2 Question 4 Page 20
(!3+ h)2! (!3)
2
h
Chapter 1 Section 2 Question 5 Page 20
(5+ h)3! (5)
3
h
Chapter 1 Section 2 Question 6 Page 20
(!1+ h)3! (!1)
3
h
Chapter 1 Section 2 Question 7 Page 20
a) True. For
f (x) = 4x3
, f (1+ h)! f (1)
h=
4(1+ h)3! 4
h
b) False. The tangent point occurs at x = 1.
c) True. For
f (x) = 4x3! 4,
f (1+ h)! f (1)
h=
(4(1+ h)3! 4)! (4(1)
3! 4)
h
=4(1+ h)
3! 4
h
d) True. The difference quotient is not defined for h = 0.
Chapter 1 Section 2 Question 8 Page 20
a) h = 0.1:
(!3+ 0.1)2! (!3)
2
0.1= −5.9
h = 0.01:
(!3+ 0.01)2! (!3)
2
0.01= −5.99
h = 0.001:
(!3+ 0.001)2! (!3)
2
0.001= −5.999
MHR • Calculus and Vectors 12 Solutions 21
b)
(!3+ h)2! (!3)
2
h=
9 ! 6h + h2! 9
h= h ! 6
h = 0.1: 0.1 – 6 = −5.9
h = 0.01: 0.01 – 6 = −5.99
h = 0.001: 0.001 – 6 = −5.999
c) Answers may vary. For example: The answers from part a) and part b) are the same. This makes sense
since the expression that is used in part b) is a simplified form of the expression in part a). As the
interval h is decreased, the calculated result for the difference quotient is getting closer to −6. The
final estimate of the instantaneous rate of change is −6.
Chapter 1 Section 2 Question 9 Page 20
(4 + h)4! 4
4
h=
44
+ 4(4)3 h + 6(4)
2 h2+ 4(4)h3
+ h4! 4
4
h= 256 + 96h +16h2
+ h3
h = 0.1: 256 + 96(0.1) + 16(0.1)2 + (0.1)
3
= 265.761
h = 0.01: 256 + 96(0.01) + 16(0.01)2 + (0.01)
3
= 256.961 601
h = 0.001: 256 + 96(0.001) + 16(0.001)2 + (0.001)
3
= 256.096 016
The final estimate of the slope at x = 4 is 256.
Chapter 1 Section 2 Question 10 Page 20
a) average rate of change:
(!3)2
+ 3(!3)! (22
+ 3(2))
!3! 2=!10
!5
= 2
b) average rate of change:
2(!3)!1! (2(2)!1)
!3! 2=!10
!5
= 2
c) average rate of change:
7(!3)2! (!3)
4! (7(2)
2! 2
4)
!3! 2=!30
!5
= 6
d) average rate of change:
!3! 2(!3)3! (2 ! 2(2)
3)
!3! 2=
65
!5
= !13
MHR • Calculus and Vectors 12 Solutions 22
Chapter 1 Section 2 Question 11 Page 20
a)
(2 + h)2
+ 3(2 + h)! (22
+ 3(2))
h=
4 + 4h + h2+ 6 + 3h ! 4 ! 6
h= 7 + h
The instantaneous rate of change at x = 2 is 7.
b)
2(2 + h)!1! (2(2)!1)
h=
4 + 2h !1! 3
h= 2
The instantaneous rate of change at x = 2 is 2.
c)
7(2 + h)2! (2 + h)
4! (7(2)
2! (2)
4)
h=
7(h2+ 4h + 4)! (h4
+ 4(2)h3+ 6(2)
2 h2+ 4(2)
3 h + 24)!12
h= !h3
! 8h2!17h ! 4
The instantaneous rate of change at x = 2 is −4.
d)
(2 + h)! 2(2 + h)3! (2 ! 2(2)
3)
h=
2 + h ! 2(h3+ 3h2
(2) + 3h(2)2
+ 23)! (!14)
h= !23!12h ! 2h2
The instantaneous rate of change at x = 2 is −23.
Chapter 1 Section 2 Question 12 Page 20
a) i)
2(a + h)2! 2a2
h=
2a2+ 4ah + 2h2
! 2a2
h= 4a + 2h
4(–3) + 2(0.01) = –11.98
ii)
(a + h)3! a3
h=
a3+ 3a2h + 3ah2
+ h3! a3
h= 3a2
+ 3ah + h2
3(–3)
2 + 3(–3)(0.01) + (0.01)
2 = 26.9101
iii)
(a + h)4! a4
h=
a4+ 4a3h + 6a2h2
+ 4ah3+ h4
! a4
h= 4a3
+ 6a2h + 4ah2+ h3
4(–3)
3 + 6(–3)
2(0.01) + 4(–3)(0.01)
2 + (0.01)
3 = –107.461 199
MHR • Calculus and Vectors 12 Solutions 23
b) Answers may vary. For example:
Each answer represents the estimate of the slope of the tangent line to the function at the point
where x = −3.
Chapter 1 Section 2 Question 13 Page 20 a) i)
f (x) = x2
ii) a = 4
iii) h = 0.01
iv) (a, f (a)) = (4, 16)
b) i) f (x) = x3
ii) a = 6
iii) h = 0.0001
iv) (a, f (a)) = (6, 216)
c) i) f (x) = 3x4
ii) a = −1
iii) h = 0.1
iv) (a, f (a)) = (−1, 3)
d) i) f (x) = !2x
ii) a = 8
iii) h = 0.1
iv) (a, f (a)) = (8, −16)
Chapter 1 Section 2 Question 14 Page 21
a)
!4.9(1+ h)2
+15(1+ h) +1! (!4.9(1)2
+15(1) +1)
h
b) Answers may vary. For example:
The expression is not valid for h = 0. Division by zero is not defined in the real number system.
MHR • Calculus and Vectors 12 Solutions 24
c) i)
!4.9(1+ 0.1)2
+15(1+ 0.1) +1!11.1
0.1= 4.71
ii)
!4.9(1+ 0.01)2
+15(1+ 0.01) +1!11.1
0.01= 5.151
iii)
!4.9(1+ 0.001)2
+15(1+ 0.001) +1!11.1
0.001= 5.1951
iv)
!4.9(1+ 0.0001)2
+15(1+ 0.0001) +1!11.1
0.0001= 5.199 51
d) Answers may vary. For example:
The instantaneous rate of change of the height of the soccer ball after 1 s is 5.2 m/s because as h gets
smaller, the rate of change of the height gets closer to 5.2 m/s.
e) Answers may vary. For example:
At time t = 1 s, the ball is moving upwards at a speed of 5.2 m/s.
f)
Chapter 1 Section 2 Question 15 Page 21
a) first 10 min:
60(25!10)2! 60(25! 0)
2
10 ! 0= !2400
The average rate of change of volume during the first 10 min is –2400 L/min.
last 10 min:
60(25! 25)2! 60(25!15)
2
25!15= !600
The average rate of change of volume during the last 10 min is –600 L/min.
Both rates of change are negative, but the rate of change of volume during the first 10 min is much
more negative since the oil is draining more quickly during this time.
MHR • Calculus and Vectors 12 Solutions 25
b) i)
60(25! (5+ h))2! 60(25! 5)
2
h=
60(h2! 40h + 400)! 60(400)
h= 60h ! 2400
The instantaneous rate of change of volume at t = 5 min is –2400 L/min.
ii)
60(25! (10 + h))2! 60(25!10)
2
h=
60(h2! 30h + 225)! 60(225)
h= 60h !1800
The instantaneous rate of change of volume at t = 10 min is –1800 L/min.
iii)
60(25! (15+ h))2! 60(25!15)
2
h=
60(h2! 20h +100)! 60(100)
h= 60h !1200
The instantaneous rate of change of volume at t = 15 min is –1200 L/min.
iv)
60(25! (20 + h))2! 60(25! 20)
2
h=
60(h2!10h + 25)! 60(25)
h= 60h ! 600
The instantaneous rate of change of volume at t = 20 min is –600 L/min.
Answers may vary. For example:
The rate of change in the flow of oil may be slowing because of the shape of the tank and a
lessening of pressure.
c)
Chapter 1 Section 2 Question 16 Page 21
a) S.A.:
4! (252)" 4! (20
2)
25" 20 or approximately 565.5 cm
2/cm
V:
4
3! (25
3)"
4
3! (20
3)
25" 20 or approximatley 6387.9 cm
3/cm
Since the surface area and volume are decreasing, the average rate of change of the surface area is
–565.5 cm2/cm and the average rate of change of the volume is –6387.9 cm
3/cm when r decreases
from 25 cm to 20 cm.
MHR • Calculus and Vectors 12 Solutions 26
b) S.A.:
4! (10 + h)2" 4! (10)
2
h=
40!h2+ 80!h + 400! " 400!
h= 40!h + 80!
V:
4
3! (10 + h)
3"
4
3! (10)
3
h=
4
3! (h3
+ 3h2(10) + 3h(10)
2+10
3)"
4000
3!
h
=4
3!h2
+ 40!h + 400!
Since the surface area and volume are decreasing, the instantaneous rate of change of the surface area
is –80π = –251.3 cm2/cm and the instantaneous rate of change of the volume is
–400π = –1256.6 cm3/cm when r = 10 cm.
c) Answers may vary. For example:
The rate of change of the volume is greater than the rate of change of the surface area because it is a
h(x) does not exist, since the left- and right-hand limits are not equal.
d) h(!1) = 1
e) limx!3
"
h(x) = 2
f) limx!3
+
h(x) = 3
x = 2
MHR • Calculus and Vectors 12 Solutions 51
g) limx!3
h(x) does not exist, since the left- and right-hand limits are not equal.
h) h(3) = 2
Chapter 1 Section 4 Question 10 Page 45 Refer to page 35 for the list of limit properties.
a)
limx!6
8 = 8
Use property 1.
b)
limx!"3
x2" 9
x + 5=
limx!"3
x( )2
" limx!"3
9
limx!"3
x + limx!"3
5
=("3)
2" 9
"3+ 5
= 0
Use properties 1, 2, 3, 4, 7 and 8.
c)
limx!"5
6x + 2
x + 5=
6 limx!"5
x( ) + limx!"5
2
limx!"5
x + limx!"5
5
=6("5) + 2
"5+ 5
="28
0
The limit does not exist. Use properties 1, 2, 3, 5 and 7.
d)
limx!0
8" x3= lim
x!0
8" limx!0
x3
= 8" 03
= 2
Use properties 1, 2, 4 and 9.
Chapter 1 Section 4 Question 11 Page 46
a)
limx!2
(2 + x)2"16
x " 2= lim
x!2
x2+ 4x "12
x " 2
= limx!2
(x " 2)(x + 6)
x " 2
= limx!2
x + 6
= 8
MHR • Calculus and Vectors 12 Solutions 52
b)
limx!6
(3" x)2" 9
x " 6= lim
x!6
x2" 6x
x " 6
= limx!6
x(x " 6)
x " 6
= limx!6
x
= 6
c)
limx!2
49 " (5+ x)2
x " 2= lim
x!2
"(x2+10x " 24)
x " 2
= limx!2
"(x " 2)(x +12)
x " 2
= limx!2
"(x +12)
= "14
d)
limx!3
1
3"
1
xx " 3
= limx!3
x " 3
3xx " 3
= limx!3
1
3x
=1
9
e)
limx!"2
x4"16
x + 2= lim
x!"2
(x2+ 4)(x2
" 4)
x + 2
= limx!"2
(x2+ 4)(x + 2)(x " 2)
x + 2
= limx!"2
(x2+ 4)(x " 2)
= "32
f)
limx!1
x2"1
x3" x2
" 3x + 3= lim
x!1
(x +1)(x "1)
(x "1)(x2" 3)
= limx!1
(x +1)
(x2" 3)
= "1
MHR • Calculus and Vectors 12 Solutions 53
Chapter 1 Section 4 Question 12 Page 46
a)
limx!0
9 + x " 3
x= lim
x!0
9 + x " 3( ) 9 + x + 3( )
x 9 + x + 3( )
= limx!0
9 + x " 9
x 9 + x + 3( )
= limx!0
1
9 + x + 3
=1
6
b)
limx!25
5" xx " 25
= limx!25
5" x
x " 5( ) x + 5( )
= limx!25
"1
x + 5
= "1
10
c)
limx!4
x " 4
x " 2
= limx!4
x " 2( ) x + 2( )x " 2
= limx!4
x + 2
= 4
d)
limx!0
1" x "1
3x= lim
x!0
1" x "1( ) 1" x +1( )
3x 1" x +1( )
= limx!0
1" x "1
3x 1" x +1( )
= limx!0
"1
3 1" x +1( )
= "1
6
MHR • Calculus and Vectors 12 Solutions 54
e)
limx!0
3" x " x + 3
x= lim
x!0
3" x " x + 3( ) 3" x + x + 3( )
x 3" x + x + 3( )
= limx!0
3" x " (x + 3)
x 3" x + x + 3( )
= limx!0
"2
3" x + x + 3
= "1
3
Chapter 1 Section 4 Question 13 Page 46
a)
limx!"2
x2" 4
x + 2= lim
x!"2
(x " 2)(x + 2)
x + 2
= limx!"2
x " 2
= "4
b)
limx!0
3x2" x
x + 5x2= lim
x!0
x(3x "1)
x(1+ 5x)
= limx!0
3x "1
1+ 5x= "1
c)
limx!"3
x2" 9
x + 3= lim
x!"3
(x + 3)(x " 3)
x + 3
= limx!"3
x " 3
= "6
d)
limx!0
"2xx2
" 4x= lim
x!0
"2xx(x " 4)
= limx!0
"2
x " 4
=1
2
=
1
9
MHR • Calculus and Vectors 12 Solutions 55
e)
limx!"5
x2+ 4x " 5
25" x2= lim
x!"5
(x "1)(x + 5)
(5+ x)(5" x)
= limx!"5
x "1
5" x
="6
10
= "3
5
f)
limx!3
2x2" 5x " 3
x2" x " 6
= limx!3
(x " 3)(2x +1)
(x " 3)(x + 2)
= limx!3
2x +1
x + 2
=7
5
= −1
g)
limx!"4
3x2+11x " 4
x2+ 3x " 4
= limx!"4
(x + 4)(3x "1)
(x + 4)(x "1)
= limx!"4
3x "1
x "1
="13
"5
=13
5
Chapter 1 Section 4 Question 14 Page 46
a) lim
x!"1+
f (x) = "2
b) lim
x!"1"
f (x) = 1
c) limx!"1
f (x) does not exist since the left and right limits are not equal.
d) limx!0
f (x) = "1
MHR • Calculus and Vectors 12 Solutions 56
Chapter 1 Section 4 Question 15 Page 46 a)
b) Answers may vary. For example: The function is a piecewise linear function.
c) The graph is discontinuous at the following distances: 1 km, 2 km, 3 km, 4 km… In general, the graph
is discontinuous for all integer values of the distance. The graph has jump discontinuities.
Chapter 1 Section 4 Question 16 Page 46 a)
b) Answers may vary. For example:
The function is a piecewise linear function.
c) The graph is discontinuous for weights of 100 g, 200 g, and 500 g. The graph has jump
discontinuities.
Chapter 1 Section 4 Question 17 Page 46 a)
MHR • Calculus and Vectors 12 Solutions 57
b)
c)
Chapter 1 Section 4 Question 18 Page 46 a)
Explanations may vary. For example: The function is discontinuous at x = –4 since
lim
x!"4"
f (x) # limx!"4
+
f (x) .
b)
Explanations may vary. For example: The function is discontinuous at x = –1 since
lim
x!"1"
f (x) # limx!"1
+
f (x) .
Chapter 1 Section 4 Question 19 Page 47 a)
limx!0
[4 f (x)"1] = 4 limx!0
f (x)"1
= 4("1)"1
= "5
MHR • Calculus and Vectors 12 Solutions 58
b)
limx!0
[ f (x)]3
= [limx!0
f (x)]3
= ("1)3
= "1
c)
limx!0
[ f (x)]2
3" f (x)
=
[limx!0
f (x)]2
3" limx!0
f (x)
=("1)
2
3" ("1)
=1
2
Chapter 1 Section 4 Question 20 Page 47 a) Answers may vary. For example:
lim
x!"1"
f (x) = a " ("1)2
= a "1; limx!"1
+
f (x) = "1" b
Setting a – 1 = –1 – b gives a = –b. However, f is discontinuous when the left and right limits are not
equal so when a ! "b .
For instance, the values a = 2 and b = 2 will make the function discontinuous at x = –1.
b) Answers may vary. For example:
c) i)
limx!"1
+
f (x) = "1" 2
= "3
ii)
limx!"1
"
f (x) = 2 " ("1)2
= 1
iii) limx!"1
f (x) does not exist since
lim
x!"1"
f (x) # limx!"1
+
f (x) .
iv)
f (!1) = 2 ! (!1)2
= 1
MHR • Calculus and Vectors 12 Solutions 59
Chapter 1 Section 4 Question 21 Page 47 a) Answers may vary. For example:
From the previous question, f was found to be discontinuous when a ! "b , thus f is continuous when
a = –b.
For instance, the values a = 4 and b = –4 will make the function continuous at x = –1.
b)
c) i)
limx!"1
+
f (x) = "1+ 4
= 3
ii)
limx!"1
"
f (x) = 4 " ("1)2
= 3
iii) limx!"1
f (x) = 3
iv)
f (!1) = 4 ! (!1)2
= 3
Chapter 1 Section 4 Question 22 Page 47 Answers may vary. For example:
a)
b)
MHR • Calculus and Vectors 12 Solutions 60
Chapter 1 Section 4 Question 23 Page 47 a)
b)
limx!4
"
16 " x2= 16 " 4
2
= 0
c) Answers may vary. For example:
The function is not defined in the real number system for x-values that are greater than 4.
Therefore the
limx!4
+
16 " x2 does not exist.
d)
limx!4
16 " x2 does not exist since
limx!4
+
16 " x2 does not exist.
Chapter 1 Section 4 Question 24 Page 47
a)
limx!8
2 " x3
8" x= lim
x!8
2 " x3
2 " x3
( ) 4 + 2 x3+ x23
( )
= limx!8
1
(4 + 2 x3+ x23
)
=1
12
b)
limx!2
x5" 32
x " 2= lim
x!2
(x " 2)(x4+ 2x3
+ 4x2+ 8x +16)
x " 2
= limx!2
x4+ 2x3
+ 4x2+ 8x +16
= 80
c)
limx!2
6x3"13x2
+ x + 2
x " 2= lim
x!2
(x " 2)(2x "1)(3x +1)
x " 2
= limx!2
(2x "1)(3x +1)
= 21
MHR • Calculus and Vectors 12 Solutions 61
Chapter 1 Section 4 Question 25 Page 47 a)
b) Answers may vary. For example:
A possible equation for this function is
f (x) =
2x !11
x ! 5. The vertical asymptote is x = 5, the
horizontal asymptote is y = 2, f (4) = 3 , and f (6) = 1.
The function satisfies all of the required conditions when graphed using a graphing calculator.
Chapter 1 Section 4 Question 26 Page 47
6x(6 !1) = 3
x(3
4!1)
6x(5) = 3
x(80)
6x
= 3x(16)
2x3
x= 3
x(16)
2x
= 16
x = 4
Chapter 1 Section 4 Question 27 Page 47
(6x2! 3x)(2x2
!13x ! 7)(3x + 5) = (2x2!15x + 7)(6x2
+13x + 5)
L.S. = 3x(2x !1)(2x +1)(x ! 7)(3x + 5)
R.S. = (2x !1)(2x +1)(x ! 7)(3x + 5)
Both sides of the equations are zero when
x =
1
2, x = !
1
2, x = 7, and x = !
5
3.
When the factors are cancelled, the equation remaining is 3x = 1, so
x =
1
3 is also a solution.
MHR • Calculus and Vectors 12 Solutions 62
Chapter 1 Section 4 Question 28 Page 47
log(2cos x )
6 + log(2cos x )
sin x = 2
log(2cos x )
6sin x = 2
(2cos x)2
= 6sin x4cos
2 x = 6sin x4(1! sin
2 x) = 6sin x4sin
2 x + 6sin x ! 4 = 0
(2sin x !1)(2sin x + 4) = 0
sin x =1
2 ( sin x ! ! 2)
x ="
6 or 30
!
MHR • Calculus and Vectors 12 Solutions 63
Chapter 1 Section 5 Introduction to Derivatives
Chapter 1 Section 5 Question 1 Page 58 a) The derivative is C, since y is a quadratic function and the slope goes from negative to positive.
b) The derivative is A, since y is a cubic function and the slope is not negative anywhere. c) The derivative is B, since y is a linear function and the slope of the graph is negative and constant. Chapter 1 Section 5 Question 2 Page 58 a)
!f (x) = 3x2 b) i)
!f ("6) = 3("6)2
= 108
ii)
!f ("0.5) = 3("0.5)2
= 0.75
iii)
!f 2
3
"
#$%
&'= 3
4
9
"
#$%
&'
=4
3
iv)
!f (2) = 3(2)2
= 12
c) i) When x = –6, f (!6) = !216
Use the point (–6, –216) and m = 108 in the equation of the tangent y = mx + b to find b.
–216 = 108(–6) + b
b = 432
The equation of the tangent is y = 108x + 432.
ii) When x = –0.5, f (!0.5) = !0.125
Use the point (–0.5, –0.125) and m = 0.75 in the equation of the tangent y = mx + b to find b.
–0.125 = 0.75(–0.5) + b
b = 0.25
The equation of the tangent is y = 0.75x + 0.25.
MHR • Calculus and Vectors 12 Solutions 64
iii) When x =
2
3,
f 2
3
!
"#$
%&=
8
27
Use the point
2
3,
8
27
!
"#$
%& and m =
4
3 in the equation of the tangent y = mx + b to find b.
8
27=
4
3
2
3
!
"#$
%&+ b
b = '16
27
The equation of the tangent is
y =
4
3x ! 16
27.
iv) When x = 2, f (2) = 8
Use the point (2, 8) and m = 12 in the equation of the tangent y = mx + b to find b.
8 = 12(2) + b
b = –16
The equation of the tangent is y = 12x – 16.
Chapter 1 Section 5 Question 3 Page 58 Answers may vary. For example:
The function is not differentiable at the given x-value, possibly because it is discontinuous or the function
makes an abrupt change.
Chapter 1 Section 5 Question 4 Page 58 a)
!f (x) = 1 b) i) !f ("6) = 1
ii) !f ("0.5) = 1
iii) !f 2
3
"
#$%
&'= 1
iv) !f (2) = 1
MHR • Calculus and Vectors 12 Solutions 65
Chapter 1 Section 5 Question 5 Page 58 a)
f (x) = 3x
b) f (x) = x2
c) f (x) = 4x3
d) f (x) = !6x2
e) f (x) =
5
x
f) f (x) = x
Chapter 1 Section 5 Question 6 Page 58
a) !f (x) = "
1
x2
b) i) !f ("6) = "
1
36
ii) !f ("0.5) = "4
iii) !f 2
3
"
#$%
&'= (
9
4
iv) !f (2) = "
1
4
c) i) When x = –6,
f (!6) = !
1
6.
Use the point
!6, !1
6
"
#$%
&' and m =
!1
36 in the equation of the tangent y = mx + b to find b.
!1
6= !
1
36(!6) + b
b = !1
3
The equation of the tangent is
y = !
1
36x ! 1
3.
MHR • Calculus and Vectors 12 Solutions 66
ii) When x = –0.5, f (!0.5) = !2 .
Use the point (–0.5, –2) and m = –4 in the equation of the tangent y = mx + b to find b.
–2 = –4(–0.5) + b
b = –4
The equation of the tangent is y = –4x – 4.
iii) When x =
2
3,
f 2
3
!
"#$
%&=
3
2.
Use the point
2
3,
3
2
!
"#$
%& and m =
!9
4 in the equation of the tangent y = mx + b to find b.
3
2= !
9
4
2
3
"
#$%
&'+ b
b = 3
The equation of the tangent is
y = !
9
4x + 3 .
iv) When x = 2,
f (2) =
1
2.
Use the point
2, 1
2
!
"#$
%& and m =
!1
4 in the equation of the tangent y = mx + b to find b.
1
2= !
1
4(2) + b
b = 1
The equation of the tangent is
y = !
1
4x +1 .
Chapter 1 Section 5 Question 7 Page 59 Explanations may vary. For example:
a) x ! (−∞, −1) or (−1, ∞)
The function is differentiable for all values of x except at x = –1 since there is an abrupt change in
slope at x = –1.
b) x ! (−∞, ∞)
The function is differentiable for all values of x since there are no discontinuities or abrupt changes.
c) x ! (3, ∞)
The function is differentiable for all values of x > 3 since the function is not defined for x < 3 and x = 3
is an endpoint.
MHR • Calculus and Vectors 12 Solutions 67
d) x ! (−∞, −1) or (−1, ∞)
The function is differentiable for all values of x except at x = –1 since this point is an infinite