Chapters 1 to 8 Course Review
Chapters 1 to 8 Course Review
a) i)
ii)
iii)
Question 1 Page 509
[2(16) ! 12 + 4] ! [2 ! 3 + 4] 21
=
4 !1
3
=7
[2(2.25) ! 4.5 + 4] ! [2 ! 3 + 4] 1
=
1.5 ! 1
0.5
=2
[2(1.21) ! 3.3 + 4] ! [2 ! 3 + 4] 0.12
=
0.1
1.1! 1
= 1.2
b) The trend is for the average slope at x = 1 to decrease. The slope at x = 1 is 1.
Chapters 1 to 8 Course Review
Question 2 Page 509
a) Average rate, since the speed is over the period of time that Ali drove.
b) Instantaneous rate, since the velocity was measured at a specific moment.
c) Average rate, since the temperature dropped over the hours of the night.
d) Instantaneous rate, since the leakage was measured at a specific moment.
Chapters 1 to 8 Course Review
a) i)
Question 3 Page 509
3(a + h)2 ! 3a 2 3a 2 + 6ah + 3h2 ! 3a 2
=
h
h
2
6ah + 3h
=
h
= 6a + 3h
6(2) + 3(0.01) = 12.03
ii)
(a + h)3 ! a 3 a 3 + 3a 2 h + 3ah 2 + h3 ! a 3 3a 2 h + 3ah 2 + h3
=
=
h
h
h
= 3a 2 + 3ah + h 2
3(2)2 +3(2)(0.01) + (0.01)2 = 12.0601
b) i) This is an approximation of the value of the slope of the tangent to f (x) = 3x 2 at x = 2.
MHR • Calculus and Vectors 12 Solutions 1020
ii) This is an approximation of the value of the slope of the tangent to f (x) = x 3 at x = 2.
Chapters 1 to 8 Course Review
a)
Question 4 Page 509
!4.9[(2 + h)2 ! 22 ] + 6h !4.9(4h + h2 ) + 6h
=
h
h
= !19.6 ! 4.9h + 6
= !13.6 ! 4.9h
The average rate of change is (–13.6 – 4.9h) m/s.
b) Choose the interval 1.9 ≤ t ≤ 2.1.
!4.9(2.12 ! 1.92 ) + 6(0.2) !4.9(0.2)(4) + 6(0.2)
=
0.2
0.2
= !19.6 + 6
= !13.6
The instantaneous rate of change is –13.6 m/s.
Alternatively, let h = 0 in the expression in part a).
c)
Chapters 1 to 8 Course Review
Question 5 Page 509
a) No limit; the sequence does not converge.
b) The limit is 5.
c) No limit; the sequence does not converge.
d) The limit is 0.
MHR • Calculus and Vectors 12 Solutions 1021
Chapters 1 to 8 Course Review
Question 6 Page 509
a) lim(3x 2 " 4x + 1) = 5
x!2
b) lim
x!"8
5x + 40
5(x + 8)
= lim
x!"8 x + 8
x+8
=5
c) lim x " 6 = 0
x!6+
d) The limit does not exist. The graph of the function has a vertical asymptote at x = 3.
Chapters 1 to 8 Course Review
Question 7 Page 509
a)
b) i) lim f (x) = 17
x!3
ii) The limit does not exist. The limits on the left and right sides are unequal.
Chapters 1 to 8 Course Review
a)
Question 8 Page 509
dy
f (x + h) " f (x)
= lim
dx h!0
h
4(x + h)2 " 3 " 4x 2 " 3
= lim
h!0
h
2
8xh + h
= lim
h!0
h
= lim (8x + h)
(
) (
)
h!0
= 8x
MHR • Calculus and Vectors 12 Solutions 1022
At x = 2,
dy
= 16 and y = 13.
dx
The equation of the tangent is:
y ! 13 = 16( x ! 2)
y = 16 x ! 19
f (x + h) # f (x)
h
(x + h)3 # 2(x + h)2 # x 3 # 2x 2
b) f !(x) = lim
h"0
= lim
(
) (
)
h
3x 2 h + 3xh2 + h3 # 4xh # 2h2
= lim
h"0
h
2
2
= lim 3x + 3xh + h # 4x # 2h
h"0
h"0
= 3x 2 # 4x
At x = 2, f !(x) = 4 and f(x) = 0.
The equation of the tangent is:
y ! 0 = 4(x ! 2)
y = 4x ! 8
f (x + h) # f (x)
h
$ 3
3'
&% x + h # x )(
= lim
h"0
h
$ 3x # 3x # 3h '
&% x(x + h) )(
= lim
h"0
h
#3h
= lim
h"0 xh(x + h)
c) g !(x) = lim
h"0
= lim
h"0
=#
#3
x(x + h)
3
x2
At x = 2, g !(x) = !
3
3
and g(x) = .
4
2
The equation of the tangent is:
MHR • Calculus and Vectors 12 Solutions 1023
y!
3
3
= ! (x ! 2)
4
2
3
y = ! x+3
4
f (x + h) # f (x)
h"0
h
d) h!(x) = lim
(2 x + h # 2 x )
h"0
h
= lim
= lim
(2 x + h # 2 x )(2 x + h + 2 x )
h(2 x + h + 2 x )
h"0
= lim =
h"0
= lim
h"0
= lim
h"0
=
(4(x + h) # 4x)
h(2 x + h + 2 x )
4h
h(2 x + h + 2 x )
4
(2 x + h + 2 x )
1
x
At x = 2, h!(x) =
1
and h(x) = 2 2 .
2
The equation of the tangent is:
y!2 2 =
y=
1
2
1
(x ! 2)
x+ 2
2
Chapters 1 to 8 Course Review
Question 9 Page 509
a)
MHR • Calculus and Vectors 12 Solutions 1024
b)
Chapters 1 to 8 Course Review
a)
Question 10 Page 509
dy
= !6x + 4
dx
b) f !(x) = "6x "2 + 10x "3
c) f !(x) = 2x
=
"
1
2
2
x
d)
e)
" 1 !1 %
dy
= (3x 2 ! 4x) $ x 2 ' +
dx
&
#2
( x ! 1) (6x ! 4)
dy 3x(2x) ! (x 2 + 4)(3)
=
dx
9x 2
6x 2 ! 3x 2 ! 12
=
9x 2
(x ! 2)(x + 2)
=
3x 2
Chapters 1 to 8 Course Review
Question 11 Page 510
Answers may vary. For example:
a) Let f(x) = x2 and g(x) = 2.
f !(x) " g !(x) = 2x " 0
( f (x) ! g(x))" = (x 2 ! 2)
= 4x
L.S. ≠ R.S.
=0
Therefore, the statement is false.
MHR • Calculus and Vectors 12 Solutions 1025
b) Let f(x) = 2x3 + x2 and g(x) = x.
! f (x) $ ' ! 2x 3 + x 2 $ '
&
#" g(x) &% = #"
x
%
(
= 2x 2 + x
= 4x + 1
f !(x) 6x 2 + 2x
=
g !(x)
1
= 6x 2 + 2x
)'
L.S. ≠ R.S
Therefore, the statement is false.
Chapters 1 to 8 Course Review
Question 12 Page 510
a) f !(x) = 4x
At x = –2, f !(x) = –8 and f(x) = 7.
Find b.
y = mx + b
b = 7 ! 16
b = !9
The equation of the tangent is y = –8x – 9.
1 " 12
b) g !(x) = x
2
At x = 4, g !(x) =
1
and g(x) = 7.
4
Find b.
y = mx + b
b = 7 !1
b=6
The equation of the tangent is y =
Chapters 1 to 8 Course Review
1
x – 6.
4
Question 13 Page 510
a) v(t) = s!(t)
= 6t 2 " 14t + 4
b) v(5) = 150 ! 70 + 4
= 84
MHR • Calculus and Vectors 12 Solutions 1026
The velocity is 84 m/s.
Chapters 1 to 8 Course Review
Question 14 Page 510
The slope is 0 when f !(x) = 0 .
Set f(x) = 0.
3x2 – 4x – 4 = 0
x=2
or
x= !
2
3
Evaluate for these values of x.
" 2%
8 8 8
f $! ' = ! ! + + 4
27 9 3
# 3&
f (2) = 8 ! 8 ! 8 + 4
= !4
40
+4
27
13
=5
27
=
" 2
# 3
The points are (2, –4) and $ ! , 5
Chapters 1 to 8 Course Review
13 %
.
27 '&
Question 15 Page 510
a) h(t) = !4.9t 2 + 28t + 2.5
b) v(t) = h!(t)
= "9.8t + 28
a(t) = v !(t)
= "9.8
c) h(2) = !4.9(4) + 28(2) + 2.5
= 38.9
v(2) = !19.6 + 28
= 8.4
a(2) = !9.8
Height is 38.9 m, velocity is 8.4 m/s, and acceleration is –9.8 m/s2.
MHR • Calculus and Vectors 12 Solutions 1027
d) When t = 2, v = 8.4. Set v = 8.4 and solve for t.
!9.8t + 28 = !8.4
!36.4
!9.8
t = 3.7
t=
After 3.7 s, the shell will have the same velocity and be falling downward.
Chapters 1 to 8 Course Review
Question 16 Page 510
Chapters 1 to 8 Course Review
Question 17 Page 510
a)
dy
= 4(3x ! x !1 )(3 + x !2 )
dx
1
2
"
1
b) g !(x) = (2x + 5) 2 (2)
1
=
c)
2x + 5
3
!
1
dy
= ! (3x ! 1) 2 (3)
2
dx
3
=!
3
2 3x ! 1
(
d) y =
)
!1
3
x 2 + 3x
!
1
= !(x 2 + 3x) 3
4
!
dy 1 2
= (x + 3x) 3 (2x + 3)
dx 3
(2x + 3)
=
4
3
3 x 2 + 3x
(
)
MHR • Calculus and Vectors 12 Solutions 1028
Chapters 1 to 8 Course Review
Question 18 Page 510
f !(x) = x 2 (3(x 3 " 3x)2 )(3x 2 " 3) + 2x(x 3 " 3x)3
= (x 3 " 3x)2 (3x 2 )(3x 2 " 3) + 2x(x 3 " 3x)
At x = –1, f !(x) = –16 and f(x) = 8.
Find b.
y = mx + b
b = 8 ! 16
b = !8
The equation of the tangent is y = –16x – 8.
Chapters 1 to 8 Course Review
Question 19 Page 510
C(2000) = 15 + 0.1 2000
= 19.47
The cost is $19.47.
" 1% (1
C !(w) = 0.1$ ' w 2
# 2&
1
(
" 1%
C !(2000) = 0.1$ ' (2000) 2
# 2&
= 0.0011
The rate of change of the cost is $0.0011/L.
Chapters 1 to 8 Course Review
Question 20 Page 510
a) The demand function is the number of orders that can be sold.
D(x) = 425 + 20x, where x is the number of $0.10 decreases and D(x) is the demand.
b) Revenue = sales × price of one order
R(x) = (2.75 ! 0.1x)(425 + 20x)
c) R!(x) = (2.75 " 0.1x)(20) + ("0.1)(425 + 20x)
= 55 " 2x " 42.5 " 2x
= 12.5 " 4x.
MHR • Calculus and Vectors 12 Solutions 1029
d) Set R!(x) = 0 and solve for x.
0 = 12.5 ! 4x
4x = 12.5
x = 3.125
Find the price when x = 1.125.
2.75 – (0.1)(3.125) = 2.44
The price is $2.44. The total revenue will be maximized after 3.125 decreases of $0.10, and at
a cost of $2.44 per order.
Chapters 1 to 8 Course Review
Question 21 Page 510
Let s = gt 2 + v0t + s0 , where t is time in seconds and g, v0, and s0 and constants.
v(t) = 2gt + v0
a(t) = 2g
100
m/s and s0 = 0.
3
25
25
a = –10 km/h/s or !
m/s2 and g = ! .
9
18
At t = 0, v0 = 120 km/h or
Find the value of t when v(t) = 0.
25 100
t+
9
3
t = 12
0=!
After 12 s, the car has stopped.
25
100
(144) +
(12)
18
3
= 200
s=!
The car's stopping distance is 200 m.
Chapters 1 to 8 Course Review
Question 22 Page 512
a) A polynomial will have the limit ±! .
Test a large value.
f (100) = 1 950 892 .
Therefore, lim (2x 3 # 5x 2 + 9x # 8) = " .
x!"
MHR • Calculus and Vectors 12 Solutions 1030
b) Use the laws of limits.
1
1+
x +1
x
lim
= lim
x!" x # 1 x!"
1
1#
x
$ 1'
lim 1+ )
x!" &
x(
%
=
$ 1'
lim & 1+ )
x!" %
x(
Divide the numerator and denominator by x.
1
1
=1
=
c) Use the laws of limits.
3 1
+
x " 3x + 1
x x2
= lim
lim 2
x!"# x + 4x + 8
x!"#
4 8
1+ + 2
x x
$
3 1'
lim & 1" + 2 )
x!"# %
x x (
=
$ 4 8'
lim & 1+ + 2 )
x!"# %
x x (
1"
2
Divide the numerator and denominator by x2.
1
1
=1
=
Chapters 1 to 8 Course Review
Question 23 Page 512
Answers may vary. For example:
The graph must have a local minimum at (–2, –5), a point of inflection at (–1, –1) and a local maximum at
(0, 3). A possible graph is shown below.
(x ! [–4, 2], y ! [–6, 6])
MHR • Calculus and Vectors 12 Solutions 1031
Chapters 1 to 8 Course Review
a)
Question 24 Page 512
f (x) = x 3 + 2x 2 ! 4x + 1
f "(x) = 3x 2 + 4x ! 4
f ""(x) = 6x + 4
Find the critical numbers.
3x 2 + 4x ! 4 = 0
(3x ! 2)(x + 2) = 0
x =–2
or
x=
2
3
Use the second derivative test to classify the critical points.
f !!("2) = "8
# 2&
f !! % ( = 8
$ 3'
! 2 59 "
Therefore, (–2, 9) is a local maximum point and # ,
$ is a local minimum.
% 3 27 &
Find the possible points of inflection.
6x + 4 = 0
x=!
2
3
Have already tested the interval around this value.
! 2 115 "
Therefore, $ # ,
% is a point of inflection.
& 3 27 '
2
2
Therefore, f is increasing for x < !2 and x > and decreasing for !2 < x < .
3
3
2
2
Also, f is concave down for x < ! and concave up for x > ! .
3
3
b)
f (x) = !3x 4 ! 2x 3 + 15x 2 ! 12x + 2
f "(x) = !12x 3 ! 6x 2 + 30x ! 12
f ""(x) = !36x 2 ! 12x + 30
Find the critical numbers.
!12x 3 ! 6x 2 + 30x ! 12 = 0
!6(2x 3 + x 2 ! 5x + 2) = 0
!6(x ! 1)(2x 2 + 3x ! 2) = 0
!6(x ! 1)(2x ! 1)(x + 2) = 0
MHR • Calculus and Vectors 12 Solutions 1032
x=1
or
x = –2
or
x=
1
2
Use the second derivative test to classify the critical points.
f !!("2) = "90
# 1&
f !! % ( = 15
$ 2'
f !!(1) = "18
!1
"
Therefore, (–2, 54) and (1, 0) are local maximum points and $ , # 0.6875 % is a local minimum.
&2
'
Find the possible points of inflection.
!36x 2 ! 12x + 30 = 0
!6(6x 2 + 2x ! 5) = 0
x ! –1.09
x=
!2 ± 124
12
x=
!1 ± 31
6
or
x ! 0.76
Have already tested the interval around these values.
Therefore, (–1.09, 31.26) and (0.76, –0.33) are points of inflection.
Therefore, f is increasing for x < !2 and 0.5 < x < 1 and decreasing for !2 < x < 0.5 and x > 1 .
Also, f is concave down for x < !1.09 and x > 0.76 and concave up for !1.09 < x < 0.76 .
3
x +1
= 3(x 2 + 1)!1
c) f (x) =
2
f !(x) = "3(x 2 + 1)"2 (2x)
= "6x(x 2 + 1)"2
f !!(x) = "6(x 2 + 1)"2 + "6x(–2)(x 2 + 1)"3 (2x)
= ("6x 2 " 6 + 24x 2 )(x 2 + 1)"3
= 6(3x 2 " 1)(x 2 + 1)"3
Find the critical numbers.
!6x(x 2 + 1)!2 = 0
x=0
Use the second derivative test to classify the critical points.
f !!(0) = "6
MHR • Calculus and Vectors 12 Solutions 1033
Therefore, (0, 3) is a local maximum.
Find the possible points of inflection.
6(3x 2 ! 1)(x 2 + 1)!3 = 0
x=±
1
3
! 1 9"
! 1 9"
, % and $
, % are points of inflection.
Therefore, $ #
3 4'
&
& 3 4'
Therefore, f is increasing for x < 0 and decreasing for x > 0 .
1
1
1
1
<x<
and x >
Also, f is concave down for !
and concave up for x < !
.
3
3
3
3
Chapters 1 to 8 Course Review
Question 25 Page 512
a) Follow the six step plan.
f (x) = 3x 4 ! 8x 3 + 6x 2
f "(x) = 12x 3 ! 24x 2 + 12x
f ""(x) = 36x 2 ! 48x + 12
Step 1. Since this is a polynomial function, the domain is {x !!} .
Step 2. f (0) = 0
The y-intercept is 0.
For x-intercepts, let y = 0 .
3x 4 ! 8x 3 + 6x 2 = 0 .
x 2 (3x 2 ! 8x + 6) = 0
x=0
The second factor has not real roots. The only x-intercept is 0.
Step 3. Find the critical numbers.
12x 3 ! 24x 2 + 12x = 0
12x(x 2 ! 2x + 1) = 0
12x(x ! 1)(x ! 1) = 0
x = 0, x =1
Use the second derivative test to classify the critical points.
MHR • Calculus and Vectors 12 Solutions 1034
f !!(0) = 12
f !!(1) = 0
Check further.
f !!(1.1) = 2.76
f !!(0.9) = "2.04
Therefore, (0, 0) local minimum point and (1, 1) is a point of inflection.
Step 4. Find all possible points of inflection.
36x 2 ! 48x + 12 = 0
12(3x 2 ! 4x + 1) = 0
12(3x ! 1)(x ! 1) = 0
x=1
or
x=
1
3
Test the intervals. Have already tested the intervals for these values.
Therefore, (0.33, 0.41) is also a point of inflection.
Step 5. From Step 3, f is increasing for 0 < x < 1 and x > 1 and decreasing for x < 0 .
1
1
From Step 4, f is concave down for < x < 1 and concave up for x < and x > 1 .
3
3
Step 6. Sketch the graph.
b) Follow the six step plan.
f (x ) = ! x3 + x 2 + 8 x ! 3
f " (x ) = !3 x 2 + 2 x + 8
f "" (x ) = !6 x + 2
Step 1. Since this is a polynomial function, the domain is {x !!} .
Step 2. f (0) = !3
The y-intercept is –3.
For x-intercepts, let y = 0 .
!x 3 + x 2 + 8x ! 3 = 0 .
This expression is not factorable. There may be up to three x-intercepts.
MHR • Calculus and Vectors 12 Solutions 1035
Step 3. Find the critical numbers.
!3x 2 + 2x + 8 = 0
(!3x ! 4)(x ! 2) = 0
x=2
x= !
or
4
3
Use the second derivative test to classify the critical points.
f !!(2) = "10
# 4&
f !! % " ( = 10
$ 3'
Therefore, (–1.33, –9.5) is a local minimum point and (2, 9) is a maximum.
Step 4. Find possible points of inflection.
!6x + 2 = 0
x=
1
3
Test the intervals. Have already tested the intervals for these values.
Therefore (0.33, –0.26) is a point of inflection.
4
4
Step 4. From Step 3, f is increasing for ! < x < 2 and decreasing for x < ! and x > 2 .
3
3
1
1
From Step 4, f is concave down for x > and concave up for x < .
3
3
Step 5. Sketch the graph.
(x ! [–5, 10], Xscl = 2, y ! [–6, 6], Yscl = 20)
c) Follow the six step plan.
x
f (x) = 2
x +1
= x(x 2 + 1)!1
MHR • Calculus and Vectors 12 Solutions 1036
f !(x) = 1(x 2 + 1)"1 + x(–1)(x 2 + 1)"2 (2x)
= (x 2 + 1" 2x 2 )(x 2 + 1)"2
= ("x 2 + 1)(x 2 + 1)"2
f !!(x) = "2x("x 2 + 1)"2 + ("x 2 + 1)(–2)(x 2 + 1)"3 (2x)
= ("x 3 " 2x + 4x 3 " 4x)(x 2 + 1)"3
= (3x 3 " 6x)(x 2 + 1)"3
Step 1. Since this is a rational function, look for asymptotes. The denominator cannot equal zero.
There are no asymptotes. Therefore, the domain is {x !!} .
Step 2. f (0) = 0
The y-intercept is 0.
For x-intercepts, let y = 0 .
x
=0
x +1
x=0
The y-intercept is 0.
2
Step 3. Find the critical numbers.
(!x 2 + 1)(x 2 + 1)!2 = 0
!x 2 + 1 = 0
x = ±1
Use the second derivative test to classify the critical points.
3
f !!(1) = "
8
3
f !!("1) =
8
Therefore, (–0.375, –0.329) is a local minimum point and (0.375, 0.329) is a maximum.
Step 4. Find possible points of inflection.
(3x 3 ! 6x)(x 2 + 1)!3 = 0
3x 3 ! 6x = 0
3x(x 2 ! 2) = 0
x=0
or
x=2
or
x = –2
Test the intervals.
12
f !!("2) =
729
12
f !!(2) = "
729
MHR • Calculus and Vectors 12 Solutions 1037
Therefore, (0, 0), (–1.41, –0.47), and (1.41, 0.47) are points of inflection.
Step 5. From Step 3, f is increasing for !0.375 < x < 0.375 and decreasing for x < –0.375 and
x > 0.375.
From Step 4, f is concave down for x < 2 and 0 < x < 2 and concave up for
! 2 < x < 0 and x > 2 .
Step 6. Sketch the graph.
(x ! [–5, 5], y ! [–1, 1], Yscl = 0.25)
Chapters 1 to 8 Course Review
Question 26 Page 512
3 3
v
16
9
P"(v) = 100 ! v 2
16
a) P(v) = 100v !
To find the maximum, let P!(v) = 0 .
100 !
9 2
v =0
16
1600
v2 =
9
40
v=±
3
Since linear velocity cannot be negative, the maximum occurs when v = 13.3 m/s.
! 40 $
! 40 $ 3 ! 40 $
b) P # & = 100 # & ' # &
" 3%
" 3 % 16 " 3 %
3
! 888.9
The maximum power is about 889 A.
MHR • Calculus and Vectors 12 Solutions 1038
Chapters 1 to 8 Course Review
Question 27 Page 512
Draw a diagram.
Let t be the time beginning when they first observe the second ship. The ship travelling north is
(15 – 12t) km away from where they first observe the second ship. The second ship is 9t km away from
this same location. The distances form a right triangle with d representing the distance between the two
ships.
9t
15 – 12t
d
1
(
" 1%
d !(t) = $ ' (225t 2 ( 360t + 225) 2 (450t ( 360)
# 2&
d(t) = (9t)2 + (15 ! 12t)2
1
= (225t 2 ! 360t + 225) 2
(
1
= (225t ( 180)(225t 2 ( 360t + 225) 2
For a minimum, the derivative must equal zero.
!
1
(225t ! 180)(225t 2 ! 360t + 225) 2 = 0
225t ! 180 = 0
t = 0.8
The ships are closest when t = 0.8 h.
d(0.8) = (9(0.8))2 + (15 ! 12(0.8))2
=9
The closest distance of approach of the two ships is 9 km.
Chapters 1 to 8 Course Review
0.000 25x 2 + 8x + 10
x
= 0.000 25x + 8 + 10x !1
C(x) =
Question 28 Page 512
C '(x) = 0.000 25 ! 10x !2
a) C(50) gives the cost per pizza at the 50 pizza per day production level.
0.000 25(50)2 + 8(50) + 10
C(50) =
50
= 8.2125
MHR • Calculus and Vectors 12 Solutions 1039
For total cost:
50C(50) = 50(8.2125)
= 410.63
The total cost of production is $410.63.
b) For a minimum cost, let C !(x) = 0 .
0.000 25 ! 10x !2 = 0
0.000 25x 2 ! 10 = 0
x 2 = 40 000
x = ±200
The negative answer has no meaning for this situation.
The minimum average daily cost per pizza occurs at a production level of 200 pizzas per day.
0.000 25(200)2 + 8(200) + 10
200
= 8.1
c) C(200) =
The minimum average daily cost per pizza is $8.10.
Chapters 1 to 8 Course Review
a)
b)
Question 29 Page 511
dy
= 3cos 2 x(! sin x)
dx
= !3(cos 2 x)(sin x)
dy
= 3x 2 cos (x 3 )
dx
c) f !(x) = "5sin (5x " 3)
#
# x & # 1& &
#
# x&&
d) f !(x) = sin 2 x % " sin % ( % ( ( + 2(sin x)(cos x) % cos % ( (
$ 2' $ 2' '
$ 2''
$
$
#
#
# x & & # 1&
# x&&
= 2(sin x)(cos x) % cos % ( ( " % ( sin 2 x % sin % ( (
$ 2' ' $ 2'
$ 2''
$
$
e) f !(x) = 2(cos 4x 2 )(" sin (4x 2 )(8x))
= "16x(cos 4x 2 )(sin 4x 2 )
MHR • Calculus and Vectors 12 Solutions 1040
f) g !(x) =
(cos x " sin x)(" sin x) " cos x(" sin x " cos x)
(cos x " sin x)2
=
– sin xcos x + sin 2 x + cos x sin x + cos 2 x)
(cos x " sin x)2
=
1
(cos x " sin x)2
Chapters 1 to 8 Course Review
y = 2 + cos 2x at x =
Question 30 Page 511
5!
.
6
dy
= –2sin 2x.
dx
dy
5!
when x =
.
dx
6
"
dy
3%
= !2 $ !
'
dx
# 2 &
Find
= 3
When x =
5! dy
5
,
= 3 and y = .
6 dx
2
Solve for b.
y = mx + b
b=
5 5 3!
!
2
6
5
2
The equation of the tangent is y = 3x + !
Chapters 1 to 8 Course Review
y = sec x
= (cos x)!1
5 3"
.
6
Question 31 Page 511
dy
= !(cos x)!2 (! sin x)
dx
sin x
=
cos 2 x
= sec x tan x
MHR • Calculus and Vectors 12 Solutions 1041
y = csc x
= (sin x)!1
y = tan x
sin x
=
cos x
dy
= !(sin x)!2 (cos x)
dx
cos x
=! 2
sin x
= – csc x cot x
dy cos x(cos x) ! sin x(! sin x)
=
dx
cos 2 x
cos 2 x + sin 2 x
=
cos 2 x
1
=
cos 2 x
= sec 2 x
Chapters 1 to 8 Course Review
Question 32 Page 511
dy
1
= 2sin x cos x !
2
dx
Set
dy
= 0.
dx
1
=0
2
1
2sin x cos x =
2
1
sin 2x =
2
2sin x cos x !
!
2x = 2k! + , for k !!
6
!
!
and 2x = (2k " 1)! " for k !!
x = k! +
6
12
7!
x = k! "
12
d2 y
! 13! 25!
!
and negative at
= 2cos 2x is positive at x = ,
,
, ..., k! +
2
12 12 12
12
dx
5! 17! 29! 41!
7!
.
x= ,
,
,
, ..., k! !
12 12
12
12
12
There are local minima at x =
! 13! 25!
!
,
,
, ..., k! + .
12 12 12
12
MHR • Calculus and Vectors 12 Solutions 1042
There are local maxima at x =
5! 17! 29! 41!
7!
.
,
,
,
, ...k! !
12 12
12
12
12
The points of inflection are given by:
d2 y
= 2cos 2x
dx 2
=0
Points of inflection are 2x =
(2k + 1)!
(2k + 1)!
or x =
for k !! .
2
4
Chapters 1 to 8 Course Review
h!(t) =
Question 33 Page 511
" 2! %
2!
cos $ t '
3
# 30 &
At the maximum height, h!(t) = 0. Solve for t.
0=
" 2! %
2!
cos $
t
3
# 30 '&
!
2!
t=
30
2
t = 7.5
" 2! % " 30 %
h(7.5) = 10sin $ ' $ ' + 12
# 30 & # 4 &
=10sin
!
+ 12
2
= 22
The maximum height is 22 m and it first occurs at 7.5 s.
Chapters 1 to 8 Course Review
Question 34 Page 511
a) d !(t) = 6cos 6t + 24sin 6t
d !(1) = 6cos 6 + 24sin 6
= "0.945
The rate of change is –0.945 cm/s.
MHR • Calculus and Vectors 12 Solutions 1043
b) For the maximum and minimum, let d !(t) = 0.
6cos 6t + 24sin 6t = 0
6cos 6t = !24sin 6t
1
4
6t = !0.245
tan 6t = !
! " 0.245, 2! " 0.245,...
2! " 0.245
! " 0.245
= 0.48,
= 1.01 ,... since t > 0.
t=
6
6
At t = 0.48, d !!(t) = "36sin 6t + 144cos 6t is negative.
So there is a maximum displacement at t = 0.48.
d(0.48) = sin 6(0.48) ! 4cos 6(0.48)
= 4.123
The maximum displacement is 4.123 cm at 0.48 s.
At t = 1.01, d !!(t) = "36sin 6t + 144cos 6t is positive.
So there is a minimum displacement at t = 1.01.
d(1.01) = sin6(1.01) ! 4cos6(1.01)
= !4.12
The minimum displacement is –4.123 cm at 1.01 s.
Chapters 1 to 8 Course Review
Question 35 Page 511
a)
y = e x increases faster as x increases. Both graphs have the same horizontal asymptote and
pass through (0, 1).
b)
MHR • Calculus and Vectors 12 Solutions 1044
The graphs are f !(x) = e x and g !(x) = ln 2(2 x ) . f !(x) increases faster as x increases and f !(x)
passes through (0, 1) while g !(x) passes through (0, ln2). Both graphs have the same
horizontal asymptote.
c)
y = e x increases faster as x increases. y = e x has a horizontal asymptote and passes through
(0, 1), while the graph of y = 1nx has a vertical asymptote and passes through (1, 0).
Chapters 1 to 8 Course Review
Question 36 Page 511
a) ln 5 = 1.61
b) ln e 2 = 2
c) (ln e) 2 = 1
Chapters 1 to 8 Course Review
Question 37 Page 511
a) ln (e x ) = x
b) eln x = x
c) eln x = x 2
2
Chapters 1 to 8 Course Review
a)
Question 38 Page 511
dy
= !2e x
dx
b) g !(x) = 5(ln 10)(10 x )
c) h(x) = !e x sin (e x )
d) f !(x) = "xe" x + e" x
= e" x (1" x)
MHR • Calculus and Vectors 12 Solutions 1045
Chapters 1 to 8 Course Review
f !(x) =
Question 39 Page 512
1 x+1
e
2
At x = 0, f (0) =
e
e
and f !(0) = .
2
2
1
2
The equation of the line perpendicular to f (x) = e x+1 has a slope !
2
and passes through
e
! e$
#" 0, 2 &% :
y!
2
e
=! x
e
2
e
2
y=! x+
2
e
Chapters 1 to 8 Course Review
Question 40 Page 512
The equation for exponential decay is N = N 0 e " !t where N 0 = 20, is the initial amount, and ! the
rate of decay.
Since the half life is 1590:
10 = 20e! " (1590)
ln(0.5) = ! " (1590)
"=!
ln(0.5)
1590
When N = 2,
2 = 20e! "t
ln(0.1) = ! "t
t=
ln(0.1)(1590)
ln(0.5)
= 5282
It will take 5282 years for 90% of the mass to decay.
MHR • Calculus and Vectors 12 Solutions 1046
Chapters 1 to 8 Course Review
Question 41 Page 512
f !(x) = x 2 e x + 2xe x
Let f !(x) = 0.
x 2 e x + 2xe x = 0
e x (x 2 + 2x) = 0
x = 0 or x = –2 since ex > 0.
Chapters 1 to 8 Course Review
Question 42 Page 512
a) P!(t) = 200("0.001)e"0.001t
= "0.2e"0.001t
b) P!(200) = "0.2e"0.001(200)
= "0.164
The rate of change of power is –0.164 Watts/day.
Chapters 1 to 8 Course Review
a)
Question 43 Page 512
" e0.0329 x ! e!0.0329 x %
dy
= !20.96(0.0329) $
'
dx
2
#
&
= !0.344 792(e0.0329 x ! e!0.0329 x )
b)
dy
= !0.344 792(e0.0329(2) ! e!0.0329(2) )
dx
= !0.0454
MHR • Calculus and Vectors 12 Solutions 1047
c) The width can be found by letting y = 0.
" e0.0329 x + e!0.0329 x
%
!20.96 $
! 10.06' = 0
2
#
&
e0.0329 x + e!0.0329 x
= 10.06
2
e0.0329 x + e!0.0329 x = 20.12
Solve this equation using CAS:
x = –91.1622 and x = 91.1622.
Therefore, the width of the arch is about 182.3 m.
The height can be found by letting
dy
= 0.
dx
!0.344792(e0.0329 x ! e!0.0329 x ) = 0
e0.0329 x = e!0.0329 x
x=0
Find the value of y when x = 0.
" e0.0329(0) + e!0.0329(0)
%
y = !20.96 $
! 10.06'
2
#
&
= 20.96(9.06)
= 189.9
The height of the arch is 189.9 m.
Chapters 1 to 8 Course Review
Question 44 Page 512
MHR • Calculus and Vectors 12 Solutions 1048
a)
dy
= e! x cos x ! e! x sin x
dx
dy
= 0.
dx
cos x = sin x
tan x = 1
! 5! 9!
,
, ...
x= ,
4
4 4
Let
d2 y
= !e! x cos x ! e! x sin x + e! x sin x ! e! x cos x
2
dx
= !2e! x cos x
d2 y
!
5!
is negative at x = and positive at x =
.
2
4
4
dx
!
5!
There are maxima at x = 2k! + , k !! and minima at x = 2k! + , k !! .
4
4
b) The maximum and minimum points rapidly get closer and closer to zero as x increases.
!
:
4
"
! # 1 &
y=e 4%
$ 2 ('
= 1.55
When x =
9!
:
4
9"
# 1 &
!
y=e 4 %
$ 2 ('
=! 0.0006.
When x =
MHR • Calculus and Vectors 12 Solutions 1049
Chapters 1 to 8 Course Review
Question 45 Page 512
Answers may vary. For example:
a)
b)
Chapters 1 to 8 Course Review
Question 46 Page 512
Since BD||AE and BD = AE, ABDE is a parallelogram and opposite sides are equal and parallel.
MHR • Calculus and Vectors 12 Solutions 1050
!
a) !u
!
b) !v
c)
1!
u
5
!!!" !!!" !!!"
d) AD = AE + ED
" "
=u+v
!!!" !!!" !!!"
e) CD = CB + BD
1" "
= ! u+ v
5
!!!" !!!" !!!"
f) CE = CD + DE
"
1" "
= ! u + v + !u
5
6" "
= ! u+v
5
( )
Chapters 1 to 8 Course Review
Question 47 Page 512
! !
! !
!
a) 3u + 5u ! 7u ! 6u = (3 + 5 ! 7 ! 6)u
!
= !5u
! "!
! "!
! "! ! "!
b) !5 c + d ! 8 c ! d = !5c ! 5d ! 8c + 8d
! ! "! "!
= !5c ! 8c ! 5d + 8d
! "!
= !13c + 3d
( ) ( )
Chapters 1 to 8 Course Review
Question 48 Page 512
a) Draw a diagram. Use the Pythagorean theorem and trigonometry.
MHR • Calculus and Vectors 12 Solutions 1051
!"
R = 122 + 152
# 19.2
tan ! =
15
12
# 15 &
! = tan "1 % (
$ 12 '
! ! 51.3o
The resultant displacement is about 19.2 km in a direction N51.3ºE.
b) Draw a diagram. Use the Pythagorean theorem and trigonometry.
!"
R = 602 + 402
# 72.1
tan ! =
40
60
# 40 &
! = tan "1 % (
$ 60 '
! ! 33.7 o
The resultant force is about 72.1 N in a direction 56.3° up from the horizontal.
MHR • Calculus and Vectors 12 Solutions 1052
Chapters 1 to 8 Course Review
Question 49 Page 512
a) Draw a diagram. Use the Pythagorean theorem and trigonometry.
!"
R = 4502 + 152
# 450.2
It is not necessary to calculate angle θ since the ground velocity only involves the projection on the
ground of this motion.
The resultant ground velocity is about 450.2 km/h in the direction of the wind.
Chapters 1 to 8 Course Review
Question 50 Page 512
a) Draw a diagram of the situation.
b) The angle between the forces in the diagram is 160º.
Use the sine law to find the angle the second force makes with the resultant.
sin ! sin 160o
=
200
340
200sin 160o
sin ! =
340
# 200sin 160o &
! = sin "1 %
(
340
$
'
! ! 11.6o
MHR • Calculus and Vectors 12 Solutions 1053
To find the magnitude of the second force, use the sine law again.
!"
F
340
=
o
sin 160o
sin 8.4
!" 340sin 8.4o
F =
sin 160o
!"
F # 145.2
The second force has a magnitude of 145.2 N at an angle of 11.6o from the resultant force.
Chapters 1 to 8 Course Review
Question 51 Page 512
Draw the vector diagram involving only two cables. They will support half of the scoreboard.
250(9.8) = 2450
The downward force will be 2450 N.
It is useful to show the vectors forming an addition triangle since their sum must be the zero vector.
!"
!!"
Label the tension vectors as T1 and T2 as shown.
!"
To find T1 , use the sine law.
!"
T1
2450
=
o
sin 140o
sin 20
!" 2450sin 20o
T1 =
sin 140o
!"
T1 # 1303.6
By symmetry the two tensions are identical.
Each of the four tensions is approximately 1303.6 N at 70º to the horizontal.
MHR • Calculus and Vectors 12 Solutions 1054
Chapters 1 to 8 Course Review
Question 52 Page 512
Draw a diagram. Use the Pythagorean theorem.
!"
F v = 582 ! 402
= 42
The vertical component of the force is 42 N.
Chapters 1 to 8 Course Review
Question 53 Page 512
The force makes a 40º with the east direction.
120cos 40o ! 91.9
120sin 40o ! 77.1
The rectangular components along the east-west and north-south lines are 91.9 km/h to the east and
77.1 km/h to the south.
Chapters 1 to 8 Course Review
Question 54 Page 512
Draw a diagram.
!"
F ramp = 100cos 48o
!"
F perp. = 100sin 48o
# 66.9
# 74.3
The component parallel to the ramp (normal force by the surface) is 66.9 N.
The component perpendicular to the ramp (friction) is 74.3 N.
MHR • Calculus and Vectors 12 Solutions 1055
Chapters 1 to 8 Course Review
Question 55 Page 513
! !
!
a) 5i + 6 j ! 4k
!
!
b) !8 j + 7 k
Chapters 1 to 8 Course Review
Question 56 Page 513
!!!"
PQ =
!!!" !!!" !!!"
a) PQ = OQ ! OP
= "# !6, 3$% ! "# 2, 4 $%
= "# !8, ! 1$%
( !8) + ( !1)
2
2
= 65
!!!" !!!" !!!"
b) PR = OR ! OP
= "# 4, ! 10 $% ! "# 2, 4 $%
= "# 2, ! 14 $%
c) 3 "# !8, ! 1$% ! 2 "# 2, ! 14 $% = "# !24, ! 3$% ! "# 4, ! 28 $%
= "# !28, 25$%
d) "# !8, ! 1$% & "# 2, ! 14 $% = !8(2) + (–1)(–14)
= !16 + 14
= !2
Chapters 1 to 8 Course Review
Question 57 Page 513
a) Assume the axes to be the east-west and north-south axes.
!
u = !"30cos 70o , 30sin 70o #$
" !"10.3, 28.2 #$
b) Assume standard x- and y-axes.
!
v = !" 40cos 80o , 40sin 80o #$
" !"6.9, 39.4 #$
MHR • Calculus and Vectors 12 Solutions 1056
Chapters 1 to 8 Course Review
Question 58 Page 513
The ship’s vector is "# !20cos 56o , ! 20sin 56o $% .
The current’s vector is "# !11cos 3o , + 11sin 3o $% .
The resultant velocity is "# !20cos 56o ! 11cos 3o , ! 20sin 56o + 11sin 3o $% ! "# !22.2, ! 16.0 $% .
!"
R = (–22.2)2 + (–16)2
# 27.4
# "16 &
! = tan "1 %
$ "22.2 ('
! 35.8o
For the bearing, 270o ! 35.8o = 234.2o .
The resultant velocity is about 27 km on a bearing of 234.2°
Chapters 1 to 8 Course Review
Question 59 Page 513
Answers may vary. For example:
!
!
"!
a) Let u = !"1, 1, 1#$ , v = !"3, 2, 1#$ , and w = !"0, % 1, 4 #$ .
! ! "!
! ! "!
LS = u + v + w
RS = u + v + w
( )
(
(
)
)
(
= !"1, 1, 1#$ + !"3, 2, 1#$ + !"0, % 1, 4 #$
= !"1, 1, 1#$ + !"3, 2, 1#$ + !"0, % 1, 4 #$
= !" 4, 3, 2 #$ + !"0, % 1, 4 #$
= !"1, 1, 1#$ + !"3, 1, 5#$
= !" 4, 2, 6 #$
= !" 4, 2, 6 #$
)
L.S. = R.S.
MHR • Calculus and Vectors 12 Solutions 1057
!
!
b) Let a = !"1, 1, 1#$ , b = !"3, 2, 1#$ , and k = %2 .
! !
! !
R.S. = a ! kb
L.S. = k a ! b
( )
( )
= (–2)(1(3) + 1(2) + 1(1))
= "#1, 1, 1$% ! "# &6, & 4, & 2 $%
= (–2)(6)
= 1(–6) + 1(–4) + 1(–2)
= "12
= &12
L.S. = R.S.
Chapters 1 to 8 Course Review
Question 60 Page 513
! ! ! !
a) u ! v = u v cos "
= (12)(21)cos 20o
" 236.8
! ! ! !
b) s ! t = s t cos "
= (115)(150)cos 42o
" 12 819.2
Chapters 1 to 8 Course Review
Question 61 Page 513
! ! "!
a) u ! v + w = #$3, " 4 %& ! #$6, 1%& + #$ "9, 6 %&
(
)
(
)
= #$3, " 4 %& ! #$ "3, 7 %&
= 3(–3) + (–4)(7)
= "37
b) This is not possible because the result of the expression in brackets is a scalar. You cannot find the dot
product of a vector and a scalar.
! ! "!
c) u v ! w = #$3, " 4 %& #$6, 1%& ! #$ "9, 6 %&
( )
(
)
= #$3, " 4 %& (6(–9) + 1(6))
= #$3, " 4 %& (–48)
= "48 #$3, " 4 %&
= #$ "144, 192 %&
! ! ! !
d) u ! v " u + v = #$3, ! 4 %& + #$6, 1%& " #$3, ! 4 %& ! #$6, 1%&
( )( ) (
)(
)
= #$9, ! 3%& " #$ !3, ! 5%&
= 9(–3) + (–3)(–5)
= !12
MHR • Calculus and Vectors 12 Solutions 1058
Chapters 1 to 8 Course Review
Question 62 Page 513
!
!
! !
u and v are perpendicular if and only if u ! v = 0 .
"# !3, 7 $% & "#6, k $% = 0
!18 + 7k = 0
18
k=
7
Chapters 1 to 8 Course Review
Question 63 Page 513
!!!"
!!!"
!!!"
AB = "# !4, ! 1$% ; BC = "#7, ! 2 $% ;AC = "#3, ! 3$%
!!!" !!!"
AB ! BC = #$ "4, " 1%& ! #$7, " 2 %&
= "26
'0
!!!" !!!"
AB ! AC = #$ "4, " 1%& ! #$3, " 3%&
= "9
'0
!!!" !!!"
BC ! AC = #$7, " 2 %& ! #$3, " 3%&
= 27
'0
None of the vectors are perpendicular. The triangle is not a right triangle.
Chapters 1 to 8 Course Review
Question 64 Page 513
! !
u"v
cos ! = ! !
u v
cos ! =
cos ! =
$% 4, 10, # 2 &' " $%1, 7, # 1&'
42 + 102 + (–2)2 12 + 7 2 + (–1)2
76
120 51
cos ! " 0.9715
! = cos #1 (0.9175)
! = 13.7 o
The angle between the vectors is about 14º.
MHR • Calculus and Vectors 12 Solutions 1059
Chapters 1 to 8 Course Review
Question 65 Page 513
!" !"
a) W = F ! d
= "#1, 4 $% ! "#6, 3$%
= 1(6) + 4(3)
= 18 J
!" !"
b) W = F ! d
= "#320, 145$% ! "#32, 15$%
= 320(32) + 145(15)
= 12 415 J
Chapters 1 to 8 Course Review
Question 66 Page 513
! !
projv! u = u cos !
= 25cos 38o
" 19.7
The projection has magnitude 19.7 and has the same direction as v .
Chapters 1 to 8 Course Review
Question 67 Page 513
!" "
W = F !s
!" "
= F s cos "
!"
100 = F (3)cos 10o
!"
100
F =
3cos10o
!"
F = 33.8
Roni applies a force with a magnitude of about 33.8 N.
MHR • Calculus and Vectors 12 Solutions 1060
Chapters 1 to 8 Course Review
Question 68 Page 513
Answers may vary. For example:
a)
b)
c)
MHR • Calculus and Vectors 12 Solutions 1061
Chapters 1 to 8 Course Review
Question 69 Page 513
!!!" !!!" !!!"
AB = OB ! OA
!!!" !!!" !!!"
OB = AB + OA
= "#6, 3, ! 2 $% + "#1, 4, 5$%
= "#7, 7, 3$%
The terminal point is B (7, 7, 3).
Chapters 1 to 8 Course Review
Question 70 Page 513
! !
a"b
cos ! = ! !
a b
cos ! =
cos ! =
#$7, 2, 4 %& " #$ '6, 3, 0 %&
7 2 + 22 + 42 (–6)2 + 32 + 02
'36
69 45
( '36 +
! = cos '1 *
) 69 45 -,
! " 130.2o
! !
a"c
cos ! = ! !
a c
cos ! =
cos ! =
#$7, 2, 4 %& " #$ 4, 8, 6 %&
7 2 + 2 2 + 4 2 4 2 + 82 + 6 2
68
69 116
(
+
68
! = cos '1 *
) 69 116 -,
! " 40.5o
! !
b"c
cos ! = ! !
b c
cos ! =
$% #6, 3, 0 &' " $% 4, 8, 6 &'
(–6)2 + 32 + 02 42 + 82 + 62
MHR • Calculus and Vectors 12 Solutions 1062
0
cos ! =
45 116
! = cos "1 0
()
! = 90
o
!
!
The angle between a and b is 130.2°.
!
!
The angle between a and c is 40.5°.
!
!
The angle between b and c is 90°.
Chapters 1 to 8 Course Review
! ! "!
L.S. = u ! v + w
(
Question 71 Page 513
! ! ! "!
R.S. = u ! v + u ! w
)
(
= "#u1 , u2 , u3 $% ! "# v1 , v2 , v3 $% + "# w1 , w2 , w3 $%
)
= "#u1 ,u2 ,u3 $% ! "# v1 + w1 , v2 + w2 , v3 + w3 $%
(
)
(
)
(
= u1 v1 + w1 + u2 v2 + w2 + u3 v3 + w3
)
= "#u1 , u2 , u3 $% ! "# v1 , v2 , v3 $% + "#u1 , u2 , u3 $% ! "# w1 , w2 , w3 $%
= u1v1 + u2 v2 + u3v3 + u1 w1 + u2 w2 + u3 w3
= u1v1 + u1 w1 + u2 v2 + u2 w2 + u3v3 + u3 w3
= u1v1 + u1 w1 + u2 v2 + u2 w2 + u3v3 + u3 w3
L.S. = R.S.
! ! "! ! ! ! "!
Therefore, u ! v + w = u ! v + u ! w .
(
)
Chapters 1 to 8 Course Review
Question 72 Page 513
Answers may vary. For example:
Choose vectors that make a zero dot product with each of the vectors.
a) [5, 1, 0] and [0, –4, 5] are two possibilities.
b) [0, –1, 3] and [6, –5, 0] are two possibilities.
Chapters 1 to 8 Course Review
Question 73 Page 513
Ignoring wind the ground speed is 180cos 14o ! 174.7 km/h.
A top-view sketch shows the velocities.
The vector for the plane is [174.7, 0].
MHR • Calculus and Vectors 12 Solutions 1063
The wind vector is !"15cos 45o , 15sin 45o #$ ! !"10.61, 10.61#$ .
The resultant vector is [185.31, 10.61].
!"
R =
(185.31) + (10.61)
2
2
# 185.6
# 10.61 &
! = tan "1 %
$ 185.31('
! 3.3o
The ground velocity is about 185.6 km/h on a bearing of 86.7º.
Chapters 1 to 8 Course Review
Question 74 Page 513
! !
a ! b = #$ 4, " 3, 5%& ! #$ 2, 7, 2 %&
= #$ "3(2) " 7(5), 5(2) " 2(4), 4(7) " 2(–3) %&
= #$ "41, 2, 34 %&
! !
b ! a = "# 2, 7, 2 $% ! "# 4, & 3, 5$%
= "# 7(5) & (–3)(2), 2(4) & 5(2), 2(–3) & 4(7) $%
= "# 41, & 2, & 34 $%
For the vectors to be orthogonal, the dot products must be zero.
! ! !
a ! a " b = $% 4, # 3, 5&' ! $% #41, 2, 34 &'
! ! !
b ! a " b = #$ 2, 7, 2 %& ! #$ '41, 2, 34 %&
= 4(–41) + (–3)(2) + 5(34)
= 2(–41) + 7(2) + 2(34)
= #164 # 6 + 170
=0
= '82 + 14 + 68
=0
( )
( )
! !
! !
You do not have to check b ! a since this vector is the opposite of a ! b and has the same direction.
Chapters 1 to 8 Course Review
The area of a triangle is given by A =
!!!" !!!" !!!"
PQ = OQ ! OP
Question 75 Page 513
1 !!!" !!!"
PQ ! QR
2
!!!" !!!" !!!"
QR = OR ! OQ
= "# !2, 5, 7 $% ! "#3, 4, 8 $%
= "# !5, ! 1, 6 $% ! "# !2, 5, 7 $%
= "# !5, 1, ! 1$%
= "# !3, ! 6, ! 1$%
MHR • Calculus and Vectors 12 Solutions 1064
1
" !5, 1, ! 1$% & "# !3, ! 6, ! 1$%
2 #
1
= 1(–1) ! (–6)(–1), (–1)(–3) ! (–1)(–5), (–5)(–6) ! (–3) 1
2
1
= "# !7 , ! 2, 33$%
2
1
=
(–7)2 + (–2)2 + 332
2
! 16.9
A=
()
( )
The area of the triangle is approximately 16.9 units2.
Chapters 1 to 8 Course Review
Question 76 Page 514
Answers may vary. For example:
!
!
Let u = [1, 0, 1], v = [0, ! 1, ! 2], and k = 3 .
! !
L.S. = ku ! v
( )
(
)
= 3 "#1, 0, 1$% ! "#0, & 1, & 2 $%
= "#3, 0, 3$% ! "#0, & 1, & 2 $%
= "# 0(–2) & (–1)(3), 3(0) & (–2)(3), 3(–1) & 0(0) $%
= "#3, 6, & 3$%
!
!
R.S. = u ! kv
( )
(
= "#1, 0, 1$% ! 3 "#0, & 1, & 2 $%
)
= "#1, 0, 1$% ! "#0, & 3, & 6 $%
= "# 0(–6) & (–3)(1), 1(0) & (–6)(1), 1(–3) & 0(0) $%
= "#3, 6, & 3$%
L.S. = R.S.
MHR • Calculus and Vectors 12 Solutions 1065
Chapters 1 to 8 Course Review
Question 77 Page 514
The cross product is orthogonal to both vectors.
! !
u ! v = #$ "1, 6, 5%& ! #$ 4, 9, 10 %&
= #$ 6(10) " 9(5), 5(4) " 10(–1), " 1(9) " 4(6) %&
= #$15, 30, " 33%& or #$5, 10, –11%&
Two orthogonal vectors are "#5, 10, ! 11$% and "# !5, ! 10, 11$% .
Note that any scalar multiple of these vectors will also answer the question.
Chapters 1 to 8 Course Review
Question 78 Page 514
!" " "
V = w!u " v
= #$5, 0, 1%& ! #$ '2, 3, 6 %& " #$6, 7, ' 4 %&
= #$5, 0, 1%& ! #$ 3(–4) ' 7(6), 6(6) ' (–4)(–2), (–2)(7) ' 6(3) %&
= #$5, 0, 1%& ! #$ '54, 28, ' 32 %&
= '302
= 302
The volume is 302 units3.
Chapters 1 to 8 Course Review
Question 79 Page 514
! ! "!
! = r F sin "
26 = (0.35)95sin "
sin " =
26
(0.35) 95
" # 51.4o
The force is applied to the wrench at about a 51.4º angle.
Chapters 1 to 8 Course Review
Question 80 Page 514
Answers may vary. For example:
!!!"
AB = !"7, 1, 17 #$
A vector equation is !" x, y, z #$ = !"5, 0, 10 #$ + t !"7, 1, 17 #$ , t %! .
MHR • Calculus and Vectors 12 Solutions 1066
Possible parametric equations are:
x = 5 + 7t
y=t
z = 10 + 17t, t !!
Chapters 1 to 8 Course Review
Question 81 Page 514
a)
b)
Chapters 1 to 8 Course Review
Question 82 Page 514
a) The equation is of the form 4 x + 8 y + C = 0 . Substitute the coordinates of P0 to determine C.
4(3) + 8(–1) + C = 0
C = !4
The scalar equation is 4x + 8y – 4 = 0 or x + 2y – 1 = 0.
b) The equation is of the form !6 x ! 7 y + C = 0 . Substitute the coordinates of P0 to determine C.
!6(–5) ! 7(10) + C = 0
C = 40
The scalar equation is –6x – 7y + 40 = 0 or 6x + 7y – 40 = 0.
Chapters 1 to 8 Course Review
Question 83 Page 514
a) The y-axis has direction vector [0, 1].
A possible vector equation for the line is !" x, y #$ = !"6, % 3#$ + t !"0, 1#$ , t &! .
MHR • Calculus and Vectors 12 Solutions 1067
b) The given line has normal vector [2, 7].
A line perpendicular to this line will have direction vector [2, 7]
A possible vector equation is !" x, y #$ = !"1, 6 #$ + t !" 2, 7 #$ , t %! .
Chapters 1 to 8 Course Review
Question 84 Page 514
Answers may vary. For example:
Choose arbitrary values for two of the coordinates and solve for the third.
If x = 0 and y = 0, then z = !2.
If x = 2 and y = !1, then z = 0.
Two possible points are (0, 0, –2) and (2, –1, 0).
Chapters 1 to 8 Course Review
Question 85 Page 514
Parametric equations are:
x = 4 + 2s + 6t
y = 3 + s + 6t
z = !5 + 4s ! 3t, t "!
For
"! equation, need to find the normal.
! the
"! scalar
n = m1 ! m2
= "# 2, 1, 4 $% ! "#6, 6, & 3$%
= "#1(–3) & 6(4), 4(6) & (–3)(2), 2(6) & 6(1) $%
= "# &27, 30, 6 $% or "#9, & 10, & 2 $%
The scalar equation is of the form 9 x ! 10 y ! 2 z + D = 0 .
Use the point (4, 3, –5) to determine C.
9(4) ! 10(3) ! 2(–5) + D = 0
D = !16
The scalar equation is 9x – 10y – 2z – 16 = 0.
Chapters 1 to 8 Course Review
Question 86 Page 514
a) To find the x-intercept, let y = 0 and z = 0 and solve for s and t.
x = 1+ s + 2t
!
0 = 8 ! 12s + 4t
0 = 6 ! 12s ! 3t
"
#
MHR • Calculus and Vectors 12 Solutions 1068
Solve  and  for s and t.
! 8 = !12s + 4t
!
! 6 = !12s ! 3t
"
! 2 = 7t
!!"
t=!
2
7
! 6 = !12s +
s=
6
7
4
7
Now substitute t = –
x = 1+
"
2
4
and s =
in equation .
7
7
" 2%
4
+ 2$ ! '
7
# 7&
=1
The x-intercept is 1.
To find the y-intercept, let x = 0 and z = 0, and solve for s and t.
0 = 1+ s + 2t
!
y = 8 ! 12s + 4t
0 = 6 ! 12s ! 3t
"
#
Solve  and  for s and t.
! 1 = s + 2t
!
! 6 = !12s ! 3t
"
!18 = 21t
6
t=!
7
12!+"
! 6 = !12s +
s=
18
7
"
5
7
6
5
and s =
in equation .
7
7
" 5%
" 6%
y = 8 ! 12 $ ' + 4 $ ! '
# 7&
# 7&
Now substitute t = –
= !4
MHR • Calculus and Vectors 12 Solutions 1069
The y-intercept is –4.
To find the z-intercept, let x = 0 and y = 0, and solve for s and t.
0 = 1+ s + 2t
!
0 = 8 ! 12s + 4t
z = 6 ! 12s ! 3t
"
#
Solve  and  for s and t.
!1 = s + 2t
!
!8 = !12s + 4t
"
6 = 14s
3
s=
7
2!!"
3
+ 2t
7
5
t=!
7
!
!1 =
3
5
and t = – in equation .
7
7
" 3%
" 5%
z = 6 ! 12 $ ' ! 3 $ ! '
# 7&
# 7&
Now substitute s =
=3
The z-intercept is 3.
b) To find the x-intercept, let y = 0 and z = 0, and solve for x.
2x ! 6(0) + 9(0) = 18
x=9
To find the y-intercept, let x = 0 and z = 0, and solve for y.
2(0) ! 6 y + 9(0) = 18
y = !3
To find the z-intercept, let x = 0 and y = 0, and solve for z.
2(0) ! 6(0) + 9z = 18
z=2
The x-intercept is 9, the y-intercept is –3, and the z-intercept is 2.
MHR • Calculus and Vectors 12 Solutions 1070
Chapters 1 to 8 Course Review
Question 87 Page 514
!!!"
a) AB = "# !9 ! 5, 10 ! 2, 3 ! 8 $%
!!!"
BC = "# !2 ! (–9), (–6) ! 10, 5 ! 3$%
= "# !14, 8, ! 5$%
= "#7, ! 16, 2 $%
! """! """!
n = AB ! BC
= #$ "14, 8, " 5%& ! #$7, " 16, 2 %&
= #$8(2) " (–16)(–5), (–5)(7) " 2(–14), (–14)(–16) " 7(8) %&
= #$ "64, " 7, 168 %&
The scalar equation is of the form 64 x + 7 y ! 168 z + D = 0 .
Use the point A(5, 2, 8) to determine C.
64(5) + 7(2) ! 168(8) + D = 0
D = 1010
The scalar equation is 64 x + 7 y ! 168 z + 1010 = 0 .
b) Planes parallel to both the x- and y-axes have scalar equations of the form z = D .
Since the plane passes through P(–4, 5, 6), the equation is z = 6.
Chapters 1 to 8 Course Review
Question 88 Page 514
Answers may vary. For example;
If the line is perpendicular to the plane, its direction vector will be the same as the normal to the plane.
!
n = [7, ! 8, 5]
The vector equation of the line is !" x, y, z #$ = !"6, 1, % 2 #$ + t !"7, % 8, 5#$ , t &! .
MHR • Calculus and Vectors 12 Solutions 1071
Chapters 1 to 8 Course Review
Question 89 Page 514
The angle between two planes is defined to be the angle between their normal vectors.
! !
n1 " n2
cos ! = ! !
n1 n2
#8, 10, 3%& " #$ 2, ' 4, 6 %&
cos ! = $
#$8, 10, 3%& #$ 2, ' 4, 6 %&
8(2) + 10(–4) + 3(6)
cos ! =
2
8 + 102 + 32 22 + (–4)2 + (6)2
'6
98.4276
( '6 +
! = cos '1 *
) 98.4276 -,
cos ! =
! " 93.5o
The angle between the planes is about 93.5º.
Chapters 1 to 8 Course Review
Question 90 Page 514
Find the intersection points, if possible.
Equate the expressions for like coordinates.
2 + 3s = 3 ! 4t
!3 + 4s = 4 + 5t
3s + 4t = 1 !
4s – 5t = 7 !
2 ! 10s = !2 + 3t
10s + 3t = 4 !
Solve equations  and  for s and t.
12s + 16t = 4
4!
12s – 15t = 21
31t = –17
17
t=–
31
3"
4!!3"
MHR • Calculus and Vectors 12 Solutions 1072
Substitute t = –
17
into equation ! to solve for s.
31
! 17 $
3s + 4 # – & = 1
" 31 %
3s –
68
=1
31
99
3s =
31
33
s=
31
! 33 17 $
The lines intersect at # , – & .
" 31 31 %
Chapters 1 to 8 Course Review
Question 91 Page 514
!"
!"
Let P1(2, –7, 3), P2(3, 4,–2), m1 = [3, ! 10, 1], and m 2 = [1, 6, 1] .
!!!!"
P1 P2 = "#1, 11, ! 5$%
! "! "!
n = m1 ! m2
= #$ "16, " 2, 28 %&
d=
d=
"#1, 11, ! 5$% & "# !16, ! 2, 28 $%
"# !16, ! 2, 28 $%
178
1044
d ! 5.51
Chapters 1 to 8 Course Review
Question 92 Page 514
! """!
n ! AQ
The distance between a point A and a plane is given by d =
where Q is any point on the plane
!
n
!
with normal n .
Choose
Q(3, 0, 0) on the plane.
!!!"
AQ = !"3, 0, 0 #$ % !" 2, 5, % 7 #$
= !"1, % 5, 7 #$
MHR • Calculus and Vectors 12 Solutions 1073
d=
=
"#3, ! 5, 6 $% & "#1, ! 5, 7 $%
32 + (–5)2 + 62
70
70
! 8.37
The distance is approximately 8.37 units.
Chapters 1 to 8 Course Review
Question 93 Page 514
The normal to the plane is:
!
n = !"1, 3, 4 #$ % !" &5, 4, 7 #$ .
= !"5, & 27, 19 #$
! "!
n ! m = #$5, " 27, 19 %& ! #$1, 6, 5%&
= "62
Therefore, the line and the plane are not parallel and they intersect at one point.
Chapters 1 to 8 Course Review
Question 94 Page 514
Answers may vary. For example:
a) The normals must be parallel, but the equations of at least two planes cannot be multiples of each
other.
x + y + 2z = 3
2 x + 2 y + 4 z = 17
x + y + 2z = 0
b) The planes must have identical equations (except for a scalar multiple).
x + y + 2z = 3
2x + 2 y + 4z = 6
! x ! y ! 2 z = !3
!
!
!
c) The normals must be coplanar (e.g., n1 = 5n 2 + 3n3 ) and the constant terms must satisfy the same
relationship as the normals (in this case, D1 = 5 D2 + 3D3 ).
4x ! 6 y + z + 2 = 0
3x ! 5 y + 2 z + 4 = 0
x! y+ z!2=0
MHR • Calculus and Vectors 12 Solutions 1074
! ! !
d) The normals must be non-coplanar. (i.e., n1 ! n 2 " n3 # 0 )
!
!
!
Choose n1 = "#1, 3, ! 2 $% , n2 = "#1, 0, 0 $% , n3 = "#0, 1, 0 $% .
! !
n2 ! n3 = "#0, 0, 1$%
! ! !
n1 ! n2 " n3 = $%1, 3, # 2 &' ! $%0, 0, 1&'
= #2
(0
x + 3y ! 2 = 8
x = 14 and y = !12
Chapters 1 to 8 Course Review
Question 95 Page 514
a) First examine the normals. None are scalar multiples of each other.
Check if the normals are coplanar.
! ! !
n1 ! n 2 " n3 = [2, # 4, 5]! [3, 2, # 1]" [4, 3, # 3]
= [2, # 4, 5]! [#3, 5, 3]
= #11
The normals are not coplanar; there is an intersection point.
Eliminate one of the variables from two pairs of equations. Eliminate z.
2x ! 4 y + 5z = !4
!
15x + 10 y ! 5z = 5
17x + 6 y
=1
5"
#
9x + 6 y ! 3z = 3
4x + 3y ! 3z = !1
3!
"
5x + 3y
#
=4
!+5"
3!!"
Solve equations  and  for x and y.
17x + 6 y = 1
!
10x + 6 y = 8
7x
= !7
x = !1
2"
!!2"
Substitute x = –1 into equation ! .
5(–1) + 3y = 4
!
y=3
MHR • Calculus and Vectors 12 Solutions 1075
Substitute x = !1 and y = 3 into equation !.
3(!1) + 2(3) ! z = 1
z=2
The three planes intersect at the point (!1, 3, 2) .
b) First examine the normals. None are scalar multiples of each other.
Check if the normals are coplanar.
! ! !
n1 ! n 2 " n3 = [5, 4, 2]! [3, 1, # 3]" [7, 7, 7 ]
= [5, 4, 2]! [28, # 42, 14]
=0
Since the triple scalar product is zero, the normals are coplanar and the planes intersect either in a line
or not at all.
Eliminate one of the variables from two pairs of equations. Eliminate y.
5x + 4 y + 2z = 7
!
12x + 4 y ! 12z = 8
4"
!7x +
+ 14z = !1
#
21x + 7 y ! 21z = 14
7x + 7 y + 7z = 12
7"
$
14x
!7x
! 28z = 2
+ 14z = !1
!!4"
7"!$
%
Equations  and  are identical.
Let z = t.
!7x + 14t = !1 !
1
x = + 2t
7
Substitute x =
1
+ 2t and z = t into equation .
7
!1
$
3 # + 2t & + y ' 3t = 2
"7
%
y=
11
' 3t
7
! 1 11 #
The planes intersect in the line !" x, y, z #$ = % , , 0 & + t !" 2, ' 3, 1#$ , t (! .
"7 7 $
MHR • Calculus and Vectors 12 Solutions 1076