MHR Calculus and Vectors 12 Solutions 462 Chapter 4 Review ...

Chapter 4 Review
Chapter 4 Review Question 1 Page 244
a) i) (0, −1), ( ! , 1), (2 ! , −1)
b)
ii) (!, 1)
iii) (2!, !1) or (0, –1)
Chapter 4 Review Question 2 Page 244
a)
b)
MHR Calculus and Vectors 12 Solutions 462

Chapter 4 Review Question 3 Page 244
dy
a) sin x
dx = !
b) f !( x) = " 2cos x
dy
c) sin x cos x
dx = ! !
d) f !(x) = 3cos x + !sin x
Chapter 4 Review Question 4 Page 244
The slope of a function at a point is given by the value of its first derivative at that point.
y = 4sin x
dy
dx x= ! 3
dy
dx = 4cos x
= 4cos ! 3
= 2
Therefore, the slope of the function
y = 4sin x at x = ! 3
is 2.
Chapter 4 Review Question 5 Page 244
a) To find the equation of the tangent line, find the slope and the y-coordinates at ! = ! 4 .
y = 2sin! + 4cos!
dy
= 2cos! " 4sin!
d!
dy
d! != ! 4
= 2 2 " 4 2
= " 2 2
= " 2
At ! = ! 4 , y = 3 2 .
MHR Calculus and Vectors 12 Solutions 463

Use the slope-point form of a line to find the equation of the tangent line.
#
y ! 3 2 = ! 2 " ! ! &
$
% 4'
(
Therefore, the equation of the tangent is y = ! 2" + 2
4 ! + 3 2 .
b) Use the same approach as in part a).
y = 2cos! " 1 2 sin!
dy
d! = "2sin! " 1 2 cos!
dy
= "2("1) " 1 d! != 3! 2 (0)
2
= 2
At ! = 3! 2 , 1
y = .
2
Use the slope-point form of a line to find the equation of the tangent.
1 3!
y $ = 2
" !
#
2
% $
2
&
' (
Therefore, the equation of the tangent is y = 2! " 3! + 1 2 .
c)
MHR Calculus and Vectors 12 Solutions 464

Chapter 4 Review Question 6 Page 244
a)
dy
= !2cos x(! sin x)
dx
= sin 2x
dy
b) 2cos 2! 4sin 2!
d! = +
c) f !(") = #!cos(2" # !)
d) f !(x) = (cos(sin x))(cos x)
e) f !(x) = #
$ " sin(cos x) %
&(" sin x)
= #
$ sin(cos x) %
&(sin x)
f) f !(") = #7sin(7") + 5sin(5")
Chapter 4 Review Question 7 Page 244
dy
a) 3xcos x 3sin x
dy = +
b) f !(t) = "4t 2 sin 2t + 4t cos 2t
c)
d)
dy
dt = !2 t(cos(!t " 6)) + !sin(!t " 6)
dy
d! = " # $ sin(sin!) %
&(cos!) – # $ cos(cos!) %
& (sin!)
e) f !(x) = "2 #
$ cos(sin x) %
& # $ sin(sin x) %
&(cos x)
f) f !(") = #7(sin7") +10(cos5")(sin5")
MHR Calculus and Vectors 12 Solutions 465

Chapter 4 Review Question 8 Page 244
a) Answers may vary. For example:
To find the slope, f '( x ) , first find all the critical points at f ''( x ) = 0 .
f (x) = 2cos3x
f '(x) = !6sin3x
f ''(x) = !18cos3x
Let f ''( x ) = 0
cos3x = 0
3x = ..., ! 3 2 !, ! ! 2 , ! 2 , 3! 2 , ...
x = ..., ! 3 6 !, ! ! 6 , ! 6 , 3! 6 , ...
x =
(2k +1)!
, k "!
6
At, x = ..., ! 5! 6 , ! ! 6 , 3! 6 , 7! 6
, ..., the maximum value of the slope is obtained f '( x ) = 6 .
Need only one equation of the tangent, so let x = ! 2 .
! ! $
f
"
# 2%
& = 2cos 3! 2
= 0
Use the slope-point form of a line to find the equation of the tangent line.
"
y ! 0 = 6 x ! ! %
#
$ 2&
'
Therefore, the equation of the tangent is y = 6x ! 3! .
b) No, there are an infinite number of tangent lines to the curve y = 2cos3x
whose slope is a maximum.
From part a), at x = ..., ! 5! 6 , ! ! 6 , 3! 6 , 7! , ..., the maximum value of the slope is
6
obtained f '( x ) = 6 . In more general terms, the set of all such x’s can be expressed as
(4k + 3)!
x = , k !! . For each value of k, you will obtain a different equation for the tangent line with
6
maximised slope.
MHR Calculus and Vectors 12 Solutions 466

Examples using a graphing calculator and the tangent function are shown below.
Chapter 4 Section Review Question 9 Page 244
a) V (t) = 130sin(5t) +18
V !(t) = 650cos(5t)
The critical points are given by V '( t ) = 0 .
0 = 650cos(5t)
0 = cos(5t)
5t = ! 2 , 3! 2 , 5! 2 ,...
t = !
10 , 3!
10 , 5!
10 ,...
!
V ! $ ! ! $
"
# 10%
& = 130sin 5'
"
# 10%
& +18
= 130 +18
= 148
!
V 3! $ ! 3! $
"
# 10 %
& = 130sin 5'
"
# 10 %
& +18
= 130((1) +18
= (112
#
% ( 4k +1)!
'
%
Maximum voltage: 148 V at time, in seconds, $ t t = , k !!, k " 0(
.
10
&%
)%
#
% (
Minimum voltage: –112 V at time, in seconds, t t = 4k + 3 )!
'
%
$
, k !!, k " 0(
.
10
&%
)%
b) Period: T = 2! 5 s
f = 1 T
= 5
2!
Frequency:
5
2! Hz
MHR Calculus and Vectors 12 Solutions 467

A = 1 [148 ! (!112)]
2
= 130
Amplitude: 130 V
Chapter 4 Section Review Question 10 Page 245
a) i) sin! = 1
! = ! 2
Maximum: The force
F = mg sin!
has a maximum value when ! = ! 2 .
ii) sin! = 0
! = 0
Minimum: The force
b) Answers may vary. For example:
The formula for force is
F = mg sin!
has a minimum value when θ = 0.
F = mg sin!
. The force will be a maximum at an angle where sin! is
maximized i.e., ! = 90°, since sine has a maximum value at 90°. The force will be a minimum at an
2
angle where sin! is minimized i.e., 0°, since sine has a minimum value at 0°.
Chapter 4 Review Question 11 Page 245
a) Answers may vary. For example:
Given:
p = mv (p is the momentum of the body.)
Differentiate with respect to time
dp dv
= m
dt dt
dv
= a
dt
(Acceleration is the rate of change of velocity.)
Therefore, dp ma
dt = .
Combined with Newton’s second law of motion:
dp
F = , this gives F = ma .
dt
MHR Calculus and Vectors 12 Solutions 468

) F = ma
= m dv
dt
= m d dt (2cos3t)
= !6msin3t
Therefore F = 0 when sin3t = 0 .
t = 0, ! 3 , 2! 3 , 3! 3 , 4! 3 , ...
t = k! 3
, k !!, k " 0.
c) v( t) = 2cos3t
!
v k! $
"
# 3 %
& = 2cos(k!)
2cos(k!) is 2 m/s when k is even and –2 m/s when k is odd.
Therefore the speed is | v | = 2 m/s.
MHR Calculus and Vectors 12 Solutions 469

Chapter 4 Practice Test
Chapter 4 Practice Test Question 1 Page 246
The correct answer is B.
Chapter 4 Practice Test Question 2 Page 246
The correct answer is C.
Chapter 4 Practice Test Question 3 Page 246
The correct answer is C.
Chapter 4 Practice Test Question 4 Page 246
The correct answer is D.
Chapter 4 Practice Test Question 5 Page 246
The correct answer is C.
Chapter 4 Practice Test Question 6 Page 246
The correct answer is D.
dy ! 1 "
= 2
dx # 2 $
% &
= 1.
Chapter 4 Practice Test Question 7 Page 246
The correct answer is B.
dy
Use a graphing calculator to graph = cos x ! sin x for the given window.
dx
Chapter 4 Practice Test Question 8 Page 246
The correct answer is B.
MHR Calculus and Vectors 12 Solutions 470

Chapter 4 Practice Test Question 9 Page 247
dy
a) sin x cos x
dx = ! !
dy
b) 6cos 2!
d! =
c) f !(x) = !sin x cos x
d) f !(t) = 3t 2 cost + 6t sint
Chapter 4 Practice Test Question 10 Page 247
a)
b)
dy
dx = cos "
! + ! %
#
$ 4&
'
dy
d! = " sin #
! " ! &
$
% 4'
(
dy
d! =
3
c) 4sin ! ( cos!
)
dy
d! =
3 4
d) 4! cos( ! )
Chapter 4 Practice Test Question 11 Page 247
The slope of the tangent is the same as that of the curve, i.e., dy
dx at ! 4 .
dy
2 2
2cos x 2sin x
dx = !
At x = ! 4 ,
dy ! 1 " ! 1 "
= 2$ % # 2$ %
dx & 2 ' & 2 '
= 0
Therefore, the slope of the line tangent to the curve y = 2sin xcos
x at x = ! 4
is 0.
MHR Calculus and Vectors 12 Solutions 471

Chapter 4 Practice Test Question 12 Page 247
dy
6cos
dx = !
2
xsin
x
At x = ! 3 ,
dy ! 1 "! 3 "
= # 6$ $ dx & 4
%$ 2 %
'& '
3 3
= #
4
= m
Also at x = ! 3 ,
! 1 "
y = 2 # $
% 2 &
1
=
4
3
Substituting for m, y, and x in y = mx + b gives:
b = 1 4 + 3 3
4
= 1 4 + 3!
4
! ! $
"
# 3%
&
y = ! 3 3
4 x + 3!
4 + 1 4
is the equation of the tangent line.
Use a graphing calculator and the tangent function to confirm this result.
MHR Calculus and Vectors 12 Solutions 472

Chapter 4 Practice Test Question 13 Page 247
a) V (t) = 325sin(100!t)
V "(t) = 325(100!)cos(100!t)
The critical points are given by V '( t ) = 0 .
0 = 325(100!)cos(100!t)
0 = cos(100!t)
100!t = ! 2 , 3! 2 , 5! 2 ,...
t = 1
200 , 3
200 , 5
200 ,...
! 1 $
V
"
# 200%
& = 325sin !
100' ! 1 $ $
#
"
# 200%
& &
"
%
= 325
! 3 $
V
"
# 200%
& = 325sin !
100' ! 3 $ $
#
"
# 200%
& &
"
%
= (325
4 1
Maximum voltage: 325 V at time, in seconds,
! k +
% t t = , k #! , k $
" 0 & .
' 200
(
4 3
Minimum voltage: –325 V at time, in seconds,
! k +
% t t = , k #! , k $
" 0 & .
' 200
(
b) i) T = 2!
100!
= 1
50
The period is
1
50 s.
ii)
1
f =
T
= 50 Hz
The frequency is 50 Hz.
1 325 ( 325)
2
= 325
iii) A = [ ! ! ]
The amplitude is 325 V.
MHR Calculus and Vectors 12 Solutions 473

Chapter 4 Practice Test Question 14 Page 247
a) V (t) = 170sin(120!t)
V "(t) = 170(120!)cos(120!t)
The critical points are given by V '( t ) = 0 .
0 = 170(120!)cos(120!t)
0 = cos(120!t)
120!t = ! 2 , 3! 2 , 5! 2 ,...
t = 1
240 , 3
240 , 5
240 ,...
! 1 $
V
"
# 240%
& = 170sin !
120' ! 1 $ $
#
"
# 240%
& &
"
%
= 170
! 3 $
V
"
# 240%
& = 170sin !
120' ! 3 $ $
#
"
# 240%
& &
"
%
= (170
4 1
Maximum voltage: 170 V at time, in seconds,
! k +
% t t = , k #! , k $
" 0 & .
' 200
(
4 3
Minimum voltage: –170 V at time, in seconds,
! k +
% t t = , k #! , k $
" 0 & .
' 200
(
2!
i) T =
120!
= 1
60
The period is
1
60 s.
ii)
1
f =
T
= 60 Hz
The frequency is 60 Hz.
1 170 ( 170)
2
= 170 V
iii) A = [ ! ! ]
The amplitude is 170 V.
MHR Calculus and Vectors 12 Solutions 474

) Answers may vary. For example;
Similarities: Both functions are sinusoidal functions and both functions pass through the origin (0, 0).
Differences: The functions have different periods, frequencies, and amplitudes.
Chapter 4 Practice Test Question 15 Page 247
L.S. = d (sin 2x)
dx
= 2cos 2x
R.S. = d (2sin x cos x)
dx
= 2cos 2 x ! 2sin 2 x
Recall:
sin 2x = 2sin x cos x
2cos 2x = 2cos 2 x ! 2sin 2 x
Therefore, differentiating both sides of the given equation gives the identity cos 2x = cos 2 x – sin 2 x.
Chapter 4 Practice Test Question 16 Page 247
a) f !(x) = " sin 2 x + cos 2 x
f !!(x) = "4sin x cos x
b) f !!! (x) = "4 #
$
" sin 2 x + cos 2 x%
&
f ( 4)
(x) = 16sin x cos x
f ( 5)
(x) = 16 "
#
! sin 2 x + cos 2 x$
%
f ( 6)
(x) = !64sin x cos x
Answers may vary. For example:
2 2
The first, third, and fifth derivatives all have the expression ! sin x + cos x in the derivative.
The second, fourth, and sixth derivative all have the expression sin x cos x in the derivative.
The second derivative is the original function multiplied by −4.
The third derivative is the first derivative multiplied by −4.
This pattern continues for fourth to sixth derivatives so that the n derivative is the (n – 2) derivative
multiplied by –4.
c) Answers may vary. For example:
My prediction for the seventh derivative is f ( 7)
(x) = !64 "
#
! sin 2 x + cos 2 x$
% .
My prediction for the eighth derivative is f ( 8)
(x) = 256sin x cos x .
When the sixth derivative is differentiated, the seventh derivative is f ( 7)
(x) = !64 "
#
! sin 2 x + cos 2 x$
% as
predicted. When the seventh derivative is differentiated, the eighth derivative is found to be
( ) (x) = 256sin x cos x as predicted.
f 8
MHR Calculus and Vectors 12 Solutions 475

d) Answers may vary. For example:
i) f ( 2n)
(x) = (!4) n sin x cos x
ii) f (2n+1) (x) = (!4) n (! sin 2 x + cos 2 x)
e) i) f (12) (x) = 4096sin x cos x
ii) f (15) (x) = !16384(! sin 2 x + cos 2 x)
Chapter 4 Practice Test Question 17 Page 247
Answers will vary. For example:
a) y = sin x
dy
dx = cos x
d 2 y
dx = ! sin x
2
d 3 y
dx = ! cos x
3
d 4 y
dx = sin x
4
d 5 y
dx = cos x
5
b) y = cos x , y = ! cos x
c) There are four functions that satisfy this differential equation. The fourth function is y = ! sin x . The
functions are sinusoidal and the derivatives of each of the functions are shifted horizontally to the left,
or to the right, ! 2
units. The graph of the fifth derivative of each of the functions will be the same as
the graph of the first derivative of each of the functions.
MHR Calculus and Vectors 12 Solutions 476