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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)
! 3! 5!
, , ,...
2 2 2
! 3! 5!
t = , , ,...
10 10 10
5t =
! !$
!
!$
V # & = 130sin # 5 ' & + 18
" 10 %
" 10 %
! 3! $
!
3! $
V # & = 130sin # 5 ' & + 18
10 %
" 10 %
"
= 130 + 18
= 148
= 130((1) + 18
= (112
(
)
#%
'%
4k + 1 !
Maximum voltage: 148 V at time, in seconds, $t t =
, k !!, k " 0 ( .
10
&%
)%
(
)
#%
'%
4k + 3 !
Minimum voltage: –112 V at time, in seconds, $t t =
, k !!, k " 0 ( .
10
%&
%)
b) Period: T =
2!
s
5
1
T
5
=
2!
f =
Frequency:
5
Hz
2!
MHR Calculus and Vectors 12 Solutions 467
1
A = [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 F = mg sin ! has a minimum value when θ = 0.
b) Answers may vary. For example:
The formula for force is F = mg sin ! . The force will be a maximum at an angle where sin ! is
!
= 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°.
maximized i.e.,
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

(Acceleration is the rate of change of velocity.)
=a
dt
dp
Therefore,
= ma .
dt
Combined with Newton’s second law of motion: F =
dp
, this gives F = ma .
dt
MHR Calculus and Vectors 12 Solutions 468
b) F = ma
dv
dt
d
= m (2cos 3t)
dt
= !6msin 3t
=m
Therefore F = 0 when sin 3t = 0 .
! 2! 3! 4!
t = 0, ,
,
,
, ...
3 3
3
3
k!
t=
, k !!, k " 0.
3
c) v(t ) = 2cos3t
! k! $
v # & = 2cos(k!)
" 3%
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.
Use a graphing calculator to graph
Chapter 4 Practice Test
dy
= cos x ! sin x for the given window.
dx
Question 8 Page 246
The correct answer is B.
MHR Calculus and Vectors 12 Solutions 470