Applications:
Derivatives of
Trigonometric
Functions
Consider a simple pendulum that has a length of 50
cm and a maximum horizontal displacement of 8 cm.
a) Find the period of the pendulum.
- is the time necessary for one complete oscillation,
is measured in time units and is given by
T
=2𝑢
l
where l is the length of the
g is the
pendulum
andacceleration due to gravity,
in meters per square seconds.
Convert 50 cm and 8 cm into
0.5 m and 0.08 m
respectively.
0.
T =2𝑢 95.
T  1.428
l
g
b) Determine a function that gives the horizontal position
of the bob as a function of time.
h
t
 
h(t)
A is the amplitude
2𝑢
t
=A
t is time in sec
2𝑢
T
cos
t
h
tT
1.42


h=8 cos
t
=8 cos1.4𝑢 t
is the period, in sec
 
- the horizontal position is
measured in cm
OR
h
t
=0.08 cos1.4𝑢 t
c) Determine a function that gives the velocity of the bob
as a function of time.
d) Determine a function that gives the acceleration
of the bob as a function of time.
A marble is placed on the end of a horizontal oscillating spring.
If you ignore the effect of friction and treat this situation as an instance of simple
harmonic motion, the horizontal position of the marble as a function of time is
given by the function
ht = A
cos2𝑢 f t 
where A is the maximum displacement from rest position, in cm,
f
is the frequency, in hertz, and t
is time, in sec.
In the given situation the spring oscillates every 1 sec and has a maximum displacemen
10 cm.
a) What is the frequency of the oscillating spring?
f =
T
1
Frequency of the oscillating spring is given by:
f
=1is 1 Hz
The frequency
The period of each oscillation is 1 sec.
b) Write the simplified equation that expresses the position
of the marble as a function of time.
c) Determine a function that expresses velocity of the marble
as a function of time.
d)
Determine a function that expresses the
acceleration of the marble as a function of time.
A differential equation is an equation involving a function
and one or more of its derivatives.
.
Determine whether the function
y =𝑢 sin 𝜃 + 2𝑢
cos
is a 𝜃
solution to the differential equation
d𝜃 2
y =𝑢 sin𝜃 + 2𝑢
cos
dy 𝜃
one
d =𝑢 cos𝜃 - 2𝑢 sin𝜃
possible 𝜃 2
d
y 2 =𝑢
- sin𝜃
- 2𝑢
solution:


d𝜃
sin𝜃
2
d y =- 𝑢 sin𝜃 - 2𝑢
d𝜃 sin𝜃
2
d 2y
+ y =0
.
d 2y
d𝜃 2
cos
d 2 y𝜃
d𝜃 2
=0
+ y =- 𝑢 sin𝜃 - 2𝑢 sin𝜃 + 𝑢 sin𝜃 + 2𝑢
+y
y =𝑢 sin 𝜃 + 2𝑢
The function
cos
𝜃
is a solution
t2o the differential
equation: d y
d𝜃 2 + y
.