EGR 1101: Unit 11 Lecture #1
Applications of Integrals in Dynamics:
Position, Velocity, & Acceleration
(Section 9.5 of Rattan/Klingbeil text)
Differentiation and Integration


Recall that differentiation and integration
are inverse operations.
Therefore, any relationship between two
quantities that can be expressed in terms
of derivatives can also be expressed in
terms of integrals.
Position, Velocity, & Acceleration
Position x(t)
Derivative
Integral
Velocity v(t)
Derivative
Integral
Acceleration a(t)
Today’s Examples
1.
2.
3.
Ball dropped from rest
Ball thrown upward from ground level
Position & velocity from acceleration
(graphical)
Graphical derivatives & integrals

Recall that:




Differentiating a parabola gives a slant line.
Differentiating a slant line gives a horizontal line
(constant).
Differentiating a horizontal line (constant) gives
zero.
Therefore:



Integrating zero gives a horizontal line
(constant).
Integrating a horizontal line (constant) gives a
slant line.
Integrating a slant line gives a parabola.
Change in velocity = Area under
acceleration curve

The change in velocity between times t1
and t2 is equal to the area under the
acceleration curve between t1 and t2:
t2
v2  v1   a(t )dt
t1
Change in position = Area under
velocity curve

The change in position between times t1
and t2 is equal to the area under the
velocity curve between t1 and t2:
t2
x2  x1   v(t )dt
t1
EGR 1101: Unit 11 Lecture #2
Applications of Integrals in
Electric Circuits
(Sections 9.6, 9.7 of Rattan/Klingbeil text)
Review


Any relationship between quantities that can
be expressed using derivatives can also be
expressed using integrals.
Example: For position x(t), velocity v(t), and
acceleration a(t),
t
dx
v (t ) 
dt
 x(t )   v(t )dt  x(0)
dv
a (t ) 
dt
 v(t )   a(t )dt  v(0)
0
t
Energy and Power

We saw in Week 6 that power is the
derivative with respect to time of energy:
dw
p (t ) 
dt

Therefore energy is the integral with
respect to time of power (plus the initial
energy):
t
w(t )   p(t )dt  w(0)
0
Current and Voltage in a Capacitor

We saw in Week 6 that, for a capacitor,
dv
i (t )  C
dt

Therefore, for a capacitor,
t
1
v(t )   i(t )dt  v(0)
C0
Current and Voltage in an Inductor

We saw in Week 6 that, for an inductor,
di
v(t )  L
dt

Therefore, for an inductor,
t
1
i(t )   v(t )dt  i(0)
L0
Today’s Examples
1.
2.
3.
4.
Current, voltage & energy in a capacitor
Current & voltage in an inductor
(graphical)
Current & voltage in a capacitor
(graphical)
Current & voltage in a capacitor
(graphical)
Review: Graphical Derivatives &
Integrals

Recall that:




Differentiating a parabola gives a slant line.
Differentiating a slant line gives a horizontal line
(constant).
Differentiating a horizontal line (constant) gives
zero.
Therefore:



Integrating zero gives a horizontal line
(constant).
Integrating a horizontal line (constant) gives a
slant line.
Integrating a slant line gives a parabola.
Review: Change in position = Area
under velocity curve

The change in position between times t1
and t2 is equal to the area under the
velocity curve between t1 and t2:
t2
x2  x1   v(t )dt
t1
Applying Graphical Interpretation
to Inductors

For an inductor, the change in current
between times t1 and t2 is equal to 1/L
times the area under the voltage curve
between t1 and t2:
t2
1
i2  i1   v(t )dt
L t1
Applying Graphical Interpretation
to Capacitors

For a capacitor, the change in voltage
between times t1 and t2 is equal to 1/C
times the area under the current curve
between t1 and t2:
t2
1
v2  v1   i (t )dt
C t1