1. (Section 9.5 of Rattan/Klingbeil text)
EGR 1101: Unit 11 Lecture #1
Applications of Integrals in Dynamics: Position, Velocity, & Acceleration
(Section 9.5 of Rattan/Klingbeil text)

2. 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.

3. Position, Velocity, & Acceleration
Position x(t)
Derivative
Integral
Velocity v(t)
Derivative
Integral
Recall that:
Differentiating position with respect to time gives velocity.
Differentiating velocity with respect to time gives acceleration.
Therefore:
Integrating acceleration with respect to time gives velocity.
Integrating velocity with respect to time gives position.
-Show these relationships in a diagram.
-Do first page and a half of Week 11 notes here.
Acceleration a(t)

4. Today’s Examples Ball dropped from rest
Ball thrown upward from ground level
Position & velocity from acceleration (graphical)
The examples are virtually identical to examples we did in Week 5 (derivatives).

5. 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.
-Do this slide and next two slides after Example #2.
-In addition to knowing what shape curve you’ll get when you integrate a particular shape, we also want to put numbers on the resulting curve. Doing so requires us to remember that the definite integral can be understood as the area under a curve.

6. 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:

7. 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:
Do Example #3 in tabular form: To find velocity, one column for each interval, 6 rows labeled “Shape of accel curve” => “Shape of vel curve,” “Area under accel curve” => “Change in Vel,” “Initial vel” => “Final vel.” Similarly to find position.

8. Applications of Integrals in Electric Circuits
EGR 1101: Unit 11 Lecture #2
Applications of Integrals in Electric Circuits
(Sections 9.6, 9.7 of Rattan/Klingbeil text)

9. 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),

10. Energy and Power
We saw in Week 6 that power is the derivative with respect to time of energy:
Therefore energy is the integral with respect to time of power (plus the initial energy):

11. Current and Voltage in a Capacitor
We saw in Week 6 that, for a capacitor,
Therefore, for a capacitor,

12. Current and Voltage in an Inductor
We saw in Week 6 that, for an inductor,
Therefore, for an inductor,

13. Today’s Examples Current, voltage & energy in a capacitor
Current & voltage in an inductor (graphical)
Current & voltage in a capacitor (graphical)

14. 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.

15. 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:

16. 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:

17. 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: