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EGR 1101: Unit 11 Lecture # 1PowerPoint Presentation

EGR 1101: Unit 11 Lecture # 1

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### EGR 1101: Unit 11 Lecture #1

### EGR 1101: Unit 11 Lecture #2

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

- 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:

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:

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),

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):

Current and Voltage in a Capacitor

- We saw in Week 6 that, for a capacitor,
- Therefore, for a capacitor,

Current and Voltage in an Inductor

- We saw in Week 6 that, for an inductor,
- Therefore, for an inductor,

Today’s Examples

- 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:

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:

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:

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