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EGR  1101: EGR  1101:

EGR 1101: - PowerPoint Presentation

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EGR 1101: - PPT Presentation

Unit 11 Lecture 1 Applications of Integrals in Dynamics Position Velocity amp Acceleration Section 95 of RattanKlingbeil text Differentiation and Integration Recall that differentiation and integration are inverse operations ID: 526515

velocity line amp position line velocity position amp voltage current graphical change area slant horizontal constant curve acceleration times

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Slide1

EGR 1101: Unit 11 Lecture #1

Applications of Integrals in Dynamics: Position, Velocity, &

Acceleration

(Section 9.5 of Rattan/Klingbeil text)Slide2

Differentiation and IntegrationRecall 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.Slide3

Position, Velocity, & Acceleration

Position

x(t)

Derivative

Velocity

v(t)

Acceleration

a(t)

Derivative

Integral

IntegralSlide4

Today’s ExamplesBall dropped from restBall thrown upward from ground level

Position & velocity from acceleration (graphical)Slide5

Graphical derivatives & integralsRecall 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.Slide6

Change in velocity = Area under acceleration curveThe change in velocity between times t1 and

t

2

is equal to the area under the acceleration curve between

t

1 and t2

:Slide7

Change in position = Area under velocity curveThe change in position between times t1 and

t

2

is equal to the area under the velocity curve between

t

1 and t2

:Slide8

EGR 1101: Unit 11 Lecture #2

Applications of Integrals in Electric

Circuits

(Sections 9.6, 9.7 of Rattan/Klingbeil text)Slide9

ReviewAny 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), Slide10

Energy and PowerWe 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):Slide11

Current and Voltage in a CapacitorWe saw in Week 6 that, for a capacitor,

Therefore

, for a capacitor,Slide12

Current and Voltage in an InductorWe saw in Week 6 that, for an inductor,

Therefore

, for an inductor,Slide13

Today’s ExamplesCurrent, voltage & energy in a capacitorCurrent & voltage in an inductor (graphical)

Current & voltage in a capacitor (graphical)

Current & voltage in a capacitor (graphical)Slide14

Review: Graphical Derivatives & IntegralsRecall 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.Slide15

Review: Change in position = Area under velocity curveThe change in position between times t1 and

t

2

is equal to the area under the velocity curve between

t

1 and t2

:Slide16

Applying Graphical Interpretation to InductorsFor an inductor, the change in current between times t1 and

t

2

is equal to 1/

L

times the area under the voltage curve between t1 and

t2:Slide17

Applying Graphical Interpretation to CapacitorsFor a capacitor, the change in voltage between times t1 and

t

2

is equal to 1/

C

times the area under the current curve between t1 and

t2: