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

Velocity v(t)

Acceleration a(t)

Derivative Integral

Integral

Today’s Examples

1. Ball dropped from rest

2. Ball thrown upward from ground level

3. 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:

2

1

)(12

t

t

dttavv

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:

2

1

)(12

t

t

dttvxx

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),

dt

dxtv )(

dt

dvta )(

t

xdttvtx0

)0()()(

t

vdttatv0

)0()()(

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):

dt

dwtp )(

t

wdttptw0

)0()()(

Current and Voltage in a Capacitor

We saw in Week 6 that, for a capacitor,

Therefore, for a capacitor,

dt

dvCti )(

t

vdttiC

tv0

)0()(1

)(

Current and Voltage in an Inductor

We saw in Week 6 that, for an inductor,

Therefore, for an inductor,

dt

diLtv )(

t

idttvL

ti0

)0()(1

)(

Today’s Examples

1. Current, voltage & energy in a capacitor

2. Current & voltage in an inductor (graphical)

3. Current & voltage in a capacitor (graphical)

4. 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:

2

1

)(12

t

t

dttvxx

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:

2

1

)(1

12

t

t

dttvL

ii

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:

2

1

)(1

12

t

t

dttiC

vv