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3.4 Rates of change 3.4 Rates of change

3.4 Rates of change - PowerPoint Presentation

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3.4 Rates of change - PPT Presentation

Rates of change Example 1 Find the rate of change of the Area of a circle with respect to its radius Evaluate the rate of change of A at r 5 and r 10 If r is measured in inches and ID: 569350

rock velocity time seconds velocity rock seconds time measured particle motion position function acc find height amp heavy acceleration

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Slide1

3.4 Rates of changeSlide2

Rates of change

Example 1:

Find the rate of change of the Area of a circle with respect to its radius.

Evaluate

the rate of change of A at

r = 5

and

r = 10.

If

r

is measured in inches and

A

is measured in square inches, what units would be appropriate for dA/dr?

in2/in

These are units, so we do not cancel them!!!Slide3

Motion

Displacement

of an object is how far an object has moved over time.

Average velocity

is the slope of a displacement vs. time graph.Slide4

Motion

Instantaneous Velocity

is the velocity at a certain point.Instantaneous Velocity is the first derivative of the position function.

Speed is the absolute value of velocity.Slide5

Motion

Acceleration

is the rate at which a particle’s velocity changes.Measures how quickly the body picks up or loses speed.

2nd derivative of the position function!!!Slide6

Example

A particle moves along a line so that its position at any time

t ≥ 0 is given by the function

s(t) = 2t2 – 5t + 3 where s is measured in meters and t

is measured in seconds.

a.) Find the displacement of the particle during the first 2 seconds.

b.) Find the average velocity of the particle during the first 6 seconds

.Slide7

Example

A particle moves along a line so that its position at any time

t ≥ 0 is given by the function

s(t) = 2t2 – 5t + 3 where s is measured in meters and t

is measured in seconds.

c.) Find the instantaneous velocity of the particle at 6 seconds.

d.) Find the acceleration of the particle when

t = 6

.Slide8

Example

A particle moves along a line so that its position at any time

t ≥ 0 is given by the function

s(t) = 2t2 – 5t + 3 where s is measured in meters and t

is measured in seconds.

e.) When does

the particle change directions?Slide9

time

distance

acc pos

vel pos &

increasing

acc zero

vel pos &

constant

acc neg

vel pos &

decreasing

velocity

zeroacc negvel neg &decreasingacc zerovel neg &constantacc posvel neg &increasingacc zero,velocity zeroIt is important to understand the relationship between a position graph, velocity and acceleration:Slide10

Free-Fall

Gravitational

Constants:

Free-fall equation:

s is the position at any time t during the fall

g is the acceleration due to Earth’s gravity (gravitational constant)Slide11

Vertical motion

Example:

A dynamite blast propels a heavy rock straight up with a launch velocity of 160 ft/sec. It reaches a height of

s = 160t – 16t2 ft after t seconds.

a.) How high does the rock go?

Find when position = 0 and divide by 2 (symmetric path)

Since it takes 10 seconds

for the rock to hit the ground, it takes it 5 seconds to reach it max height. Slide12

Vertical motion

Example:

A dynamite blast propels a heavy rock straight up with a launch velocity of 160 ft/sec. It reaches a height of

s = 160t – 16t2 ft after t seconds.

a.) How high does the rock go?

Find when velocity = 0 (this is when

the rock changes direction)Slide13

Vertical motion

Example:

A dynamite blast propels a heavy rock straight up with a launch velocity of 160 ft/sec. It reaches a height of

s = 160t – 16t

2 ft after t seconds.

b.) What is the velocity

and speed of the rock when it is 256 ft above the ground on the way up?

At what time is the rock 256 ft above the ground on the way up?Slide14

Vertical motion

Example:

A dynamite blast propels a heavy rock straight up with a launch velocity of 160 ft/sec. It reaches a height of

s = 160t – 16t

2 ft after t seconds.

b.) What is the velocity

and speed of the rock when it is 256 ft above the ground on the way down?

At what time is the rock 256 ft above the ground on the way down?Slide15

Vertical motion

Example:

A dynamite blast propels a heavy rock straight up with a launch velocity of 160 ft/sec. It reaches a height of

s = 160t – 16t2 ft after t seconds.

c.) What is the acceleration

of the rock at any time t

at any time t during its flight?Slide16

from Economics:

Marginal cost

is the first derivative of the cost function, and represents an approximation of the cost of producing one more unit.

Marginal

revenue

is the first derivative of the revenue function

, and represents an approximation of the revenue of selling one more unit.