3 4 rates of change
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3.4 Rates of change. Example 1: Find the rate of change of the Area of a circle with respect to its radius. Rates of change. If r is measured in inches and A is measured in square inches, what units would be appropriate for dA / dr ?. in 2 /in.

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3.4 Rates of change

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3 4 rates of change

3.4 Rates of change


Rates of change

  • Example 1:

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

Rates of change

  • 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!!!

  • Evaluate the rate of change of A at r = 5 and r = 10.


Motion

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

Motion

  • Average velocity is the slope of a displacement vs. time graph.


Motion1

  • 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.

Motion


Motion2

  • 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!!!

Motion


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.

Example

  • 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.


Example1

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.

Example

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

  • d.) Find the acceleration of the particle when t = 6.


Example2

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.

Example

  • e.) When does the particle change directions?


3 4 rates of change

distance

time

It is important to understand the relationship between a position graph, velocity and acceleration:

acc neg

vel pos &

decreasing

acc neg

vel neg &

decreasing

acc zero

vel neg &

constant

acc zero

vel pos &

constant

acc pos

vel neg &

increasing

velocity

zero

acc pos

vel pos &

increasing

acc zero,

velocity zero


Free fall

Gravitational

Constants:

Free-Fall

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)


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.

Vertical motion

  • 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.


Vertical motion1

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.

Vertical motion

  • a.) How high does the rock go?

  • Find when velocity = 0 (this is when the rock changes direction)


Vertical motion2

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.

Vertical motion

  • 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?


Vertical motion3

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.

Vertical motion

  • 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?


Vertical motion4

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.

Vertical motion

  • c.) What is the acceleration of the rock at any time t at any time t during its flight?


3 4 rates of change

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.


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