Aim: Motion Problems Course: Calculus Do Now: Aim: What do these derivatives do for us anyway? Find the average rate of change of f(x) = x 2 on the interval

Aim: Motion Problems Course: Calculus

Do Now:

Aim: What do these derivatives do for us anyway?

Find the average rate of change of f(x) = x2 on the interval [-1, 2]

Average rate of change is the slope of the secant line connecting the

two endpoints of the interval


Average Rate of Change

The average rate of change of a function f(x) on an interval [a, b] is found by computing the slope of the line joining the end points of the function on that interval.

average rate of change

of ( ) on interval [ , ]

( ) ( )

f x a b

f b f a

b a

Find avg. rate of change f(x) = x2 for [-1, 2]

(2) ( 1)1

2 ( 1)

f f

6

5

4

3

2

1

-1

2

f x = x2

2 – (-1)

f(2) – f(-1)

-1, f(-1)

2, f(2)


Velocity Function

t

s st t tv tt0

( ) lim

Instantaneous Rate of Change of distance

with respect to time.

Displacement, Velocity & Acceleration

Average velocity = st

Change in distanceChange in time

( ) '( )v t s t 1st Derivative

Position Function s t gt v t s20 0

1( )

2

g = -32 ft/sec2Freely Falling Object

Acceleration Function ( ) '( )a t v t 1st Derivative of Velocity

Instantaneous Rate of Change in Velocity

with respect to time.

ds

dt

( ) ''( )a t s t2nd Derivative of Distance or Displacement

dv

dt2

2

d s

dt

Velocity – speed and direction

Speed – Absolute Value of velocity


Displacement vs. Distance

You drive 100 miles and then return 70 of those miles. What is your total distance traveled? What is your displacement?

100 miles: + displacementstart

return – 70 miles: – displacement

finish

30 miles: net displacement

100 + 70 = 170 miles total distance


Do Now:


Academic rigor is teaching, learning, and assessment which promotes student growth in knowledge of the discipline and the ability to any analyze, synthesize

and critically evaluate the content under study.

What does it look like?

“Rigor” in the context of intellectual work refers to thoroughness, carefulness, and right understanding of the material learned. Rigor is to academic work what careful practice and nuanced performance is to musical performance, and what intense and committed play is to athletic performance. When we talk about a ‘rigorous course’ in something, it’s a course that examines details, insists on diligent and scrupulous study and performance, and doesn’t settle for a mild or informal contact with the key ideas. Robert Talbert


Do Now:


Find dy/dx given sin x + 2cos(2y) = 1


Motion of Freely Falling Object

A ball is dropped from a window 20 feet above the ground.

a. Find the height of ball after 1 sec.

b. Find the velocity of ball after 1 sec.

c. When will the ball hit the ground?

d. What is its speed at that moment? Position Function s t gt v t s2

0 0

1( )

2

g = -32 ft/sec2Freely Falling Object

21( ) ( 32) 0 20

2s t t t


A ball is dropped from a window 20 feet above the ground.

Find the height of ball after 1 sec.

Find the velocity of ball after 1 sec.

When will the ball hit the ground?

What is its speed at that moment?

Motion of Freely Falling Object

21( ) ( 32) 1 20 4 ft.

2s t

( ) 32v t t ( ) 32 1 32 ft/secv t

( ) 0s t 2 5( ) 16 20 0;

2s t t t

( ) 32v t t5

( ) 32 35.776 ft/sec.2

v t


A football is punted into the air. Its displacement (directed distance) from the ground is a function of time t in seconds and is described by the equation y = -16t2 + 37t + 3.

a. Find the velocity at t = 1 at t = 2

b. Find the acceleration

at t = 1 at t = 2

Model Problem

30

25

20

15

10

5

1 2

f x = -16x2+37x+3


30

25

20

15

10

5

1 2

a. Find the velocity at t = 1 at t = 2

Model Problem: y = -16t2 + 37t + 3.

( ) '( )v t s t

' 32 37ds

v t s t tdt

1 32 1 37 5ft / sec

2 32 2 37 27ft / sec

v

v

@ 1 sec

@ 2 sec

velocity is directed speed

speed is faster a 2

sec. than at 1 sec. but velocity is

less


30

25

20

15

10

5

1 2

b. Find the acceleration at t = 1 at t = 2

Model Problem: y = -16t2 + 37t + 3.

( ) '( )a t v t

' 32dv

a t v tdt

2

2

1 32ft / sec

2 32ft / sec

a

a

@ 1 sec

@ 2 sec

negative acceleration

means velocity is decreasing


Speeding Up or Slowing Down?

If velocity and acceleration have the same sign, the object is speeding up

If velocity and acceleration have the opposite signs, the object is slowing down.

positive velocity

positive acceleration

negative velocity


Same signs

positive velocity


negative velocity

positive acceleration

Opposite signs


Rectilinear Motion

Rectilinear motion is motion along a straight line.

Particle Velocity

v(t) = 0 & a(t) isn’t = 0

v(t) > 0

v(t) < 0

Sign of v(t) changes

particle moves right or up

particle moves left or down

particle at rest

particle changes directions


Do Now:


Find the equation of the normal to 3x2 – 4y2 + y = 9 when x = 2 in QI.


Model Problem

A particle moves along a line with its position at time t given by s(t) = t3 – 3t + 2, t in seconds, s in feet.

a. Find velocity function.

b. Find v(0) and v(2).

c. When is velocity zero? Where is the particle at that time?

d. Draw a number line indicating the position and velocity of particle at t = 0, t = 1 and t = 2.

e. Refer to the number line to write a description of the motion of the particle from t = 0 to t = 2. How far does it travel in from t = 0 to t = 4?


Model Problem


a. Find velocity function.

b. Find v(0) and v(2).

c. When is velocity zero? Where is the particle at that time?

v(t) = s’(t) = 3t2 – 3

v(0) = -3 ft/sec and v(2) = 9 ft/sec

3t2 – 3 = 0: t = 1; s(1) = 0


Model Problem


d. Draw a number line indicating the position and velocity of particle at t = 0, t = 1 and t = 2.

0 2 4 s(t)

Particle Motion

Time t

Position

Velocity

0

s(t) = t3 – 3t + 2

v(t) = s’(t) = 3t2 – 3

1 2

2 0 4

-3 ft/s 0 ft/s 9 ft/s


Model Problem

A particle moves along a line with its position at time t given by s(t) = t3 - 3t2 + 2, t in seconds, s in feet.

e. Refer to the number line to write a description of the motion of the particle from t = 0 to t = 2.

0 2 4 s(t)

Particle Motion

Time t

Position

Velocity

0

s(t) = t3 – 3t + 2

v(t) = s’(t) = 3t2 – 3

1 2

2 0 4


At t = 0, the particle is at s = 2 and is moving to the left at 3 ft/s. One second later at t = 1, the particle is at s = 0 and is at rest. The particle then turns and at t = 2, the particle is a s = 4 and moving to the right at 9 ft/s.


Model Problem

e. How far does it travel from t = 0 to t = 4?

0 2 4 s(t)

Particle Motion

Time t

Position

Velocity

0

s(t) = t3 – 3t + 2

v(t) = s’(t) = 3t2 – 3

1 2

2 0 4


We must divide the time interval into the time when the velocity is negative and positive and add the absolute values of each distance.


Model Problem

e. How far does it travel from t = 0 to t = 4?

We must divide the time interval into the time when the velocity is negative and positive and add the absolute values of each distance during those intervals.

Interval [0, 1]; velocity negative

Interval [1, 4]; velocity positive

s(t) = t3 – 3t + 2

(1) (0) (4) (1)s s s s

3 3

3 3

1 3 1 2 0 3 0 2

4 3 4 2 1 3 1 2

56


Model Problem

A particle moves in the x-direction (miles) in such a way that its displacement from the y-axis is x = 3t3 - 30t2 + 64t + 57, for t > 0 in hours.

a. Find equations for its velocity and accel.

b. Find velocity and accel. at t = 2, t = 4, and t = 6. At each time, state

• Whether x is increasing or decreasing and at what rate

• Whether the object is speeding up or slowing down

c. At what times in the interval [0, 8] is x at a maximum? Is x ever negative in [0, 8]?


Model Problem

x = 3t3 - 30t2 + 64t + 57, for t > 0

a. Find equations for its velocity and accel.

2' 9 60 64dx

v t x t t tdt

2' 18 60dv dx

a t v t tdt dt

Find velocity and accel. at t = 2, t = 4, and t = 6.

t v(t) a(t) x is … Speeding/Slowing

2

4

6

-20

-32

28

-24

12

48

decreasing at 20mph

decreasing at 32mph

increasing at 28mph

Speeding up

Slowing down

Speeding up


180

160

140

120

100

80

60

40

20

5 10

Model Problem

x = 3t3 - 30t2 + 64t + 57, for t > 0

c. At what times in the interval [0, 8] is x at a maximum? Is x ever negative in [0, 8]?

x

t

29 60 64v t t t

29 60 64 0v t t t 1 1

5 or 13 3

t

at t = - relative maximum; velocity changes from positive to negative

11 3

at t = 8 - absolute maximum;

x is never negative in interval


Do Now:


Given the position function x(t) = t4 – 8t2 find the distance that the particle travels from t = 0 to t = 4.


Model Problem

Use implicit differentiation to find the derivative of 2 2 33 10x xy y


Model Problem

Use implicit differentiation to find

2

2 cos 3sin 1d

y xdx


Model Problem

If the position function of a particle is

find when the particle is changing direction.

2( ) , 09

tx t t

t


Model Problem

If the position function of a particle is

find the distance that the particle travels from t = 2 to t = 5.

2( ) 3 2 4, 0x t t t t


Model Problem

A particle moves along a line with its position at time t given by s(t) = tsint, t in seconds, s in feet.

a. Find velocity function and acceleration function

b. When is velocity zero? Where is the particle at that time?

• Draw the path of the particle on a number line and show the direction of increasing t.

• Is the distance traveled by the particle greater than 10?

Documents

Aim: Motion Problems Course: Calculus Do Now: Aim: What do these derivatives do for us anyway? Find the average rate of change of f(x) = x 2 on the interval