aim: motion problems course: calculus do now: aim: what do these derivatives do for us anyway? find...
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![Page 1: 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](https://reader035.vdocument.in/reader035/viewer/2022062519/5697c0291a28abf838cd73f0/html5/thumbnails/1.jpg)
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
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Aim: Motion Problems Course: Calculus
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)
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Aim: Motion Problems Course: Calculus
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
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Aim: Motion Problems Course: Calculus
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
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Aim: Motion Problems Course: Calculus
Do Now:
Aim: What do these derivatives do for us anyway?
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
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Aim: Motion Problems Course: Calculus
Do Now:
Aim: What do these derivatives do for us anyway?
Find dy/dx given sin x + 2cos(2y) = 1
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Aim: Motion Problems Course: Calculus
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
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Aim: Motion Problems Course: Calculus
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
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Aim: Motion Problems Course: Calculus
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
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Aim: Motion Problems Course: Calculus
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
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Aim: Motion Problems Course: Calculus
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
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Aim: Motion Problems Course: Calculus
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
negative acceleration
Same signs
positive velocity
negative acceleration
negative velocity
positive acceleration
Opposite signs
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Aim: Motion Problems Course: Calculus
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
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Aim: Motion Problems Course: Calculus
Do Now:
Aim: What do these derivatives do for us anyway?
Find the equation of the normal to 3x2 – 4y2 + y = 9 when x = 2 in QI.
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Aim: Motion Problems Course: Calculus
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?
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Aim: Motion Problems Course: Calculus
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?
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
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Aim: Motion Problems Course: Calculus
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.
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
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Aim: Motion Problems Course: Calculus
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
-3 ft/s 0 ft/s 9 ft/s
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.
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Aim: Motion Problems Course: Calculus
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
-3 ft/s 0 ft/s 9 ft/s
We must divide the time interval into the time when the velocity is negative and positive and add the absolute values of each distance.
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Aim: Motion Problems Course: Calculus
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
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Aim: Motion Problems Course: Calculus
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]?
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Aim: Motion Problems Course: Calculus
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
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Aim: Motion Problems Course: Calculus
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
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Aim: Motion Problems Course: Calculus
Do Now:
Aim: What do these derivatives do for us anyway?
Given the position function x(t) = t4 – 8t2 find the distance that the particle travels from t = 0 to t = 4.
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Aim: Motion Problems Course: Calculus
Model Problem
Use implicit differentiation to find the derivative of 2 2 33 10x xy y
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Aim: Motion Problems Course: Calculus
Model Problem
Use implicit differentiation to find
2
2 cos 3sin 1d
y xdx
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Aim: Motion Problems Course: Calculus
Model Problem
If the position function of a particle is
find when the particle is changing direction.
2( ) , 09
tx t t
t
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Aim: Motion Problems Course: Calculus
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
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Aim: Motion Problems Course: Calculus
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?