Post on 28-Dec-2015
Calculus
6.3 Rectilinear Motion
6.3 Rectilinear Motion8. Apply derivatives
4 Solve a problem that involves applications and combinations of these skills.
3 Use implicit differentiation to find the derivative of an inverse function
Use derivatives to solve problems involving rates of change with velocity, speed and acceleration Use derivatives to find absolute and relative extrema in real-life context (Unit 4 Ch 5-6)Use derivatives to model rates of change (Unit 4 Ch 5-6)
2 Uses implicit differentiation to find the derivative of an inverse but makes errors in writing the inverses or in computing the derivativeCan state the relationship between position, velocity, speed and acceleration, and can write the velocity function given position, or acceleration function given the velocity functionShows understanding of using the derivatives to find extrema, but makes mistakes, or finds local extrema when absolute are required.Writes an equation to describe a situation but does not write a differential equation, or makes mistakes in evaluating the function.
1 With help, partial success at 2.0 & 3.0 content
0 Even with help, no success
Vocabulary
• Day 1:– Rectilinear Motion
• Position function– Velocity function
• Instantaneous velocity
• Day 2:– Speed function
• Instantaneous speed
• Day 3– Acceleration Function
• Speeding up• Slowing down
Rectilinear Motion
• Motion on a line
Moving in a positive direction from the origin
Moving in a negative direction from the origin
Position Function
• Horizontal axis:– time
• Vertical Axis:– position on a line
Moving in a positive direction from the origin
time
position
Moving in a negative direction from the origin
Position function: s(t)s = position (sposition duh!)t = times(t)= position changes as time changes
Sketchpad Example
Example
• Use the position and time graph to describe how the puppy was moving
time
position
Velocity• Rate
– position change vs time change
– Velocity can be positive or negative
• positive: going in a positive direction
• negative: going in a negative direction
18
16
14
12
10
8
6
4
2
-2
-4
-6
-8
-10
p
1 2 3 4 5 6 7 8 9 10 11 12
t
position
time
A A
18
16
14
12
10
8
6
4
2
-2
-4
-6
-8
-10
-12
p
-1 1 2 3 4 5 6 7 8 9 10 11
t
v(t) x = 3x2+-34x+76
4
time
Animate Points
Vel
ocity
Pos
ition
Velocity function
• Velocity is the slope of the position function (change in position /change in time)
• velocity =
– Technically this is instantaneous velocity
dt
dstv )( )(ts
Position Velocity Meaning
Positive Slope Positive y’s moving in a positive direction
Negative slope Negative y’s
Moving in a negative direction
Practice
• Let s(t)= t3-6t2 be the position function of a particle moving along an s-axis were s is in meters and t is in seconds. – Graph the position function– On a number line, trace the path that the particle
took. – Where will the velocity be positive? Negative?– Graph the instantaneous velocity. – Identify on the velocity function when the particle was
heading in a positive direction and when it was heading in a negative direction.
Velocity or Speed
• Speed change in position with respect to time in any direction
• Velocity is the change in position with respect to time in a particular direction– Thus – Speed cannot be negative – because
going backwards or forwards is just a distance– Thus – Velocity can be negative – because
we care if we go backwards
Speed
• Absolute Value of Velocity–
dt
dstv
)(
speed
ousinstantane
example: • if two particles are moving on the same coordinate line • with velocity of v=5 m/s and v=-5 m/s,• then they are going in opposite directions• but both have a speed of |v|=5 m/s
Example - s(t)= t3-6t2 position
time
23 6)( ttts
time
velocity
tttv 123)( 2
tttv 123)( 2
time
speed
Practice
• Graph the velocity function
• What will the speed function look like?
• At what time(s) was the particle heading in a negative direction? Positive direction?
19163)( 2 tttv
Acceleration
• the rate at which the velocity of a particle changes with respect to time.– If s(t) is the position function of a particle
moving on a coordinate line, then the instantaneous acceleration of the particle at time t is
•
• or
dt
dvta )(
2
2
)(")(dt
sdtsta
Example
• Let s(t) = t3 – 6t2 be the position function of a particle moving along an s-axis where s is in meters and t is in seconds. Find the instantaneous acceleration a(t) and shows the graph of acceleration verses time
tttstv 123)(')( 2 126)('')(')( ttstvta
Day 3: Speeding Up & Slowing down
• Speeding up when slope of speed is positive
• Slowing down when slope of speed is negative
Example
• When is s(t) speeding up and slowing down?
position
time
23 6)( ttts
time
velocity
tttv 123)( 2 speed
acceleration
Velocity & Acceleration function20
18
16
14
12
10
8
6
4
2
-2
-4
-6
-8
-10
-12
p
-1 1 2 3 4 5 6 7 8 9 10 11
t
Animate Points
A AB
Slowing down
Velocity +
Acceleration -
Speeding up
Velocity -
Acceleration -
Slowing down
Velocity -
Acceleration +
Speeding up
Velocity +
Acceleration +
Analyzing MotionGraphically Algebraically Meaning
Position
Velocity
Acceleration
Positive “s” values Positive side of the number line
Negative side of the number line
Negative “s” values
s(t)=velocity.
Look for Critical Pts
Postive “v” values
0 “v” values (CP)
Negative “v” values
Moving in + direction
Turning/stopped
Moving in a – direction
v(t)=accelerationLook for Critical Pts
+ a, + v = speeding up- a, - v = speeding up+ a, - v = slowing down- a, + v = slowing down
Example
Suppose that the position function of a particle moving on a coordinate line is given by s(t) = 2t3-21t2+60t+3 Analyze the motion of the particle for t>0
Graphically Algebraically Meaning
Pos
ition
Vel
ocity
Acc
eler
atio
n
0360212)( 23 tttts
Never 0 (t>0), always postive
Always on postive side of number line
060426)()( 2 tttvts0)107(6 2 tt0)5)(2(6 tt
0 2 5
+ - +0 0
0<t<2 going pos direction
t=2 turning
2<t<5 going neg. directiont=5 turning
t>5 going pos. direction
t=0 t=2t=5
04212)()( ttatv4212 t 5.3t
+ - - +
0 2 53.5
va - - + +
0<t<2 slowing down
2<t<3.5 speeding up
3.5<t<5 slowing down
5<t speeding up
Day 4: Applications; Gravity•
• s = position (height)
• s0= initial height
• v0= initial velocity
• t = time• g= acceleration due to gravity
– g=9.8 m/s2 (meters and seconds)– g=32 ft/s2 (feet and seconds)
200 2
1gttvss
s0
Example• Nolan Ryan was capable of throwing a baseball at 150ft/s (more
than 102 miles/hour). Could Nolan Ryan have hit the 208 ft ceiling of the Houston Astrodome if he were capable of giving the baseball an upward velocity of 100 ft/s from a height of 7 ft?
2161007 tts tv 32100 the maximum height occurs when velocity = 0
t=100/32=25/8 seconds
s(25/8)=163.25 feet