1. Aim: What do these derivatives do for us anyway?
Do Now:
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

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

3. Displacement, Velocity & Acceleration
Average velocity =
Velocity – speed and direction
Position Function
g = -32 ft/sec2
Freely Falling Object
Speed – Absolute Value of velocity
Velocity Function
Instantaneous Rate of Change of distance with respect to time.
1st Derivative
Acceleration Function
1st Derivative of Velocity
Instantaneous Rate of Change in Velocity with respect to time.
2nd Derivative of Distance or Displacement

4. Displacement vs. Distance
You drive 100 miles and then return 70 of those miles. What is your total distance traveled? What is your displacement?
= 170 miles total distance
100 miles: + displacement
start
return – 70 miles: – displacement
finish
30 miles: net displacement

5. Aim: What do these derivatives do for us anyway?
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

6. Aim: What do these derivatives do for us anyway?
Do Now:
Find dy/dx given sin x + 2cos(2y) = 1

7. Motion of Freely Falling Object
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?
Position Function
g = -32 ft/sec2
Freely Falling Object

8. Motion of Freely Falling Object
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?

9. Model Problem
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.
Find the velocity at t = at t = 2
Find the acceleration
at t = at t = 2

10. Model Problem: y = -16t2 + 37t + 3.
Find the velocity at t = at t = 2
@ 1 sec
@ 2 sec
speed is faster a 2 sec. than at 1 sec. but velocity is less
velocity is directed speed

11. Model Problem: y = -16t2 + 37t + 3.
b. Find the acceleration at t = at t = 2
@ 1 sec
@ 2 sec
negative acceleration means velocity is decreasing

12. 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

13. 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 at rest
particle moves right or up
particle moves left or down
particle changes directions

14. Aim: What do these derivatives do for us anyway?
Do Now:
Find the equation of the normal to 3x2 – 4y2 + y = 9 when x = 2 in QI.

15. 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.
Find velocity function.
Find v(0) and v(2).
When is velocity zero? Where is the particle at that time?
Draw a number line indicating the position and velocity of particle at t = 0, t = 1 and t = 2.
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?

16. 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.
Find velocity function.
Find v(0) and v(2).
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

17. 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. 2 4
s(t)
Particle Motion
Time t
Position
Velocity 1 2
s(t) = t3 – 3t + 2 2 4
v(t) = s’(t) = 3t2 – 3
-3 ft/s
0 ft/s
9 ft/s

18. 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.
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. 2 4
s(t)
Particle Motion
Time t
Position
Velocity 1 2
s(t) = t3 – 3t + 2 2 4
v(t) = s’(t) = 3t2 – 3
-3 ft/s
0 ft/s
9 ft/s

19. 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. 2 4
s(t)
Particle Motion
Time t
Position
Velocity 1 2
s(t) = t3 – 3t + 2 2 4
v(t) = s’(t) = 3t2 – 3
-3 ft/s
0 ft/s
9 ft/s

20. e. How far does it travel from t = 0 to t = 4?
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.
s(t) = t3 – 3t + 2
Interval [0, 1]; velocity negative
Parametric graphing: Mode: t-range [0, 10]; increments of .1; x-range [0, 50]; y-range [0, 2]; x = t3 – 3t + 2; y = 1
compare with Function graph: x-min: -1; max: 5; y min -10, y max 60
Interval [1, 4]; velocity positive 56

21. 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.
Find equations for its velocity and accel.
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]?

22. Model Problem
x = 3t3 - 30t2 + 64t + 57, for t > 0
Find equations for its velocity and accel.
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
-24
decreasing at 20mph
Speeding up
-32 12
decreasing at 32mph
Slowing down 28 48
increasing at 28mph
Speeding up

23. x t 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
at t = relative maximum; velocity changes from positive to negative
at t = absolute maximum;
x is never negative in interval t

24. Aim: What do these derivatives do for us anyway?
Do Now:
Given the position function x(t) = t4 – 8t2 find the distance that the particle travels from t = 0 to t = 4.

25. Model Problem
Use implicit differentiation to find the derivative of

26. Model Problem
Use implicit differentiation to find

27. Model Problem
If the position function of a particle is
find when the particle is changing direction.

28. Model Problem
If the position function of a particle is
find the distance that the particle travels from t = 2 to t = 5.

29. 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.
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?