AP CALCULUS AB
CHAPTER 3 REVIEW
A particle in motion moves along
the x-axis so its position at time, t,
is given by: s(t)=t3-8t2+17t-10.
(time in seconds and s(t) in feet)
For example: s(0)=-10 means at time t=0, the
particle is 10 units to the left of the origin.
A particle in motion moves along
the x-axis so its position at time, t,
is given by: s(t)=t3-8t2+17t-10
A graphical look at s(t):
s(t)=t3-8t2+17t-10
1. Determine the average velocity on [1,5].
s(t)=t3-8t2+17t-10
1. Determine the average velocity on [1,8].
Average velocity= changein position
change in time
or
s(t)=t3-8t2+17t-10
1. Determine the average velocity on [1,8].
Average velocity= changein position
change in time
or
s(8) − s(1) 126 − 0
=
= 18 feet per second
8 −1
8 −1
*Change in position can also be refered to as “displacement”
Determine the instantaneous
velocity at t=8.
Determine the instantaneous
velocity at t=8.
Instantaneous velocity is s’(8)
s(t)=t3-8t2+17t-10
Determine the instantaneous
velocity at t=8.
Instantaneous velocity is s’(8)
s’(t)=3t2-16t+17
s(t)=t3-8t2+17t-10
Determine the instantaneous
velocity at t=8.
Instantaneous velocity is s’(8)
s’(t)=3t2-16t+17
s’(8)=382-168+17=81 ft/sec
s(t)=t3-8t2+17t-10
Determine the instantaneous
velocity at t=8.
Instantaneous velocity is s’(8)
Graphically:
s(t)=t3-8t2+17t-10
When does the particle change
directions?
When does the particle change
directions?
When s’(t)=0 and changes signs:
+ to – or – to +
When does the particle change
directions?
When s’(t)=0 and changes signs:
+ to – or – to +
When does the particle change
directions?
When s’(t)=0 and changes signs:
+ to – or – to +
When does the particle change
directions?
When s’(t)=0 and changes signs:
+ to – or – to +
Use nDeriv or algebraically find the derivative!!
What is the acceleration at t=4?
What is the acceleration at t=4?
Acceleration at t=4 is s”(4)
The second derivative of position
What is the acceleration at t=4?
Acceleration at t=4 is s”(4)
s(t)=t3-8t2+17t-10
What is the acceleration at t=4?
Acceleration at t=4 is s”(4)
s(t)=t3-8t2+17t-10
s’(t)=3t2-16t+17
What is the acceleration at t=4?
Acceleration at t=4 is s”(4)
s(t)=t3-8t2+17t-10
s’(t)=3t2-16t+17
s”(t)=6t-16
What is the acceleration at t=4?
Acceleration at t=4 is s”(4)
s(t)=t3-8t2+17t-10
s’(t)=3t2-16t+17
s”(t)=6t-16
s”(4)=64-16=8 ft per sec2
Write the equation of the line
through the origin and the local
max on f(x)=4sin(2x)
Write the equation of the line
through the origin and the local
max on f(x)=4sin(2x)
Write the equation of the line
through the origin and the local
max on f(x)=4sin(2x)
Use your calculator to
find the local max.
Write the equation of the line
through the origin and the local
max on f(x)=4sin(2x)
(.785,4)
(0,0)
Use your calculator to
find the local max.
Write the equation of the line
through the origin and the local
max on f(x)=4sin(2x)
(.785,4)
(0,0)
Use your calculator to
find the local max.
4−0
slope =
.785 − 0
Answer: y-0=4/.785(x-0) or y-4=4/.785(x-.785)
Find where f(x)=6x2-3x is parallel to
the line 8x+4y=12.
Find where f(x)=6x2-3x is parallel to
the line 8x+4y=12.
Find where f(x)=6x2-3x is parallel to
the line 8x+4y=12.
Locate where f’(x)=the slope of line 8x+4y=12
Find where f(x)=6x2-3x is parallel to
the line 8x+4y=12.
Locate where f’(x)=the slope of line 8x+4y=12
f ’(x)=12x-3
Find where f(x)=6x2-3x is parallel to
the line 8x+4y=12.
Locate where f’(x)=the slope of line 8x+4y=12
f ’(x)=12x-3
8x+4y=12
Find where f(x)=6x2-3x is parallel to
the line 8x+4y=12.
Locate where f’(x)=the slope of line 8x+4y=12
f ’(x)=12x-3
8x+4y=12
4y=-8x+12
Find where f(x)=6x2-3x is parallel to
the line 8x+4y=12.
Locate where f’(x)=the slope of line 8x+4y=12
f ’(x)=12x-3
8x+4y=12
4y=-8x+12
y=-2x+3
Find where f(x)=6x2-3x is parallel to
the line 8x+4y=12.
Locate where f’(x)=the slope of line 8x+4y=12
f ’(x)=12x-3
8x+4y=12
4y=-8x+12
y=-2x+3
Slope=-2
Find where f(x)=6x2-3x is parallel to
the line 8x+4y=12.
Locate where f’(x)=the slope of line 8x+4y=12
f ’(x)=12x-3
Let 12x-3=-2
8x+4y=12
4y=-8x+12
y=-2x+3
Slope=-2
Find where f(x)=6x2-3x is parallel to
the line 8x+4y=12.
Locate where f’(x)=the slope of line 8x+4y=12
f ’(x)=12x-3
Let 12x-3=-2
12x=1
8x+4y=12
4y=-8x+12
y=-2x+3
Slope=-2
Find where f(x)=6x2-3x is parallel to
the line 8x+4y=12.
Locate where f’(x)=the slope of line 8x+4y=12
f ’(x)=12x-3
Let 12x-3=-2
12x=1
1
x=
12
8x+4y=12
4y=-8x+12
y=-2x+3
Slope=-2
Find where f(x)=6x2-3x is parallel to
the line 8x+4y=12.
A graphical look:
#47 page170: The amount A in
grams of radioacitve plutonium
remaining in a 20-gram sample
after t days is given by the formula:
A=20(1/2)(t/140)
At what rate is the plutonium decaying when
t=2 days?
Rate is refering to the derivative, A’ ,at t=2.
Find the derivative algebraically or use dy/dx
on your calculator: Answer: -.098 grams per day