Calculus
For AP Physics C
• The derivative:
Velocity vs. Time
3.5
3
v (m/s)
2.5
2
1.5
1
0.5
0
0
1
2
3
4
5
t (s)
6
7
8
9
10
• The derivative:
• The derivative is the slope of a line at a
particular point.
• The line is the graph of a function which we
will call f(x).
• To graphically determine the slope of a curved
line you must first draw a tangent line to the
curve at the point of interest.
• The derivative:
• The derivative:
• We can mathematically determine the slope by
taking the derivative of the function and then
evaluating it at the point of interest.
• There are two primary ways of denoting the
derivative of a function.
f '(x)
• and
• The derivative:
• The power rule:
• Where f(x) = xn
• f '(x) = nxn–1
• The derivative:
• The derivative of a polynomial is also very
simple, as each term can be evaluated
separately.
• Where f(x) = 5x3 + 2x2 – 8x +4
• f '(x) = 15x2 +4x – 8
• The derivative:
• Derivatives of trigonometric functions:
• The derivative:
• Special derivatives:
• The derivative:
• The Product Rule:
• When a function consists of the product of two
terms, each of which contains the variable, the
product rule must be used to determine the
derivative. The general form of the product
rule is
• (fg)' = f 'g + fg'
• The derivative:
•
•
•
•
•
•
•
Here is an example.
f(x) = (4x2 + 9x)(2x4 + 6x2 – 3)
8x + 9
(8x + 9)(2x4 + 6x2 – 3)
8x3 + 12x
(4x2 + 9x)(8x3 + 12x)
f '(x) = (8x + 9)(2x4 + 6x2 – 3) + (4x2 + 9x)(8x3 + 12x)
• The derivative:
• Quotient Rule:
• When a function consists of the quotient of two
terms, each of which contains the variable, the
quotient rule must be used to determine the
derivative. The general form of the product rule
is:
• The derivative:
•
•
•
•
The derivative of the numerator is
12x2 + 4x
The derivative of the denominator is
10x + 8
• Lo d hi – hi d lo over lo2
• The derivative:
• The Chain Rule:
• The chain rule is necessary in order to find the
derivative of a function raised to a power.
• If we let u be a function of x, then the derivative
can be found by
• 𝑢𝑛 ′ = 𝑛 𝑢𝑛−1 𝑑𝑢
•
•
•
•
•
•
•
The derivative:
f(x) = (4x2 + 9x)4
n=4
n–1=3
du = 8x + 9
Putting it all together our answer is
f '(x) = 4(4x2 + 9x)3(8x + 9)
• Integration:
Velocity vs. Time
10
v (m/s)
8
6
4
2
0
0
1
2
3
4
5
t (s)
6
7
8
9
10
• Integration:
Velocity vs. Time
35
30
v (m/s)
25
20
15
10
5
0
0
1
2
3
4
5
t (s)
6
7
8
9
10
• Integration:
• Integration is in essence the opposite of the
derivative.
• An integral gives the area between a curve on a
graph and the x-axis.
• Integration:
• Integration:
• Integration:
• The integral of a simple function can be
computed as follows:
• Integration:
• f(x) = 6x2 + 2x – 4
f '(x) = 12x + 2
• If we integrate the derivative we will have
• Differential Equations:
• Using a differential equation, write an expression for
the position of a particle with a velocity given by: v =
2t2 – 5. The position x at t = 3 s is 5 m.
2 3
x  t  5t  C
3
dx
 2t 2  5
dt


dx  2t 2  6 dt


2
dx

2
t
   5 dt
2 3
5  3  53  C
3
2 3
x  t  5t  2
3
• Kinematic Equations:
• Velocity is change in position.
x
average velocity 
v
t
dx
instantane ous velocity 
v
dt
• Kinematic Equations:
• Acceleration is change in velocity.
v
average accelerati on 
a
t
dv
instantane ous accelerati on 
a
dt
• Deriving Kinematic Equations:
dv
a
dt
at  0  v v
adt  dv
at  0   v f  v0
t
vf
0
 a dt  
t
vf
0
v0
dv
If accelerati on is constant
t
vf
0
v0
a  dt   dv
at  v f  v0
v f  v0  at
• Deriving Kinematic Equations:
dx
v
dt
t
1 2
xf

 v0t  at   x x0
2

0
vdt  dx
 v dt  
t
xf
0
x0
1 2
v0t  at  x f  x0
2
1 2
x f  x0  v0t  at
2
dx
v f  v0  at
 v
t
0
0
 at dt   dx
xf
x0
• Deriving Kinematic Equations:
v f  v0  at
a
v f  v0
t
1 2
x f  x0  v0t  at
2
1  v f  v0  2
t
x f  x0  v0t  
2 t 
vf
v0
x f  x0  v0t  t  t
2
2
v0t v f
x f  x0 
 t
2
2
1
x f  x0  v0  v f t
2
• Deriving Kinematic Equations:
v f  v0  at
t
v f  v0
a
1 2
x f  x0  v0t  at
2
 v f  v0  1  v f  v0 
  a

x f  x0  v0 
 a  2  a 
2x f  x0 a  2v0 v f  v0   v f  v0 
2
2
• Deriving Kinematic Equations:
2x f  x0 a  2v0 v f  v0   v f  v0 
2
2x f  x0 a  2v0v f  2v02  v 2f  2v0v f  v02
2x f  x0 a  v02  v 2f
v 2f  v02  2ax f  x0 
• Position, velocity, and acceleration:
dx
v
dt
x   vdt
dv
a
dt
v   adt
• For a case of constant, but non zero, acceleration,
draw a graph of (a) acceleration vs. time, (b) velocity
vs. time, and (c) position vs. time.
• Kinematics terms:
1) Distance – how far something moves
2) Displacement – how far for initial to final location

Symbols for 1 & 2: d, x, y, h, l, r, s
3) Speed – distance per time
4) Velocity – displacement per time

Symbol for 3 & 4: v
5) Acceleration – rate of change of velocity

Symbol for 5: a
6) Scalar – magnitude (amount or size) only
7) Vector – magnitude and direction
• Types of acceleration:
1) Positive – speed increases in the positive direction or
decreases in the negative
2) Negative – speed decreases in the positive direction or
increases in the negative
3) Centripetal – direction changes
• Using the kinematic equations:
vf
v0
a
t
x
vf = v0 + at




X
xf = x0 + v0t + ½at2
X




vf2 = v02 + 2a(xf – x0)



X

xf = x0 + ½(vf + v0)t


X


xf = x0 + vft – ½at2

X



• Determine the time for an object to fall a distance d
near Earth’s surface.
1 2
x f  x0  v0t  at
2
1 2
d  0  0t  gt
2
2d
2
t
g
d
t
5
• Determine the time for a projectile’s flight on level
ground.
v fy  v0 y  at
v0 y  v0 y  gt
2v0 y
g
t
v0 y
5
t
• Trig functions of common angles:
0
30
45
60
90
sin
0
0
2
1
2
2
2
3
2
1
cos
1
4
2
3
2
2
2
1
2
0
tan
0
0
4
11
3
1
2
2
3
11
-
4
2
0
2
4
0
• Projectile motion
• Projectiles undergo unpowered flight, typically
moving horizontally and vertically.
• Horizontal and vertical motion for a projectile is
completely independent!
• For projectiles, horizontal acceleration is ALWAYS
zero
• Horizontal motion
v f  v0  at
v f  v0
1 2
x f  x0  v0t  at
2
x  vt
v  v  2ax f  x0 
v f  v0
1
x f  x0  v0  v f t
2
x  vt
2
f
2
0
• Projectile motion
• The velocity of a projectile launched at an angle must
be separated into horizontal and vertical components!
• A softball is hit at 41.3 m/s at an angle of 35.0° above
the horizontal. Determine (a) how long it is in the air,
(b) how high it goes, and (c) how far it travels.
vx = (41.3)(cos 35) = 33.8 m/s
vy = (41.3)(sin 35) = 23.7 m/s
a)
vyf = vy0 + at
-vy0 = vy0 + gt
-23.7 = 23.7 + (-10)t
t = 4.74 s
• A softball is hit at 41.3 m/s at an angle of 35.0° above
the horizontal. Determine (a) how long it is in the air,
(b) how high it goes, and (c) how far it travels.
vx = (41.3)(cos 35) = 33.8 m/s
vy = (41.3)(sin 35) = 23.7 m/s
b)
vyf2 = vy02 + 2ad
0 = (23.7)2 + 2(-10)d
d = 28.1 m
• A softball is hit at 41.3 m/s at an angle of 35.0° above
the horizontal. Determine (a) how long it is in the air,
(b) how high it goes, and (c) how far it travels.
vx = (41.3)(cos 35) = 33.8 m/s
vy = (41.3)(sin 35) = 23.7 m/s
c)
x = vxt
x = (33.8)(4.74)
d = 160. m
• A football is thrown horizontally at 15 m/s from a
height of 20. m. With what velocity does it strike the
ground?
vyf2 = vy02 + 2ad
vyf2 = 0 + 2(10)(20)
vyf = 20. m/s
vf = 25 m/s at 37° WRT vertical
• An accelerating car emerges from behind a building
and observer #1 notes that the car travels 50. m in 3.0
s. Observer #2 notes that the car travels 75 m in 4.0 s.
What is the acceleration of the car?
d = v0t + ½at2
50 = v0(3) + ½a(3)2
16.7 = v0 + 1.5a
v0 = 16.7 – 1.5a
16.7 – 1.5a = 18.8 – 2.0a
a = 4.16 m/s2
75 = v0(4) + ½a(4)2
18.8 = v0 + 2a
v0 = 18.8 – 2.0a
• Vectors
• Vector quantities can be represented by arrows called
vectors.
• The length of a vector should be proportional to its
size.
• Vectors are added head to tail
r2
R?
r1
R
r2
r1
• Unit vectors
• Unit vectors have a length of 1 and are used only for
direction
Direction Unit Vector
+x
–x
+y
–y
+z
–z
iˆ
 iˆ
ĵ
 ĵ
k̂
 k̂
• Determine the acceleration of a 2.0 kg particle with
the velocity vector v  3t 2 iˆ  4t ˆj m/s




a  6t iˆ  4 ˆj m/s 2
• Determine the force acting on the particle


F  12t iˆ  8 ˆj N
• Determine the magnitude of the acceleration at t = 2s


a  62 iˆ  4 ˆj m/s 2
a  12 2  4 2  160  12.6 m/s 2
• When a high-speed passenger train travelling at 161
km/h rounds a bend, the engineer is shocked to see
that a locomotive has improperly entered onto the
track from a siding and is a distance D = 676 m
ahead. The locomotive is moving at 29.0 km/h. The
engineer of the high-speed train immediately applies
the brakes. What must be the magnitude of the
resulting deceleration if a collision is to be just
avoided?
𝑣𝑓 = 𝑣0 + 𝑎𝑡
𝑥 = 𝑣𝑥𝑡
8.06 = 44.72 + 𝑎𝑡
𝑥 = 8.06𝑡
𝑣𝑓 2 = 𝑣0 2 + 2𝑎𝑑
8.06
2
= 44.72
1 2
𝑑 = 𝑣0 𝑡 + 𝑎𝑡
2
2
+ 2𝑎 676 + 𝑥
1 2
676 + 𝑥 = 44.72𝑡 + 𝑎𝑡
2
𝑥
𝑡=
8.06
𝑥
8.06 = 44.72 + 𝑎
8.06
−295.5 = 𝑎𝑥
8.06
2
= 44.72
2
+ 2𝑎 676 + 𝑥
8.06
2
= 44.72
2
+ 1352𝑎 + 2𝑎𝑥
2
+ 1352𝑎 + 2 −295.5
−295.5 = 𝑎𝑥
8.06
2
= 44.72
−1344 = 1352𝑎
𝑎 = −0.994 m/s2
• Relative motion
• A car is moving at 50 km/h on a rainy day. Raindrops
that hit the side window make an angle on the glass of
60° relative to the vertical. Determine the speed of the
raindrop relative to (a) the car and (b) the earth.
28.9 km/h
• (a) 57.7 km/h
• (b) 28.9 km/h
50 km/h
30°
60°
60°
• When a high-speed passenger train travelling at 161
km/h rounds a bend, the engineer is shocked to see
that a locomotive has improperly entered onto the
track from a siding and is a distance D = 676 m
ahead. The locomotive is moving at 29.0 km/h. The
engineer of the high-speed train immediately apples
the brakes. What must be the magnitude of the
resulting deceleration if a collision is to be just
avoided?
• v0 = 44.72 – 8.06 = 36.66 m/s
• vf = 0 m/s
2
2
𝑣𝑓 =𝑣0 + 2𝑎𝑑
• d = 676 m
02 =36.662 + 2𝑎 676
• a=?
𝑎 = −0.99 m/s2
• Centripetal motion
• Centripetal means center seeking
2
v
ac 
r
• r is the radius of the circle or curve
• Centripetal acceleration is also known as radial
• “Normal” acceleration is known as linear or
tangential
• A car travelling at 90.0 km/h slows as it enters a sharp
curve (r = 150. m). After 15.0 seconds, the car has
slowed to 50.0 km/h. Determine the acceleration of the
car at this point.
v2
ac 
v f  v0  at t
13.9  25.0  at 15.0
at  0.74 m/s 2
a  a a
2
t
2
c
a
r
2

13.9 
ac 
150
ac  1.29 m/s 2
0.74
2

 1.29
2

a  1.49 m/s inward and 29.8 backward
2