calculus for ap physics c. the derivative: the derivative is the slope of a line at a particular...
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Calculus
For AP Physics C
• The derivative:
Velocity vs. Time
0
0.5
1
1.5
2
2.5
3
3.5
0 1 2 3 4 5 6 7 8 9 10
t (s)
v (m
/s)
• 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:
yslope m
x
• 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
dx
dt
• 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:
sincos
d xx
dx
cossin
d xx
dx
• The derivative:
• Special derivatives:
ln 1d x
dx x
xx
d ee
dx
• 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:
2
'' 'f gf fg
g g
• 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
3 2
2
4 2 6
5 8
x xf x
x x
2 2 3 2
22
5 8 12 4 4 2 6 10 8'
5 8
x x x x x x xf x
x x
• 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
• 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
0
2
4
6
8
10
0 1 2 3 4 5 6 7 8 9 10
t (s)
v (m
/s)
• Integration:
Velocity vs. Time
0
5
10
15
20
25
30
35
0 1 2 3 4 5 6 7 8 9 10
t (s)
v (m
/s)
• 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:
1
1
nn x
x dx Cn
• Integration:
• f(x) = 6x2 + 2x – 4 f '(x) = 12x + 2
• If we integrate the derivative we will have
212 2 6 2x dx x x
• 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.
52 2 tdt
dx
dttdx 52 2
Cttx 53
2 3
dttdx 62 2 C 3533
25 3
253
2 3 ttx
• Kinematic Equations:
• Velocity is change in position.
velocityaverage t
xv
velocityousinstantane dt
dxv
• Kinematic Equations:
• Acceleration is change in velocity.
onaccelerati average t
va
onaccelerati ousinstantane dt
dva
• Deriving Kinematic Equations:
dt
dva
dvdta
fv
v
tdvdta
00
fv
v
tdvdta
00
constant ison accelerati If
fv
v
tvta
00
00 vvta f
0vvat f
atvv f 0
• Deriving Kinematic Equations:
dt
dxv
dxdtv
fx
x
tdxdtv
00
fx
x
t
xattv0
0
20 2
1
atvv f 0
fx
x
tdxdtatv
00 0
02
0 2
1xxattv f
200 2
1attvxx f
• Deriving Kinematic Equations:
atvv f 0
2000 2
1t
t
vvtvxx f
f
t
vva f 0
tv
tv
tvxx ff 22
000
tvtv
xx ff 22
00
tvvxx ff 00 2
1
200 2
1attvxx f
• Deriving Kinematic Equations:
atvv f 0
2
0000 2
1
a
vva
a
vvvxx ff
f
a
vvt f 0
20000 22 vvvvvaxx fff
200 2
1attvxx f
• Deriving Kinematic Equations:
20000 22 vvvvvaxx fff
200
22000 2222 vvvvvvvaxx ffff
22002 ff vvaxx
020
2 2 xxavv ff
• Position, velocity, and acceleration:
dt
dxv
dt
dva
vdtx
adtv
• 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 = v0
2 + 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.
2
00 2
1attvxx f
2
2
100 gttd
22t
g
d
5
dt
• Determine the time for a projectile’s flight on level ground.
atvv yfy 0
gtvv yy 00
tg
v y 02
50 yv
t
• Trig functions of common angles:
0 30 45 60 90
sin 0 1
cos 1 0
tan 0 1 -
2
1
2
3
3
1
2
2
2
2
2
3
2
1
1
3
2
0
2
0
2
4
2
4
4
0
2
2
0
4
3
1
1
3
• 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
atvv f 0
200 2
1attvxx f
tvvxx ff 00 2
1
020
2 2 xxavv ff
0vv f
vtx
0vv f
vtx
x = vt
• 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 = vy0
2 + 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 = vy0
2 + 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 75 = v0(4) + ½a(4)2
16.7 = v0 + 1.5a 18.8 = v0 + 2a
v0 = 16.7 – 1.5a v0 = 18.8 – 2.0a
16.7 – 1.5a = 18.8 – 2.0a
a = 4.16 m/s2
• 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
r1
r2R ?
r1
r2
R
• Unit vectors
• Unit vectors have a length of 1 and are used only for direction
Direction Unit Vector
+x
–x
+y
–y
+z
–z
i
ij
j
k
k
• Determine the acceleration of a 2.0 kg particle with the velocity vector
• Determine the force acting on the particle
• Determine the magnitude of the acceleration at t = 2s
m/s ˆ 4ˆ 3 2 jtitv
2m/s ˆ 4ˆ 6 jita
N ˆ 8ˆ 12 jitF
2m/s ˆ 4ˆ 26 jia
222 m/s 12.6160412 a
• 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?
𝑣 𝑓=𝑣0+𝑎𝑡
𝑣 𝑓2=𝑣0
2+2𝑎𝑑
𝑑=𝑣0 𝑡+12𝑎𝑡
2
8.06=44.72+𝑎𝑡
(8.06 )2=( 44.72 )2+2𝑎 ( 676+𝑥 )
(676+𝑥 )=44.72 𝑡+ 12𝑎𝑡
2
𝑥=𝑣𝑥𝑡𝑥=8.06 𝑡
𝑡=𝑥
8.06
8.06=44.72+𝑎𝑥
8.06
−295.5=𝑎𝑥
(8.06 )2=( 44.72 )2+2𝑎 ( 676+𝑥 )
−295.5=𝑎𝑥
(8.06 )2=( 44.72 )2+1352𝑎+2𝑎𝑥
(8.06 )2=( 44.72 )2+1352𝑎+2 (−295.5 )
−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.
• (a) 57.7 km/h• (b) 28.9 km/h
60°
50 km/h
60°
30°
57.7 km/h
28.9 km/h
• 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
• d = 676 m• a = ?
=
=
𝑎=− 0.99m / s❑2
• Centripetal motion
• Centripetal means center seeking
• r is the radius of the circle or curve
• Centripetal acceleration is also known as radial
• “Normal” acceleration is known as linear or tangential
r
vac
2
• 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.
r
vac
2
tavv tf 0
0.150.259.13 ta2m/s 74.0ta
150
9.13 2
ca
2m/s 29.1ca
22ct aaa 22 29.174.0 a
backward 29.8 and inward m/s 49.1 2 a
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