calculus for ap physics c. the derivative: the derivative is the slope of a line at a particular...

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:

• 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


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


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


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


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


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


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

calculus for ap physics c. the derivative: the derivative is the slope of a line at a particular...

Documents