math 34b lecture 7

Math 34B Lecture 7

January 31, 2012

Click as you doA d

dx

(2e3x

)B d

dx (2x)C∫

(4e2x + 6x2 + a) dx

D∫ 21 x−2 dx

E What value of x minimizes f (x) = −x2 + 2bx + c

A = 6e3x B = (ln 2)2x C = 2e2x + 2x3 + ax + C D = 1/2 E = bHow many did you get correct ? A = 5 B = 4 C = 3 D = 2 E =

B 2x = ex ln 2 so ddx (2x) = d

dx

(ex ln 2

)= (ln 2)ex ln 2 = (ln 2)2x

D∫ 21 x−2 dx =

[x−1/(−1)

]21

= (2−1/(−1))− (1−1/(−1)) = 1/2 UPOW

E for min f ′(x) = 0 so −2x + 2b = 0 so x = b

Section (12.1) Sine Waves.

Sine waves arise in many situations:geometry

circles, triangles, navigationperiodic phenomena

child on a swing,waves ( sound waves , radio waves, sea waves, light waves)oscillations: economic, populationoscilloscope

It all comes down to

x = cos(θ) y = sin(θ)

Circles and angleCircles and angle

circle of radius 1

circumfrence =

( cos( ) , sin( ) )

Equation of unit circle is x2 + y2 = 1 because this says the point (x , y) isdistance 1 from the origin. This comes from Pythagoras (section 1.7).Therefore cos2 θ + sin2 θ = 1. Mathematica demo + I get around

sin(θ) = b/c= opposite

hypotenuse

sin(45o) = 1√2

sin(30o) = 12

sin(60o) =√

32

y = Sin( ) one cycleone cycle

periodperiod = time taken for one cycle

frequencyfrequency = number of cycles in one unit of time

= 1 / period

Example

period = 0.2 seconds

frequency = 1/0.2 = 5 hertz hertz or cpscps (ccycles pper ssecond)

Vanilla sine wave y = sin(t). Period = 2π ∴ frequency = 12π .

Speeded up sine waves

For y = sin(Kt) time passes K times faster.Thus period is 1/K that of vanilla sine wave.

y = sin(Kt)period = 2π/Kfrequency = K/(2π)

y = sin(5t)period =A = 5 B = 10π C = 1/5 D = 2π/5 E = 1/(10π)

D = 2π/5 Reason: vanilla sine wave has period 2π. This one is 5 times

faster so has period 1/5 of vanilla sine wave.

The blue sine wave is y = sin(t). Find the equation of the red sine waveA = sin(Kt) B = sin(4t) C = sin(4πt) D = sin(t/4) E = sin(t/(4π))give a hint ??

Is the red one faster or slower than the blue one ?, How manytimes faster or slower ? DReason: the blue one does 4 cycles in the time taken for the red one to do 1cycle. So the red one is (1/4) as fast as blue one. Thus K = 1/4.

The blue sine wave is y = sin(t). Find the equation of the red sine waveA = sin(Kt) B = sin(4t) C = sin(4πt) D = sin(t/4) E = sin(t/(4π))give a hint ?? Is the red one faster or slower than the blue one ?, How manytimes faster or slower ?

DReason: the blue one does 4 cycles in the time taken for the red one to do 1cycle. So the red one is (1/4) as fast as blue one. Thus K = 1/4.

The blue sine wave is y = sin(t). Find the equation of the red sine waveA = sin(Kt) B = sin(4t) C = sin(4πt) D = sin(t/4) E = sin(t/(4π))give a hint ?? Is the red one faster or slower than the blue one ?, How manytimes faster or slower ? DReason: the blue one does 4 cycles in the time taken for the red one to do 1cycle. So the red one is (1/4) as fast as blue one. Thus K = 1/4.

What is the amplitude of y = 5 + 7 sin(3t)A = 7 B = 5 C = 3 D = 12 E = 15

A=7 Because 5+ just shifts whole thing upwards 5 units, but does notchange difference in height between the top and bottom of sine wave. Sosame amplitude as 7 sin(3t) namely 7.

What is the amplitude of y = 5 + 7 sin(3t)A = 7 B = 5 C = 3 D = 12 E = 15A=7 Because 5+ just shifts whole thing upwards 5 units, but does not

change difference in height between the top and bottom of sine wave. Sosame amplitude as 7 sin(3t) namely 7.

y = A sin(Kt)period = 2π/Kamplitude = A

t is time in hours. Which of the following is a sine wave with period 24hours ?A = sin(t/24) B = sin(24t) C = sin(πt/12) D = sin(πt/24)

C = sin(πt/12) vanilla sine wave sin(t) has period 2π. Need to slow this

down. sin(Kt) has period 2π/K . So 2π/K = 24 and K = π/12.

y = Sin(t) y = Sin(t+2)

sin(t + 2) is shifted 2 seconds into the future compared to y = sin(t).Everything for sin(t + 2) happens 2 seconds earlier than it does for sin(t).This is called phase shift.

The height of water in a harbor goes up and down with the tide. The timebetween two successive high tides is about 12 hours. The average height ofwater in the harbor is 20 feet and high tide is 3 feet higher. Let t be thenumber of hours after high tide. Which function describes height of water?A = 3 + 20 sin(πt/6) B = 20 + 3 sin(πt/6)C = 20 + 3 sin(π(t + 3)/6) D = 20 + 3 sin(π(t − 3)/6)) Give a hint ?

Draw a pretty little picture showing height of water plotted against time.C



The height of water in a harbor goes up and down with the tide. The timebetween two successive high tides is about 12 hours. The average height ofwater in the harbor is 20 feet and high tide is 3 feet higher. Let t be thenumber of hours after high tide. Which function describes height of water?A = 3 + 20 sin(πt/6) B = 20 + 3 sin(πt/6)C = 20 + 3 sin(π(t + 3)/6) D = 20 + 3 sin(π(t − 3)/6)) Give a hint ?Draw a pretty little picture showing height of water plotted against time.

C



The height of water in a harbor goes up and down with the tide. The timebetween two successive high tides is about 12 hours. The average height ofwater in the harbor is 20 feet and high tide is 3 feet higher. Let t be thenumber of hours after high tide. Which function describes height of water?A = 3 + 20 sin(πt/6) B = 20 + 3 sin(πt/6)C = 20 + 3 sin(π(t + 3)/6) D = 20 + 3 sin(π(t − 3)/6)) Give a hint ?Draw a pretty little picture showing height of water plotted against time.C

ddt (sin(t)) = cos(t)ddt (cos(t)) = − sin(t)

discuss

ddt (sin(Kt)) = K cos(Kt)ddt (cos(Kt)) = −K sin(Kt)

changes K times faster

∫sin(Kt)dt = − 1

K cos(Kt)∫cos(Kt)dt = 1

K sin(Kt)

check by taking derivative

The height of water in a harbor goes up and down with the tide.The time between two successive high tides is about 12 hours.The average height of water in the harbor is 20 feetand high tide is 3 feet higher. Let t be the number of hours after high tide.Which function describes height of water?

A = 20 + 3 sin(π(t − 3)/6)B = 20 + 3 sin(π(t + 3)/6)C = 20 + 3 sin(πt/6)D = 3 + 20 sin(πt/6)

B

y = sin(x) y = cos(x)

cos does everything π/2 sooner than sin cos(x) = sin(x + π/2)

cos has a phase shift of +π/2 compared to sin

θ 0 π/2 π 3π/2

sin(θ) 0 1 0 −1

cos(θ) 1 0 −1 0

You will find it easier toremember this if you canpicture the circle and theangles 0, π/2, π, 3π/2.

(cos(π/2) , sin(π/2)) = (0,1)

(cos(π) , sin(π)) = (0,-1)

(cos(3π/2) , sin(3π/2)) = (0,-1)

(cos(0) , sin(0)) = (1,0)θ = 0

θ = π

θ = π/2

θ = 3π/2

(cos(π) , sin(π)) = (-1,0)

(12.1) Sine waves

d

dtsin(Kt) = K cos(Kt)

d

dtcos(Kt) = −K sin(Kt)

Remember d(42f (t))/dt = 42 f ′(t).

d

dt(7 sin(3t) + 3 cos(4t)) = ?

A = 21 sin(3t) + 12 cos(4t)B = 21 cos(3t) + 12 sin(4t)C = −21 cos(3t)− 12 sin(4t)D = 21 cos(3t)− 12 sin(4t)E = OTHER

D

(12.1) Sine waves

d


d

dtcos(Kt) = −K sin(Kt)

Remember d(42f (t))/dt = 42 f ′(t).

d

dt(7 sin(3t) + 3 cos(4t)) = ?

A = 21 sin(3t) + 12 cos(4t)B = 21 cos(3t) + 12 sin(4t)C = −21 cos(3t)− 12 sin(4t)D = 21 cos(3t)− 12 sin(4t)E = OTHERD

An explanation of why the derivatives of sin and cos are what they are.Don’t worry if you don’t follow this, it won’t be on the test

Once you know d sin(t)/dt = cos(t) then it is easy to convince yourself thatd sin(Kt)/dt = K cos(Kt). Basically because time is running K times faster.

+C

Integration is opposite of differentiation.Differentiating swaps sine and cosine, with some funny sign business.So integration does the reverse.It still swaps sine and cosine, still does the sign business. But this time youdivide by K instead of multiplying.∫

sin(Kt) dt = − 1

Kcos(Kt)+C

∫cos(Kt) dt =

1

Ksin(Kt)+C

The sign business is a pain in the *** . Most people remember the formulafor derivatives then check their integrals by doing the derivative in their headto see if they got the sign right (and did the 1

K thing right too)

Sin(x) + C+ C Cos(x)

differentiate

integrate

Cos(x) + C+ C -Sin(x)

differentiate

integrate

d


d

dtcos(Kt) = −K sin(Kt)∫

sin(Kt) dt = − 1

Kcos(Kt)+C

∫cos(Kt) dt =

1

Ksin(Kt)+C

What is the slope of the graph y = 3 + 7 sin(2t) when t = 0 ?You may want to know sin(0) = 0 and cos(0) = 1.

A = 3 B = 7 C = 14 D = −14 E = OTHER

Cslope = derivatived(3 + 7 sin(2t))/dt = 14 cos(2t).Plug in t = 0 get 14 cos(0) = 14.

d


d


sin(Kt) dt = − 1

Kcos(Kt)+C

∫cos(Kt) dt =

1

Ksin(Kt)+C

What is the slope of the graph y = 3 + 7 sin(2t) when t = 0 ?You may want to know sin(0) = 0 and cos(0) = 1.

A = 3 B = 7 C = 14 D = −14 E = OTHERC

slope = derivatived(3 + 7 sin(2t))/dt = 14 cos(2t).Plug in t = 0 get 14 cos(0) = 14.

d


d


sin(Kt) dt = − 1

Kcos(Kt)+C

∫cos(Kt) dt =

1

Ksin(Kt)+C

What is the area under one arch of y = sin(t). Draw a diagram.First arch starts at t = 0 and ends t = π. Told cos(0) = 1 and cos(π) = −1.A = 1 B = 2 C = π D = −π E = 0.

B Notice how I lay out my work.

area =∫ π0 sin(t) dt

= [ − cos(t) ]π0 CTBA= (− cos(π))− (− cos(0)) UPOW= −(−1)− (−1)= 2

(12.2) Product RuleThe derivative of the sum of two functions is simple:

d

dt(f (t) + g(t)) = f ′(t) + g ′(t)

But the derivative of a product of two functions is tricky

d

dt(f (t) · g(t)) 6= f ′(t) · g ′(t) WrongWrongWrong

Exampled

dt

((t2)(t3)

)6= (2t)(3t2) = 6t3 Wrong

The correct method gives

d

dt

((t2)(t3)

)=

d

dt

(t5)

= 5t4 Right

Product Rule

d

dx(f (x) · g(x)) = f ′(x)g(x) + f (x)g ′(x)

“derivative of first term times second plus first term times derivative ofsecond.”“eff prime gee plus eff gee prime”

Exampled

dx

(x3e3x

)= 3x2e3x + x33e3x

Find derivative of (x2 + 2)(x3 − 1)A = 6x3 B = 2x · 3x2 C = 2x(x3 − 1)D = 2x(3x2 − 1) E = 5x4 + 6x2 − 2x .

E

d

dx

((x2 + 2)(x3 − 1)

)= 2x(x3 − 1) + (x2 + 2)3x2 = 5x4 + 6x2 − 2x

product rule ddx (f (x) · g(x)) = f ′(x)g(x) + f (x)g ′(x)

ddx (sin(Kx)) = K cos(Kx) d

dx (cos(Kx)) = −K sin(Kx)

d

dx

(x3sin(2x)

)=

A = 6x2 cos(2x)B = 3x2 sin(2x) + x3 cos(2x)C = 3x2 sin(2x) + x3 cos(x)D = 3x2 sin(2x) + 2x3 cos(2x)

D

Explanation of product ruleFarmer sells milk.t = time in days.p(t) = price of one gallon of milk on day tn(t) = number of gallons sold on day t

p(t)n(t) = daily income =amount of money farmer earns on day t

How fast daily income changes depends on two things: how quickly the priceof milk changes and how quickly the amount of milk produced changes

d

dt(p(t)n(t))︸︷︷︸ = p′(t)n(t)︸︷︷︸ + p(t)n′(t)︸︷︷︸

how quickly contribution from contribution fromfarmer’s earnings changing price of increasing number

increase one gallon of milk of gallons of milk

Explanation of product ruleFarmer sells milk.t = time in days.p(t) = price of one gallon of milk on day tn(t) = number of gallons sold on day tp(t)n(t) = daily income =amount of money farmer earns on day t


d

dt(p(t)n(t))︸︷︷︸ = p′(t)n(t)︸︷︷︸ + p(t)n′(t)︸︷︷︸




How fast daily income changes depends on two things:

how quickly the priceof milk changes and how quickly the amount of milk produced changes

d

dt(p(t)n(t))︸︷︷︸ = p′(t)n(t)︸︷︷︸ + p(t)n′(t)︸︷︷︸




How fast daily income changes depends on two things: how quickly the priceof milk changes

and how quickly the amount of milk produced changes

d

dt(p(t)n(t))︸︷︷︸ = p′(t)n(t)︸︷︷︸ + p(t)n′(t)︸︷︷︸





d

dt(p(t)n(t))︸︷︷︸ = p′(t)n(t)︸︷︷︸ + p(t)n′(t)︸︷︷︸



Explanation of product rule part two

d

dt(7f (t)) = 7f ′(t)

Expresses the idea that 7f (t) changes 7 times as fast as f (t).One might say that the rate of change of 7f (t) is the rate of change of f (t)speeded up by a factor of 7.

General situation

ddt (f (t)g(t)) = f ′(t)g(t)︸︷︷︸ + f (t)g ′(t)︸︷︷︸

contribution due to contribution due tof(t) changing g(t) changing

Look at f (t)g ′(t). This is the rate of change g ′(t) but speeded up by afactor of f (t) This happens because any change in g(t) is multiplied by f (t).Other term mutatis mutandi (= with necessary changes)




Find derivative of sin(2x)e3x

A = 2 cos(2x)e3x + 3 sin(2x)e2x

B = cos(2x)e3x + 3 cos(2x)e3x

C = 2 cos(2x)e3x + sin(2x)e3x

D = 2 cos(2x)e3x + 3 sin(2x)e3x

E = OTHER

D




Find derivative of sin(2x)e3x

A = 2 cos(2x)e3x + 3 sin(2x)e2x

B = cos(2x)e3x + 3 cos(2x)e3x

C = 2 cos(2x)e3x + sin(2x)e3x

D = 2 cos(2x)e3x + 3 sin(2x)e3x

E = OTHERD

A rectangular hole in the ozone layer is 100 miles wide and 300 miles long.The width increases at a rate of 20 miles per year. The length increases at arate of 10 miles per year. How fast is the area of the hole increasing ?Answer in square miles per year.A = 30 B = 200 C = 7000 D = 30000 E = OTHERHint: PLP. Label it. Area is product of length and width. NUN.

10

20

100

200

L(0) = 200W(0) = 100L’(0) = 10W’(0)= 20

A = LWA’ = L’W + LW’A’(0) = L’(0) W(0) + L(0)W’(0) = (10)(100) + (200)(20) = 5000

Told L(t) = length in year t in miles of holeW(t) = width in year t in miles of holeA(t) = area in sqaure miles in year t

L(t)

W(t)

NUN A(t) = area of hole in square miles in year t.L(t) = length of hole in year tW (t) = width of hole in year tarea is A(t) = L(t)×W (t)

C

so A′(0) = L′(0)×W (0) + L(0)×W ′(0)= 10× 100 + 300× 20 = 1000 + 6000 = 7000


10

20

100

200

L(0) = 200W(0) = 100L’(0) = 10W’(0)= 20

A = LWA’ = L’W + LW’A’(0) = L’(0) W(0) + L(0)W’(0) = (10)(100) + (200)(20) = 5000


L(t)

W(t)


C

so A′(0) = L′(0)×W (0) + L(0)×W ′(0)

= 10× 100 + 300× 20 = 1000 + 6000 = 7000


10

20

100

200

L(0) = 200W(0) = 100L’(0) = 10W’(0)= 20

A = LWA’ = L’W + LW’A’(0) = L’(0) W(0) + L(0)W’(0) = (10)(100) + (200)(20) = 5000


L(t)

W(t)


C

so A′(0) = L′(0)×W (0) + L(0)×W ′(0)= 10× 100 + 300× 20 = 1000 + 6000 = 7000

math 34b lecture 7

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