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Applications of Sinusoidal derivatives. Tuesday, April 8 th. Sinusoidal derivatives. Questions from homework? . Sinusoidal derivatives. F(x) = sin 3 ( x). What is F’(x)? . 3sin 2 (x) 3cos 2 (x ) 3sin 2 (x ) cos (x) 3cos 2 (x )sin( x). Sinusoidal derivatives. F(x) = sin 3 ( x).

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sinusoidal derivatives
Sinusoidal derivatives
  • Questions from homework?
sinusoidal derivatives1
Sinusoidal derivatives
  • F(x) = sin3(x)

What is F’(x)?

3sin2(x)

3cos2(x)

3sin2(x)cos(x)

3cos2(x)sin(x)

sinusoidal derivatives2
Sinusoidal derivatives
  • F(x) = sin3(x)

What is F’(x)?

3sin2(x)

3cos2(x)

3sin2(x)cos(x)

3cos2(x)sin(x)

sinusoidal derivatives3
Sinusoidal derivatives
  • F’(x) = cos(2x)

What is F(x)?

2sin(2x)

-2sin(2x)

½sin(2x)

-½sin(2x)

sinusoidal derivatives4
Sinusoidal derivatives
  • F’(x) = cos(2x)

What is F(x)?

2sin(2x)

-2sin(2x)

½sin(2x)

-½sin(2x)

sinusoidal derivatives5
Sinusoidal derivatives
  • F(x) = cos(2x)sin(2x)

What is F’(x)?

2cos2(2x) + 2sin2(2x)

2cos2(2x) – 2sin2(2x)

4sin2(2x) + 4cos2(2x)

4sin2(2x) – 4cos2(2x)

sinusoidal derivatives6
Sinusoidal derivatives
  • F(x) = cos(2x)sin(2x)

What is F’(x)?

2cos2(2x) + 2sin2(2x)

2cos2(2x) – 2sin2(2x)

4sin2(2x) + 4cos2(2x)

4sin2(2x) – 4cos2(2x)

ac circuit application
AC circuit application
  • This morning, I plugged my kettle into our wall, and somehow delicious tea appeared. How?
  • Our electricity grid uses AC (alternating current). This means that voltage flips from + to – every 0.0167s, and the current changes direction in turn.

Visualizing AC

ac circuit application1
AC circuit application
  • The voltage in your wall changes from –170V to +170V and back again every 0.0167s. (You may have heard that your wall has “120V” but that’s just the root mean square or “average voltage”).
  • Create an equation for the voltage as a function of time.
ac circuit application2
AC circuit application
  • The voltage in your wall changes from –170V to +170V and back again every 0.0167s. (You may have heard that your wall has “120V” but that’s just the root mean square or “average voltage”).
  • Create an equation for the voltage as a function of time.

V(t) = Acos(ωt)

V(t) = Asin(ωt)

ac circuit application3
AC circuit application
  • The voltage in your wall changes from –170V to +170V and back again every 0.0167s. (You may have heard that your wall has “120V” but that’s just the root mean square or “average voltage”).
  • Create an equation for the voltage as a function of time.

V(t) = Acos(ωt)

V(t) = Asin(ωt)

Amplitude

Amplitude is

(max θ + min θ)/2

ac circuit application4
AC circuit application
  • The voltage in your wall changes from –170V to +170V and back again every 0.0167s. (You may have heard that your wall has “120V” but that’s just the root mean square or “average voltage”).
  • Create an equation for the voltage as a function of time.

V(t) = Acos(ωt)

V(t) = Asin(ωt)

Amplitude

Amplitude = 170V

ac circuit application5
AC circuit application
  • The voltage in your wall changes from –170V to +170V and back again every 0.0167s. (You may have heard that your wall has “120V” but that’s just the root mean square or “average voltage”).
  • Create an equation for the voltage as a function of time.

V(t) = Acos(ωt)

V(t) = Asin(ωt)

Angular frequency

ω = 2πf

ac circuit application6
AC circuit application
  • The voltage in your wall changes from –170V to +170V and back again every 0.0167s. (You may have heard that your wall has “120V” but that’s just the root mean square or “average voltage”).
  • Create an equation for the voltage as a function of time.

V(t) = Acos(ωt)

V(t) = Asin(ωt)

Angular frequency

ω = 2π / T

ω = 2πf

ac circuit application7
AC circuit application
  • The voltage in your wall changes from –170V to +170V and back again every 0.0167s. (You may have heard that your wall has “120V” but that’s just the root mean square or “average voltage”).
  • Create an equation for the voltage as a function of time.

V(t) = Acos(ωt)

V(t) = Asin(ωt)

Angular frequency

ω = 380 rad/s

ac circuit application8
AC circuit application
  • The voltage in your wall changes from –170V to +170V and back again every 0.0167s. (You may have heard that your wall has “120V” but that’s just the root mean square or “average voltage”).
  • Create an equation for the voltage as a function of time.

V(t) = 170sin(380t)

V(t) = 170cos(380t)

ac circuit application9
AC circuit application
  • If V(t) = 170sin(380t), find where the voltage would be at a maximum.

V(t) = 170sin(380t)

ac circuit application10
AC circuit application
  • If V(t) = 170sin(380t), find where the voltage would be at a maximum.

V(t) = 170sin(380t)

V’(t) = 64600cos(380t)

ac circuit application11
AC circuit application
  • If V(t) = 170sin(380t), find where the voltage would be at a maximum.

V(t) = 170sin(380t)

V’(t) = 64600cos(380t)

0 = 64600cos(380t)

0 = cos(380t)

ac circuit application12
AC circuit application
  • If V(t) = 170sin(380t), find where the voltage would be at a maximum.

cos(380t)

V(t) = 170sin(380t)

V’(t) = 64600cos(380t)

0 = 64600cos(380t)

0 = cos(380t)

time

ac circuit application13
AC circuit application
  • If V(t) = 170sin(380t), find where the voltage would be at a maximum.

cos(380t)

V(t) = 170sin(380t)

V’(t) = 64600cos(380t)

0 = 64600cos(380t)

0 = cos(380t)

time

ac circuit application14
AC circuit application
  • If V(t) = 170sin(380t), find where the voltage would be at a maximum.

cos(380t)

V(t) = 170sin(380t)

V’(t) = 64600cos(380t)

0 = 64600cos(380t)

0 = cos(380t)

T = 0.0167

time

0.0041s

0.012s

applications of sinusoidal derivatives1
Applications of sinusoidal derivatives
  • Recall:

x(t) = a function that describes position as a function of time

v(t) = rate of change of position as a function of time (aka. velocity)

a(t) = rate of change of velocity as a function of time (aka. acceleration)

v(t) = x’(t)

a(t) = v’(t) = x’’(t)

applications of sinusoidal derivatives2
Applications of sinusoidal derivatives
  • New:

θ(t) = a function that describes angle as a function of time

ω(t) = rate of change of angle as a function of time. This is called the “angular velocity”

α(t) = rate of change of angular velocity as a function of time. This is called the “angular acceleration”

ω(t) = θ’(t)

α(t) = ω’(t) = θ’’(t)

applications of sinusoidal derivatives3
Applications of sinusoidal derivatives
  • As a calculus student, you see derivatives everywhere you go. On this occasion, you are visiting the swings at Erieau’s beach and you notice that the angle of your swing varies from +30° to -30° (where θ is the angle measured from the vertical). It takes you 4 seconds to complete one swing.
  • a. Construct an equation for the angle you are relative to the vertical as a function of time.
applications of sinusoidal derivatives4
Applications of sinusoidal derivatives
  • As a calculus student, you see derivatives everywhere you go. On this occasion, you are visiting the swings at Erieau’s beach and you notice that the angle of your swing varies from +30° to -30° (where θ is the angle measured from the vertical). It takes you 4 seconds to complete one swing.
  • a. Construct an equation for the angle (in radians) you are relative to the vertical as a function of time.

θ

applications of sinusoidal derivatives5
Applications of sinusoidal derivatives
  • As a calculus student, you see derivatives everywhere you go. On this occasion, you are visiting the swings at Erieau’s beach and you notice that the angle of your swing varies from +30° to -30° (where θ is the angle measured from the vertical). It takes you 4 seconds to complete one swing.
  • a. Construct an equation for the angle (in radians) you are relative to the vertical as a function of time.

In general:

θ(t) = Asin(ωt)

θ

applications of sinusoidal derivatives6
Applications of sinusoidal derivatives
  • As a calculus student, you see derivatives everywhere you go. On this occasion, you are visiting the swings at Erieau’s beach and you notice that the angle of your swing varies from +30° to -30° (where θ is the angle measured from the vertical). It takes you 4 seconds to complete one swing.
  • a. Construct an equation for the angle (in radians) you are relative to the vertical as a function of time.

In general:

θ(t) = Asin(ωt)

θ

Amplitude

applications of sinusoidal derivatives7
Applications of sinusoidal derivatives
  • As a calculus student, you see derivatives everywhere you go. On this occasion, you are visiting the swings at Erieau’s beach and you notice that the angle of your swing varies from +30° to -30° (where θ is the angle measured from the vertical). It takes you 4 seconds to complete one swing.
  • a. Construct an equation for the angle (in radians) you are relative to the vertical as a function of time.

In general:

θ(t) = Asin(ωt)

θ

Amplitude

Amplitude is

(max θ + min θ)/2

applications of sinusoidal derivatives8
Applications of sinusoidal derivatives
  • As a calculus student, you see derivatives everywhere you go. On this occasion, you are visiting the swings at Erieau’s beach and you notice that the angle of your swing varies from +30° to -30° (where θ is the angle measured from the vertical). It takes you 4 seconds to complete one swing.
  • a. Construct an equation for the angle (in radians) you are relative to the vertical as a function of time.

In general:

θ(t) = Asin(ωt)

θ

Amplitude

Amplitude is 30° or ⅙π radians

applications of sinusoidal derivatives9
Applications of sinusoidal derivatives
  • As a calculus student, you see derivatives everywhere you go. On this occasion, you are visiting the swings at Erieau’s beach and you notice that the angle of your swing varies from +30° to -30° (where θ is the angle measured from the vertical). It takes you 4 seconds to complete one swing.
  • a. Construct an equation for the angle (in radians) you are relative to the vertical as a function of time.

In general:

θ(t) = Asin(ωt)

Angular frequency

θ

Amplitude

applications of sinusoidal derivatives10
Applications of sinusoidal derivatives
  • As a calculus student, you see derivatives everywhere you go. On this occasion, you are visiting the swings at Erieau’s beach and you notice that the angle of your swing varies from +30° to -30° (where θ is the angle measured from the vertical). It takes you 4 seconds to complete one swing.
  • a. Construct an equation for the angle (in radians) you are relative to the vertical as a function of time.

In general:

θ(t) = Asin(ωt)

θ

Angular frequency

Alert: Angular frequency has the same letter as (and is related to) angular velocity, but these are not the same thing!

applications of sinusoidal derivatives11
Applications of sinusoidal derivatives
  • As a calculus student, you see derivatives everywhere you go. On this occasion, you are visiting the swings at Erieau’s beach and you notice that the angle of your swing varies from +30° to -30° (where θ is the angle measured from the vertical). It takes you 4 seconds to complete one swing.
  • a. Construct an equation for the angle (in radians) you are relative to the vertical as a function of time.

In general:

θ(t) = Asin(ωt)

θ

Angular frequency

Angular frequency is 2π times the number of swings that occur in one second. ω = 2πf

applications of sinusoidal derivatives12
Applications of sinusoidal derivatives
  • As a calculus student, you see derivatives everywhere you go. On this occasion, you are visiting the swings at Erieau’s beach and you notice that the angle of your swing varies from +30° to -30° (where θ is the angle measured from the vertical). It takes you 4 seconds to complete one swing.
  • a. Construct an equation for the angle (in radians) you are relative to the vertical as a function of time.

In general:

θ(t) = Asin(ωt)

θ

Angular frequency

Angular frequency:

ω = 2π / T

applications of sinusoidal derivatives13
Applications of sinusoidal derivatives
  • As a calculus student, you see derivatives everywhere you go. On this occasion, you are visiting the swings at Erieau’s beach and you notice that the angle of your swing varies from +30° to -30° (where θ is the angle measured from the vertical). It takes you 4 seconds to complete one swing.
  • a. Construct an equation for the angle (in radians) you are relative to the vertical as a function of time.

In general:

θ(t) = Asin(ωt)

θ

Angular frequency

Angular frequency:

ω = 2π / 4 = ½π

applications of sinusoidal derivatives14
Applications of sinusoidal derivatives
  • As a calculus student, you see derivatives everywhere you go. On this occasion, you are visiting the swings at Erieau’s beach and you notice that the angle of your swing varies from +30° to -30° (where θ is the angle measured from the vertical). It takes you 4 seconds to complete one swing.
  • a. Construct an equation for the angle (in radians) you are relative to the vertical as a function of time.

Therefore our equation is:

θ(t) = ⅙π sin(½π t)

θ

applications of sinusoidal derivatives15
Applications of sinusoidal derivatives
  • As a calculus student, you see derivatives everywhere you go. On this occasion, you are visiting the swings at Erieau’s beach and you notice that the angle of your swing varies from +30° to -30° (where θ is the angle measured from the vertical). It takes you 4 seconds to complete one swing.
  • b. Determine the angular velocity by taking the derivative of θ with respect to time: dθ/dt.

θ(t) = Asin(ωt)

θ’(t) = Aωcos(ωt)

θ

applications of sinusoidal derivatives16
Applications of sinusoidal derivatives
  • As a calculus student, you see derivatives everywhere you go. On this occasion, you are visiting the swings at Erieau’s beach and you notice that the angle of your swing varies from +30° to -30° (where θ is the angle measured from the vertical). It takes you 4 seconds to complete one swing.
  • b. Determine the angular velocity by taking the derivative of θ with respect to time: dθ/dt.

θ(t) = ⅙π sin(½π t)

θ’(t) = (π2/12)cos(½π t)

This is the angular velocity, ω(t)

θ

applications of sinusoidal derivatives17
Applications of sinusoidal derivatives
  • As a calculus student, you see derivatives everywhere you go. On this occasion, you are visiting the swings at Erieau’s beach and you notice that the angle of your swing varies from +30° to -30° (where θ is the angle measured from the vertical). It takes you 4 seconds to complete one swing.
  • b. Determine the angular acceleration.

θ(t) = Asin(ωt)

θ’(t) = Aωcos(ωt)

θ

applications of sinusoidal derivatives18
Applications of sinusoidal derivatives
  • As a calculus student, you see derivatives everywhere you go. On this occasion, you are visiting the swings at Erieau’s beach and you notice that the angle of your swing varies from +30° to -30° (where θ is the angle measured from the vertical). It takes you 4 seconds to complete one swing.
  • b. Determine the angular acceleration.

θ(t) = Asin(ωt)

θ’(t) = Aωcos(ωt)

θ’’(t) = −Aω2sin(ωt)

This is the angular acceleration, α(t)

θ

epic swing design
Epic Swing design
  • You are designing a giant swing for our elementary school playground, and you are tasked with keeping it relatively safe for children. You decide that you should keep the maximum speed under 30m/s.
  • If the children will start by being pulled up by 70o (1.22 rad), what is the maximum length of the swing?
epic swing design1
Epic Swing design
  • 1. Should we model the equation for angle as a cosine, sine or something in between?
epic swing design2
Epic Swing design
  • Should we model the angle as a cosine, sine or something in between?
  • cosine – because the children will start at their maximum angle.
  • θ(t) = Acos(ωt)
epic swing design3
Epic Swing design
  • Should we model the angle as a cosine, sine or something in between?
  • cosine – because the children will start at their maximum angle.
  • θ(t) = Acos(ωt)
  • θ(t) = 1.22cos(ωt)

The amplitude is set by our initial angle

epic swing design4
Epic Swing design
  • Should we model the angle as a cosine, sine or something in between?
  • cosine – because the children will start at their maximum angle.
  • θ(t) = Acos(ωt)
  • θ(t) = 1.22cos(ωt)

The angular frequency will be determined by the length of the swing…

epic swing design5
Epic Swing design
  • Should we model the angle as a cosine, sine or something in between?
  • cosine – because the children will start at their maximum angle.
  • θ(t) = Acos(ωt)
  • θ(t) = 1.22cos(ωt)

The angular frequency will be determined by the length of the swing…

For a pendulum length L where all the weight is at the end, the angular frequency is:

ω = √(g / L)

epic swing design6
Epic Swing design
  • Should we model the angle as a cosine, sine or something in between?
  • cosine – because the children will start at their maximum angle.
  • θ(t) = Acos(ωt)
  • θ(t) = 1.22cos(√[g/L] t)

The angular frequency will be determined by the length of the swing…

For a pendulum length L where all the weight is at the end, the angular frequency is:

ω = √(g / L)

epic swing design7
Epic Swing design
  • We can use this model to figure out the angular velocity as a function of time:
  • θ(t) = 1.22cos(√[g/L] t)
epic swing design8
Epic Swing design
  • We can use this model to figure out the angular velocity as a function of time:
  • θ(t) = 1.22 cos(√[g/L] t)
  • θ’(t) = –1.22√[g/L] sin(√[g/L] t)
epic swing design9
Epic Swing design
  • Then we can use the angular velocity to figure out the linear velocity:
  • θ(t) = 1.22 cos(√[g/L] t)
  • θ’(t) = –1.22√[g/L] sin(√[g/L] t)
  • V(t) = L θ’(t)
epic swing design10
Epic Swing design
  • Then we can use the angular velocity to figure out the linear velocity:
  • θ(t) = 1.22 cos(√[g/L] t)
  • θ’(t) = –1.22√[g/L] sin(√[g/L] t)
  • V(t) = L θ’(t)
  • V(t) = L(–1.22√[g/L] sin(√[g/L] t))
  • = –1.22√[gL] sin(√[g/L] t))
epic swing design11
Epic Swing design
  • Finally, we can figure out what L will ensure that we never exceed 30m/s.
  • 30 = –1.22√[gL] sin(√[g/L] t))

sin(√[g/L] t)) reaches a maximum of 1

epic swing design12
Epic Swing design
  • Finally, we can figure out what L will ensure that we never exceed 30m/s.
  • 30 = –1.22√[gL] sin(√[g/L] t))

sin(√[g/L] t)) reaches a maximum of 1

Therefore as long as this amplitude is 30 or under, kids will be happy

epic swing design13
Epic Swing design
  • Finally, we can figure out what L will ensure that we never exceed 30m/s.
  • 30 = –1.22√[9.8L]
epic swing design14
Epic Swing design
  • Finally, we can figure out what L will ensure that we never exceed 30m/s.
  • 30 = –1.22√[9.8L]
  • 30/–1.22= √[9.8L]
epic swing design15
Epic Swing design
  • Finally, we can figure out what L will ensure that we never exceed 30m/s.
  • 30 = –1.22√[9.8L]
  • 30/–1.22= √[9.8L]
  • 24.6 = √[9.8L]
epic swing design16
Epic Swing design
  • Finally, we can figure out what L will ensure that we never exceed 30m/s.
  • 30 = –1.22√[9.8L]
  • 30/–1.22= √[9.8L]
  • 24.6 = √[9.8L]
  • 604.7 = 9.8L
epic swing design17
Epic Swing design
  • Finally, we can figure out what L will ensure that we never exceed 30m/s.
  • 30 = –1.22√[9.8L]
  • 30/–1.22= √[9.8L]
  • 24.6 = √[9.8L]
  • 604.7 = 9.8L
  • L = 61.7m
applications of sinusoidal derivatives19
Applications of sinusoidal derivatives
  • For the unique case where:
  • Amplitude of the swing is 1
  • Angular frequency is also 1
  • Graph:
  • the angle from the vertical vstime
  • the angular velocity vstime
  • the angular acceleration vs time  

θ(t) = Asin(ωt)

θ’(t) = Aωcos(ωt)

θ’’(t) = −Aω2sin(ωt)

applications of sinusoidal derivatives20
Applications of sinusoidal derivatives

θ(t) = sin(t)

θ’(t) = cos(t)

θ’’(t) = −sin(t)

homework
Homework
  • Page 241 #1, 3 – 7
  • Assignment #8