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### Applications of Sinusoidal derivatives

Tuesday, April 8th

Sinusoidal derivatives

- Questions from homework?

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 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

- 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 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 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 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 application

- Create an equation for the voltage as a function of time.

V(t) = Acos(ωt)

V(t) = Asin(ωt)

Amplitude

Amplitude = 170V

AC circuit application

- 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 application

- 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 application

- 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 application

- Create an equation for the voltage as a function of time.

V(t) = 170sin(380t)

V(t) = 170cos(380t)

AC circuit application

- If V(t) = 170sin(380t), find where the voltage would be at a maximum.

V(t) = 170sin(380t)

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 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 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 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 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 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 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 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 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 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 derivatives

- 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 derivatives

In general:

θ(t) = Asin(ωt)

θ

Amplitude

Amplitude is

(max θ + min θ)/2

Applications of sinusoidal derivatives

In general:

θ(t) = Asin(ωt)

θ

Amplitude

Amplitude is 30° or ⅙π radians

Applications of sinusoidal derivatives

In general:

θ(t) = Asin(ωt)

Angular frequency

θ

Amplitude

Applications of sinusoidal derivatives

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 derivatives

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 derivatives

In general:

θ(t) = Asin(ωt)

θ

Angular frequency

Angular frequency:

ω = 2π / T

Applications of sinusoidal derivatives

In general:

θ(t) = Asin(ωt)

θ

Angular frequency

Angular frequency:

ω = 2π / 4 = ½π

Applications of sinusoidal derivatives

Therefore our equation is:

θ(t) = ⅙π sin(½π t)

θ

Applications of sinusoidal derivatives

- 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 derivatives

- 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 derivatives

- b. Determine the angular acceleration.

θ(t) = Asin(ωt)

θ’(t) = Aωcos(ωt)

θ

Applications of sinusoidal derivatives

- 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

- 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 design

- 1. Should we model the equation for angle as a cosine, sine or something in between?

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 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 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 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 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 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 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 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 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 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 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 design

- Finally, we can figure out what L will ensure that we never exceed 30m/s.
- 30 = –1.22√[9.8L]

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 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 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 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 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)

Homework

- Page 241 #1, 3 – 7
- Assignment #8

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