Section 5.4: Modeling with CircularFunctions

Circular Motion

Example

A ferris wheel with radius 25 feet is rotating at a rate of 3revolutions per minute, When t = 0, a chair starts at its lowestpoint on the wheel, which is 5 feet above ground. Write amodel for the height h (in feet) of the chair as a function of thetime t (in seconds).

What kind of curve should we use - sine or cosine?

First, let’s find the amplitude. Thoughts?

Since the radius is 25 feet, the amplitude is 25 feet.

Circular Motion

Example

A ferris wheel with radius 25 feet is rotating at a rate of 3revolutions per minute, When t = 0, a chair starts at its lowestpoint on the wheel, which is 5 feet above ground. Write amodel for the height h (in feet) of the chair as a function of thetime t (in seconds).

What kind of curve should we use - sine or cosine?

First, let’s find the amplitude. Thoughts?

Since the radius is 25 feet, the amplitude is 25 feet.

Circular Motion

Example

A ferris wheel with radius 25 feet is rotating at a rate of 3revolutions per minute, When t = 0, a chair starts at its lowestpoint on the wheel, which is 5 feet above ground. Write amodel for the height h (in feet) of the chair as a function of thetime t (in seconds).

What kind of curve should we use - sine or cosine?

First, let’s find the amplitude. Thoughts?

Since the radius is 25 feet, the amplitude is 25 feet.

Circular Motion

Example

A ferris wheel with radius 25 feet is rotating at a rate of 3revolutions per minute, When t = 0, a chair starts at its lowestpoint on the wheel, which is 5 feet above ground. Write amodel for the height h (in feet) of the chair as a function of thetime t (in seconds).

What kind of curve should we use - sine or cosine?

First, let’s find the amplitude. Thoughts?

Since the radius is 25 feet, the amplitude is 25 feet.

Circular Motion

What is the maximum of the chair?

Minimum?

What is the midline? y = 30

Now the period ... we are looking for t to be in seconds, so howlong is a revolution here?

20 seconds.

Circular Motion

What is the maximum of the chair? Minimum?

What is the midline? y = 30

Now the period ... we are looking for t to be in seconds, so howlong is a revolution here?

20 seconds.

Circular Motion

What is the maximum of the chair? Minimum?

What is the midline?

y = 30

Now the period ... we are looking for t to be in seconds, so howlong is a revolution here?

20 seconds.

Circular Motion

What is the maximum of the chair? Minimum?

What is the midline? y = 30

Now the period ... we are looking for t to be in seconds, so howlong is a revolution here?

20 seconds.

Circular Motion

What is the maximum of the chair? Minimum?

What is the midline? y = 30

Now the period ... we are looking for t to be in seconds, so howlong is a revolution here?

20 seconds.

Circular Motion

What is the maximum of the chair? Minimum?

What is the midline? y = 30

Now the period ... we are looking for t to be in seconds, so howlong is a revolution here?

20 seconds.

Circular Motion

Parameters:

A =

−25B = 2π

20 = π10

C = 0D = 30

Putting this together, we have

h(t) = −25cos( π

10t)+ 30

Circular Motion

Parameters:

A = −25

B = 2π20 = π

10

C = 0D = 30

Putting this together, we have

h(t) = −25cos( π

10t)+ 30

Circular Motion

Parameters:

A = −25B =

2π20 = π

10

C = 0D = 30

Putting this together, we have

h(t) = −25cos( π

10t)+ 30

Circular Motion

Parameters:

A = −25B = 2π

20 = π10

C = 0D = 30

Putting this together, we have

h(t) = −25cos( π

10t)+ 30

Circular Motion

Parameters:

A = −25B = 2π

20 = π10

C =

0D = 30

Putting this together, we have

h(t) = −25cos( π

10t)+ 30

Circular Motion

Parameters:

A = −25B = 2π

20 = π10

C = 0

D = 30

Putting this together, we have

h(t) = −25cos( π

10t)+ 30

Circular Motion

Parameters:

A = −25B = 2π

20 = π10

C = 0D =

30

Putting this together, we have

h(t) = −25cos( π

10t)+ 30

Circular Motion

Parameters:

A = −25B = 2π

20 = π10

C = 0D = 30

Putting this together, we have

h(t) = −25cos( π

10t)+ 30

Circular Motion

Parameters:

A = −25B = 2π

20 = π10

C = 0D = 30

Putting this together, we have

h(t) = −25cos( π

10t)+ 30

A Modeling Example

Example

A team of biologists have discovered a new creature in the rainforest. They note the temperature of the animal appears to varysinusoidally over time. A maximum temperature of 125◦

occurs 15 minutes after they start their examination. Aminimum temperature of 99◦ occurs 28 minutes later. The teamwould like to find a way to predict the animal’s temperatureover time in minutes. Your task is to help them by creating agraph of one full period and an equation of temperature as afunction over time in minutes

A Modeling Example

•(15,125)

•(43,99)

So this looks like what kind of curve? A sine curve ...

A Modeling Example

•(15,125)

•(43,99)

So this looks like what kind of curve? A sine curve ...

A Modeling Example

•(15,125)

•(43,99)

So this looks like what kind of curve? A sine curve ...

A Modeling Example

•(15,125)

•(43,99)

So this looks like what kind of curve? A sine curve ...

A Modeling Example

•(15,125)

•(43,99)

So this looks like what kind of curve?

A sine curve ...

A Modeling Example

•(15,125)

•(43,99)

So this looks like what kind of curve? A sine curve ...

A Modeling Example

We need to find the midline, amplitude and period first, whichwe can do from the information here. Then we will worryabout the horizontal shift.

How can we find the amplitude?

125 − 992

=262

= 13

So, if the amplitude is 13, what is the midline?

y = 112

A Modeling Example

We need to find the midline, amplitude and period first, whichwe can do from the information here. Then we will worryabout the horizontal shift.

How can we find the amplitude?

125 − 992

=262

= 13

So, if the amplitude is 13, what is the midline?

y = 112

A Modeling Example

We need to find the midline, amplitude and period first, whichwe can do from the information here. Then we will worryabout the horizontal shift.

How can we find the amplitude?

125 − 992

=262

= 13

So, if the amplitude is 13, what is the midline?

y = 112

A Modeling Example

We need to find the midline, amplitude and period first, whichwe can do from the information here. Then we will worryabout the horizontal shift.

How can we find the amplitude?

125 − 992

=262

= 13

So, if the amplitude is 13, what is the midline?

y = 112

A Modeling Example

We need to find the midline, amplitude and period first, whichwe can do from the information here. Then we will worryabout the horizontal shift.

How can we find the amplitude?

125 − 992

=262

= 13

So, if the amplitude is 13, what is the midline?

y = 112

A Modeling Example

Now, the period ... thoughts?

It takes 28 minutes from the highest to the lowest temperature,so the period therefore is 56 minutes.

So at this point we knowA = 13B = π

28

D = 112

A Modeling Example

Now, the period ... thoughts?

It takes 28 minutes from the highest to the lowest temperature,so the period therefore is

56 minutes.

So at this point we knowA = 13B = π

28

D = 112

A Modeling Example

Now, the period ... thoughts?

It takes 28 minutes from the highest to the lowest temperature,so the period therefore is 56 minutes.

So at this point we knowA = 13B = π

28

D = 112

A Modeling Example

Now, the period ... thoughts?

It takes 28 minutes from the highest to the lowest temperature,so the period therefore is 56 minutes.

So at this point we knowA =

13B = π

28

D = 112

A Modeling Example

Now, the period ... thoughts?

It takes 28 minutes from the highest to the lowest temperature,so the period therefore is 56 minutes.

So at this point we knowA = 13B =

π28

D = 112

A Modeling Example

Now, the period ... thoughts?

It takes 28 minutes from the highest to the lowest temperature,so the period therefore is 56 minutes.

So at this point we knowA = 13B = π

28

D =

112

A Modeling Example

Now, the period ... thoughts?

It takes 28 minutes from the highest to the lowest temperature,so the period therefore is 56 minutes.

So at this point we knowA = 13B = π

28

D = 112

A Modeling Example

Now the horizontal shift. Notice the curve doesn’t cross themidline on the y-axis, so there is a horizontal shift.

•(15,125)

•(43,99)

How can we find how big this shift needs to be?

Based on length of period and where peaks are, the shift is oneto the right.

A Modeling Example

Now the horizontal shift. Notice the curve doesn’t cross themidline on the y-axis, so there is a horizontal shift.

•(15,125)

•(43,99)

How can we find how big this shift needs to be?

Based on length of period and where peaks are, the shift is oneto the right.

A Modeling Example

So, our final equation therefore is ...

y = 13sin( π

28(x − 1)

)+ 112

A Modeling Example

So, our final equation therefore is ...

y = 13sin( π

28(x − 1)

)+ 112

Another Example

Example

On February 10, 1990, high tide in Boston was at midnight. Thewater level at high tide was 9.9 feet; later, at low tide, it was 0.1feet. Assuming the next high tide is exactly 12 hours later andthat the height of the water is given by a sine or cosine curve,find a formula for water level in Boston as a function of time t.

y = .1

y = 9.9

12 24

Another Example

Example

On February 10, 1990, high tide in Boston was at midnight. Thewater level at high tide was 9.9 feet; later, at low tide, it was 0.1feet. Assuming the next high tide is exactly 12 hours later andthat the height of the water is given by a sine or cosine curve,find a formula for water level in Boston as a function of time t.

y = .1

y = 9.9

12 24

Another Example

Now let’s determine what we need.

What type of curve should we use?

CosineAmplitude? 4.9Midline? 5Period? 12 hours

And the parameters ...

A = 4.9B = π

6

C = 0D = 5

Which gives y = 4.9cos(

π6 t)+ 5

Another Example

Now let’s determine what we need.

What type of curve should we use? Cosine

Amplitude? 4.9Midline? 5Period? 12 hours

And the parameters ...

A = 4.9B = π

6

C = 0D = 5

Which gives y = 4.9cos(

π6 t)+ 5

Another Example

Now let’s determine what we need.

What type of curve should we use? CosineAmplitude?

4.9Midline? 5Period? 12 hours

And the parameters ...

A = 4.9B = π

6

C = 0D = 5

Which gives y = 4.9cos(

π6 t)+ 5

Another Example

Now let’s determine what we need.

What type of curve should we use? CosineAmplitude? 4.9

Midline? 5Period? 12 hours

And the parameters ...

A = 4.9B = π

6

C = 0D = 5

Which gives y = 4.9cos(

π6 t)+ 5

Another Example

Now let’s determine what we need.

What type of curve should we use? CosineAmplitude? 4.9Midline?

5Period? 12 hours

And the parameters ...

A = 4.9B = π

6

C = 0D = 5

Which gives y = 4.9cos(

π6 t)+ 5

Another Example

Now let’s determine what we need.

What type of curve should we use? CosineAmplitude? 4.9Midline? 5

Period? 12 hours

And the parameters ...

A = 4.9B = π

6

C = 0D = 5

Which gives y = 4.9cos(

π6 t)+ 5

Another Example

Now let’s determine what we need.

What type of curve should we use? CosineAmplitude? 4.9Midline? 5Period?

12 hours

And the parameters ...

A = 4.9B = π

6

C = 0D = 5

Which gives y = 4.9cos(

π6 t)+ 5

Another Example

Now let’s determine what we need.

What type of curve should we use? CosineAmplitude? 4.9Midline? 5Period? 12 hours

And the parameters ...

A = 4.9B = π

6

C = 0D = 5

Which gives y = 4.9cos(

π6 t)+ 5

Another Example

Now let’s determine what we need.

What type of curve should we use? CosineAmplitude? 4.9Midline? 5Period? 12 hours

And the parameters ...

A = 4.9B = π

6

C = 0D = 5

Which gives y = 4.9cos(

π6 t)+ 5

Another Example

Now let’s determine what we need.

What type of curve should we use? CosineAmplitude? 4.9Midline? 5Period? 12 hours

And the parameters ...

A = 4.9

B = π6

C = 0D = 5

Which gives y = 4.9cos(

π6 t)+ 5

Another Example

Now let’s determine what we need.

What type of curve should we use? CosineAmplitude? 4.9Midline? 5Period? 12 hours

And the parameters ...

A = 4.9B = π

6

C = 0D = 5

Which gives y = 4.9cos(

π6 t)+ 5

Another Example

Now let’s determine what we need.

What type of curve should we use? CosineAmplitude? 4.9Midline? 5Period? 12 hours

And the parameters ...

A = 4.9B = π

6

C = 0

D = 5

Which gives y = 4.9cos(

π6 t)+ 5

Another Example

Now let’s determine what we need.

What type of curve should we use? CosineAmplitude? 4.9Midline? 5Period? 12 hours

And the parameters ...

A = 4.9B = π

6

C = 0D = 5

Which gives y = 4.9cos(

π6 t)+ 5

Another Example

Now let’s determine what we need.

What type of curve should we use? CosineAmplitude? 4.9Midline? 5Period? 12 hours

And the parameters ...

A = 4.9B = π

6

C = 0D = 5

Which gives y = 4.9cos(

π6 t)+ 5

Another Example

Now let’s determine what we need.

What type of curve should we use? CosineAmplitude? 4.9Midline? 5Period? 12 hours

And the parameters ...

A = 4.9B = π

6

C = 0D = 5

Which gives y = 4.9cos(

π6 t)+ 5