Daylight Hours Homework: Solving for Portland, OR 2016

[Coltjb7]

Homework Statement

My teacher created this following problem:
The equation H(t)=3.3sin(0.017214t-1.369247)+12 estimates the number of daylight hours in Portland, OR for any day, t, during 2016.
1. Give the date in 2016 when the number of daylight hours is increasing the fastest. How many additional minutes of daylight do we gain on that day?
2.Give the date in 2016 when the number of daylight hours is decreasing the fastest. How many minutes of daylight do we lose on that day?
3. How many days into the year is the longest day of 2016? What is the date?

Homework Equations

H(t)=3.3sin(0.017214t-1.369247)+ 12 0<t January 1=0; December 31=365; *2016 is
H'(t)=0.0568062cos(0.017214t-1.369247) a leap year.
H''(t)=-9.77861927E^-4sin(0.017214t-1.369247)

The Attempt at a Solution

I attempted to use the first derivative test but am not sure what I am doing. Do I need to solve for t? One kid at my table thinks they figured it out and got t=(about)74 so the answer t #1 would be around March 10th. I m totally lost. Please help

 

[SammyS]

Homework Statement

My teacher created this following problem:
The equation H(t)=3.3sin(0.017214t-1.369247)+12 estimates the number of daylight hours in Portland, OR for any day, t, during 2016.
1. Give the date in 2016 when the number of daylight hours is increasing the fastest. How many additional minutes of daylight do we gain on that day?
2.Give the date in 2016 when the number of daylight hours is decreasing the fastest. How many minutes of daylight do we lose on that day?
3. How many days into the year is the longest day of 2016? What is the date?

Homework Equations

H(t)=3.3sin(0.017214t-1.369247)+ 12 0<t January 1=0; December 31=365; *2016 is
H'(t)=0.0568062cos(0.017214t-1.369247) a leap year.
H''(t)=-9.77861927E^-4sin(0.017214t-1.369247)

The Attempt at a Solution

I attempted to use the first derivative test but am not sure what I am doing. Do I need to solve for t? One kid at my table thinks they figured it out and got t=(about)74 so the answer t #1 would be around March 10th. I m totally lost. Please help

Hello Coltjb7. Welcome to PF !

I general, what does the first derivative test tell you?

 

[Coltjb7]

Hello Coltjb7. Welcome to PF !

I general, what does the first derivative test tell you?

It would tell me on what intervals the equation is increasing/decreasing.
Honestly I am having trouble finding the Critical Values.

Thank you for the welcome

 

[SteamKing]

It would tell me on what intervals the equation is increasing/decreasing.
Honestly I am having trouble finding the Critical Values.

Thank you for the welcome

Why don't you show us what you've tried to find the critical values?

 

[Coltjb7]

Why don't you show us what you've tried to find the critical values?

so what I have been trying to do is this:
H'(t)=0.0568062cos(0.017214t-1.369247)
Since anything multiplied by 0 is equal to zero I ignored the constant leaving me with this:
0=cos(0.017214t-1.369247).
I know that cos=0 at pi/2 and 3pi/2 but am having trouble figuring out how to work this out.

 

[SteamKing]

so what I have been trying to do is this:
H'(t)=0.0568062cos(0.017214t-1.369247)
Since anything multiplied by 0 is equal to zero I ignored the constant leaving me with this:
0=cos(0.017214t-1.369247).
I know that cos=0 at pi/2 and 3pi/2 but am having trouble figuring out how to work this out.

Do you know about inverse trigonometric functions?

 

[Coltjb7]

Haha, yes I know about them but have not used them in a while. Would I do this:
0=cos(0.017214t-1.369247)
cos^-1(0)=0.017214t-1.369247
[(cos^-1(0))+1.369247]/0.017214=t
t=170.79

 

[Coltjb7]

So would my intervals go from (-infinity,170.79) and (170.79, infinity)?

 

[SteamKing]

Haha, yes I know about them but have not used them in a while. Would I do this:
0=cos(0.017214t-1.369247)
cos^-1(0)=0.017214t-1.369247
[(cos^-1(0))+1.369247]/0.017214=t
t=170.79

That's one solution. What's the other one?

So would my intervals go from (-infinity,170.79) and (170.79, infinity)?

It's not clear what these intervals are for.

 

[Coltjb7]

That's one solution. What's the other one?

It's not clear what these intervals are for.

I honestly am not sure how to find the other solution. Mind helping me out? I would be using the intervals for my first derivative test. our teacher has us create a table like this:
Interval ( we gather this from the critical values)
Test Value (any number you want within each interval)
Sign of f'(x) (test the value you chose as a test value to see if it is >0 or <0)
Conclusion (where we say if it is increasing/decreasing)

 

[SammyS]

$H(t)\$ gives the hours of daylight on day $\ t\,.$

The derivative, $\ H'(t)\$ gives the rate at which the $\ H(t)\$ is changing on day $\ t\,.$

If you solve for the derivative being zero, that gives the day on which the hours of daylight is minimum or maximum. That's what you're asked to find in part 3.

To find out when the hours of daylight is changing at a maximum or minimum rate, you need to look at the derivative of the function that gives the rate of change.

Added in Edit:
The following solution is for part 3,
3. How many days into the year is the longest day of 2016?

Haha, yes I know about them but have not used them in a while. Would I do this:
0=cos(0.017214t-1.369247)
cos^-1(0)=0.017214t-1.369247
[(cos^-1(0))+1.369247]/0.017214=t
t=170.79

That's some day in late June, close to the day with most sunlight hours.

 

[Coltjb7]

I understand I need to look at the derivative but how do I find a specific date where it is increasing/decreasing the most. She said we could use our calculators to do it, but I cannot remember how to do this on a calculator. I have the TI-84 Plus.

 

[SammyS]

I understand I need to look at the derivative but how do I find a specific date where it is increasing/decreasing the most. She said we could use our calculators to do it, but I cannot remember how to do this on a calculator. I have the TI-84 Plus.

Look at the derivative of WHICH function?

 

[SteamKing]

I honestly am not sure how to find the other solution. Mind helping me out?

You already told me the solutions:

I know that cos=0 at pi/2 and 3pi/2 but am having trouble figuring out how to work this out.

You have apparently worked out the value of t using π/2 as the argument for the cosine where its value is zero. What do you get for t if the argument is 3π/2?

 

[Coltjb7]

You already told me the solutions:
You have apparently worked out the value of t using π/2 as the argument for the cosine where its value is zero. What do you get for t if the argument is 3π/2?

See that's what I am not understanding. I did nothing with pi/2. I ignored that information and solved it using cos^-1

 

[Coltjb7]

Look at the derivative of WHICH function?

The original function. H(t)=3.3sin(0.017214t-1.369247)+12

 

[SammyS]

The original function. H(t)=3.3sin(0.017214t-1.369247)+12

No. Setting the derivative of that to zero would give the date of max or min hours of daylight.

You already have the function and the first and second derivatives!

...

H(t)=3.3sin(0.017214t-1.369247)+ 12 where 0<t January 1=0; December 31=365; *2016 is a leap year.
H'(t)=0.0568062cos(0.017214t-1.369247)
H''(t)=-9.77861927E^-4sin(0.017214t-1.369247)

$H'(t)\$ is the rate of change in the hours of daylight.

You need to find the max and min for this rate of change, not for the number of hours (not until part 3 anyway).

rate of change: $\ H'(t)\$

number of hours: $\ H(t)\$

 

[SteamKing]

See that's what I am not understanding. I did nothing with pi/2. I ignored that information and solved it using cos^-1

But what value did you do the cos^-1 of?

 

[Coltjb7]

See that's what I am not understanding. I did nothing with pi/2. I ignored that information and solved it using cos^-1

So now i solved H'(t) for zero when t=pi/2 and 3pi/2. These are my new answers:
when t=pi/2 there is a critical value at 108.7306
When t=3pi/2 there is a critical value at 36.24.

So, is day number 108.7306 (do i round to day 109 or keep it at 108?) the day where the most daylight is being added? And, is day 36.2435 the day where the the daylight is decreasing the fastest?

 

[SteamKing]

So now i solved H'(t) for zero when t=pi/2 and 3pi/2. These are my new answers:
when t=pi/2 there is a critical value at 108.7306
When t=3pi/2 there is a critical value at 36.24.

So, is day number 108.7306 (do i round to day 109 or keep it at 108?) the day where the most daylight is being added? And, is day 36.2435 the day where the the daylight is decreasing the fastest?

I think you have things askew here.

If H'(t) = cos(0.017214t-1.369247), then you want to find t such that it makes H'(t) = 0. Since H'(t) is a cosine function, cosine is zero when its argument is π/2 or 3π/2, which means that (0.017214t-1.369247) = π/2 or 3π/2. You have to solve for t to get those values where H'(t) = 0.

 

[SammyS]

If $\ H'(t)=0\,,\$ then $\ H(t)\$ is at a max or min.

If $\ H''(t)=0\,,\$ then $\ H'(t)\$ is at a max or min.

 

[Coltjb7]

I think you have things askew here.

If H'(t) = cos(0.017214t-1.369247), then you want to find t such that it makes H'(t) = 0. Since H'(t) is a cosine function, cosine is zero when its argument is π/2 or 3π/2, which means that (0.017214t-1.369247) = π/2 or 3π/2. You have to solve for t to get those values where H'(t) = 0.

Ohhh. Okay. Thank you. I was going about it wrong and set t=pi/2 and 3pi/2 then plugged that value into H'(t). Now I realize that I made a stupid mistake. I worked out those values and came up with:

When pi/2=0.017214t-1.369247 Then following the same premise for 3pi/2
0.017214t=pi/2 +1.369247 t=353.2959
0.017214t=(pi+2.73849)/2
t=(pi+2.73849)/(2(0.017214))
t=170.7937

*I got the value 2.73849 by creating a common denominator of 2*

Now, are those my intervals for my first derivative test, or are they the values I need to answer questions 1&2?

 

[SammyS]

Now, are those my intervals for my first derivative test, or are they the values I need to answer questions 1&2?

Go back & review all of the posts.

 

[Coltjb7]

Go back & review all of the posts.

I think I have got it. Thank you guys very much!