# Daylight Hours

1. Jan 5, 2016

### Coltjb7

1. The problem statement, all variables and given/known data
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?

2. Relevant 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)

3. 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

2. Jan 5, 2016

### SammyS

Staff Emeritus
Hello Coltjb7. Welcome to PF !

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

3. Jan 5, 2016

### Coltjb7

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

4. Jan 5, 2016

### SteamKing

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

5. Jan 5, 2016

### Coltjb7

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.

6. Jan 5, 2016

### SteamKing

Staff Emeritus
Do you know about inverse trigonometric functions?

7. Jan 5, 2016

### 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

8. Jan 5, 2016

### Coltjb7

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

9. Jan 5, 2016

### SteamKing

Staff Emeritus
That's one solution. What's the other one?
It's not clear what these intervals are for.

10. Jan 5, 2016

### Coltjb7

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)

11. Jan 5, 2016

### SammyS

Staff Emeritus
$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.

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

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

Last edited: Jan 5, 2016
12. Jan 5, 2016

### 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.

13. Jan 5, 2016

### SammyS

Staff Emeritus
Look at the derivative of WHICH function?

14. Jan 5, 2016

### SteamKing

Staff Emeritus
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?

15. Jan 5, 2016

### Coltjb7

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

16. Jan 5, 2016

### Coltjb7

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

17. Jan 5, 2016

### SammyS

Staff Emeritus
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)\$ 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)\$

18. Jan 5, 2016

### SteamKing

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

19. Jan 5, 2016

### Coltjb7

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?

20. Jan 5, 2016

### SteamKing

Staff Emeritus
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.