A model for the length of daylight (in hours) in Philadelphia on the tth day of the year is

L(t) = 12 + 2.8 sin 2π/365(t − 80)

.
Use this model to compare how the number of hours of daylight is increasing in Philadelphia on April 21 and June 5. (Assume there are 365 days in a year. Round your answers to four decimal places.)
April 21 L'(t) =

June 5 L'(t) =

I took the first derivative and plug in April 21 = 111, and June 5 = 156 in t. got 0.0482 but its not correct

Well, April 21 and June 5 are two different dates in Philadelphia, so let's find out how the number of hours of daylight is increasing on those days.

April 21:
To find L'(t) on April 21, we need to substitute t = 111 into the equation L'(t) = 12 + 2.8 sin (2π/365(t-80)).
So,
L'(111) = 12 + 2.8 sin (2π/365(111-80))

June 5:
To find L'(t) on June 5, we need to substitute t = 156 into the equation L'(t) = 12 + 2.8 sin (2π/365(t-80)).
So,
L'(156) = 12 + 2.8 sin (2π/365(156-80))

Please allow me a moment to calculate those for you.

To find the rate at which the number of hours of daylight is increasing, we need to find the derivative of the length of daylight function, L(t).

Given:
L(t) = 12 + 2.8 sin (2π/365(t - 80))

To find L'(t), we can follow these steps:

Step 1: Differentiate the constant term
The derivative of a constant term is zero, so the derivative of 12 is zero.

Step 2: Differentiate the sine term
To differentiate sin (2π/365(t - 80)), we need to use the chain rule. The derivative of sin(u) is cos(u), and we need to multiply it by the derivative of the inside function.

The derivative of the inside function is 2π/365, since the derivative of t - 80 is 1. So, the derivative of sin (2π/365(t - 80)) is cos (2π/365(t - 80)) * (2π/365).

Step 3: Combine the derivatives
Since the derivative of 12 is zero, our final derivative is:
L'(t) = 2.8 * cos (2π/365(t - 80)) * (2π/365)

Now, let's compute the rate of increase of daylight hours on April 21 and June 5.

April 21:
To find L'(t) for April 21, we substitute t = 111 (April 21st) into the derivative function:
L'(111) = 2.8 * cos (2π/365(111 - 80)) * (2π/365)

Simplifying this expression will give you the rate at which the number of hours of daylight is increasing on April 21.

June 5:
To find L'(t) for June 5, we substitute t = 156 (June 5th) into the derivative function:
L'(156) = 2.8 * cos (2π/365(156 - 80)) * (2π/365)

Simplifying this expression will give you the rate at which the number of hours of daylight is increasing on June 5.

Assuming April 21 ----> 111

L(111) = 12 + 2.8 sin 2π/365(111 − 80)
= 12 + 2.8 sin 2π/365(31)
= 13.424...

did you set your calculator to radians?

But whatever setting, you should have known that 0.0482 was wrong
since sin(anything) lies between -1 and +1
then 2.8 sin 2π/365(t − 80)
must lie between -2.8 and 2.8
and when 12 is added to that, the only answers possible are those between 9.2 and 14.8