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 March 21 and April 21. (Assume there are 365 days in a year. Round your answers to four decimal places.)
To compare the increase in the number of hours of daylight in Philadelphia on March 21 and April 21, we need to find the difference in L(t) values for those dates.
First, we need to convert the dates to the corresponding day of the year:
- March 21st is the 81st day of the year (since t = 81)
- April 21st is the 112th day of the year (since t = 112)
Now we can calculate L(t) for each date:
- For March 21st: L(81) = 12 + 2.8 * sin[(2π/365)(81 - 80)]
- For April 21st: L(112) = 12 + 2.8 * sin[(2π/365)(112 - 80)]
Let's calculate these values:
For March 21st:
L(81) = 12 + 2.8 * sin[(2π/365)(1)]
= 12 + 2.8 * sin[(2π/365)]
≈ 12 + 2.8 * 0.1710
≈ 12 + 0.479
≈ 12.479
For April 21st:
L(112) = 12 + 2.8 * sin[(2π/365)(32)]
= 12 + 2.8 * sin[(2π/365)(32)]
≈ 12 + 2.8 * 0.386
≈ 12 + 1.0808
≈ 13.081
Therefore, the number of hours of daylight in Philadelphia on March 21st is approximately 12.479 hours, while on April 21st it is approximately 13.081 hours.
To compare the increase, we subtract the number of hours of daylight on March 21st from the number of hours on April 21st:
Increase = 13.081 - 12.479 ≈ 0.602
Thus, the number of hours of daylight in Philadelphia increases by approximately 0.602 hours from March 21st to April 21st.
To compare the rate at which the number of hours of daylight is increasing in Philadelphia on March 21 and April 21, we need to calculate the derivative of the function L(t) with respect to time (t).
The derivative of L(t) with respect to t can be found by applying the chain rule. First, we differentiate the outer function (12 + 2.8 sin[(2π/365)(t − 80)]) with respect to the inner function [(2π/365)(t − 80)]. Then, we multiply this by the derivative of the inner function with respect to t.
Let's calculate the derivative:
dL(t)/dt = (2.8) * (2π/365) * cos[(2π/365)(t − 80)]
Now, let's evaluate the derivative at t = 21 (March 21) and t = 21 + 31 = 52 (April 21):
For March 21 (t = 21):
dL(21)/dt = (2.8) * (2π/365) * cos[(2π/365)(21 − 80)]
For April 21 (t = 52):
dL(52)/dt = (2.8) * (2π/365) * cos[(2π/365)(52 − 80)]
Now, we can substitute the values into the formula and calculate the derivatives:
dL(21)/dt ≈ (2.8) * (2π/365) * cos[(2π/365)(21 − 80)]
dL(52)/dt ≈ (2.8) * (2π/365) * cos[(2π/365)(52 − 80)]
Finally, round the answers to four decimal places to compare the rate of change in the number of hours of daylight between March 21 and April 21 in Philadelphia.
3/31: t=90
4/21: t=111
L'(t) = 2.8cos[(2π/365)(t − 80)](2π/365)
now just plug in t=90 and t=111
The answer, of course, will be in hr/day.