A model for the length of daylight (in hours) in Philadelphia on the t th day of the year is
L(t)= 12+2.8 sin[2pi/365(t-80)]
Use this model to compare how the number of hours of daylight is increasing in Philadelphia on March 21 and May 21. (Assume there are 365 days in a year. Round your answers to four decimal places.
March 21 L'(t)= ?
May 21 L'(t)=?
I found;
L'(t)=2.8cos(2pi/365(t-80))(2pi/365)
I get the same derivative.
You can post your numerical answer for a check if you wish.
To find the rate of change of the number of hours of daylight in Philadelphia on March 21 and May 21, we need to find the derivative of the equation L(t) with respect to t.
For March 21 (t = 80 + 31 + 21 = 132), we have:
L'(132) = 2.8 * cos(2π/365 * (132 - 80)) * (2π/365)
= 2.8 * cos(2π/365 * 52) * (2π/365)
Now let's calculate this value.
L'(132) ≈ 2.8 * cos(0.2857) * 0.0173
≈ 2.8 * 0.9659 * 0.0173
≈ 0.0477 hours
So on March 21, the number of hours of daylight in Philadelphia is increasing at a rate of approximately 0.0477 hours per day.
Now let's calculate the rate for May 21 (t = 80 + 31 + 21 + 30 + 21 = 183):
L'(183) = 2.8 * cos(2π/365 * (183 - 80)) * (2π/365)
= 2.8 * cos(2π/365 * 103) * (2π/365)
Now let's calculate this value.
L'(183) ≈ 2.8 * cos(0.5644) * 0.0173
≈ 2.8 * 0.8253 * 0.0173
≈ 0.0328 hours
So on May 21, the number of hours of daylight in Philadelphia is increasing at a rate of approximately 0.0328 hours per day.
To find the rate at which the number of hours of daylight is increasing, we need to differentiate the given function, L(t), with respect to time (t).
The derivative of L(t) with respect to t, denoted as L'(t), can be calculated using the chain rule. Recall that the derivative of sin(u) with respect to u is cos(u). Here's how we can find L'(t):
L(t) = 12 + 2.8 * sin[2π/365(t - 80)]
Using the chain rule, we differentiate each term separately. The derivative of 12 with respect to t is zero since it is a constant. For the second term, we apply the chain rule:
L'(t) = 0 + 2.8 * cos[2π/365(t - 80)] * [d/dt(2π/365(t - 80))]
The derivative of (2π/365(t - 80)) with respect to t can be calculated as follows:
d/dt(2π/365(t - 80)) = (2π/365) * d/dt(t - 80)
= (2π/365) * 1
= 2π/365
Substituting this back into our previous equation:
L'(t) = 2.8 * cos[2π/365(t - 80)] * (2π/365)
= 5.6π/365 * cos[2π/365(t - 80)]
Now, let's use this derivative to find the rate at which the number of hours of daylight is increasing in Philadelphia on March 21 and May 21.
To find L'(t) for March 21, we substitute t = 81 into our derivative:
L'(81) = 5.6π/365 * cos[2π/365(81 - 80)]
= 5.6π/365 * cos[2π/365]
Rounding this value to four decimal places, we get L'(81) ≈ 0.4372
Therefore, the rate at which the number of hours of daylight is increasing in Philadelphia on March 21 is approximately 0.4372 hours per day.
To find L'(t) for May 21, we substitute t = 141 into our derivative:
L'(141) = 5.6π/365 * cos[2π/365(141 - 80)]
= 5.6π/365 * cos[2π/365(61)]
Rounding this value to four decimal places, we get L'(141) ≈ 0.2210
Therefore, the rate at which the number of hours of daylight is increasing in Philadelphia on May 21 is approximately 0.2210 hours per day.