In Philadelphia the number of hours of daylight on day t (where t is the number of days after Jan. 1) is modeled by the function
L(t)= 12+2.83sin(2pi/365(t-365))
A) Which days of the year have about 10 hours of daylight?
B) Which days of the year have more than 10 hours of daylight?
(A) Solve
10 = 12 + 2.83 sin[(2*pi/365*(t-365)]
-2 = 2.83 sin[(2*pi*/365)(t-365)]
-0.7067 = sin[(2*pi*/365)(t-365)]
sin[(2*pi*/365)(t-365)] = -0.7848
Solve for t.
[(2*pi*/365)(t-365)] = -0.90244
Use the first value
t-365 = -52
t = 313 days
Oct 1 is day 303. So Oct 11 is one answer. The other day will be 52 days after the winter solstice, or about Feb 11.
To find the days of the year with about 10 hours of daylight, we need to solve the equation:
L(t) = 10
Substituting the given function, we have:
12 + 2.83sin(2π/365(t-365)) = 10
Simplifying the equation:
2.83sin(2π/365(t-365)) = 10 - 12
2.83sin(2π/365(t-365)) = -2
Dividing both sides by 2.83:
sin(2π/365(t-365)) ≈ -2/2.83
sin(2π/365(t-365)) ≈ -0.707
Taking the inverse sine of both sides:
2π/365(t-365) ≈ arcsin(-0.707)
Solving for t:
t - 365 ≈ (365/2π) * arcsin(-0.707)
t ≈ (365/2π) * arcsin(-0.707) + 365
Using a calculator to find the approximate value of arcsin(-0.707):
arcsin(-0.707) ≈ -0.775
Substituting this back into the equation for t:
t ≈ (365/2π) * (-0.775) + 365
Calculating this expression:
t ≈ -2.11 + 365
t ≈ 362.89
Since t represents the number of days after January 1, the value of t ≈ 362.89 means that there are about 10 hours of daylight on day 362 or 363 of the year.
Therefore, the days of the year with about 10 hours of daylight are around December 28th and 29th.
To find the days of the year with more than 10 hours of daylight, we can simply observe that any days before or after December 28th and 29th will have more than 10 hours of daylight.
Thus, the days of the year with more than 10 hours of daylight are from January 1st until December 27th and from December 30th onwards.
To determine which days of the year have about 10 hours of daylight, we need to find the values of t for which the function L(t) is approximately equal to 10.
A) To solve this, we'll set up the equation L(t) = 10 and solve for t:
12 + 2.83sin(2π/365(t - 365)) = 10
Subtracting 12 from both sides of the equation, we get:
2.83sin(2π/365(t - 365)) = -2
Dividing both sides of the equation by 2.83, we get:
sin(2π/365(t - 365)) ≈ -0.706
To find the values of t that satisfy this equation, we need to find the values of the angle that have a sine value of approximately -0.706. We'll use the inverse sine function (sin^-1) to find these angles.
sin^-1(-0.706) ≈ -43.06 degrees
Since the sine function is periodic, we can add or subtract multiples of 360 degrees or 2π radians to find all the angles that satisfy the equation. In this case, since we're dealing with days, we'll use degrees:
t = (365 ± 43.06) / (2π/365)
Calculating the values of t, we get:
t ≈ 57.76 or t ≈ 348.24
Therefore, the days of the year with about 10 hours of daylight are approximately day 58 (around February 27th) and day 348 (around December 14th).
B) To determine which days of the year have more than 10 hours of daylight, we need to find the values of t for which the function L(t) is greater than 10.
We can use the same equation L(t) > 10, but this time we'll solve it without the approximation.
12 + 2.83sin(2π/365(t - 365)) > 10
Subtracting 12 from both sides, we get:
2.83sin(2π/365(t - 365)) > -2
Dividing by 2.83 (a positive number), the inequality remains the same:
sin(2π/365(t - 365)) > -0.706
Now, we can proceed to solve the inequality similarly to before:
t = (365 ± sin^-1(-0.706) / (2π/365)
Calculating the values of t, we get:
t > 58.24 or t < 348.76
Therefore, the days of the year with more than 10 hours of daylight are approximately day 59 (around February 28th) to day 348 (around December 14th).