In Philadelphia the number of hours of daylight on day t (t is the number of days after january 1) is modeled by the function

L(t)= 12+2.83sin((2*Pi)/365(t-800)
a. Which day have about 10 hours of daylight?
b. How many days of the year have more than 10 hours of day light?
the back of the book gave the answer
a. 34th day
b. 308th day
I want to know how they got this answer?

Well, sometimes the answers in the back of the book are like a secret club where only a few people know the secret handshake. But fear not, my friend, I can help you unravel this mystery!

To find the day when there are about 10 hours of daylight, we need to plug in L(t) = 10 and solve for t. Let's do some math!

10 = 12 + 2.83sin((2*Pi)/365(t-800)

First, we need to isolate the sine term:

2.83sin((2*Pi)/365(t-800)) = -2

Now, divide both sides by 2.83 (approximately):

sin((2*Pi)/365(t-800)) = -0.7

To find the angle that has a sin of -0.7, we can use a calculator or a handy dandy table of values. The angle is approximately -44 degrees. But, remember that the calculator gives us the angle in radians. So, let's convert -44 degrees to radians:

-44 degrees = -0.766 radians

Now, we can write our equation in terms of radians:

sin((2*Pi)/365(t-800)) = -0.766

To find t, we need to take the inverse sine (also known as arcsine) of both sides:

(2*Pi)/365(t-800) = arcsin(-0.766)

Now, let's solve for t:

(t-800) = (365/(2*Pi)) * arcsin(-0.766)

t = 800 + (365/(2*Pi)) * arcsin(-0.766)

Using a calculator, we find t ≈ 34.1. So, the 34th day (approximately) has about 10 hours of daylight.

Now, for part b, we need to find the number of days in the year that have more than 10 hours of daylight. The pattern of daylight repeats every 365 days, so we can find the number of days with more than 10 hours in one complete cycle.

Since we found that the 34th day has about 10 hours, we can subtract this from the total number of days in a year:

365 - 34 ≈ 331

So, there are approximately 331 days in a year that have more than 10 hours of daylight. Therefore, the answer in the back of the book is approximateloy correct with the 34th day having about 10 hours of daylight and approximately 331 days in the year that have more than 10 hours of daylight.

Hope that clears things up for you!

To find the day with about 10 hours of daylight and the number of days with more than 10 hours of daylight, you need to solve the equation L(t) = 10, where L(t) is the function that models the number of hours of daylight.

a. To find the day with about 10 hours of daylight, solve the equation L(t) = 10:
12 + 2.83sin((2*Pi)/365(t-800)) = 10

Subtract 12 from both sides:
2.83sin((2*Pi)/365(t-800)) = -2

Divide by 2.83:
sin((2*Pi)/365(t-800)) = -2/2.83

Taking the inverse sine (sin^(-1)), we get:
(2*Pi)/365(t-800) = sin^(-1)(-2/2.83)

Now solve for t by multiplying both sides by 365/(2*Pi):
t-800 = (365/(2*Pi)) * sin^(-1)(-2/2.83)

Add 800 to both sides:
t = 800 + (365/(2*Pi)) * sin^(-1)(-2/2.83)

Calculate the value on the right-hand side to find the day with about 10 hours of daylight.

b. To find the number of days with more than 10 hours of daylight, substitute L(t) > 10 into the equation:

12 + 2.83sin((2*Pi)/365(t-800)) > 10

Subtract 12 from both sides:
2.83sin((2*Pi)/365(t-800)) > -2

Divide by 2.83:
sin((2*Pi)/365(t-800)) > -2/2.83

Again, taking the inverse sine (sin^(-1)), we get:
(2*Pi)/365(t-800) > sin^(-1)(-2/2.83)

Now solve for t by multiplying both sides by 365/(2*Pi):
t-800 > (365/(2*Pi)) * sin^(-1)(-2/2.83)

Add 800 to both sides:
t > 800 + (365/(2*Pi)) * sin^(-1)(-2/2.83)

Calculate the value on the right-hand side to find the number of days with more than 10 hours of daylight.

To find the day with about 10 hours of daylight, we need to solve the equation L(t) = 10.

L(t) = 12 + 2.83sin((2π/365)(t-800)

Substituting in 10 for L(t), we have:

10 = 12 + 2.83sin((2π/365)(t-800)

To solve this equation, first subtract 12 from both sides:

-2 = 2.83sin((2π/365)(t-800)

Next, divide both sides by 2.83:

-2/2.83 = sin((2π/365)(t-800)

Using a calculator, find the inverse sin of -2/2.83:

sin^(-1)(-2/2.83) ≈ -41.81 degrees

Now, we can solve for (2π/365)(t-800):

(2π/365)(t-800) = -41.81 degrees

Since the sine function is negative in the 3rd and 4th quadrants, we know that the angle must be in either the 3rd or 4th quadrant. Let's choose the 4th quadrant.

The reference angle (the corresponding angle in the 1st quadrant) can be found by subtracting the angle from 180 degrees:

Reference angle = 180 - (-41.81) = 221.81 degrees

Since (2π/365)(t-800) is negative in the 4th quadrant, we use the reference angle and subtract it from 360 degrees:

(2π/365)(t-800) = 360 - 221.81 ≈ 138.19 degrees

To solve for t, we divide by (2π/365):

(t-800) ≈ (138.19 degrees) * (365/(2π)) ≈ 1077.408

Lastly, we add 800 to both sides to isolate t:

t ≈ 1077.408 + 800 ≈ 1877.408

Therefore, the approximate 34th day has about 10 hours of daylight.

To find how many days of the year have more than 10 hours of daylight, we need to count the number of days that give values of L(t) larger than 10.

Since the function L(t) = 12 + 2.83sin((2π/365)(t-800)) represents the number of hours of daylight, any value of t that satisfies L(t) > 10 indicates a day with more than 10 hours of daylight.

To find these days, we need to evaluate L(t) for each day of the year and count the number of days that give values greater than 10.

Starting from t = 0 (January 1st), we can calculate L(t) for each day of the year incrementing t by 1 each time until we find the last day with more than 10 hours of daylight.

By counting the number of days with L(t) > 10, we can determine how many days of the year have more than 10 hours of daylight. In this case, the answer provided is 308th day.

just set it equal to 10

10 = 12+2.83sin((2?)/365(t-800)
-2 = 2.83sin((2?)/365(t-800)
-2/2.83 = sin((2?)/365(t-800)
-.706713... = sin((2?)/365(t-800)
using my calculator set to radians, taking the inverse sine of +.706713, I get a reference angle of .78484 radians
but for the sine to be negative, it must be in III or IV

first answer = ? + .78484 = 3.926435...
so ((2?)/365(t-800) = 3.926435...
t-800 = 3.926435...(365)/(2?)
= 228.09
t = appr 1028
but the period of the curve is 365, so adding or subtracting multiples of 365 will produce more answers.
possible answer in a year:
1028-2(365) = 298.1 day

second answer:
2? - .78484 = 5.498345..
((2?)/365(t-800) = 5.498345..
t-800 = 319.4
t = 1119.4

again, subtracting periods of 365
t = 24 th day

So it reaches 10 hrs of sunlight on day 24 until day 298
that is,
there is daylight for more than 10 hours for
298-24 or 274 days


I know your answer given is 34
but both of my answers work and are confirmed by Wolfram:
http://www.wolframalpha.com/input/?i=plot+y+%3D+12%2B2.83sin((2%CF%80)%2F365(x-800)

hover your mouse at y = 10 and you will see the x value at close to 20 and 300