I'm having a little trouble with this problem...it would be great if you could point out where I'm going wrong.
Hours of Daylight as a Function of Latitude.
Let S(x) be the number of sunlight hours on a cloudless June 21st, as a function of latitude, x, measured in degrees.
a.) What is S(0)?
(b) Let x0 be the latitude of the Arctic Circle. In the northern hemisphere, S (x)
is given, for some constants a and b, by the formula:
S (x) = bracket (piecewise function)
a + b arcsin*(tan x/tan x0)
for 0 ≤ x < x0
24 for x0 ≤ x ≤ 90.
Find a and b so that S (x) is continuous.
(c) Calculate S (x) for Tucson, Arizona (x = 32◦ 13′ ) and Walla Walla, Washington (46◦ 4′ ).
(d) Graph S (x), for 0 ≤ x ≤ 90.
(e) Does S (x) appear to be differentiable?
My answers:
a.) @ 0 degrees latitude, you're at the equator, so S(0) = 12 hours daylight
b.) Since S(0)=12,
12 = a + b arcsin * (tan0/tan x0)
12 = a + b arcsin 0
12 = a + b0
12 = a,
and for b, because the function must be continuous,
24 = 12 + b arcsin* (tan 66/tan 66)
12 = b arcsin* 1
7.639 (rounded) = b.
But now when I try to find the answer to part C,
latitude of Tucson= 32 degrees
S(x)= 12 + 7.639 arcsin* (tan 32/tan 66)
...I get a nonreal answer for the arcsin of (tan 32/tan 66).
Can anyone please tell me where I'm going wrong? Thanks!
"I get a nonreal answer for the arcsin of (tan 32/tan 66)"
"arcsin of (tan 32/tan 66)" exists and equals 16.15338661° as long as the angles 32 and 66 are in degrees. I guess that your calculator has been set to radians or grad. Please check.
Post if you have other queries.
I am in the process of going through your other answers, so unable to comment for now.
Oh, I am sorry! I had my calculator on radian mode instead of degree mode. Silly me. Thanks so much for the help!
I think I have managed to do the remainder of the problem, except for part E. How would I be able to tell if that function were differentiable or not?
How would you draw the graph of this problem?
It seems like you're on the right track with your calculations, but there are a couple of mistakes in your approach. Let's go through each part of the problem to identify where the issues arise:
a) You correctly identified that at the equator (0 degrees latitude), there are 12 hours of daylight. So, S(0) = 12 hours is correct.
b) To find the values of 'a' and 'b' such that S(x) is continuous, you need to set the two parts of the piecewise function equal to each other at x0. In this case, x0 is the latitude of the Arctic Circle. So, you should solve the following equation:
a + b * arcsin(tan(x0)/tan(x0)) = 24
However, tan(x0)/tan(x0) simplifies to 1, so the equation becomes:
a + b * arcsin(1) = 24
Since arcsin(1) is equal to 90 degrees, the equation can be simplified further:
a + b * 90 = 24
a + 90b = 24
This equation will allow you to find the values of 'a' and 'b' simultaneously. By substituting the value you found for 'a' in part (b) (a = 12), you can solve for 'b'. It seems like you made an error when solving this equation, leading to the incorrect value of 'b'.
c) To calculate S(x) for Tucson, Arizona (32 degrees latitude) and Walla Walla, Washington (46.067 degrees latitude), you should plug these values into the equation you found in part (b) with the correct values of 'a' and 'b'.
For Tucson: S(32) = 12 + 7.639 * arcsin(tan(32)/tan(66))
For Walla Walla: S(46.067) = 12 + 7.639 * arcsin(tan(46.067)/tan(66))
Keep in mind that these calculations involve trigonometric functions, and it's important to ensure your calculator is set to the correct angle mode (degrees in this case).
d) To graph S(x) for 0 ≤ x ≤ 90, you can plot the equation obtained in part (b) for the corresponding range of x values.
e) Whether S(x) appears to be differentiable can be determined by examining the graph in part (d). If the graph has no sharp corners, cusps, or vertical tangent lines, it suggests that S(x) is differentiable in the given range.
I hope this helps you identify and correct your mistakes. Let me know if you have any further questions!