need help with this modeling problem (trig functions)

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abel muroi

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I was given this word problem..

you want to model the day length on an asteroid with year length of 100 days, where the first day is the shortest, with 250 minutes of daylight, while the longest day has 760 minutes of daylight. Build a trig function to model this situation.

So im not entirely sure how to start here. I know the trig function is in the form y = a + b cos/sin ( c ( d + x))

x = angle

how can i find the values of a, b , c and d?
 
abel muroi said:
I was given this word problem..

you want to model the day length on an asteroid with year length of 100 days, where the first day is the shortest, with 250 minutes of daylight, while the longest day has 760 minutes of daylight. Build a trig function to model this situation.

So im not entirely sure how to start here. I know the trig function is in the form y = a + b cos/sin ( c ( d + x))

x = angle

how can i find the values of a, b , c and d?
Click to expand...
You need to assign some values to x. Since cos(x-\(\displaystyle \frac{\pi}{2}\)) = sin(x), lets just use the sine function for our form. So given day length DL as
DL = a + b sin( c ( d + x ))
what value would x [or maybe one should say c (d + x)] be if DL is the smallest? What value is x when DL is a a maximum. Hint, the sine is periodic and has values between -1 and 1 with min and max at \(\displaystyle x\, =\, \frac{3 \pi}{2}\, and\, x\, =\, \frac{\pi}{2}\), respectively. The min and max will repeat every 2\(\displaystyle \pi\) for the argument of the sine.
 
Ishuda said:
You need to assign some values to x. Since cos(x-\(\displaystyle \frac{\pi}{2}\)) = sin(x), lets just use the sine function for our form. So given day length DL as
DL = a + b sin( c ( d + x ))
what value would x [or maybe one should say c (d + x)] be if DL is the smallest? What value is x when DL is a a maximum. Hint, the sine is periodic and has values between -1 and 1 with min and max at \(\displaystyle x\, =\, \frac{3 \pi}{2}\, and\, x\, =\, \frac{\pi}{2}\), respectively. The min and max will repeat every 2\(\displaystyle \pi\) for the argument of the sine.
Click to expand...

So I am assuming that a = 250 since it is the first day and also the shortest day

and i think b = 510 because.... well i just subtracted 760 (which is the amount of minutes for the longest day) and 250 (which is the amount of minutes for the shortest day)

and c must be 100 because that is the number of days in a year for the asteroid

(this is pretty much all i understand from this problem)

so i guess the function is y = 250 + 510 sin (100 (d + x)

so yeah... am i doing this correctly?
 
abel muroi said:
So I am assuming that a = 250 since it is the first day and also the shortest day

and i think b = 510 because.... well i just subtracted 760 (which is the amount of minutes for the longest day) and 250 (which is the amount of minutes for the shortest day)

and c must be 100 because that is the number of days in a year for the asteroid

(this is pretty much all i understand from this problem)

so i guess the function is y = 250 + 510 sin (100 (d + x)

so yeah... am i doing this correctly?
Click to expand...
The most sin(c(d+x)) can be is 1 and the least it can be is -1. So a+b=760 and a-b=250. Now solve for a and b.
 
Jomo said:
The most sin(c(d+x)) can be is 1 and the least it can be is -1. So a+b=760 and a-b=250. Now solve for a and b.
Click to expand...

ok so the two numbers that i got for a and b are

a = 505

b = 255

are these the correct numbers? if so, what do i do next?
 
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abel muroi said:
ok so the two numbers that i got for a and b are

a = 505

b = 255

are these the correct numbers? if so, what do i do next?
Click to expand...
Let x represent the beginning of the day, so the beginning of the first day is x=0, and the end of the first day [beginning of the second day] is x=1.
So, for x=0 the day is the minimum length of time, i.e.
sin(c (d + 0)) = sin (c d) = -1
So, what is cd? [note: there are cycles of the sine function involved here, so it depends on how you are going to count] You should now have d in terms of c. At the end of day 100 [beginning of day 101], you have the completion of one cycle of the sine wave [the start of a new year] and x=100 and again
sin(c (d + 100)) = sin (c d + 100 c) = -1
So, what is cd+100 c? [make sure you define you cycles well and you are 'in the same cycle'] Given that, since you know cd, what is c? Given c, what is d?
 
Ishuda said:
Let x represent the beginning of the day, so the beginning of the first day is x=0, and the end of the first day [beginning of the second day] is x=1.
So, for x=0 the day is the minimum length of time, i.e.
sin(c (d + 0)) = sin (c d) = -1
So, what is cd? [note: there are cycles of the sine function involved here, so it depends on how you are going to count] You should now have d in terms of c. At the end of day 100 [beginning of day 101], you have the completion of one cycle of the sine wave [the start of a new year] and x=100 and again
sin(c (d + 100)) = sin (c d + 100 c) = -1
So, what is cd+100 c? [make sure you define you cycles well and you are 'in the same cycle'] Given that, since you know cd, what is c? Given c, what is d?
Click to expand...

i have a few questions to ask..

why does cd = -1?

I'm a little confused on how you explained this. so is c = -pi? if so, why does it equal to -pi
 
abel muroi said:
i have a few questions to ask..

why does cd = -1?

I'm a little confused on how you explained this. so is c = -pi? if so, why does it equal to -pi
Click to expand...

Do you mean "why does sin(cd)=-1"? If so, look at the problem: "...where the first day is the shortest...".

Where did you get c = -pi. Show your work.
 
Ishuda said:
Do you mean "why does sin(cd)=-1"? If so, look at the problem: "...where the first day is the shortest...".

Where did you get c = -pi. Show your work.
Click to expand...

hmm so you are saying that sin (cd) = -1 because the shortest day of the asteroid is 250 minutes.

Using the unit circle (instead of 360 degrees/2 pi radians, an entire cycle of the unit circle will give 360 minutes) and since 250 minutes is on the third quadrant of the unit circle and the y-axis is -1.... that means that sin (cd) = -1 (since sin represents the y-axis)

am i correct so far?
 
Unfortunately, no. For this one particular part of the problem, the length of the day in minutes is irrelevant. Instead, it's the fact that the first day is the shortest that's important. Think about what you know about the sine function - it cycles between 1 and -1. Since the function, we'll call it f(x), for finding the length of day x is a modified sine function, we'll want the shortest day to have the lowest function value, right?

For right now, we're starting by finding the values of c and d in order to work up to finding a and b. So, sin(c[d+x]) is the function we're concerned with. On the first day, call it day 0 so that x = 0, sin(c[d+x]) = sin(c[d+0]) = sin(cd). And because the problem states that the first day is the shortest, we arrive at sin(cd) = -1.

Does that make sense? Hopefully I didn't confuse you further...
 
ksdhart said:
Unfortunately, no. For this one particular part of the problem, the length of the day in minutes is irrelevant. Instead, it's the fact that the first day is the shortest that's important. Think about what you know about the sine function - it cycles between 1 and -1. Since the function, we'll call it f(x), for finding the length of day x is a modified sine function, we'll want the shortest day to have the lowest function value, right?

For right now, we're starting by finding the values of c and d in order to work up to finding a and b. So, sin(c[d+x]) is the function we're concerned with. On the first day, call it day 0 so that x = 0, sin(c[d+x]) = sin(c[d+0]) = sin(cd). And because the problem states that the first day is the shortest, we arrive at sin(cd) = -1.

Does that make sense? Hopefully I didn't confuse you further...
Click to expand...

ok so everytime a problem gives out information about the shortest day, does that mean that sin(cd)/cos(cd) will always equal to -1? since its the lowest value of the wave.

OK i understand that much. so what do i do next?
 
abel muroi said:
ok so everytime a problem gives out information about the shortest day, does that mean that sin(cd)/cos(cd) will always equal to -1? since its the lowest value of the wave.

OK i understand that much. so what do i do next?
Click to expand...
a and b are fixed. You have y=a +b*(some function of x) Assume b>0. To make y the smallest you want to make the function of x the least, which will make b*function as small as possible. Adding a to b*function will not change the fact that y is still the least. If we let the function be the most it can be, then y = a + b*function will be the most y can be. You need to see this concept crystal clear. It really isn't that hard. For the record, this function in your case is the sine function. And it has a max of 1 and a min of -1.
 
Jomo said:
a and b are fixed. You have y=a +b*(some function of x) Assume b>0. To make y the smallest you want to make the function of x the least, which will make b*function as small as possible. Adding a to b*function will not change the fact that y is still the least. If we let the function be the most it can be, then y = a + b*function will be the most y can be. You need to see this concept crystal clear. It really isn't that hard. For the record, this function in your case is the sine function. And it has a max of 1 and a min of -1.
Click to expand...

I dont think i'm understanding this at all. Can you give me a list of steps on how to do this?

i usually learn better when i am given a list of step by step instructions on how to solve this problem
 
abel muroi said:
I dont think i'm understanding this at all. Can you give me a list of steps on how to do this?

i usually learn better when i am given a list of step by step instructions on how to solve this problem
Click to expand...
Suppose x can ONLY be 3,-4,2,-1 or 6. Let y= 3+5x (a=3 and b=5) What is the largest and smallest value y can be?
What if -2<=f(x)<=5 and y= 9+7f(x)?
 
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Jomo said:
Both values are wrong! y= 3+5x. 6=3+5x so 5x=3 (after all 3 + 3 =6) then x=3/5. But this value was not given. What is the largest and smallest value which y can be??!!
Click to expand...

can you just tell me what the trig function of the problem is? im still not getting this.

maybe i'll understand better once i know what the trig function is.
 
abel muroi said:
can you just tell me what the trig function of the problem is? im still not getting this.

maybe i'll understand better once i know what the trig function is.
Click to expand...

OK - let's start all over. However, you will need to work through the following and tell us where you are stuck. That is, what question can you not answer in order for us to help you.

We will model the day length DL by

DL(x)= a + b * sin(c x + d) [slightly different than before but equivalent]

where x = j is the end of the jth day, j = 1, 2, 3, .... Since x=j is the end of the jth day, j-1 is the start of the jth day. We will designate the day by the value of x for the start of the day. So
x=0 is the start of the first day
x=1 is the start of the second day and end of the first day
x=2 is the start of the second day and end of the second day
....
Note: You can do the count differently and/or use the cosine and/or use a different form [say c(x+d) instead of cx+d] but it would have to be consistent with the above.

(1) Now b is going to be the amplitude of the sine function. What is the min and max of the day length? Given that, what is the amplitude of the sine function? If you can't answer that question I'm afraid you will need to go back and do some more studying/review about definitions. Maybe this
http://www.purplemath.com/modules/grphtrig.htm
will help

(2) Now that we know b, we need to find a. The value of a is the midpoint (average) of the function. What is the min and max of the day length? Given that, what is the mid point (average of those two numbers)? Again, if you can't answer that question I'm afraid you will need to go back and do some more review.

(3) Now that we know a and b, lets find d. The first day is the shortest and the beginning of the day is x=0, so
DLshortest = a + b sin( c*0 + d) = a + b sin(d)
Since you know DLshortest, a, and b you should be able to compute sin(d), Given sin(d), you should be able to compute d. Again, if you can't answer that question I'm afraid you will need to go back and do some more review.

(4) Now that you know a, b, and d, you can compute c. The value of c determines the period of the sine function for DL(x). The argument of the sine function at the start of the first day is c*0+d = d. The argument at the start of the second year is c*Year_Length + d. So the difference between these two is period of the sine function. Given that you should be able to compute c. Again, if you can't answer that question I'm afraid you will need to go back and do some more review.
 
Ishuda said:
OK - let's start all over. However, you will need to work through the following and tell us where you are stuck. That is, what question can you not answer in order for us to help you.

We will model the day length DL by

DL(x)= a + b * sin(c x + d) [slightly different than before but equivalent]

where x = j is the end of the jth day, j = 1, 2, 3, .... Since x=j is the end of the jth day, j-1 is the start of the jth day. We will designate the day by the value of x for the start of the day. So
x=0 is the start of the first day
x=1 is the start of the second day and end of the first day
x=2 is the start of the second day and end of the second day
....
Note: You can do the count differently and/or use the cosine and/or use a different form [say c(x+d) instead of cx+d] but it would have to be consistent with the above.

(1) Now b is going to be the amplitude of the sine function. What is the min and max of the day length? Given that, what is the amplitude of the sine function? If you can't answer that question I'm afraid you will need to go back and do some more studying/review about definitions. Maybe this
http://www.purplemath.com/modules/grphtrig.htm
will help

(2) Now that we know b, we need to find a. The value of a is the midpoint (average) of the function. What is the min and max of the day length? Given that, what is the mid point (average of those two numbers)? Again, if you can't answer that question I'm afraid you will need to go back and do some more review.

(3) Now that we know a and b, lets find d. The first day is the shortest and the beginning of the day is x=0, so
DLshortest = a + b sin( c*0 + d) = a + b sin(d)
Since you know DLshortest, a, and b you should be able to compute sin(d), Given sin(d), you should be able to compute d. Again, if you can't answer that question I'm afraid you will need to go back and do some more review.

(4) Now that you know a, b, and d, you can compute c. The value of c determines the period of the sine function for DL(x). The argument of the sine function at the start of the first day is c*0+d = d. The argument at the start of the second year is c*Year_Length + d. So the difference between these two is period of the sine function. Given that you should be able to compute c. Again, if you can't answer that question I'm afraid you will need to go back and do some more review.
Click to expand...

1) the amplitude of the function is 255, the maximum day length is 760 minutes, the minimum day length is 250

2) the midpoint is 505,

3) sin (d) = -1

4) the period of the function is pi, but using the formula

2pi/b = period,

i get 2 for the period.

with all this information i got this trig function

y = 505 + 255 sin (2*x - 1)

is this the right equation?
 
As far as testing to see if the equation is right, you need only see if it matches the points you know. Since you only know the exact value of the first day, try f(0). If:

f(x) = 505 + 255 * sin(2x - 1)

then

\(\displaystyle f(0) = 505 + 255 sin(2(0)-1) = 505 + 255 sin(-1) \approx 505 + 255(-0.84) \approx 290.8\)

And since that's not 250, as it should be, that means the equation isn't right. You're close to the right answer, and I'll give you two hints to hopefully get you the rest of the way.

First, consider the period. Why do you say the period \(\displaystyle \pi\)? Remember that the period is how often your sine function repeats itself. And second, you put d = -1 into your equation. But is that the correct value? If sin(d) = -1, then what is d?
 
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