need help with this modeling problem (trig functions)

[abel muroi]

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?

d must be -1/2 pi right?

the thing i dont fully understand is finding the period here. what two values do i have to use to find the period?

 

[ksdhart]

First off, yes, d = sin^-1​(-1) or \(\displaystyle -{\pi \over 2}\), good job. And you actually already know the period - it was given in the problem. As I said, the period is how often the function repeats itself. The sine function is modeling one year on the asteroid, so how often does the function repeat? Given the period then, what you want to find is the value of b. The period is always equal to \(\displaystyle {2 \pi} \over b\), so it's simply a matter of solving for b.

 

[abel muroi]

First off, yes, d = sin^-1​(-1) or \(\displaystyle -{\pi \over 2}\), good job. And you actually already know the period - it was given in the problem. As I said, the period is how often the function repeats itself. The sine function is modeling one year on the asteroid, so how often does the function repeat? Given the period then, what you want to find is the value of b. The period is always equal to \(\displaystyle {2 \pi} \over b\), so it's simply a matter of solving for b.

so the period is 100 then?

using the 2pi/b = period... i get 1/50pi

is 1/50pi the correct period?

 

[ksdhart]

\(\displaystyle \pi \over 50\) is indeed the correct value for b. So, if you put it all together, your sine function looks like:

\(\displaystyle 505+255sin\left(\frac{x\pi }{50}-\frac{\pi }{2}\right)\)

If you graph that function, you'll see it oscillates over 100 days, with a minimum at 250 at x=0 and a maximum of 760 at x=50. That meets all the criteria, so you're finally done. Hopefully the next time you have a problem like this, you'll know what to do and it won't give you so much grief.

 

[abel muroi]

\(\displaystyle \pi \over 50\) is indeed the correct value for b. So, if you put it all together, your sine function looks like:

\(\displaystyle 505+255sin\left(\frac{x\pi }{50}-\frac{\pi }{2}\right)\)

If you graph that function, you'll see it oscillates over 100 days, with a minimum at 250 at x=0 and a maximum of 760 at x=50. That meets all the criteria, so you're finally done. Hopefully the next time you have a problem like this, you'll know what to do and it won't give you so much grief.

btw I have a few questions that i need to ask..

is this equation, y = 505 + 255 * sin (1/50 pi(x - 1/2)), the same thing as this equation y = 505 + 255 * sin (x/50pi - 1/2))?

and also, how can i make sure if my amplitude will be negative or positive?

 

[ksdhart]

No, the two equations you posted are not the same. You can see that by multiplying and comparing. The basic forms you have written are:

a + b*sin(c[x+d]) and a + b*sin(cx+d)

If you multiply the first form out, it becomes a + b * sin(cx + cd), which is not the same as the second form Either form is acceptable, but I'd say since you've solved it using the second form, I'd stick with that rather than complicating things further.

Concerning the amplitude, the amplitude will always be positive. The amplitude just measures how far the value ranges from the average value. For sin(x), the function ranges from -1 to 1, so the average value is 0. The function is never more than 1 unit above or below 0, so the amplitude is 1. And in the modified function, the values range from 250 to 760, for an average is 505. Both extremes are 255 units away from 505.

 

[abel muroi]

No, the two equations you posted are not the same. You can see that by multiplying and comparing. The basic forms you have written are:

a + b*sin(c[x+d]) and a + b*sin(cx+d)

If you multiply the first form out, it becomes a + b * sin(cx + cd), which is not the same as the second form Either form is acceptable, but I'd say since you've solved it using the second form, I'd stick with that rather than complicating things further.

Concerning the amplitude, the amplitude will always be positive. The amplitude just measures how far the value ranges from the average value. For sin(x), the function ranges from -1 to 1, so the average value is 0. The function is never more than 1 unit above or below 0, so the amplitude is 1. And in the modified function, the values range from 250 to 760, for an average is 505. Both extremes are 255 units away from 505.

Do you think that this method..

y = 255 * sin (0 (0) + d) + 505
250 = 255 * sin(d) + 505
-255 = 255 * sin(d)
-1 = sin(d)

d = -1/2 pi

.. is the best method to find the period of a trig function? (as long as y represents the minimum)

 

[Steven G]

so the period is 100 then?

using the 2pi/b = period... i get 1/50pi

is 1/50pi the correct period?

you wrote 3 lines and contradicted yourself. You wrote so the period is 100. Then you wrote period... i get 1/50pi
You need to stop confusing yourself. Think and write real mathenmatics.

 

[Steven G]

so the period is 100 then?

using the 2pi/b = period... i get 1/50pi

is 1/50pi the correct period?

Do you know why the period is 2pi/b? Do you know what b represents?
Here is why it works.
Let y = a*sin(bx+c).
All sine (and cosine) graphs start a cycle when the angle equals 0: so bx+c=0 implies x=-c/b
All sine (and cosine) graphs finish that cycle when the angle equals 2pi: so bx+c=2pi implies x=(2pi-c)/b.
The period is the length of a single cycle (cycle is the portion that repeats) so it is (2pi-c)/b - (-c/b) which equals 2pi/b.
I wondered from the start of this thread why you wanted y=a +bsin(c(x+d)). Note that c(x+d)=cx+cd=cx+d'
Why not y = a + bsin(cx+d)?
Now try to continue.