Outside temperature over a day can be modeled as a sinusoidal function.

pkdroid777

New member
Joined
Feb 15, 2019
Messages
3
Outside temperature over a day can be modeled as a sinusoidal function. Suppose you know the temperature varies between 64 and 86 degrees during the day and the average daily temperature first occurs at 11 AM. How many hours after midnight does the temperature first reach 80.5 degrees? Round your answer to the nearest full hour.

Hello, I have been having issues with this homework problem, I know there are two things I need to do in order to complete the problem 1) figure out the equation of the word problem, and 2) make it equal to the equation and then solve.

The equation I got, was -11cos(pi/12(x-11))+75 so then I had 80.5=-11cos(pi/12(x-11))+75 after solving everything I ended up with 12/pi*cos-1​(5.5/-11)+11 I then put it in my calculator I got 19 and it told me I was wrong. Im guessing its probably something to do with my phase shift.

Thanks.
 
Last edited:
Outside temperature over a day can be modeled as a sinusoidal function. Suppose you know the temperature varies between 64 and 86 degrees during the day and the average daily temperature first occurs at 11 AM. How many hours after midnight does the temperature first reach 80.5 degrees? Round your answer to the nearest full hour.

Hello, I have been having issues with this homework problem, I know there are two things I need to do in order to complete the problem 1) figure out the equation of the word problem, and 2) make it equal to the equation and then solve.

The equation I got, was -11cos(pi/12(x-11))+75 so then I had 80.5=-11cos(pi/12(x-11))+75 after solving everything I ended up with 12/pi*cos-1​(5.5/-11)+11 I then put it in my calculator I got 19 and it told me I was wrong. Im guessing its probably something to do with my phase shift.

I think you misread the problem. It says "the average daily temperature first occurs at 11 AM"; your use of the cosine suggests that you thought it said "the minimum daily temperature first occurs at 11 AM". Try fixing that.
 
… The [function] I got [is] -11cos(pi/12(x-11))+75 so then I [wrote the equation] 80.5=-11cos(pi/12(x-11))+75 … I got 19 and [was told it is] wrong.

… something to do with my phase shift …
Yes, the graph is not shifted correctly.

x = the number of hours after midnight

The average temperature is 75º, so you need f(11) = 75.


To correct the horizontal shift (which is not the same as the phase shift, by the way), you could temporarily remove that shift -- solving the following for x.

-11 cos(pi/12*x) + 75 = 75

That will yield the number of hours after midnight (x) when 75º is first reached without a shift. The difference between that value of x and 11 is the amount of shift needed.

Alternatively, you could solve the following, and get the horizontal shift directly. :cool:

-11 cos(pi/12(11-C)) + 75 = 75
 
Last edited:
Yes, the graph is not shifted correctly.

x = the number of hours after midnight

The average temperature is 75º, so you need f(11) = 75.


To correct the horizontal shift (not the same as phase shift), you could remove the current shift, and solve the following for x.

-11 cos(pi/12*x) + 75 = 75

That will yield the number of hours after midnight when 75º is first reached (without a shift). The difference between that value of x and 11 is the amount of shift your function needs.

Alternatively, you could solve the following, and get the shift directly. :cool:

-11 cos(pi/12(11-C)) + 75 = 80.5

Sorry for the late reply, so when you say f(11)=75 does that mean I need my y value to equal 75 if I was to plug in a number for x (like in the calculator table). I am still a little confused on the equations I don't think I am solving them right the second one in particular because I end up with 19 and I believe that couldn't be the shift since it doesn't equal 75 when I make x=11. Hopefully what I am asking makes sense.

Thanks.
 
… when you say f(11)=75 does that mean I need my y value to equal 75 …
Yes! f(x) is y. That is, symbol f(x) is another way of expressing the y variable (called function notation).


… I don't think I am solving … the second [equation right] because I end up with 19 …
No, that's my fault. :oops: Post #3 contained a math typo. I had substituted 11 for x, but then I set y equal to 80.5 instead of 75. (I've edited that post.)

So, the correct equation to find the horizontal shift C is:

-11 cos(Pi/12*(11 - C)) + 75 = 75
 
Top