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# highest temperature of 80 degrees occurs at 5 pm...

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help

felrose13  Feb 17, 2017

#1
+91262
+10

A sign wave is of the form  (I am developing the graph in radians)

y=sinx

This has a amplitude of 1 a centre y=0  and a wavelength of  2pi

$$y=sin(x) \qquad \text{ has a wavelength of }2\pi\\\ y=sin(nx) \qquad \text{ has a wave length of } 2\pi\div n\ so\\ y=sin(\frac{2\pi }{24} *x) \qquad \text{ has a wave length of }2\pi\div \frac{2\pi }{24}=24 \text{ and this will represent hours.}\\$$

So so far we have

$$y=sin[\frac{2\pi}{24}(x)]$$

Now the amplitude of the wave is 15 degrees (80-65=15)

$$y=15sin[\frac{2\pi}{24}(x)]$$

The centre of the wave is y=65 so the wave is 'lifted' by 64

$$y=15sin[\frac{2\pi}{24}(x)]+65$$

The hottest part of the day is 5pm=17:00 in 24 hour time, so the average temps must be 6 hours either side,  11:00 and 23:00

So at 11:00 the is average temp but getting hotter.  So the horizonal shift is 11 in a positive direction

$$y=15sin[\frac{2\pi}{24}(x-11)]+65$$

So at 3am it will be

15*sin(2pi/24(3-11))+65 = 52 degrees

Here is the graph:

Full interactive version:

https://www.desmos.com/calculator/uri6zjnpns

picture version:

Melody  Feb 18, 2017
Sort:

#1
+91262
+10

A sign wave is of the form  (I am developing the graph in radians)

y=sinx

This has a amplitude of 1 a centre y=0  and a wavelength of  2pi

$$y=sin(x) \qquad \text{ has a wavelength of }2\pi\\\ y=sin(nx) \qquad \text{ has a wave length of } 2\pi\div n\ so\\ y=sin(\frac{2\pi }{24} *x) \qquad \text{ has a wave length of }2\pi\div \frac{2\pi }{24}=24 \text{ and this will represent hours.}\\$$

So so far we have

$$y=sin[\frac{2\pi}{24}(x)]$$

Now the amplitude of the wave is 15 degrees (80-65=15)

$$y=15sin[\frac{2\pi}{24}(x)]$$

The centre of the wave is y=65 so the wave is 'lifted' by 64

$$y=15sin[\frac{2\pi}{24}(x)]+65$$

The hottest part of the day is 5pm=17:00 in 24 hour time, so the average temps must be 6 hours either side,  11:00 and 23:00

So at 11:00 the is average temp but getting hotter.  So the horizonal shift is 11 in a positive direction

$$y=15sin[\frac{2\pi}{24}(x-11)]+65$$

So at 3am it will be

15*sin(2pi/24(3-11))+65 = 52 degrees

Here is the graph:

Full interactive version:

https://www.desmos.com/calculator/uri6zjnpns

picture version:

Melody  Feb 18, 2017
#2
+91262
0

typo error :

"The centre of the wave is y=65 so the wave is 'lifted' by 64"    that should have been lifted by 65.

Melody  Feb 18, 2017
#3
+79903
0

Nice, Melody....!!!!

CPhill  Feb 18, 2017
#4
+91262
0

Thanks Chris,

I am sure it can be done much easier, the time is not a difficult one, but I just decided to do it like this.

I suppose if I was good at it, and did not have to think at all, it would be a dead simple method but at present I still have to think about what is happening.

Melody  Feb 18, 2017
#5
+79903
0

It makes perfect sense..however...I still have to wrap my  mind around the  2pi/24  part....!!!!

CPhill  Feb 18, 2017

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