Question

The desert temperature oscillates daily between 4 degrees celsius at 5am and 32 degrees celsius at 5pm.
Write a formula for the temperature as a function of time "t" measured in hours from 5am. I know at t = 0, it is 5am.
Therefore, at t = 12, it is 5pm. So, (0,4) is a minimum, and (5,32) is a maximum. Drawing just those points out, I can see that it is going to end up as a negative cosine function. I figured that by taking the min and the max y-values and dividing by 2, I get the axis (or the vertical shift), which is 18.

Now I am stuck. I have (so far), f(x) = -cos(x) + 18

Answer 1

no that is not right

Answer 2

32 - 4 = 24 so you need an amplitude of 12

Answer 3

if you use
\[-12\cos(x)\] then it will fluctuate between -12 and 12

Answer 4

Wait so a vertical shift of 4 units up is its translation right?

Answer 5

am i an idiot or what?

Answer 6

32 - 4 = 28
so you need an amplitude of 14

Answer 7

is that for saying i am an idiot? thanks ishaan!

Answer 8

Wait I did not call you an idiot..

Answer 9

then
\[-14\cos(x)\] will fluctuate between -14 and 14 to add 18 like you had

Answer 10

no i called myself one and ishaan gave me a medal for it!

Answer 11

so now we should have
\[-14\cos(x)+18\] that takes care of the range. now you have to take care of the period

Answer 12

as it stands if
\[x=0\] you get 4 and if
\[x=\pi\] you get 32

Answer 13

Ok so to find the amplitude, it is max - min / 2, to find the axis, it is max + min / 2

Answer 14

yeah i guess. i wasn't using a formula but it looks like you got it

Answer 15

OHHH Now I remember.

Answer 16

Ok so how would you know what the period is, given this information?

Answer 17

Here, I am confused because you have a min, and you have the ending maximum

Answer 18

yeah that works because you used
\[-\cos(x)\]

Answer 19

Ok so with the given information, you CANNOT determine a definite period?

Answer 20

we are going from 4 to 32 right?

Answer 21

sure

Answer 22

Yea

Answer 23

looks like it is 24 hours right?

Answer 24

5 am for min, 5 pm for max, and then back again

Answer 25

That is what I was thinking too, but it cannot be.

Answer 26

why not? we just have to use a 24 hour clock

Answer 27

so put the period at 24 hours and period is
\[\frac{2\pi}{b}\]

Answer 28

so
\[\frac{2\pi}{b}=24\]
\[b=\frac{\pi}{12}\]

Answer 29

and finally we have to adjust to make 5 am 0 so subtract 5

Answer 30

Ok wait so what is the formula for determining the period?

Answer 31

period of
\[\cos(bx)\] or
\[\sin(bx)\] is
\[\frac{2\pi}{b}\]

Answer 32

How do you know that this is the formula? How can you tell that it is right?

Answer 33

well i know it because i know it, but we can find it by doing an example if you like

Answer 34

Yes please.

Answer 35

ok suppose you have
\[\sin(6x)\] and you want the period

Answer 36

Ok can we use the equation that I was taught: Period = 2pi / k

Answer 37

If you don't mind.

Answer 38

when
\[x=0\]
then
\[6x=0\] and so we start there. when
\[6x=2\pi\] we have completed one full period. solve for x get
\[x=\frac{2\pi}{6}\]

Answer 39

i.e. as x goes from 0 to
\[\frac{2\pi}{6}\] 6x goes from 0 to 2pi

Answer 40

Ok I understand, so you are saying that 6x = 2pi, this is when we have completed a period. But why did you not say sin(6x) = 2pi

Answer 41

it is an easy enough formula to know and derive if you need it.

Answer 42

2pi is the input. sin(2pi)=0 is the output

Answer 43

x= 2pi/6
6x=2pi
sin(2pi)=0

Answer 44

back to the problem at hand since the period is 24 hours we know the period is 24 and therefore
\[b=\frac{\pi}{12}\]

Answer 45

so now we have
\[-14\cos(\frac{\pi}{12}x)+18\]

Answer 46

and finally we have to adjust so that 5 am becomes 0. that is easy. subtract 5 from the input

Answer 47

\[-14\cos(\frac{\pi}{12}(x-5))+18\]

Answer 48

what a nice picture! it works.
http://www.wolframalpha.com/input/?i=-14cos%28pi%2F12%28x-5%29%29%2B18

Answer 49

Wait they said STARTING FROM 5AM, so wouldn't that mean that t = 0 at 5am? We wouldn't need to do a shift horizontally..

Answer 50

i want to put in 5 and take the -cos(0) to get the lowest value

Answer 51

so i want to shift right 5 units. make 0 into 5

Answer 52

so yes, like you said, it is a horizontal shift. shift 5 to the right by subtracting by 5

Answer 53

check the graph and you see that at 5 you get 4, and at 17 (12 hours later) you get 32

Answer 54

Yea I do not think that you need horizontal shift at all. This is because:
f(12) = -14cos(pi/12(12)) + 18
f(12) = 32

Answer 55

My calculator is in radians.

Answer 56

we are working in radians. look here
http://www.wolframalpha.com/input/?i=-14cos%28pi%2F12%28x-5%29%29%2B18

Answer 57

you don't need a horizontal shift if you are shifting first in your head. but 12 is not 0 right? we arranged it so if you plug in 5 you get out 4

Answer 58

But I tried for both (0,4) and (12,32) and it worked without doing a h.shift

Answer 59

right i agree. but you shifted 5 am to 0 in your head first.

Answer 60

if you really want 5 to give you 4 you have to shift

Answer 61

Observe:
T(0) = -14cos((pi/2(0)) + 18 = 4
T(12) = -14cos((pi/2(12)) + 18 = 32

Answer 62

Ok I understand what you mean

Answer 63

yeah i see. for sure it works but you have said in your head "make 5 am 0 and start counting from there"

Answer 64

You are saying that if you start at 1am, and you want to move to 5am?

Answer 65

no i am saying that if you want to make 5 actually give you a temperature of 4, then you have to shift horizontally 5 units right.

Answer 66

you have "put in 0, get out 4, put in 12, get out 32" which is fine, but if you shift you get
"put in 5, get out 4, put in 17, get out 32"

Answer 67

i agree it is a red herring. we can alway say "start counting at 5 and make that 0"

Answer 68

OK so then x = 0 will be 12am, from what you are saying?