# Thread: Finding a Trig Equation

1. ## Finding a Trig Equation

The desert temperature, H, oscillates daily between 39ºF at 6am and 79ºF at 6pm. Write a possible formula for H in terms of t, measured in hours from 6am. Use the form:

H = A cos(Bt) + C.

What I tried was this: I subtracted 39 from 79 to get 40, then divided that in half. That should be the amplitude A, right? Then I took 12 hours (from 6am to 6pm), and set that equal to 2pi / b, the period for a cosine function. I got pi/6, which should be B. Then to get the y axis up to 40, C should be 20 since the amplitude is 20.

However, when I submitted this, all of them were wrong. Having tried everything I can think of, I'm asking for help. I'd appreciate any responses! Thank you!

2. For a start C ought to be 59: if H goes from 39 to 79 then the point in the middle where cos Bt = 0 is when C = 59.

Next, note that the time starts when H is at its lowest. Thus when t = 0, H = 59 - 20 cos (0) (note that minus sign). So A will have to be -20, not 20.

Next note that your period is 24 not 12 - 12 hours is *half* the period.

$t = 0$ means $Bt = 0$, $t = 24$ means $Bt = 2 \pi$ so $B = 2 \pi / 24$ and so $B = \pi / 12$.

Draw a graph - it may clarify things.

3. equation, H = A cos(Bt) + C.

@ t = 0, H = 39F, then
39 = A( cos 0) + C
39 = A(1) + C
39 = A + C ---- (1)
-------------------------------------

@ t = 12, H = 79F, then
79 = A cos B(12) + C ---- (2)
-------------------------------------

39 = A + C ---- (1)
79 = A cos B(12) + C ---- (2) - subtract (1) from (2)
-------------------------------------

40 = A(1 + cos 12 B) + 0;

(40)(1) = (A)(1 + cos 12 B)

equate factors, A = 40 and (1 + cos 12B) = 1.

then, A = 40

(1 + cos 12B) = 0

cos 12B = -1

12B = arccos(-1) = 180 degrees 0r pi rad

then, B = pi/12

period of the cosine function is 2pi/B = 2pi/(pi/12) = 24

What is C? For H = 40 cos (pi/12)t + C

is it 39F?

4. For H= 40cos ((pi/12)t) + 39 , for t = 0 to 24 hours . i will plot it below . . . . What is wrong with this?

5. For H = 40 cos (pi/12)t + C, for t = - 6 hours, cos pi/2 = 0, H = C = 39.

Since t = -6 hours corresponds to 12am, and pi/2 was added tp compensate for the 12 hours lag, then the new equation will be

H = 40 cos ((pi/12)t - pi/2) + 39

see my my graph.

i am still reluctant to post this, i believe there is something wrong with it. HELP

the 6pm H does not corresponds to the given value

6. Originally Posted by pacman
equation, H = A cos(Bt) + C.

@ t = 0, H = 39F, then
39 = A( cos 0) + C
39 = A(1) + C
39 = A + C ---- (1)
-------------------------------------

@ t = 12, H = 79F, then
79 = A cos B(12) + C ---- (2)
-------------------------------------

39 = A + C ---- (1)
79 = A cos B(12) + C ---- (2) - subtract (1) from (2)
-------------------------------------

etc.
I think this is where you went wrong.(2) - (1) gives:
$40 = A(cos 12 B - 1)$

7. Thanks Matt Westwood,

8. This is the plot of Matt Westwood's equation, H = 69 - 20cos (pi/12)t, for t = 0 to 24 hours.

I believe this is the right solution

9. the above post is still wrong, H = 59 - 20 cos (pi/12)t, for t = 0 to 24 hours

10. this is the graph of H = 59 - 20 cos (pi/12)t, for t = 0 to 24 hours

11. Originally Posted by pacman
This is the plot of Matt Westwood's equation, H = 69 - 20cos (pi/12)t, for t = 0 to 24 hours.

I believe this is the right solution
59 not 69, yeah?

12. Originally Posted by pacman
this is the graph of H = 59 - 20 cos (pi/12)t, for t = 0 to 24 hours
What I said.