Question

the temperature is given by T(H)=-A-Bcos(piH/12), T is temperature and H is the number of hours from midnight (0 14 years ago

Answer 1

in your formula is Capital H and lower h supposed to be the same variable?

Answer 2

oops yeah

Answer 3

Just checking.

Answer 4

Another thing the portion you have that says ( 0 <= T < 24). Are you sure that T doesn't say H?

Answer 5

Sorry, this may sound like a stupid question that part really doesn't make sense.

Answer 6

yup it's T, but now that you mentioned it it doesnt make sense

Answer 7

Yea I believe that T should be an H, because H is time and if we consider military time, which has 24 hours in a day, then that inequality makes sense... hmmm

Answer 8

i don't knowww, some one making this assignment made a mistake. let's go with it's supposed to be H

Answer 9

Is we go with the inequality as (0 <= H < 24)...

A.)

-15 = -A - B*cos(0) --> -15 = -A - B

5 = -A - B*cos(pi) --> 5 = -A + B.

you have two equations with two unknowns, all thats left is to solve.

Answer 10

how did you go from -15 to 5.

Answer 11

or did you solve for both?

Answer 12

I solved two separate times The told us that when T= -15 H =0, and T=5 and H = 12

Answer 13

and did you have to plug the 12 in for when you plugged the 5 degrees in cuz its T(H)?

Answer 14

T(H) is just telling you that t is a function of H. Sorry i assumed to much left me redo what i did early to be more explicit with what I mean.

Answer 15

ohh ok i get it. I was thinking that but for other things i've been going over i had to substitute in all the variables.

Answer 16

We are told that initially (at midnight) the temperature is -15 degrees right?

well...

T(0) = -A - B*cos (pi*0/12) --> T(0) = -A - B*cos (0) -- >where T(0) = -15

Therefore,

-15 = -A - B.

Then i did same with other information to the same equation again and got

5 = -A + B understand?

Answer 17

From there we have two equations and two unknowns.solve and tell me what you get and I'll tell you what i got.

Answer 18

ok

Answer 19

i got a=5 b=7

Answer 20

Check you B again. show me your math.

Answer 21

ohh b=10

Answer 22

YUP =)

Answer 23

Now part B....

Answer 24

yeahh... i think it has something to do with something similair to the trapizoid method..

Answer 25

Maybe, but we can do it another way. We can find all of the temperatures for the first ten hours then average them. So....

Are new equation is now T(H) =-5-10*cos(pi*H/12), since we found A and B.

T(0) = ?
T(1) = ?
T(2) = ?
.
.
.
T(10) = ?

from here [T(0) + T(1) + T(2) + ... + T(10)]/10

Does that make sense?

Answer 26

To be honest I don't remember the trapezoidal method. Not much help form here on

Answer 27

i think so.

Answer 28

haha me eitherr, learned it like three months ago but nothing sticks with migraines 24/7

Answer 29

Hmm. Well for part b linear interpolation should get you the right answer i believe. From here on I don't remember, I thought i did but my answers are coming up super wrong, not even close haha sorry.

Answer 30

its all good, thanks for the help

Answer 31

T(H)=-A-B\[T(H)=-A-B \cos (\frac{\pi H}{12})\]
At midnight:
\[T(H=0)=-A-B \cos( 0)=-A-(B \times 1)=-A-B=-15\]
At noon:
\[T(H=12)=-A-B \cos( \pi)=-A-(B \times -1)=-A+B=5\]
-A -B = -15
-A + B = 5
Adding gives:
-2A = -10
A = 5
By substitution:
B = 10

Answer 32

yepp

Answer 33

The average temperature for the first 10 hours is found by integrating the formula between the limits H=0 and H=10 and then dividing the result by 10 to find the average.
Do you follow?

Answer 34

\[\int\limits_{0}^{10}[-5-10\cos (\frac{\pi H}{12})]dH\]
\[\left[ -5H+\frac{10 \pi}{12}\sin \frac{\pi H}{12} \right]\]Between H=10 and H=0

Answer 35

The value of the definite integral = -48.7
Therefore the average temperature over the first 10 hours is:
-48.7/10 = -4.87 degrees

Answer 36

sorry, i got off here, but it makes sense using the integration, i knew i had to but wasn't sure how to do it. thanks for helping