4 Part Calc Problem Experts Only Please...really hard and challenging, i need help please! details below?

Question

The Temperature on New Year's Day in Hinterland was given by
T(H)= -A-Bcos(piH/12), where T is the temperature in degrees Fahrenheit and H is the number of hours from midnight (0 < or equal T which is less than 24).

a) The initial temperature at midnight was -15 degrees Fahrenheit, and at noon of New Year's Day was 5 degrees F. Find A and B
b) Find the average temperature for the first ten hours
C) Use the trapezoidal rule with 4 equal subdivisions to estimate the integral of T(H) dH with upper limit 8 and lower limit 6
d) Find an expression for the rate that the temperature is changing with respect to H.

Answer 1

if T(h) = - A - Bcos(piH/12)
and t(0) = -15
and t(12) = 5
then -15 = -A-B
and 5 = -A + B (cos(pi) = -1)
so, two equations, 2 unknowns. add them:
-15 + 5 = (-a-b) + (-a+b) = -2a
or -10 = -2a, so a = 5 and therefore b = 10

b) The average temperature is the integral of the function divided by the elapsed time, or 1/10*integral from 0 to 10 of (-5-10cos(pih/12)).

c) compute the value of the function at 6, 6.5, 7, 7.5, and 8. You will want to sum : .25*T(6) + .5*(t(6.5)+t(7) + t(7.5)) + .25*t(8). the 1/2 is the time increment per division. You only put half the weitght on each end point.

d) t'(h) is simply the derivative of t(h), which is +bsin(piH/2)*pi/2 (use the chain rule)

Answer 2

(a) The initial temperature at midnight was -15 degrees Fahrenheit
then, T(0) = -15
A+B = 15
at noon of New Year's Day was 5 degrees F
so, T(12) = 5
-A+B = 5
then ... A = 5 & B = 10
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(b) the average temperature for the first ten hours
Taverage = (1/H)∫T(H)dH (integration from H = 0 to H = 10)
= (1/10)∫[-A-Bcos(piH/12)]dH
= (1/10){-A(10-0) - B(12/pi)[sin(pi*10/12)-sin(pi*0/12)]}
= [-10A - 6B/pi]/10 = -50-60/pi = -6.91
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(d) rate that the temperature is changing with respect to H = (dT/dH)
= Bpi/12*sin(piH/12) = 2.618sin(piH/12)
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(c) sorry i don't know that rule

Answer 3

Just part a for now because I have to go somewhere in a minute! I'll come back to it later if no one else answers.

a) Substitute in the values T=-15 and H=0 to get -15=-A-B, and then substitute T=5 and H=12 to get 5=-A+B

Solve simultaneously:
-15=-A-B
5=-A+B
Add them together to get
-10=-2A
therefore A=5 and B=10

Answer 4

T(H)= -A-Bcos(piH/12)

a)
-15 = -A - B
5 = -A + B

Solving the above two equations for A and B gives,
A = 5 , B = 10

b) Tav = (1/10) ∫T(H) d H = -6.91 F

c) (6, 6.5), (6.5, 7), (7, 7.5), (7.5, 8)

∫T(H) d H [6..8]
= 0.5 [T(6)+T(6.5)]/2 +0.5 [T(6.5)+T(7)]/2 +0.5 [T(7)+T(7.5)]/2 +
0.5 [T(7.5)+T(8)]/2
= Can you finish this part?

d)
dT(H)/d H
= 10 [sin(piH/12)](pi/12)
= (5 pi/6)[sin(piH/12)]

Answer 5

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