One fine day in Rochester the low temperature occurs at 5am and the high temperature at 5pm. the temperature varies sinusoidally all day.

Question

the temperature t hours after midnight is: T(t) = A + Bsin(pi(t-C)/12)

where A,B, and C are certain constants.

the low temperature is 55 and the high temperature is 71.

find the average temperature during the first 7 hours after noon.

Hint: the high and low temperature can be used together to find A and B. Determine C from the fact that it is hottest at 5pm

Answer 1

The key to this problem is remembering that a definite integral of a function between a and b is the same value as the definite integral of the function's average value from a to b. In other words:

∫_a^b f(x) dx = ∫_a^b f_{avg, b-a} dx

But the integral of a constant is simple, so let's start there:

∫_a^b f_{avg, b-a} dx = f_{avg, b-a} (b) - f_{avg, b-a} (a) = f_{avg, b-a} (b - a)

Now it's easy to see that

∫_a^b f(x) dx = f_{avg, b-a} (b - a)

f_{avg, b-a} = ∫_a^b f(x) dx / (b - a)

Now all we need are an f(x), a, and b, and we can solve it.

In this problem, instead of f(x), we are using T(t) (temperature vs. time) due to the context. They give us this function, sort of:

T(t) = A + B sin( (t-C) pi / 12)

But we need to find A, B, and C to really know what's going on. To do this we use our given temp values:

The low at 5 AM (t = 5) is 55

The high at 5 PM (t = 17) is 71

When a sine function is at its max value, it equals 1. At its min value, it equals -1. This is important to remember, because even though we don't know what C is yet, we do know that the min happens at t = 5 and the max happens at t = 17 in this problem. Therefore:

T(5) = A + B sin( (5-C) pi / 12)

sine part equals -1, because t = 5 is where the low temp happens

T(5) = A + B (-1) = A - B

T(17) = A + B sin( (17-C) pi / 12)

sine part equals 1, because t = 17 is where the high temp happens

T(17) = A + B (1) = A + B

Now we have this simple set of simultaneous equations:

T(5) = A - B = 55

T (17) = A + B = 71

Solve for A and B:

A = 63

B = 8

So now we have:

T(t) = A + B sin( (t-C) pi / 12)

T(t) = 63 + 8 sin( (t-C) pi / 12)

The easiest way to find C is to use the same kinds of properties of sine as above.

We know that sine reaches its maximum when the part inside the sine function is pi/2, and we know that in this problem the maximum happens when t = 17. So we can set the part inside the sine equal to pi / 2 and plug in the associated value of t:

(t_max - C) pi / 12 = pi / 2

(17 - C) pi / 12 = pi / 2

then solve for C:

(17 - C) / 12 = 1/2

17 - C = 6

C = 11

Our function is now complete:

T(t) = 63 + 8 sin( (t-11) pi / 12)

Great! But the problem is actually asking for the average value of this function from noon to 7 PM (t = 12 to t = 19). So we go back to this:

f_{avg, b-a} = ∫_a^b T(t) dt / (b - a)

We have everything we need now. Remember that the variables a and b are just the start and end points we are calculating our integral over. In this case, that's the start and end time for our average value calculation, which are givens. Plug it all in:

f_{avg, 19-12} = ∫₁₂¹⁹ [63 + 8 sin( (t-11) pi / 12)] dt / (19 - 12)

= (1/7) ∫₁₂¹⁹ [63 + 8 sin( (t-11) pi / 12)] dt

Evaluate the integral:

(1/7) ∫₁₂¹⁹ [63 + 8 sin( (t-11) pi / 12)] dt = (1/7) [63t - (96/pi) cos( (t-11) pi / 12)]₁₂¹⁹

=(1/7) [( 63(19) - (96/pi) cos( ((19)-11) pi / 12) ) - ( 63(12) - (96/pi) cos( ((12)-11) pi / 12) )]

=(1/7) (485.79) = 69.40

So the average temperature from noon to 7 PM is 69.40°.