Question 4

 

A Calculator MAY be used to help answer this question.

Be certain to limit yourself to the four permissable calculator functions.

 

The temperature on New Year's Day in Hinterland was given by , where T is the temperature in degrees Fahrenheit and H is the number of hours from midnight (0 ≤ H < 24).

 

(a) The initial temperature at midnight was -15°F, and at noon of New Year's Day was 5°F. Find A and B.

 

(b) Find the average temperature for the first 10 hours.

 

(c) Use the Trapezoid Rule with 4 equal subdivisions to estimate .

 

(d) Find an expression for the rate that the temperature is changing with respect to H.

 

Solution

 

My name is Craig and I have decided to solve this question, considering no one else has claimed it yet.

NOTE: It is best viewed in maximum window dimensions.

 

 

(a) So reading the question, it is clear what we need to do. Find the constants A and B. Also given in the question are two points which include the time of day and the temperature at that time. (0,-15) and (12,5).

The first order of business is to isolate either of the constants, and we have now have two ways to do so using simple algebra:

 

                                   T(H) = -A -B cos(πH/12)

                                        /                    \

-15 = -A - B cos (π(0)/12)               5 = -A - B cos (π(12)/12)

-15 = -A - B cos (0)                           5 = -A - B cos (π)

-15 = -A - B                                       5 = -A + B

   A  = 15 - B                                       A = B - 5

 

Now that we have two functions that are both equal to A (could have also isolated B), we can easily set them equal to each other and solve for B:

 

15 - B = B - 5

      20 = 2B

        B = 10

 

Next take B=10 and plug it in to one of the simplified equations to solve for A:

 

A = B - 5

A = 10 - 5

A = 5

 

Therefore, the values of the two constants A and B, are 5 and 10, respectively.

 

 

(b) One way to find the average (or mean) value, is to use the Mean Value Theorem. This states that the average value of a function over a certain interval[a,b] is equivalent to the integral of the function over the same interval[a,b] divided by (b - a). Reading the question again, it asks for the average value during the first 10 hours, which is the same as over the interval [0 , 10]. So all we need to do is set up the equation and use our calculators to solve it:

 

AVG. Value = ƒ(c) = (1/(b-a))•()

T(c) = (1/(k-0))•()       (PS: due to technical difficulties, the program will not allow me

T(c) ≈ (1/10)•(-69.0986)                            to put "10" on top of theintegral notation so let 10=k.)

T(c) ≈ -6.9099 ºF

 

The average temperature for the first 10 hours of New Year's Day is -6.9099 ºF.

 

 

(c) The next part asks for an estimation of the integral from 6 to 8 of the Temperature function using a "Trapezoid Sum". This uses two Riemann Sums (method for approximating integrals of graphs). The average of the left hand sum and the right hand sum produces what we call a Trapezoid Sum.

First, I will set up the left and right hand sums seperately and then apply them to the definition of a Trapezoid Sum.

*One key point to understand is the fact that it asks for 4 sub-intervals. This means that the left and right hand sums the left and right hand sums will consist of 4 values whose x-coordinates are equidistant from each other. *

 

8 - 6 = 2          2 ÷ 4 = 0.5

Therefore ∂x = 0.5

 

 

LEFT HAND SUM:

≈ [ T(6) + T(6.5) + T(7) + T(7.5) ] • 0.5

                           ≈ [ -12.2797 ] • 0.5

                           ≈ -6.1399

 

RIGHT HAND SUM:

≈ [ T(6.5) + T(7) + T(7.5) + T(8) ] • 0.5

                           ≈ [ -7.2797 ] • 0.5

                           ≈ -3.6399

 

TRAPEZOID SUM:

≈ ( LHS + RHS ) ÷ 2

                           ≈ ( -6.1399 + -3.6399 ) ÷ 2

                           ≈ -4.8899

 

Another, quicker way to write out a Trapezoid Sum can be used as well. Expand the LHS and RHS from the following question and watch how it fits together nicely.

 

TRAPEZOID SUM:

≈ ( LHS + RHS ) ÷ 2

                           ≈ [ ( [ T(6) + T(6.5) + T(7) + T(7.5) ] • 0.5 ) + ( [ T(6.5) + T(7) + T(7.5) + T(8) ] • 0.5 ) ] ÷ 2

                           ≈ [ [ T(6) + T(6.5) + T(6.5) + T(7) + T(7) + T(7.5) + T(7.5) + T(8) ] • 0.5 ] ÷ 2

                           ≈ [ [ T(6) + 2[ T(6.5) ] + 2[ T(7) ] + 2[ T(7.5) ] + T(8) ] • 0.5 ] ÷2

                           ≈ ( -19.5594 ) ÷ 4

                           ≈ -4.8899

 

 

(d) The final part of the question is to find an equation that represents the rate at which the Temperature is changing. This is also known as the slope of the tangent line to a given point on the Temperature function, which is also known as the DERIVATIVE of the Tempertaure function ( T'(H) ). This is done by simply differentiating both sides of the equation. The derivative of a constant, trigonometric function, and the Chain Rule will be needed to solve this question.

 

∂/∂H T = ∂/∂H ( -5 - 10cos( πH/12 ) )     <--differentiate both sides

         T' =∂/∂H  ( -5 ) -  ∂/∂H ( 10cos( πH/12 ) )     <--break up the two terms

         T' =0 -  10 • ∂/∂H cos( πH/12 ) • ∂/∂H (πH/12)     <--differentiate constant, extract multiplier, set up Chain Rule

         T' = -10 • -sin( πH/12 ) • ( π/12 )     <--apply the chain rule

         T' = -( 10π/12 ) • -sin( πH/12)     <--combine the multipliers

         T' = ( 5π/6 ) sin( πH/12 )     <--simplify the multiplier, get rid of negatives

 

The rate that the Temperature is changing can be written as T'=(5π/6)sin(πH/12).

 

*Edit: Trapezoid Sum image created and added by Tim-math-y