Accumulation

Part 1. Background

The following command loads the necessary plotting package.  (No response is required in this part.)

<< Graphics`Graphics`

Part 2. Summations: Size Distribution of Particles

Here are the data for particles in the Pasadena atmosphere.  The size values p are defined as diameters and the corresponding values of Δ(N/Δ)p as rates.  The Thread command creates pairs for plotting.

diameters  =  {0.00875, 0.0125, 0.0175, 0.0250, 0.0350, 0.0500, 
          0.0700, 0.0900, 0.112, 0.137, 0.175, 0.250, 0.350, 0.440, 0.550, 
          0.660, 0.770, 0.880, 1.05, 1.27, 1.48, 1.82, 2.22, 2.75, 3.30, 
          4.12, 5.22};
rates  =  {1.57*10^7, 5.78*10^6, 2.58*10^6, 1.15*10^6, 6.01*10^5, 
          2.87*10^5, 1.39*10^5, 8.90*10^4, 7.02*10^4, 4.03*10^4, 2.57*10^4, 
          9.61*10^3, 2.15*10^3, 9.33*10^2, 2.66*10^2, 1.08*10^2, 5.17*10, 
          2.80*10, 1.36*10, 5.82, 2.88, 1.25, 4.80*10^(-1), 
          2.17*10^(-1), 1.18*10^(-1), 6.27*10^(-2), 
          3.03*10^(-2)};
diamData  =  Thread[{diameters, rates}]

Plot the data points.

ListPlot[diamData, PlotStyle -> PointSize[0.015],  
        AxesLabel -> {"p", " "}];

Make a semilog plot of the data.

LogListPlot[diamData, 
        PlotStyle -> PointSize[0.015]];

Answer to question 3:

Answer to question 4:

Do your calculations for question 5 in the input cell below.

Our task for step 6 is to add up all the products of the form (Subscript[rates, k])*(Subscript[run, k]) where (Subscript[run, k]) = (Subscript[diameters, k + 1]) - (Subscript[diameters, k]), and where k runs from 1 up to one less than the number of data points.  First we find out how many terms will be in the sum.  The Mathematica command Length counts the number of elements in a list.

n  =  Length[rates] - 1

Now we create the list of differences for run or Δp.  The first command, incidentally, suppresses the warning that Mathematica would have generated about the similarity between our variable run and the built-in function Run.

Off[General::spell1];
run  =  
        Table[diameters[[k + 1]] - diameters[[k]],  {k,  
            n}];

The Sum command is used to evaluate (∑_(k = 1)^n) f[k], where f[k] can be any expression using the index k.  Since k may have a value left over from some other statement, we clear it first.

Clear[k];
Sum[rates[[k]]*run[[k]], {k, n}]

Answer to question 7:

Part 3. Accumulation and Area

Do your calculations of the exact area in the input cells below.

b  =  ?;

triangleArea  =  ?;

sectorArea  =  ?;

area  =  triangleArea + sectorArea
N[%]

Define the function whose graph is the arc of the circle:

f[x_] :=   ?;

Set up a left-hand sum (LHS) with n = 25 terms.  Note that the interval [0, b] is divided into n subintervals, each of width Δ = (b/n).  The points of subdivision are 0, Δ, 2Δ, 3Δ, ..., nΔ.  Find the error in the LHS approximation by subtracting it from the exact area.

n  =  25
Delta  = b/n
LHS  =  Sum[f[(k - 1)*Delta]*Delta,  {k, n}];

N[LHS]
N[area - %]

Complete the formula for a right-hand sum (RHS) with the same number of terms.  Calculate RHS and the error in the approximation.

RHS  =  ?;
N[RHS]
N[area - %]

Fill in the following table by changing n in the calculations above and recalculating.

({{n, Area  -  LHS, Area  -  RHS}, {25, -0.102, \[Placeholder]}, {250, \[Placeholder], \[Placeholder]}, {2500, \[Placeholder], \[Placeholder]}})

Answer to question 5:

Answer to question 6:

Enter your NEW approximation scheme, and calculate for the same values of n.

NEW  =  ?;
N[%]
N[area - %]

Fill in the following table:

({{n, Area  -  NEW}, {25, \[Placeholder]}, {250, \[Placeholder]}, {2500, \[Placeholder]}})

Answers to questions in step 8:

Part 4. Area Investigations

Enter your functions and your area approximation formulas in the input cell below.

Part 5. Summary