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HW 1 Mathematics 503, Mathematical Modeling, CSUF June 2 2007

Nasser M. Abbasi

June 15, 2014

Contents

1 Problem 1 (section 1.1,#3, page 7)
2 Problem 2 (section 1.1, #11, page 18)

1 Problem 1 (section 1.1,#3, page 7)

problem:

PIC

solution:

The blast equation is given by

     (   2) 15
r = C  Et-
        ρ

Where r is radius of blast in meters, t is time since explosion in seconds, E energy of explosion in Joules, ρ is air density in Kg/m^3

We are given              3
ρ = 1.25 kg∕m   and C = 1  , hence the above can be rewritten as

5        2
r = 0.8Et

Let x = E,         2
A = 0.8t   ,     5
b= r   , then we have

b = Ax

And now x is found by least squares solution.  Once x is found, then E is found since E = x .

The code is shown on the next page. The least squares solutions shows that

|----------------------|
|                      |
| E = 16.6175 kilotons |
-----------------------

PIC

PIC

2 Problem 2 (section 1.1, #11, page 18)

problem:

PIC

solution:

First make a list of all the physical variables and the corresponding dimensions.








Variable P (power) l  (ship length) V ρ ν             g







meaning work rate(F*d/t) in meters ship speed water density water Viscosity gravity







Dimension ML2
-T3   L L
T  M
L3   L2
T-              L
            T2







Hence we seek a physical law of the form

f(P,l,V,ρ,ν, g)= 0

The function f is a combination of all the physical variables. Hence we write

pict

For the above combination to be dimensionless, we must have each exponent term equal to zero. Hence we obtain the following 3 equations

pict

Writing the above in matrix form

                           ⌊ ⌋

                           ||a||
                           || ||
⌊                        ⌋ ||b||   ⌊  ⌋
| 1   0  0    1    0   0 | | |   | 0|
||                        || ||c||   ||  ||
||                        || || || = ||  ||
| 2   1  1   − 3   2   1 | || ||   | 0|
|⌈                        |⌉ ||d||   |⌈  |⌉
 − 3  0  − 1  0   − 1 − 2  || ||     0
                           |e|
                           || ||
                           ⌈ ⌉
                            f

We see that the dimension matrix 3× 6  , hence its row space has dimension 6  and its column space has dimension 3  . We need now to find the basis for the Null Space of the dimension matrix. Now reduce A to its reduce row echelon form

⌊                        ⌋     ⌊                        ⌋      ⌊                       ⌋

|| 1   0   0   1    0   0 ||     || 1   0   0   1    0   0 ||      ||1  0   0    1   0    0 ||
||                        || l21=2||                        || l31=−3||                       ||
|| 2   1   1  − 3   2   1 ||  →  || 0   1   1  − 5   2   1 ||  →   ||0  1   1   − 5  2    1 ||
|                        |     |                        |      |                       |
⌈                        ⌉     ⌈                        ⌉      ⌈                       ⌉
 − 3  0  − 1  0   − 1  − 2      − 3  0  − 1  0   − 1 − 2        0  0  − 1   3  − 1  − 2

Now multiply last row by       ⌊                  ⌋

      |1  0  0   1   0  0|
      ||                  ||
− 1→  ||0  1  1  − 5  2  1||
      ||                  ||
      ⌈                  ⌉
       0  0  1  − 3  1  2 , and now subtract last row from second row

   ⌊                   ⌋
   ||1  0  0   1  0   0 ||
   |                   |
→  ||0  1  0  − 2 1  − 1||
   ||                   ||
   ⌈                   ⌉
    0  0  1  − 3 1   2 and this is the final reduce row echelon form

we have rank=3 (the first 3 columns are Linearly independent). Therefor, we use the first 3 variables as the pivot variables, which are a,b,c and use as the free variables those which correspond to the last 3 columns, which are d,e,f

Now since

                      ⌊  ⌋
                      | a|
                      |  |
⌊                    ⌋||  ||   ⌊ ⌋
                      || b||
| 1  0 0   1   0   0 |||  ||   |0|
||                    |||| c||   || ||
|| 0  1 0  − 2  1  − 1|||  | = ||0||
||                    ||||  ||   || ||
⌈                    ⌉|| d||   ⌈ ⌉
  0  0 1  − 3  1   2  ||  ||    0
                      || e||
                      |  |
                      ⌈  ⌉
                        f

By back substitution, from the 3rd row we obtain  c− 3d + e+ 2f = 0  , or c = 3d− e − 2f and from the second row we obtain b− 2d+ e − f = 0  , or b = − e+ f + 2d , and from the first row we obtain a+ d = 0  or a = − d

Hence the solution now can be written in terms of the free variables as

pict

Therefor the basis for the null space of A are

( ⌊   ⌋  ⌊   ⌋ ⌊   ⌋ )
||                    ||
|||| || 1 ||  || 0 || || 0 || ||||
||| |   |  |   | |   | |||
|||| || − 2|| || 1 || || − 1|| ||||
|||| ||   ||  ||   || ||   || ||||
||| ||   ||  ||   || ||   || |||
{ || − 3|| || 1 || || 2 || }
|| ||   || ,||   ||,||   || ||
||| | − 1| | 0 | | 0 | |||
|||| ||   ||  ||   || ||   || ||||
|||| || 0 ||  ||− 1|| || 0 || ||||
||| ||   ||  ||   || ||   || |||
|||| ⌈   ⌉  ⌈   ⌉ ⌈   ⌉ ||||
(   0      0     − 1 )

Hence

pict

Hence the complete set is

{       P        Vl      V2}
 π1 = l2V3ρ,π2 = ν-,π3 = l g

Hence the general solution is      α β  γ
π = π1 π2π3

The Pi theorem says that there is a physical law expressed in terms of the dimensionless quantities called F (π1,π2,π3)= 0  corresponding to the physical law f(P,l,V,ρ,ν,g) = 0

Now, we need to solve for P , hence we write

pict

Hence if we let  ( Vl-V2)       √ --
g  ν ,l g = π2×  π3   , then we see that

|-----------------|
|  2P3-= g(Fr,Re) |
|  lVρ            |
------------------

where

     Vl-    √-V--
Fr = ν ,Re =  l g

Let k be some constant, then the power needed is given by

pict