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

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

Nasser M. Abbasi

June 15, 2014

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:

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 | -----------------------

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

problem:

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

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

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

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

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

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