Ramakrishnan

Ramakrishnan Vaidyanathan

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7 years, 245 days

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With twenty years of Industrial experience and twenty years of teaching experience, I am now as retired Professor, using Maple to teach mathematics subject for students studying X to XII standards. Published XII Mathematics books.

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These are questions asked by Ramakrishnan

In the following problem though b and c are same (except the way denominator 2 is hanfled), command ' a-b ' readily answers zero, but a-c not so. Why? Only on condition of assumption real it gives zero!

a := (1/2)*(kappa*omega^2+omega^3)*(Y+(1/2)*(-sqrt(N)*omega^(3/2)*sin(theta[2])*cos(varphi[2])*lambda__b+sqrt(N)*omega^(3/2)*sin(theta[1])*cos(varphi[1])*lambda__a)/(kappa*omega^2+omega^3))^2/omega:

b := (1/2)*(kappa*omega^2+omega^3)*(Y+(1/2)*(-sqrt(N)*omega^(3/2)*sin(theta[2])*cos(varphi[2])*lambda__b+sqrt(N)*omega^(3/2)*sin(theta[1])*cos(varphi[1])*lambda__a)/(kappa*omega^2+omega^3))^2/omega:

a-b;

0

(1)

c := (1/2)*(kappa*omega^2+omega^3)*(Y+(-sqrt(N)*omega^(3/2)*sin(theta[2])*cos(varphi[2])*lambda__b+sqrt(N)*omega^(3/2)*sin(theta[1])*cos(varphi[1])*lambda__a)/(2*(kappa*omega^2+omega^3)))^2/omega:

a-c;

(1/2)*(kappa*omega^2+omega^3)*(Y+(1/2)*(-N^(1/2)*omega^(3/2)*sin(theta[2])*cos(varphi[2])*lambda__b+N^(1/2)*omega^(3/2)*sin(theta[1])*cos(varphi[1])*lambda__a)/(kappa*omega^2+omega^3))^2/omega-(1/2)*(kappa*omega^2+omega^3)*(Y+(-N^(1/2)*omega^(3/2)*sin(theta[2])*cos(varphi[2])*lambda__b+N^(1/2)*omega^(3/2)*sin(theta[1])*cos(varphi[1])*lambda__a)/(2*kappa*omega^2+2*omega^3))^2/omega

(2)

"(->)"

0

(3)

``

Why the answer is not given as zero?

``

``

 

Download what_is_the_difference_between_b_and_c.mw

What difference therms b and c make for Maple? Are they not same?

Ramakrishnan V

rukmini_ramki@hotmail.com

There have come unwanted lines and marks . I donot know how to remove them. Using doc.block, remove block seems to be little tough to incorporate! Please enlighten me. Modified doc. is most welcome. Thanks. Ramakrishnan V 

Gaussian Elimination Method

 

 

Given*the*equations

  restartreset:

with(Student[LinearAlgebra])``

(1)
Coefficient Tanle

Equation 1

Equation 2

Equation 3

Equations

`m__1,1` := 3:
`` 

`m__2,1` := 2:
``

`m__3,1` := 1:
``

`m__1,1`*x__1+`m__1,2`*y+`m__1,3`*z = `m__1,4`; = 3*x__1+y-z = 3

`m__2,1`*x__1+`m__2,2`*y+`m__2,3`*z = `m__2,4`; = 2*x__1-8*y+z = -5

```m__3,1`*x__1+`m__3,2`*y+`m__3,3`*z = `m__3,4`; = x__1-2*y+9*z = 8

The equations in matrix form is given by

Matrix([[3, 1, -1, 3], [2, -8, 1, -5], [1, -2, 9, 8]])

(2)

The Gaussian Elimination gives the simplified natrix equation as given below:

Matrix([[3, 1, -1, 3], [0, -26/3, 5/3, -7], [0, 0, 231/26, 231/26]])

(3)

``The equations in simplified form are:

3*x+y-z = 3

(4)

-(26/3)*y+(5/3)*z = -7

(5)

(231/26)*z = 231/26

(6)

``

The aolution ia obtained by solving the above equations in reverse order

{x = 1, y = 1, z = 1}

(7)

 

``

 

Download GausianFinal15Nov2015.mwGausianFinal15Nov2015.mw

Please_Help_for_a_better_presentation.mw
I have used conduction Fourier formula to find k or q or T. Each time i hAvE to run from restart and units. Can this duplication be eliminated?
How do i rewrite the codes?

Thanks for considering this worth answering.

Ramakrishnan Vaidyanathan

restart

with(Units[Standard])

with(Units)

UsingSystem()

SI

(1)

q := -k*(T__2-T__1)/t

-k*(T__2-T__1)/t

(2)

S := {T__1 = 550*Unit('K'), T__2 = 50*Unit(Unit('K')), k = 19.1*Unit('W')/(Unit('m')*Unit('K')), t = 2*Unit('cm')}

{T__1 = 550*Units:-Unit(K), T__2 = 50*Units:-Unit(K), k = 19.1*Units:-Unit(m*kg/(s^3*K)), t = (1/50)*Units:-Unit(m)}

(3)

eval(q, S)

477.5000000*Units:-Unit(kW/m^2)

(4)

refresh

restart

with(Units[Standard])

with(Units)

UsingSystem()

 

 

t := -k*(T__2-T__1)/qNULL

 

-k*(T__2-T__1)/q

(5)

S := {T__1 = 550*Unit('K'), T__2 = 50*Unit(Unit('K')), k = 19.1*Unit('W')/(Unit('m')*Unit('K')), q = 477.5*Unit('kW'/'m'^2)}

{T__1 = 550*Units:-Unit(K), T__2 = 50*Units:-Unit(K), k = 19.1*Units:-Unit(m*kg/(s^3*K)), q = 477500.0*Units:-Unit(kg/s^3)}

(6)

eval(t, S)

2.000000000*Units:-Unit(cm)

(7)

restart

clear

 

``NULL

NULL

 

  refresh

restart

with(Units[Standard])

with(Units)

UsingSystem()

NULL

T__2 := q*t/k+T__1

q*t/k+T__1

(8)

S := {T__1 = 50*Unit('K'), k = 19.1*Unit('W')/(Unit('m')*Unit('K')), q = 477.5*Unit('kW'/'m'^2), t = 2*Unit('cm')}

{T__1 = 50*Units:-Unit(K), k = 19.1*Units:-Unit(m*kg/(s^3*K)), q = 477500.0*Units:-Unit(kg/s^3), t = (1/50)*Units:-Unit(m)}

(9)

eval(T__2, S)

550.0000000*Units:-Unit(K)

(10)

NULL

 

NULL

 

Download Please_Help_for_a_better_presentation.mw

restart

with(Units[Standard])

with(Units)

UsingSystem()

SI

(1)

q := -k*(T__2-T__1)/t

-k*(T__2-T__1)/t

(2)

S := {T__1 = 550*Unit('K'), T__2 = 50*Unit(Unit('K')), k = 19.1*Unit('W')/(Unit('m')*Unit('K')), t = 2*Unit('cm')}

{T__1 = 550*Units:-Unit(K), T__2 = 50*Units:-Unit(K), k = 19.1*Units:-Unit(m*kg/(s^3*K)), t = (1/50)*Units:-Unit(m)}

(3)

eval(q, S)

477.5000000*Units:-Unit(kW/m^2)

(4)

refresh

restart

with(Units[Standard])

with(Units)

UsingSystem()

 

 

t := -k*(T__2-T__1)/qNULL

 

-k*(T__2-T__1)/q

(5)

S := {T__1 = 550*Unit('K'), T__2 = 50*Unit(Unit('K')), k = 19.1*Unit('W')/(Unit('m')*Unit('K')), q = 477.5*Unit('kW'/'m'^2)}

{T__1 = 550*Units:-Unit(K), T__2 = 50*Units:-Unit(K), k = 19.1*Units:-Unit(m*kg/(s^3*K)), q = 477500.0*Units:-Unit(kg/s^3)}

(6)

eval(t, S)

2.000000000*Units:-Unit(cm)

(7)

restart

clear

 

``NULL

NULL

 

  refresh

restart

with(Units[Standard])

with(Units)

UsingSystem()

NULL

T__2 := q*t/k+T__1

q*t/k+T__1

(8)

S := {T__1 = 50*Unit('K'), k = 19.1*Unit('W')/(Unit('m')*Unit('K')), q = 477.5*Unit('kW'/'m'^2), t = 2*Unit('cm')}

{T__1 = 50*Units:-Unit(K), k = 19.1*Units:-Unit(m*kg/(s^3*K)), q = 477500.0*Units:-Unit(kg/s^3), t = (1/50)*Units:-Unit(m)}

(9)

eval(T__2, S)

550.0000000*Units:-Unit(K)

(10)

NULL

 

NULL

 

Download Please_Help_for_a_better_presentation.mw

How do i proceed to solve two differential equations?

Two equations two unknowns is easy to solve in polynomial algebraic equations. Example: x+y=5; x-y=3; The solution is x=4; y=1 by adding the equations we arrive at.

The two equations are second order differential equations with two variables say temperature T (x,y) and velocity c(x,y). Assume any simple equation (one dimensional as well i.e. T(x) and c(x) which you can demonstrate with ease, I have not formulated the exact equations and boundary conditions yet for SI Engine simulation.

Thanks for comments, suggestions and answers expected eagerly.

Ramakrishnan

y(t) = _C1*exp(-1.*t)*sin(.57736*t)+_C2*exp(-1.*t)*cos(.57736*t)

The answer i got for a DE raised in mapleprime is given above.

What command do i write now to get a plot of the same?

Ramakrishnan V

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