Z4K

06-26-2004, 08:30 AM

It's time for MULTIVARIABLE CALCULUS!! Hooray!

http://www.photodump.com/direct/Z4k/z_calc001_United.jpg

I need some help with a maths problem I have - I can't do it and don't have Matlab or Maple.

If any of you are mathsy types I'd really appreciate a hand with this, or even just a solution (including a graphical one).

There are a couple of ways of stating the problem. In vector notation (where a' is the derivative of a, a'' is the 2nd derivative of a, and a is the magnitude of vector a) it can be written:

(x'', y'') = L * sqrt [ x'^2 + y'^2 ] * (-y', x') + D * sqrt [ x'^2 + y'^2 ] * (-x', -y') + ( 0, -g )

where L, D and g are constants; x and y are postion vectors; x' and y' are the rate of change of x and y with time (ie. velocities in horizontal and vertical planes); similarly for x'' & y'' (acceleration).

Or in words:

I have a particle subject to the following accelerations:

- A constant acceleration vertically down;

- An acceleration proportional to the square of the velocity (= L*v^2) perpendicular to the path of the particle upwards (ie. there are two perpendiculars, it's on the upward one);

- An acceleration proportional to the square of the velocity (= D*v^2) parallel to the path of the particle and ******ing its motion (ie. slowing it down).

I want to be able to find two of the following; x, y, or velocity, from the other one. For instance "at a y displacement of 3 metres, the particle will have an x displacement of 2 metres and a velocity of 6 m/sec."

http://www.photodump.com/direct/Z4k/z_calc003_Train.jpg

Try as I might (having forgotten all my university calculus, and probably never having done any that could solve this anyway...) I can't solve it. If you have a look at it, you'll see that the acceleration in either direction is a function of the velocity in not only that direction, but the other as well ( a(x) = F(x, y) ). I can't separate it, can't make it an "exact differential equation" and hence don't think I can use integration of partial differentials to solve it.

I've got to the following random collection of expressions:

(B: angle path makes to horizontal;

i, j: x & y unit vectors;

v: velocity vector;

v: magnitude of velocity;

v{x}: horizontal velocity vector;

v{y}: magnitude of vertical velocity;

a: acceleration;

a{t}: magnitude of tangential acceleration;

a{n}: magnitude of normal acceleration;

L, D, g, m: constants)

IN NO PARTICULAR ORDER:

dv/dt = g sin(B) - Dv^2

dB/dt = (g cos(B) - Lv^2)/v = a{n}/v

d2x/dt2 = a{x} = (v^2/m)(L sin(B) - D cos(B))

d2y/dt2 = a{y} = (v^2/m)(L cos(B) + D sin(B)) - g

v = (dx/dt)*i + (dy/dt)*j

a = (sqrt ((dx/dt)^2 + (dy/dt)^2))*(( L(dy/dt) - D(dx/dt) )*i + ( L(dx/dt) + D(dy/dt) )*j ) - g*j

http://www.photodump.com/direct/Z4k/z_calc004_CadetNurses.jpg

If you read this far, thank you.

- - - - - - - - - - - - - - -

Let's play Global Termonuclear War

[This message was edited by Tully__ on Sun June 27 2004 at 12:30 AM.]

http://www.photodump.com/direct/Z4k/z_calc001_United.jpg

I need some help with a maths problem I have - I can't do it and don't have Matlab or Maple.

If any of you are mathsy types I'd really appreciate a hand with this, or even just a solution (including a graphical one).

There are a couple of ways of stating the problem. In vector notation (where a' is the derivative of a, a'' is the 2nd derivative of a, and a is the magnitude of vector a) it can be written:

(x'', y'') = L * sqrt [ x'^2 + y'^2 ] * (-y', x') + D * sqrt [ x'^2 + y'^2 ] * (-x', -y') + ( 0, -g )

where L, D and g are constants; x and y are postion vectors; x' and y' are the rate of change of x and y with time (ie. velocities in horizontal and vertical planes); similarly for x'' & y'' (acceleration).

Or in words:

I have a particle subject to the following accelerations:

- A constant acceleration vertically down;

- An acceleration proportional to the square of the velocity (= L*v^2) perpendicular to the path of the particle upwards (ie. there are two perpendiculars, it's on the upward one);

- An acceleration proportional to the square of the velocity (= D*v^2) parallel to the path of the particle and ******ing its motion (ie. slowing it down).

I want to be able to find two of the following; x, y, or velocity, from the other one. For instance "at a y displacement of 3 metres, the particle will have an x displacement of 2 metres and a velocity of 6 m/sec."

http://www.photodump.com/direct/Z4k/z_calc003_Train.jpg

Try as I might (having forgotten all my university calculus, and probably never having done any that could solve this anyway...) I can't solve it. If you have a look at it, you'll see that the acceleration in either direction is a function of the velocity in not only that direction, but the other as well ( a(x) = F(x, y) ). I can't separate it, can't make it an "exact differential equation" and hence don't think I can use integration of partial differentials to solve it.

I've got to the following random collection of expressions:

(B: angle path makes to horizontal;

i, j: x & y unit vectors;

v: velocity vector;

v: magnitude of velocity;

v{x}: horizontal velocity vector;

v{y}: magnitude of vertical velocity;

a: acceleration;

a{t}: magnitude of tangential acceleration;

a{n}: magnitude of normal acceleration;

L, D, g, m: constants)

IN NO PARTICULAR ORDER:

dv/dt = g sin(B) - Dv^2

dB/dt = (g cos(B) - Lv^2)/v = a{n}/v

d2x/dt2 = a{x} = (v^2/m)(L sin(B) - D cos(B))

d2y/dt2 = a{y} = (v^2/m)(L cos(B) + D sin(B)) - g

v = (dx/dt)*i + (dy/dt)*j

a = (sqrt ((dx/dt)^2 + (dy/dt)^2))*(( L(dy/dt) - D(dx/dt) )*i + ( L(dx/dt) + D(dy/dt) )*j ) - g*j

http://www.photodump.com/direct/Z4k/z_calc004_CadetNurses.jpg

If you read this far, thank you.

- - - - - - - - - - - - - - -

Let's play Global Termonuclear War

[This message was edited by Tully__ on Sun June 27 2004 at 12:30 AM.]