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Some of my friends, upon finding out that I'd joined these forums, decided to give me a math challenge problem they thought couldn't be solved. Does everyone get the same answer? It goes as follows:

A knight jumps on a chess board (assume it makes a legal move) from point A to point B. Relative to an observing point O, what is the rate of change of angle of the chess piece as a function of time?

The piece makes a legal move

The piece is a knight

s (given) : the scale of the chess board. One space has width and height S

C = current position of the piece as a function of time

P = the current position of the piece projected down onto the plane of the chess board

R is a reference point which is always directly right of the center point

[itex]\theta[/itex] (given) is the launch angle

g (constant) is acceleration due to gravity

Coordinate system is defined as the chess board itself, with the origin in one of the corners.

Let theta be the launch angle of the piece.

After launch the piece is solely under the influence of gravity

A function for the measure of the angle AOP, n (angle in the plane of the board)

A function for the measure of the angle POC, a (accention)

Time rate of change of n, n'

Time rate of change of a, a'

You have access to the aTan2 function, which properly returns the angle between the reference (+X axis) and any point giving an ordered pair IE: aTan2(-2,-1) => 210degrees

-I)describe the distance between points A and B, in the plane of the board(d)

-II)derive a function for the vertical component of velocity of the piece, a function of theta, launch velocity, and time

-III)Integrate said function to get a h as a function of time, launch velocity, and launch angle

-IV)Use the definition of velocity to generate a function for s of t and d

-V)Solve the system of eqs from III and IV for initial velocity

-VI)Define the magnitude ||OA|| and ||OB||, r

-VII)Define the angle between the reference(+X axis) , and OA, OB

-VIII)Write R(t,R

-IX)Take the derivative dR/dt

-X)Realize n is given by ROA plus angle AOP; AOP is determined by creating a right triangle with AO and the component AP perpendicular to AO, and adding their associated rate of change times time.

-XI)Take the derivative with respect to time of n to get n', and one of the answers. Remember your chain rule!

-XII)go back to IX, and realize that a (COP) can be determined from the triangle COP. h = ||CP||, r = ||OP||.

-XIII)once you have a, take the derivative to find the time rate of change of a, a'
-I)describe the distance between points A and B, in the plane of the board(d)

knights move up 1 over 2, or up 2 over 1, either way, the distance between A and B is:

d = S*[itex]\sqrt{1^2+2^2}[/itex]

d = S[itex]\sqrt{5}[/itex]
-II)derive a function for the vertical component of velocity of the piece, a function of theta, launch velocity, and time

accel=g

V

-III)Integrate said function to get a h as a function of time, launch velocity, and launch angle

h=gt

(assume time zero is launch)
-IV)Use the definition of velocity to generate a function for s of t and d

distance=rate*time

d/([itex]\Delta[/itex]t) = cos([itex]\theta[/itex])V

s= cos([itex]\theta[/itex])V
-V)Solve the system of equations from III and IV for initial velocity

h=gt

0=gt+V

-gV

...

I'll finish the rest of this later...

View attachment challengeProblem.zip

A knight jumps on a chess board (assume it makes a legal move) from point A to point B. Relative to an observing point O, what is the rate of change of angle of the chess piece as a function of time?

**Let/Assume:**The piece makes a legal move

The piece is a knight

s (given) : the scale of the chess board. One space has width and height S

C = current position of the piece as a function of time

P = the current position of the piece projected down onto the plane of the chess board

R is a reference point which is always directly right of the center point

[itex]\theta[/itex] (given) is the launch angle

g (constant) is acceleration due to gravity

Coordinate system is defined as the chess board itself, with the origin in one of the corners.

Let theta be the launch angle of the piece.

After launch the piece is solely under the influence of gravity

**Find:**A function for the measure of the angle AOP, n (angle in the plane of the board)

A function for the measure of the angle POC, a (accention)

Time rate of change of n, n'

Time rate of change of a, a'

You have access to the aTan2 function, which properly returns the angle between the reference (+X axis) and any point giving an ordered pair IE: aTan2(-2,-1) => 210degrees

**Strategy(with calculus)**-II)derive a function for the vertical component of velocity of the piece, a function of theta, launch velocity, and time

-III)Integrate said function to get a h as a function of time, launch velocity, and launch angle

-IV)Use the definition of velocity to generate a function for s of t and d

-V)Solve the system of eqs from III and IV for initial velocity

-VI)Define the magnitude ||OA|| and ||OB||, r

_{I}and r

_{F}

-VII)Define the angle between the reference(+X axis) , and OA, OB

-VIII)Write R(t,R

_{I},R

_{F}), giving the length of OP at any time

-IX)Take the derivative dR/dt

-X)Realize n is given by ROA plus angle AOP; AOP is determined by creating a right triangle with AO and the component AP perpendicular to AO, and adding their associated rate of change times time.

-XI)Take the derivative with respect to time of n to get n', and one of the answers. Remember your chain rule!

-XII)go back to IX, and realize that a (COP) can be determined from the triangle COP. h = ||CP||, r = ||OP||.

-XIII)once you have a, take the derivative to find the time rate of change of a, a'

**I**knights move up 1 over 2, or up 2 over 1, either way, the distance between A and B is:

d = S*[itex]\sqrt{1^2+2^2}[/itex]

d = S[itex]\sqrt{5}[/itex]

**II**accel=g

V

_{h}=h'=gt+C=gt+V

_{0}Sin([itex]\theta[/itex])

**III**h=gt

^{2}+V

_{0}Sin([itex]\theta[/itex])t+0

(assume time zero is launch)

**IV**distance=rate*time

d/([itex]\Delta[/itex]t) = cos([itex]\theta[/itex])V

_{0}

s= cos([itex]\theta[/itex])V

_{0}[itex]\Delta[/itex]t

**V**h=gt

^{2}+V

_{0}Sin([itex]\theta[/itex])t

0=gt+V

_{0}Sin([itex]\theta[/itex])

-gV

_{0}csc([itex]\theta[/itex]) = [itex]\Delta[/itex]t

...

I'll finish the rest of this later...

**Answer**View attachment challengeProblem.zip

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