dy/d(x,z)

can Fermat's method be modified for use on 3d forms?
i've run into a number of problems, i'll see what turns up before i ask more spicific questions.

Originally posted by fearlessleader
can Fermat's method be modified for use on 3d forms?
i've run into a number of problems, i'll see what turns up before i ask more spicific questions.

What is Fermat's method?
What is a 3D form?
What do you mean 'more specific'?

Originally posted by THUDandBLUNDER
What is Fermat's method?
What is a 3D form?
What do you mean 'more specific'?

fermat's method is the basis of all derivatives.

d(f(x))/d(x) = lim(h-->0) (f(x+h)-f(x))/h

h is delta x, this keybord has no greek.

this finds the slope of a line tangent to f(x) at x. if properly executed, it always produces a function of x, denoted f'(x)

i'm looking for a similar method of finding a funtion in terms of x and z that will find planes tangent to three dimensional surfaces, ie: funtions with two independant variabels and one dependant.

i've done some work on the matter, and i've made a little progress, there are still problems. first, a line requires only two peices of information: a point and a slope. in a 2d function, we already know the point, and a single function (dy/dx) gives us the slope. however, a plane requires three peices of information, including the point of tangency, so i am starting to think that two equations will be needed. that would no be elegant.

i wont post my work so far untill i see what other approaches you might come up with on your own.

Originally posted by fearlessleader
fermat's method is the basis of all derivatives.

d(f(x))/d(x) = lim(h-->0) (f(x+h)-f(x))/h

h is delta x, this keybord has no greek.

this finds the slope of a line tangent to f(x) at x. if properly executed, it always produces a function of x, denoted f'(x)

i'm looking for a similar method of finding a funtion in terms of x and z that wi ...[text shortened]... wont post my work so far untill i see what other approaches you might come up with on your own.

I won't say the answer here, but see what happens when you differentiate your function with respect to only one variable at a time, keeping the other constant. This will give you two new functions. See if you can work out the total change in the function as the independent variables move about.

Originally posted by royalchicken
I won't say the answer here, but see what happens when you differentiate your function with respect to only one variable at a time, keeping the other constant. This will give you two new functions. See if you can work out the total change in the function as the independent variables move about.

Gallus Nobellus!! i thought i would hear from you sooner or later.

if i understand you correctly, then i've gotten that far. but all it gives me is dy/dx and a seperate function dy/dz (from which, granted, i can get the tangent plane.)

the troubel is that for most planes, you need two peices of information.

i've figured that any plane can be represented by a single line, given that for every point on that line that line is the steepest line passing through that point. unfortenetly, i can't find any better way to define these lines than as the intersection of two planes, still leaving me with a minimum of two peices of information and thus two equations.

is there any way around this? for basic practice such a unified function is unnessesary, as one can get tangent planes through the two single-variable derivatives, but i'd like to be able to graph this, and i dont see how to graph two dependant variables as a singel entity.

Originally posted by fearlessleader
Gallus Nobellus!! i thought i would hear from you sooner or later.

if i understand you correctly, then i've gotten that far. but all it gives me is dy/dx and a seperate function dy/dz (from which, granted, i can get the tangent plane.)

the troubel is that for most planes, you need two peices of information.

i've figured that any pla ...[text shortened]... o be able to graph this, and i dont see how to graph two dependant variables as a singel entity.

Have you met vectors?

Originally posted by royalchicken
Have you met vectors?

only in passing.
i see how they apply, but do they help?

Originally posted by fearlessleader

if i understand you correctly, then i've gotten that far. but all it gives me is dy/dx and a seperate function dy/dz (from which, granted, i can get the tangent plane.)

the troubel is that for most planes, you need two peices of information.

You have the extra bit of information you need... it's the point (x0,z0) on the surface at which you calculated your derivatives... With that, and your knowledge of the equation for a plane, you should be able to arrive at (in different notation from yours, but better-suited to this format)

y = f(x0,z0)+fx(x0,z0)*(x-x0)+fz(x0,z0)*(z-z0).

Originally posted by fearlessleader
only in passing.
i see how they apply, but do they help?

A plane is defined by any two non-parallel vectors. A plane is also defined by a vector which is normal to it and any point in it.

Originally posted by royalchicken
A plane is defined by any two non-parallel vectors. A plane is also defined by a vector which is normal to it and any point in it.

neight: could you explain your notation a little better, and how you arrived at this. it looks good, but i dont have a clue what it means.

RC: the line normal is prehaps a better definition for a plane than any i have tried so far, but it still leaves the problem: how do you define a line not parrele to either x=0 or z=0 using only two peices of information (including the given point of tangency)?

Originally posted by fearlessleader
neight: could you explain your notation a little better, and how you arrived at this. it looks good, but i dont have a clue what it means.

sorry, I was away from the forums for a couple days.
y = f(x0,z0)+fx(x0,z0)*(x-x0)+fz(x0,z0)*(z-z0)

assumes your original surface is y=f(x,z).
the notation fx(x,z) or fz(x,z) denote the derivatives of f with respect to x or z, respectively, ie dy/dx and dy/dz, which you said you had computed.

I suggested this form because it should be very easy to graph, which I think was your original intention?

as for how this is arrived at, recall the point-slope definition of a line. This is the natural adaptation of that for a plane: f(x0,z0) is the point,
and the slopes are (x-x0) in the fx direction, and (z-z0) in the fz direction.

Originally posted by royalchicken
Have you met vectors?

Stay away from them! They are like those creepy hobos trying to sell you roses on your way back from class. They look like nice guys, but they really deep down are just trying buy weed......crap. That started off like a great analogy, and I just blew it to hell.

Originally posted by neight
sorry, I was away from the forums for a couple days.
y = f(x0,z0)+fx(x0,z0)*(x-x0)+fz(x0,z0)*(z-z0)

assumes your original surface is y=f(x,z).
the notation fx(x,z) or fz(x,z) denote the derivatives of f with respect to x or z, respectively, ie dy/dx and dy/dz, which you said you had computed.

I suggested this form because it should be very easy to gra ...[text shortened]... s the point,
and the slopes are (x-x0) in the fx direction, and (z-z0) in the fz direction.

i'm not sure i understand.
you're saying that
d(f(x,z))/d(x,z)=f(x,z)+(d(f(x,z))/d(x))(X-x)+(d(f(x,z))/d(z))(Z-z)
caps are the variable, lowercase are the spicific value in question.

in speach, for cross-refrencing, that is
the deriviative of f of x,z with respect to x,z is equal to the functional value of f of x,z, plus ( the derivatice of f of x,z with respect to x, times x minus the spicific value of x ), plus (the derivative in terms of z times z minus the spicific value of z

this dosnt make sence to me. it may work if you have spicific values of x and z to work with, but if you're trying to make a general statment about the function, then it seems meeningless. am i misinterpreting x0 ?

Originally posted by fearlessleader
i'm not sure i understand.
you're saying that
d(f(x,z))/d(x,z)=f(x,z)+(d(f(x,z))/d(x))(X-x)+(d(f(x,z))/d(z))(Z-z)
caps are the variable, lowercase are the spicific value in question.

in speach, for cross-refrencing, that is
the deriviative of f of x,z with respect to x,z is equal to the functional value of f of x,z, plus ( the derivatice of ...[text shortened]... ake a general statment about the function, then it seems meeningless. am i misinterpreting x0 ?

look again at my post- there is no mention of d(f(x,z)/d(x,z). (Which you are correct in saying does not make sense)

(x0,z0) is the point at which your tangent plane is tangent. x0 and z0 are fixed (specific) values.

Is the source of the confusion that I called y=f(x,z) and y=f(x0,z0)+fx...(the equation in my post from 11-Mar.)? Those are not meant to be substituted, the one in the other. They are seperate curves. The first is your original surface, the second is the equation for the plane tangent to that surface at the point (x0,z0).

I like your method of avoiding confusion by writing in 'speach.' I will do the same to describe my equation for the tangent plane:

"Given a function f of two variables, x and z, write an expression for the plane tangent to f at the point (x0,z0)."

solution:
"Evaluate f at the point (x0,z0). Call this value f of (x0,z0). Compute the derivative of f with respect to x alone. Call this function fx of (x,z) (I think you called this dy/dx in your post of 11 Mar '05 16:10). Repeat for the derivative of f with respect to z alone. Call this function fz of (x,z) (This is your dy/dz). Also construct the algebraic expressions x-x0 and z-z0."

Finally, we are ready to express the tangent plane:
"f of x0,z0, plus fx of x0,z0 times the quantity x minus x0, plus fz of x0 and z0 times the quantity z-z0."

You can verify that this expression is indeed the tangent plane by checking the following:
1) At the point (x0,z0) this expression has the same value as your original function (this value is f(x0,z0), so they intersect at that point.
2) At the point (x0,z0) this expression has the same slope as your original function (this slope is fx(x0,z0) in the x direction, and fz(x0,z0) in the z direction- it may be slightly more difficult to show that the slopes are the same in any given direction, but that's why vectors are useful...)


I hope that's a bit less confusing...

Originally posted by neight
look again at my post- there is no mention of d(f(x,z)/d(x,z). (Which you are correct in saying does not make sense)

(x0,z0) is the point at which your tangent plane is tangent. x0 and z0 are fixed (specific) values.

Is the source of the confusion that I called y=f(x,z) and y=f(x0,z0)+fx...(the equation in my post from 11-Mar.)? Those are not meant to b ...[text shortened]... n direction, but that's why vectors are useful...)


I hope that's a bit less confusing...

i understand.

this is a very elegant formula for finding the tangent plane, much better than the dual-formula sysytem i had.

i think my original goal of an actual derivative is impossible, at least in three dimensions.