*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...