quietrain said:
oh so it is like an equation of the line right? rx = x + tv where v is a vector parallel to the x-direction ? so in this case it is the dx/dt?
Yes, every vector can be written as a sum of vectors parallel to the axes.
also, i meant that if i have r(x,y,z) = (x2,y2,z2)
I have no idea at all what that means. Are (a, b, c) the x,y,z coordinates or not? If so, then you are saying that x= x^2, y= y^2, z= z^2 so that your set is not a curve at all but is 8 discrete points.
, and i do
partial differentiation, to get r'(x,y,z) = (2x,2y,2z) , then this is the tangent line equation right?
NO.
so if i parameterize r(x(t),y(t),z(t)) , and do r'(t), i get the equation you wrote above right? which is the tangent line equation parameterized.
I don't know what you mean by that. Are you still saying that r(x,y,z)= (x^2, y^2, z^2)? As I said before, points satifying that do NOT form a curve and it cannot be parameterized.
so are they the same? just that one is parameterized one is not?
A vector (or point) function in an xyz-coordinate system is
always (x, y, z) by definition of "xyz-coordinate system". You have to have x, y, and z functions of some other variables in order to have a set. For example, if they are functions of one parameter, (x(t), y(t), z(t)), this is a one dimensional figure, a curve. If they are functions of two parameters, (x(u,v), y(u,v), z(u,v)) then it is a two dimensional figure, a surface. To write something like "r(x, y, z)= (f(x), g(y), h(z)) would mean that you are requiring the points (x,y,z) to satisfy x= f(x), y= g(y), z= h(z) which, typically, will reduce to a finite number of points.