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#1 2008-09-28 14:34:58

mikau
Member
Registered: 2005-08-22
Posts: 1,504

tangent plane to a curve

let f(x,y,z) = x + y^2 -z^2

find the equation of the tangent plane to f at (1,2,3).

The book then proceeds to construct a plane in 3 variables by some logic that escapes me. I don't understand this geometrically. The graph of this is in 4 dimensions. How can a plane in 3 dimensional space, be tangent to a 4 dimensional graph.

Now, if we were to set: f(x,y,z) = K, now we've got a 3d surface that kinda makes sense to me. We then know that the gradient vector is perpendicular to this at the point (x,y,z), and we construct a tangent plane. However, if we're going to let f(x,y,z) = K, then we must let K = f(1,2,3) since we're assuming the graph passes though this point.

So is this problem really asking me to consider a 4 dimensional graph, or the graph of f(x,y,z) = f(1,2,3) ???


further, consider the next problem:
Find the direction of the greatest rate of increase of f(x,y,z) = xye^z at the point (2,1,2)

again, we seem to be considering a 3 dimensional surface, whereas the graph is, I think, in 4 dimensions. So whats going on? Am I not understanding, or is this being presented in a screwy manner?

Last edited by mikau (2008-09-28 14:50:58)


A logarithm is just a misspelled algorithm.

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#2 2008-10-06 23:43:36

ojabo
Member
Registered: 2008-10-04
Posts: 1

Re: tangent plane to a curve

Differentiate the function explicitly with respect to x and make dy/dx subject of the formula,after that substitute the values of x,y,z, i.e(1,2,3) into the equation.What you get is call the gradient if the function.Equate the value you got to this, (y1-y)/(x1-x)  and cross multiply to get the equation.

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