Directional derivatives
1. The problem statement, all variables and given/known data
Find the directional derivative of w = x^3 + y^2 + 3z at the point (1, 3, 2) on the curve of intersection of the surfaces x^2 + y^2− 2xz = 6, 3x^2− y^2 + 3z = 0, in the direction of increasing z. 2. Relevant equations 3. The attempt at a solution i know how to solve the directional dirivative of a function of at a point on the given vector.but in this question I dont know how to solve the equations,and "in the direction of increasing z" does not make sense 
Re: Directional derivatives
The tangent vector along the curve is perpendicular to the gradient vectors of the two surfaces. Find them at (1,3,2). Now that you have these two vectors, how do you find one that is perpendicular to them both?

Re: Directional derivatives
cross product of these two vectors

Re: Directional derivatives
But I cant get the meaning of "in the direction of increasing z"

Re: Directional derivatives
Quote:

Re: Directional derivatives
I want to get the gradient vector of x^2 + y^2− 2xz = 6,so z=(x^2+y^26)/2x
,is the gradient vector (dz/dx dz/dy) but i should have three dimensions? 
Re: Directional derivatives
If f=x^2+y^22xz6=0 is your surface the gradient vector is (df/dx,df/dy,df/dz) (all partial derivatives). Of course you get three components.

Re: Directional derivatives
Why do we calculate the gradient vectors of the surfaces? What is the geometrical meaning of gradient vector? I think I should calculate the normals of two surfaces and get the cross product of two normals, because the vector of the intersection curve is perpendicular to both of the normals..

Re: Directional derivatives
The normal vector points in the same direction as the gradient vector. Don't you calculate the normal vector using the gradient vector??

Re: Directional derivatives
AH, finally,i figured it out , thank a lot

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