Directional Derivative of w on Intersection of Surfaces

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Homework Help Overview

The discussion revolves around finding the directional derivative of the function w = x^3 + y^2 + 3z at the point (1, 3, 2) along the curve of intersection of two surfaces defined by the equations x^2 + y^2− 2xz = 6 and 3x^2− y^2 + 3z = 0, specifically in the direction of increasing z.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the need to find tangent vectors that are perpendicular to the gradient vectors of the surfaces at the given point. There is also confusion regarding the interpretation of "in the direction of increasing z" and the dimensionality of the gradient vector.

Discussion Status

Some participants have offered guidance on calculating the tangent vector using the cross product of the gradient vectors. There is ongoing exploration of the geometrical meaning of the gradient vectors and their relationship to the normals of the surfaces. A participant expresses a breakthrough in understanding, indicating some progress in the discussion.

Contextual Notes

Participants are grappling with the implications of the problem's constraints, particularly regarding the interpretation of directional derivatives and the dimensional aspects of the gradient vectors. The discussion reflects a mix of established knowledge and points of confusion that remain to be addressed.

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Homework Statement


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.


Homework Equations





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 don't know how to solve the equations,and "in the direction of increasing z" does not make sense
 
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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?
 
cross product of these two vectors
 
But I can't get the meaning of "in the direction of increasing z"
 
alvielwj said:
But I can't get the meaning of "in the direction of increasing z"

Using your cross product you'll get a vector t for the tangent vector. But -t is also a tangent. Pick the one that has positive z component.
 
I want to get the gradient vector of x^2 + y^2− 2xz = 6,so z=(x^2+y^2-6)/2x
,is the gradient vector (dz/dx dz/dy) but i should have three dimensions?
 
If f=x^2+y^2-2xz-6=0 is your surface the gradient vector is (df/dx,df/dy,df/dz) (all partial derivatives). Of course you get three components.
 
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..
 
The normal vector points in the same direction as the gradient vector. Don't you calculate the normal vector using the gradient vector??
 
  • #10
AH, finally,i figured it out , thank a lot
 

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