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 alvielwj Oct13-08 03:13 PM

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

 Dick Oct13-08 03:24 PM

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?

 alvielwj Oct13-08 03:30 PM

Re: Directional derivatives

cross product of these two vectors

 alvielwj Oct13-08 03:31 PM

Re: Directional derivatives

But I cant get the meaning of "in the direction of increasing z"

 Dick Oct13-08 03:33 PM

Re: Directional derivatives

Quote:
 Quote by alvielwj (Post 1913923) But I cant 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.

 alvielwj Oct13-08 03:48 PM

Re: Directional derivatives

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?

 Dick Oct13-08 04:00 PM

Re: Directional derivatives

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.

 alvielwj Oct13-08 04:43 PM

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..

 Dick Oct13-08 04:48 PM

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??

 alvielwj Oct13-08 04:53 PM

Re: Directional derivatives

AH, finally,i figured it out , thank a lot

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