Directional derivative of a surface

  • Thread starter Poley
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  • #1
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Homework Statement


What is the directional derivative of the function z = x3 - y at the point (1, 2, -1) and in the direction of a vector (1,1,1)?


Homework Equations





The Attempt at a Solution


If f(x,y) = x3 - y, then ∇f = (3x2, -1) which equals (3, -1) at the given point. Now I understand I have to take the dot product of the gradient with the unit vector (1/√3, 1/√3, 1/√3) but I'm not quite sure how to...

Can a function like z = f(x,y) have a directional derivative in the direction of a three dimensional vector?

Thanks for any help!
 

Answers and Replies

  • #2
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You are right, you cannot take the directional derivative of f(x,y) in the direction of a vector in three dimensions (unless the z component is zero, in which case you sort of can).
 
  • #3
HallsofIvy
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The surface [itex]z= x^3- y[/itex] is given by [itex]x\vec{i}+ y\vec{j}+ (x^3- y)\vec{k}[/itex] the gradient of that vector gives the normal vector and you can take the dot product of that with a unit vector in the direction of [itex]\vec{i}+ \vec{j}+ \vec{k}[/itex]- but I would NOT call that a "directional derivative".
 

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