Directional derivative of a surface

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SUMMARY

The discussion centers on calculating the directional derivative of the function z = x³ - y at the point (1, 2, -1) in the direction of the vector (1, 1, 1). The gradient ∇f is determined to be (3x², -1), which evaluates to (3, -1) at the specified point. It is clarified that while the directional derivative can be computed using the dot product of the gradient and a unit vector, it is not valid to take the directional derivative of a function defined in two dimensions (f(x,y)) in the direction of a three-dimensional vector unless the z component is zero.

PREREQUISITES
  • Understanding of gradient vectors and their computation
  • Knowledge of directional derivatives in multivariable calculus
  • Familiarity with unit vectors and dot product operations
  • Basic concepts of surfaces represented by functions of two variables
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  • Explore the implications of surfaces defined by functions of two variables
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Poley
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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!
 
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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).
 
The surface z= x^3- y is given by x\vec{i}+ y\vec{j}+ (x^3- y)\vec{k} 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 \vec{i}+ \vec{j}+ \vec{k}- but I would NOT call that a "directional derivative".
 

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