Directional derivative of a surface

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".
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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