Slope of the tangent line of an intersection - Directional Derivatives

RaoulDuke
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Homework Statement


Find the slope of the tangent line to the curve of intersection of the vertical plane x - y + 1 =0 and the surface z = x2+y2 at the point (1, 2, 5)


Homework Equations


Gradients, Cross products


The Attempt at a Solution



I'm pretty lost here. I think I have to cross the two gradients of the functions that I have to find the intersection at (1, 2, 5). However, what then?
 
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how about finding the horizontal direction of the plane, then using it to find the 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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