Slope of the tangent line of an intersection - Directional Derivatives

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To find the slope of the tangent line at the intersection of the vertical plane x - y + 1 = 0 and the surface z = x² + y² at the point (1, 2, 5), the gradients of both functions must be calculated. The cross product of these gradients will yield a vector normal to the surface at the intersection. To determine the slope, the horizontal direction of the plane should be analyzed, which can then be used to compute the directional derivative. The discussion highlights confusion regarding the steps needed to solve the problem effectively. Understanding the relationship between the gradients and the directional derivative is crucial for finding the correct slope.
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...
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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