Maximizing Gradient for Steep Climb on Hill Surface

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To maximize the gradient for a steep climb on the hill surface defined by z=100exp((-x^2+3y^2)/701), one must compute the gradient vector using the partial derivatives of z with respect to x and y. The discussion clarifies that z is a function of x and y, so only these variables should be considered for the gradient calculation. There is confusion about the relevance of partial derivatives of x and y, which are unnecessary in this context. The gradient vector points in the direction of the steepest ascent, confirming the need to compute the gradient of z. Understanding these concepts is crucial for determining the optimal climbing direction.
kasse
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"You are standing at the point (30, 20, 5) on a hill with the shape of the surface z=100exp((-x^2+3y^2)/701). In what direction should you proceed in order to climb most steeply?"

SInce the grad vector allegedly points in the most steep direction of the surface, I guess I'll have to compute that one. But I'm not sure if I'm suppoesd to compute the partials of all of x, y and z or only x and y in the gradient. How can I know that?
 
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Edit: z is a function of x and y.
 
the thing is, it depends on what kind of increase you want. obviously from the question, it implies the direction that z ("climb most steeply") increases most rapidly. so, take the derivatives with respect to z.
 
No, it doesn't. You are given that z is a function of x and y so take the gradient of z using the partial derivatives of z with respcect to x and y. (That may be what tim lou meant to say.)

It would make no sense to talk about taking partial derivatives of x and y- with respect to what other variables?

SInce the grad vector allegedly points in the most steep direction of the surface, I guess I'll have to compute that one.
Why "allegedly"? And you surely, by "that one" mean the gradient of z don't you? So why was there any question?
 
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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