Maximizing Gradient for Steep Climb on Hill Surface

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Homework Help Overview

The problem involves determining the direction of steepest ascent on a hill defined by the surface equation z=100exp((-x^2+3y^2)/701) at a specific point (30, 20, 5). Participants are discussing the computation of the gradient vector to find this direction.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are considering how to compute the gradient vector, specifically whether to include partial derivatives with respect to z, x, and y. There is discussion about the implications of z being a function of x and y and the correct interpretation of the problem's requirements.

Discussion Status

The discussion is ongoing, with participants clarifying the need to compute the gradient using partial derivatives of z with respect to x and y. There is some confusion regarding the terminology and the interpretation of the problem, but guidance has been offered on focusing on the gradient of z.

Contextual Notes

There is a mention of the need to clarify what is meant by "climb most steeply," indicating that participants are exploring different interpretations of the problem's requirements.

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?
 

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