• kasse
In summary, the conversation discusses how to determine the direction in which a person should proceed in order to climb most steeply on a hill with a given surface. The gradient vector is mentioned as a way to calculate the direction, but there is confusion about which variables to take partial derivatives with respect to. Eventually, it is clarified that the gradient of z should be used.

#### kasse

"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?

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?