Maximum Slope of a 3D shape (not gradient at a point)

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Nick11234

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Apr 26, 2022
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Hi all, I'm stuck on this question.

My equation is f(x,y) = -1-x^2-6x-y^2-2y.
I'm required to find the number of points where the steepest slope of this shape occurs, where z is greater than or equal to 0, and where these points are.

My first thought was to find the where the second partial derivatives =0, but this gives me two constant values.
I then thought that as this graph expands out from a central maximum, the slope would occur at the lowest z, which in this case occurs at z=0, but i got stuck there.
 
To update, the answer was all x & y which satisfy the circular function at Z approaching 0.
 
Do you know how to compute the slope value in each point of a multi-variate function domain?
 
Yes, its the dot product of the gradient vector, and the unit gradient vector, at the point. If you are looking for max slope at a point, it always occurs in the direction of the gradient.
 
Nick11234 said:
Yes, its the dot product of the gradient vector, and the unit gradient vector, at the point. If you are looking for max slope at a point, it always occurs in the direction of the gradient.
Click to expand...
That's true. It's also equal to the norm of the gradient. I usually look for the maximum of the squared norm because it is easier to compute (no square roots please:). Can you figure them out ?
 
Hey mate, I'm unfamiliar with finding the norm of the gradient, I know the norm of a vector is its magnitude. If I take the magnitude of the gradient vector it does not give me the slope.
 
Nick11234 said:
Hey mate, I'm unfamiliar with finding the norm of the gradient, I know the norm of a vector is its magnitude. If I take the magnitude of the gradient vector it does not give me the slope.
Click to expand...
The magnitude of the gradient vector does give you the slope, and it is the same as the dot product of the gradient with its unit vector.
 
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