What is the nature of the surface at the point of partial derivative equality?

KingBigness
Messages
94
Reaction score
0

Homework Statement



Let f(x,y)=1−x^{2}−y^{2}. Find the point at which

\frac{\partial f}{\partial x} = \frac{\partial f}{\partial y} = 0
and illustrate graphically the nature of the surface z = f (x, y) at this point.

The Attempt at a Solution



Just did the partial derivatives and got 2x=2y=0 hence x=0 and y=0

So illustrating z=f(x,y) at this point give z=1

Creating a contour plot I get the plot attached below.

This question is part of a difficult assignment I got from uni but this question seems way to easy in comparison to the rest of the assignment. I just want to make sure there isn't something I missed or is it really this simple?

Thank you
 

Attachments

  • MSP241419hb99bd9ig0gh4i00004aa457092c3a6ee2.gif
    MSP241419hb99bd9ig0gh4i00004aa457092c3a6ee2.gif
    8.7 KB · Views: 401
Last edited:
Physics news on Phys.org
Any help?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top