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Gradient Vector

  • Thread starter Air
  • Start date
  • #1
Air
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Homework Statement


Calculate the gradient vector at the point [tex]S[/tex] for the function, [tex]f(x,y,z)=x-\sqrt{z^2 - y^2}; S(x,y,z)=(4, 8, -6)[/tex].


2. The attempt at a solution
[tex]\frac{\partial f}{\partial x} = 1[/tex]
[tex]\frac{\partial f}{\partial y} = \frac{y}{\sqrt{z^2-y^2}}[/tex]
[tex]\frac{\partial f}{\partial z} = -\frac{z}{\sqrt{z^2-y^2}}[/tex]
Therefore at Point [tex]S[/tex]:
[tex]\frac{\partial f}{\partial x} = 1[/tex]
[tex]\frac{\partial f}{\partial y} = \frac{-6}{\sqrt{36-64}} = \frac{-6}{\sqrt{-28}}[/tex]


3. The problem that I'm having
When I put the point values of Point [tex]S[/tex] into the equation, I get a negative square root in the denominator. Where am I going wrong? Thanks in advance.
 

Answers and Replies

  • #2
Matterwave
Science Advisor
Gold Member
3,965
326
Indeed, this is weird. Have you double checked you copied the problem down right?

We can see that the scalar field becomes imaginary at that point as well (you will also get a negative sign under the square root).
 

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