Unit Normal to a level surface

  • #1
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Homework Statement



Given $$\phi = x^{2} +y^{2}-z^{2}-1 $$

Calculate the unit normal to level surface φ = 0 at the point r = (0,1,0)

Homework Equations


  1. $$ \hat{\mathbf n} = \frac{∇\phi}{|\phi|}$$
  2. $$ z = \sqrt{x^{2}+y^{2} -1} $$
  3. $$ \mathbf n = (1,0,(\frac{\partial z}{\partial x})_{P}) \times (0,1,(\frac{\partial z}{\partial y})_{P}) $$

The Attempt at a Solution



Using equation (1) above I obtained:
$$\hat{\mathbf n} = \frac{1}{\sqrt{x^{2} + y^{2} +z^{2}}} (x\hat{\mathbf i}, y\hat{\mathbf j}, -z\hat{\mathbf k})$$

Evaluating this at (0,1,0) resulted in $$\hat{\mathbf n} = \hat{\mathbf j}$$

As far I can tell this is correct? I ran into a problem when I tried using equations (2) and (3) to check my answer;
I worked out the partial derivatives as follows:

$$\frac{\partial z}{\partial x} = \frac{x}{\sqrt{x^{2}+y^{2}-1}} $$
$$\frac{\partial z}{\partial y} = \frac{y}{\sqrt{x^{2}+y^{2}-1}} $$

Evaluating these partial derivatives at (0,1,0) gave:

$$(\frac{\partial z}{\partial x})_{P} = 0 $$

$$(\frac{\partial z}{\partial y})_{P} = \frac{1}{\sqrt{1^{2}-1}} = \frac{1}{0} $$

My problem is that I am getting an indeterminate value for the partial derivative w.r.t y evaluated at (0,1,0). I'm struggling to see where I have gone wrong, if anyone could help point out my mistake it would be greatly appreciated :)

I have also attached a wolfram alpha screenshot in relation to my second attempt's method, does this second method only work for some scalars?
 

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Answers and Replies

  • #2
LCKurtz
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It's best to use the formula ##\nabla \phi = \langle \phi_x,\phi_y, \phi_z \rangle##. Your answer is correct. The problem is probably that the surface has a vertical tangent plane at the point whose normal is in the y direction. z isn't a function of x and y at that point, but the surface still has a unit normal.
 
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  • #3
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It's best to use the formula ##\nabla \phi = \langle \phi_x,\phi_y, \phi_z \rangle##. Your answer is correct. The problem is probably that the surface has a vertical tangent plane at the point whose normal is in the y direction. z isn't a function of x and y at that point, but the surface still has a unit normal.
Ah okay thank you for that, I'll stick to the grad method
 

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