Why does the equation of the level surface not have ln?

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Discussion Overview

The discussion revolves around finding an equation of the level surface for the function g(x,y,z)=ln(x^2−y+z^2) at the point (−1,2,1). Participants explore the implications of the function being undefined at that point and how to derive an equation that does not involve the logarithm.

Discussion Character

  • Exploratory, Technical explanation, Conceptual clarification

Main Points Raised

  • One participant, bbelson01, questions the process of finding the level surface since g(−1,2,1) results in ln(0), which is undefined.
  • Another participant suggests that the level found is "undefined" and proposes that the equation x^2−y+z^2 = 0 is a valid representation that does not involve ln.
  • In response, bbelson01 seeks clarification on how to transition from ln(x^2−y+z^2) to the equation x^2−y+z^2=0, questioning the elimination of ln.
  • A later reply explains that to find the surface corresponding to a level c, one can express it as x^2−y+z^2=e^c and notes that as c approaches -∞, the equation simplifies to x^2−y+z^2=0.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the method of eliminating ln from the equation, as bbelson01 seeks further clarification while another participant provides an explanation. The discussion remains unresolved regarding the clarity of the transition between the logarithmic and non-logarithmic forms.

Contextual Notes

The discussion highlights the undefined nature of the logarithmic function at certain points and the implications for deriving level surfaces. There are assumptions about the behavior of the function as c approaches -∞ that are not fully explored.

bbelson01
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Could someone please help me with the following question:

Consider the function g(x,y,z)=ln(x^2−y+z^2). Find an equation of the level surface of the function through the point (−1,2,1) which does not have ln. First find g(−1,2,1).

When I sub in the points I get:
g(−1,2,1)=ln(1−2+1)=ln(0)=undefined.

Where am I going wrong?

Thanks, bbelson01
 
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bbelson01 said:
Could someone please help me with the following question:

Consider the function g(x,y,z)=ln(x^2−y+z^2). Find an equation of the level surface of the function through the point (−1,2,1) which does not have ln. First find g(−1,2,1).

When I sub in the points I get:
g(−1,2,1)=ln(1−2+1)=ln(0)=undefined.

Where am I going wrong?

Thanks, bbelson01

Hi bbelson01! Welcome to MHB! :)

I don't thank that you've gone wrong.
You've found the level, which is "undefined".
To get other points, you need that:
$$x^2−y+z^2 = 0$$
Hey! That is an equation that does not have $\ln$.
That piece of information actually seems to be a hint.
 
Thanks for replying Selena.

Could you please step me through how you go from ln(x^2−y+z^2) to x^2−y+z^2=0? How can you just eliminate ln?

Thanks,
bbelson01
 
bbelson01 said:
Thanks for replying Selena.

Could you please step me through how you go from ln(x^2−y+z^2) to x^2−y+z^2=0? How can you just eliminate ln?

Thanks,
bbelson01

Let's take a look at our surface. It's:
$$g(x,y,z)=c \quad\Rightarrow\quad \ln(x^2−y+z^2)=c \quad\Rightarrow\quad x^2−y+z^2=e^c$$

To get to the surface that contains $(−1,2,1)$ we need to let $c$ approach $-\infty$.
We reach that surface in the limit for $c \to -\infty$.
If we do, we're left with:
$$x^2−y+z^2=0$$
 

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