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

[bbelson01]

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

 

[I like Serena]

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.

 

[bbelson01]

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

 

[I like Serena]

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:
$$x^2−y+z^2=0$$