Here's the question,
Determine level set L(f,c) for the following
f(x,y,z)= x2-2x+y2+z2-6y+11 ,c=5
please help!!!!!
As we have three variables, the level set $\displaystyle L(f,c) = {(x,y,z}|f(x,y,z)=c$ is a level surface. It is given by the equation $\displaystyle x^2-2x+y^2-6y+z^2+11 = c= 5$
As this is quadratic in x, y, and z, the result is a quadric.
We have:
$\displaystyle x^2-2x+y^2-6y+z^2+11 = 5$
$\displaystyle x^2-2x+y^2-6y+z^2+6 = 0$
$\displaystyle x^2-2x+1+y^2-6y+9+z^2+6-1-9 = 0$ (completing the square for x and y)
$\displaystyle (x-1)^2+(y-3)^2+z^2-4 = 0$
$\displaystyle (x-1)^2+(y-3)^2+z^2 = 2^2$
The resulting geometric surface should be easy to see (hint: remember the distance formula).
--Kevin C.