Q 57 - Calculus

Question

In Exercises $53-58$, determine the level surfaces $c=-3,-2,-1,0,1,2,3$ if they exist for the specified function.

$f\left(x,y,z\right)=x^2+y^2+z^2$.

Step-by-Step Solution

Answer

The level surfaces are determined by the equation $x^2+y^2+z^2=c$. For $c=0$, $x^2+y^2+z^2=0$ represents the origin of 3-D plane.

Also the level surfaces are sphere with equation :-

$x^2+y^2+z^2=c; c=1,2,3$.

For $c=-3,-2,-1$ level surfaces are undefined.

Step 1. Given Information

We have given the following function :-

$f\left(x,y,z\right)=x^2+y^2+z^2$.

We have to determine level surfaces $c=-3,-2,-1,0,1,2,3$ for the given function. 

Step 2. Determine level surfaces

The given function is :-

$f\left(x,y,z\right)=x^2+y^2+z^2$.

We know that the level surfaces are determined as :-

$f\left(x,y,z\right)=c$.

That is we have :-

$x^2+y^2+z^2=c$.

This is the equation of sphere.

So the level surfaces are undefined for $c=-3,-2,-1$ as radius of sphere cannot be negative.

Also for $c=0$, $x^2+y^2+z^2=0$ represents the origin of 3-D plane.

So that we have :-

The level surfaces are sphere with the equation $x^2+y^2+z^2=c$ for $c=1,2,3$.

For $c=0$, level surface is origin.

For $c=-3,-2,-1$ the level surfaces are undefined.