Q 55 - 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.

$f\left(x,y,z\right)=\frac{x}{y-z}$.

Step-by-Step Solution

Answer

The level surfaces are planes with the equation $x-cy+cz=0; c=-3,-2,-1,0,1,2,3$ consisting all points on the plane except those points for $y=z$.

Step 1. Given Information

We have given the following function :-

$f\left(x,y,z\right)=\frac{x}{y-z}$.

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 :-

$f\left(x,y,z\right)=\frac{x}{y-z}$.

We know that the level surfaces are determined as :-

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

That is we have :-

$$\begin{gathered} \frac{x}{y-z}=c \\ \Rightarrow x=cy-cz \\ \Rightarrow x-cy+cz=0 \end{gathered}$$

We know that this is the equation of plane for any value of  $c$. It consists all points on the plane except those for $y=z$.

So that we can conclude that the level surfaces are the planes with equation $x-cy+cz=0$ for $c=-3,-2,-1,0,1,2,3$ consisting all points except those for $y=z$.