Functions of Several Variables — Question 5

Question 5 – Solution

Strategy Substitute first and expand carefully.

Step 1: Simplify Since $$(u+v)^2+(u-v)^2=2u^2+2v^2,$$ we get $\boxed{g(u,v)=\sqrt{1-2u^2-2v^2}}$.

Step 2: Domain The radicand must be nonnegative: $$1-2u^2-2v^2\ge 0\Longrightarrow\boxed{u^2+v^2\le\frac 12}.$$

Step 3: Levels For $0\le c\le 1$, $g=c$ implies $$u^2+v^2=\frac{1-c^2}{2},$$ circles centered at the origin. The map $(u,v)\mapsto(x,y)=(u+v,u-v)$ reflects $v$ to $-v$, then rotates by $45^\circ$ and scales lengths by $\sqrt 2$. Its inverse image of the unit disk is the disk of radius $1/\sqrt 2$. At $c=1$ the level is just the origin; levels outside $[0,1]$ are empty.