## Problem on surfaces and level curves

Sketch the three-dimensional surface and level curves of $z = y^2 + x$.
• ## Solution

#### Sketch of surface in perspective

Recall that
Because $z$'s dependence on $x$ is simpler than its dependence on $y$, we plot cross-sections with three values: $x=0$, $x$ negative, and $x$ positive.
Notice that the minimal value of each parabola is $z = x$. For visual clarity, the figure is not drawn to scale.
We observe that the surface is like a parabola in $(y,z)$ that slopes along the $x$ axis. This shape is similar to a half-pipe on a ski slope. Adding shading and lines of constant $y$:

#### Sketch of level curves

Recall that
We will sketch level curves corresponding to a couples values, such as $0, 1, -1$.
The $z=0$ level set is given by $y^2 +x= 0$, or $x = -y^2$. This is a parabola in $x$ as a function of $y$.
Now we add the $z=-1$ and $z=1$ level sets. These are given by the equations $y^2 +x=-1$ and $y^2 +x=1$, which are parabolas:
Additional level curves look similar to these.