8.05a–8.05c - Surfaces, sections and contours

Equations and surfaces

Could a flat map tell you whether you are looking at a bowl or a saddle? The surfaces $z=x^2+y^2$ and $z=x^2-y^2$ both pass through the origin and both contain the parabola $z=x^2$ when $y=0$. One view is not enough. Sections and contours will reveal the difference.

For $z=x^2+y^2$, choose two coordinates and calculate the third: $(1,2,5)$ lies on the surface because $1^2+2^2=5$. The point $(1,2,6)$ does not. Letting $x$ and $y$ vary produces the whole surface, rather than just one curve.

The sphere $x^2+y^2+z^2=9$ contains both $(0,0,3)$ and $(0,0,-3)$: the same horizontal position has two heights. Solving for $z$ gives two pieces, $z=\sqrt{9-x^2-y^2}$ and $z=-\sqrt{9-x^2-y^2}$. By contrast, $x^2+y^2-z=0$ rearranges to the single explicit surface $z=x^2+y^2$. The equation $x^2+y^2=4$ is another implicit surface: a circular cylinder, because $z$ is unrestricted.

[DIAGRAM: asset_name: A bowl, a saddle and a sphere; asset_slug: h245_8_05_surfaces; caption: A bowl has one height above each horizontal position; a sphere can have two. The saddle rises in one coordinate direction and falls in the other.; description: Three exact three-dimensional plots. The left plot is z=x squared+y squared, the middle plot is z=x squared-y squared, and the right plot is the sphere x squared+y squared+z squared=9. Coordinate axes are labelled x, y and z. The bowl and saddle include highlighted coordinate sections. The sphere marks its north and south poles at heights 3 and -3.]

What a partial derivative measures

Imagine walking along the surface while keeping $y$ fixed. You follow a curve in a vertical plane. Its gradient measures how the height changes as $x$ changes; this is the partial derivative with respect to $x$. Holding $x$ fixed instead gives the partial derivative with respect to $y$.

At $(1,2,5)$ on the bowl, the section with $y=2$ is $z=x^2+4$, so its gradient at $x=1$ is $2$. The section with $x=1$ is $z=1+y^2$, whose gradient at $y=2$ is $4$. These are two different slopes at the same point. The next lesson develops their notation and calculation.

Some equations simply do not define a point at a proposed position. For $z=x/y$, setting $y=0$ is impossible: even when $x=0$, the original expression is undefined. This is an algebraic restriction, rather than a missing line that you should fill in. The explicit and implicit descriptions, and this interpretation of partial derivatives, are not printed in the formula booklet.

The sphere provides a useful test of the difference between an equation in three variables and a function of two.

Vertical sections

Cutting a surface with a vertical coordinate plane turns the equation into an ordinary function of one variable.

For example, fixing $x=2$ in $z=x^2+y^2$ gives $z=4+y^2$. Sketch it on axes labelled $y$ and $z$, because $y$ varies and $x$ does not. This curve sits in the plane $x=2$; it is not in the plane $x=0$.

Substitution is the whole algebraic method. The geometry comes from naming the correct axes and identifying the remaining curve's intercepts, turning points and shape. These section rules are not in the formula booklet.

The next example changes only which coordinate is fixed. The same surface now produces a parabola rather than a line.

These two sketches are different cuts through the same object.

[DIAGRAM: asset_name: Two coordinate sections through one surface; asset_slug: h245_8_05_sections; caption: Fixing x gives a straight section; fixing y gives a parabolic section of the same surface.; description: Two side-by-side exact coordinate plots for z=2x squared+2xy. The left has horizontal axis y and vertical axis z, the line z=2+2y, and labelled intercepts (-1,0) and (0,2). The right has horizontal axis x and vertical axis z, the curve z=2x squared+2x, vertex (-1/2,-1/2), and intercepts (-1,0) and (0,0). The titles identify the cutting planes x=1 and y=1.]

For a sketch, show the axes and label the features you calculated; a correct equation alone is not a sketch.

Horizontal contours

A contour joins positions with the same height, so its equation comes from fixing $z$ rather than $x$ or $y$.

For the bowl $z=x^2+y^2$, the horizontal cut $z=4$ is a circle of radius $2$ at height $4$. Its contour map shows the circle $x^2+y^2=4$ on the $xy$-plane and labels it with the height $4$. The map has flattened the location of the circle, while retaining its height as a label.