Quadric surfaces

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We will get to know some basic quadric surfaces.

As we have seen, if we look at the set of points that satisfy an equation $F(x,y,z)=0$ where $F:\R ^3\to \R$ , we obtain a surface in $\R ^3$ . A basic class of surfaces are the quadric surfaces.

A quadric surface in $\R ^3$ is a surface of the form $Ax^2 + By^2 + Cz^2 + Dxy + Exz+ Fyz + Gx + Hy + I z + J = 0$ where $A$ , $B$ , $C$ , $D$ , $E$ , $F$ , $G$ , $H$ , $I$ , and $J$ are constants and at least one of $A$ , $B$ , $C$ , $D$ , $E$ , or $F$ are nonzero.

Which of the following are quadric surfaces?

$x^2 = 0$ $y=0$ $z=y^2$

Do not confuse a quadric with a quadratic, or quartic, as these are different beasts entirely.

We will be interested in a special class of quadric surfaces, those that arise naturally when computing the Taylor polynomial of a surface $z=F(x,y)$ at a point $\vec {c}$ where: $F^{(1,0)}(\vec {c}) = 0 = F^{(0,1)}(\vec {c})$ In this case, the quadric is of the form:

Understanding quadric surfaces will help us find extrema of surfaces.

In what follows, we will study each shape by considering various sections of the surface.

A section of a surface is the intersections of a surface with a plane.

Consider the following surface: $z = 6x^2 + y^2 + 5 xy$ Compute the section of the surface given by the plane $y=0$ . $z = \answer {6x^2}$

Does this parabola open “up” or “down?”

up down

Compute the section of the surface given by the plane $y=-2.5 x$ . $z = \answer {6x^2 + 6.25 x^2 - 12.5x^2}$

Does this parabola open “up” or “down?”

up down

Elliptic paraboloids

An elliptic paraboloid is a surface with graph:

and equation, after moving the vertex to the origin: $z=\pm \frac {x^2}{a^2}\pm \frac {y^2}{b^2}$ To understand this surface better consider the sections when:

$x=d$ , in this case we now have $z = \pm \frac {d^2}{a^2} \pm \frac {y^2}{b^2}$ , a parabola.
$y=d$ , in this case we now have $z = \pm \frac {x^2}{a^2} \pm \frac {d^2}{b^2}$ , a parabola.
$z=d$ , in this case we now have $d = \pm \frac {x^2}{a^2} \pm \frac {y^2}{b^2}$ , an ellipse.

$\begin {array}{cc} \textbf {Plane} & \textbf {Section} \\ \hline x=d & \text {Parabola} \\ y=d & \text {Parabola}\\ z=d & \text {Ellipse} \end {array}$

Hyperbolic paraboloids

A hyperbolic paraboloid is a surface with graph:

and equation, after moving the vertex to the origin: $z=\pm \frac {x^2}{a^2}\mp \frac {y^2}{b^2}$ To understand this surface better consider the sections when:

$x=d$ , in this case we now have $z = \pm \frac {d^2}{a^2} \mp \frac {y^2}{b^2}$ , a parabola.
$y=d$ , in this case we now have $z = \pm \frac {x^2}{a^2} \mp \frac {d^2}{b^2}$ , a parabola that opens the opposite direction as the previous one.
$z=d$ , in this case we now have $d = \pm \frac {x^2}{a^2} \mp \frac {y^2}{b^2}$ , a hyperbola.

We’ll give an additional graph to show the hyperbolas:

$\begin {array}{cc} \textbf {Plane} & \textbf {Section} \\ \hline x=d & \text {Parabola}\\ y=d & \text {Parabola}\\ z=d & \text {Hyperbola} \end {array}$

Identifying quadric surfaces

Let’s start by working a specific example.

Consider the surface: $z = x^2+4y^2-5xy$ Is this surface an elliptic paraboloid or a hyperbolic paraboloid?

We’ll work somewhat naively. Consider the plane $y= mx$ . This plane is perpendicular to the $(x,y)$ -plane. If we intersect this plane with the surface above, we will find a parabola. If we can change the direction that the parabola opens by varying $m$ , then our surface is a hyperbolic paraboloid. If we cannot change the direction that the parabola opens by varying $m$ , then our surface is an elliptic paraboloid. Start by setting $m = 0$ . In this case we find: $z = \answer [given]{x^2}$ This is a parabola that opens “up” in the $(x,z)$ -plane. Can we find a parabola that opens “down” by varying $m$ ? Intersecting the surface $z = x^2 + 4y^2 -5xy$ and $y=mx$ we find: $\begin{align*} z &= \answer[given]{x^2 + 4 m^2 x^2 -5 x^2 m}\\ &=x^2\left(\answer[given]{4 m^2-5m + 1}\right) \end{align*}$ This parabola will open “downward” when we can find $m$ such that $4 m^2-5m + 1$ is negative. The expression $4 m^2-5m + 1$ is zero when $\begin{align*} m &= 1/4\\ m &= \answer[given]{1}. \end{align*}$ Let’s draw a sign-chart:

Since the intersection of $z=x^2+4y^2-5xy$ with the plane $y=0$ is a parabola that opens “up” in the $(x,z)$ -plane, and the intersection of the surface with the plane $y = x/2$ (or $y=mx$ where $m$ is between $1/4$ and $1$ ) is a parabola that opens “down” in the $(x,z)$ -plane, we have a hyperbolic paraboloid. For your viewing pleasure, we’ve included a graph of the hyperbolic paraboloid and the plane:

Later in this course, we will be looking at quadric surfaces of the form

$\begin{align*} z = &\left(\frac{1}{2}\right)F^{(2,0)}(\vec{c})(x-c_1)^2\\ &+ \left(\frac{1}{2}\right)F^{(0,2)}(\vec{c})(y-c_2)^2 \\ &+ F^{(1,1)}(\vec{c}) (x-c_1)(y-c_2)\\ &+ F(\vec{c}). \end{align*}$ and trying to identify them as either elliptic paraboloids, or as hyperbolic paraboloids. In what follows, let $\sign (x) = \begin {cases} 1 &\text {if $x>0$,}\\ -1 &\text {if $x<0$,}\\ 0 &\text {if $x=0$.} \end {cases}$ This will aid in our analysis of the quadric surfaces.

The pure partials have opposite signs

If $\sign (F^{(2,0)}(\vec {c})) = -\sign (F^{(0,2)}(\vec {c})),$ then we can examine the following sections: $y= c_2 \quad \text {and}\quad x = c_1$ If $y=c_2$ then the surface

$\begin{align*} z = &\left(\frac{1}{2}\right)F^{(2,0)}(\vec{c})(x-c_1)^2\\ &+ \left(\frac{1}{2}\right)F^{(0,2)}(\vec{c})(y-c_2)^2 \\ &+ F^{(1,1)}(\vec{c}) (x-c_1)(y-c_2)\\ &+ F(\vec{c}). \end{align*}$ becomes $z = \left (\frac {1}{2}\right )F^{(2,0)}(\vec {c})(x-c_1)^2 + F(\vec {c}),$ and this is a parabola that opens in the $z$ -direction of the sign of $F^{(2,0)}(\vec {c})$ .

If $x=c_1$ then the surface becomes $z = \left (\frac {1}{2}\right )F^{(0,2)}(\vec {c})(y-c_2)^2 + F(\vec {c}),$ and this is a parabola that opens in the $z$ -direction of the sign of $F^{(0,2)}(\vec {c})$ . Since $\sign (F^{(2,0)}(\vec {c})) = -\sign (F^{(0,2)}(\vec {c})),$ we see that when the pure partials have opposite signs, then the quadric surface is a hyperbolic paraboloid.

The pure partials have the same sign

If $\sign (F^{(2,0)}(\vec {c})) = \sign (F^{(0,2)}(\vec {c})),$ then we start by examining the section: $y -c_2 = m (x-c_1)$ Substituting this into the surface above, we find

$\begin{align*} z = &\left(\frac{1}{2}\right)F^{(2,0)}(\vec{c})(x-c_1)^2\\ &+ m^2 \left(\frac{1}{2}\right)F^{(0,2)}(\vec{c})(x-c_1)^2\\ &+ m F^{(1,1)}(\vec{c}) (x-c_1)^2 \\ &+ F(\vec{c}). \end{align*}$ Factoring and rearranging, set $A = m^2 \left (\frac {1}{2}\right )F^{(0,2)}(\vec {c}) + m F^{(1,1)}(\vec {c}) + \left (\frac {1}{2}\right )F^{(2,0)}(\vec {c})$ and now $\begin{align*} z = (x-c_1)^2 A + F(\vec{c}). \end{align*}$ This is a parabola that opens in the $z$ -direction of the sign of $F^{(2,0)}(\vec {c})$ when $A = m^2 \left (\frac {1}{2}\right )F^{(0,2)}(\vec {c}) + m F^{(1,1)}(\vec {c}) + \left (\frac {1}{2}\right )F^{(2,0)}(\vec {c})$ has the same sign as $F^{(2,0)}(\vec {c})$ . The parabola opens in the opposite direction, when $A = m^2 \left (\frac {1}{2}\right )F^{(0,2)}(\vec {c}) + m F^{(1,1)}(\vec {c}) + \left (\frac {1}{2}\right )F^{(2,0)}(\vec {c})$ has the opposite sign as $F^{(2,0)}(\vec {c})$ . We can find a $m$ that produces this opposite sign when the quadratic equation, in the variable $m$ , $m^2 \left (\frac {1}{2}\right )F^{(0,2)}(\vec {c}) + m F^{(1,1)}(\vec {c}) + \left (\frac {1}{2}\right )F^{(2,0)}(\vec {c}) = 0$ has two real solutions. Let’s investigate using the quadratic formula: $m = \frac {- F^{(1,1)}\pm (\vec {c})\sqrt {F^{(1,1)}(\vec {c})^2-F^{(2,0)}(\vec {c})F^{(0,2)}(\vec {c})}}{F^{(0,2)}(\vec {c})}$ We see that there are two real solutions for $m$ when $F^{(1,1)}(\vec {c})^2-F^{(2,0)}(\vec {c})F^{(0,2)}(\vec {c})>0$ or equivalently when $F^{(2,0)}(\vec {c})F^{(0,2)}(\vec {c})-F^{(1,1)}(\vec {c})^2<0.$ So, we see that when the pure partials have the same sign, the quadric surface is a hyperbolic paraboloid when $F^{(2,0)}(\vec {c})F^{(0,2)}(\vec {c})-F^{(1,1)}(\vec {c})^2<0,$ and an elliptic paraboloid when $F^{(2,0)}(\vec {c})F^{(0,2)}(\vec {c})-F^{(1,1)}(\vec {c})^2>0.$

The second derivative test

Given a function $F:\R ^2\to \R$ , and a point $\vec {c}$ where $F^{(1,0)}(\vec {c}) = 0 = F^{(0,1)}(\vec {c}),$ our work above allows us to identify what a surface looks like locally. Specifically we get what is known as the second derivative test:

Second derivative test Given a function $F:\R ^2\to \R$ , and a point $\vec {c}$ where $F^{(1,0)}(\vec {c}) = 0 = F^{(0,1)}(\vec {c})$ set $D(\vec {c}) = F^{(2,0)}(\vec {c})F^{(0,2)}(\vec {c})-F^{(1,1)}(\vec {c})^2.$

If $D(\vec {c})>0$ , then $F$ locally looks like an elliptic paraboloid.
If $D(\vec {c})<0$ , then $F$ locally looks like a hyperbolic paraboloid.
If $D(\vec {c})=0$ , the test is inconclusive.

Try your hand at identifying local behavior of a surface.

Consider $F(x,y) = x^3-3x-y^2+4y$ . Does this surface locally look like an elliptic paraboloid or a hyperbolic paraboloid at the point $(-1,2)$ ? Compute: $\begin{align*} \pp[F]{x} &= \answer{3x^2-3}\\ \pp[F]{y} &= \answer{4-2y}\\ \frac{\partial^2F}{\partial x^2} &= \answer{6x}\\ \frac{\partial^2F}{\partial y^2} &= \answer{-2}\\ \frac{\partial^2F}{\partial x\partial y} &= \answer{0}\\ \end{align*}$

$D(-1,2) = \answer {12}$

An elliptic paraboloid. A hyperbolic paraboloid. We cannot tell.

Again consider $F(x,y) = x^3-3x-y^2+4y$ . Does this surface locally look like an elliptic paraboloid or a hyperbolic paraboloid at the point $(1,2)$ ? Compute: $D(1,2) = \answer {-12}$

An elliptic paraboloid. A hyperbolic paraboloid. We cannot tell.

Consider $F(x,y) = e^{x^2+y^2}$ . Does this surface locally look like an elliptic paraboloid or a hyperbolic paraboloid at the point $(0,0)$ ? Compute: $\begin{align*} \pp[F]{x} &= \answer{2x e^{x^2+y^2}}\\ \pp[F]{y} &= \answer{2y e^{x^2+y^2}}\\ \frac{\partial^2F}{\partial x^2} &= \answer{2 e^{x^2+y^2}+4x^2 e^{x^2+y^2}}\\ \frac{\partial^2F}{\partial y^2} &= \answer{2 e^{x^2+y^2}+4y^2 e^{x^2+y^2}}\\ \frac{\partial^2F}{\partial x\partial y} &= \answer{4 x y e^{x^2+y^2}}\\ \end{align*}$

$D(0,0) = \answer {4}$

An elliptic paraboloid. A hyperbolic paraboloid. We cannot tell.

Consider $F(x,y) = x y e^{-xy}$ . Does this surface locally look like an elliptic paraboloid or a hyperbolic paraboloid at the point $(0,0)$ ? Compute: $\begin{align*} \pp[F]{x} &= \answer{y e^{-x y}-xy^2e^{-x y}}\\ \pp[F]{y} &= \answer{x e^{-x y}-x^2ye^{-x y}}\\ \frac{\partial^2F}{\partial x^2} &= \answer{xy^3e^{-xy}-2y^2e^{-xy}}\\ \frac{\partial^2F}{\partial y^2} &= \answer{x^3ye^{-xy}-2x^2e^{-xy}}\\ \frac{\partial^2F}{\partial x\partial y} &= \answer{x^2 y^2 e^{-x y}-3 x y e^{-x y}+e^{-x y}}\\ \end{align*}$

$D(0,0) = \answer {-1}$

An elliptic paraboloid. A hyperbolic paraboloid. We cannot tell.