Quadric surfaces

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We will get to know some basic quadric surfaces.

Our goal is to be able to identify local maximums and minimums of a surface. When we did this in our first calculus course, we had a “second derivative test” to help us out. In this section we lay the ground work for developing the “second derivative test” for functions of two variables.

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: $F^{(1,0)}(\vec {c}) = 0 = F^{(0,1)}(\vec {c})$ When these first partials are zero, 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 cross-sections of the surface.

A cross-section of a surface is the intersection of a surface with a plane.

Consider the following surface: $z = 6x^2 + y^2 + 5 xy$ Compute the cross-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 cross-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

Now that we have the basic tool of using a cross-section, we will explore our quadric surfaces.

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 cross-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.

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

Hyperbolic paraboloids

A hyperbolic paraboloid is a surface with graph:

and equation, after moving the vertex to the origin: $z=\pm \frac {x^2}{a^2}\mp \frac {y^2}{b^2}$ Here the symbols “ $\pm$ ” and “ $\mp$ ” just mean that both signs cannot be the same. To understand this surface better consider the cross-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:

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

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 cross-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 cross-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: $m = \frac {- F^{(1,1)}(\vec {c})\pm \sqrt {F^{(1,1)}(\vec {c})^2-F^{(2,0)}(\vec {c})F^{(0,2)}(\vec {c})}}{F^{(0,2)}(\vec {c})}$ 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 $F^{(2,0)}(\vec {c})F^{(0,2)}(\vec {c})-F^{(1,1)}(\vec {c})^2> 0.$ When $F^{(2,0)}(\vec {c})F^{(0,2)}(\vec {c})-F^{(1,1)}(\vec {c})^2 =0,$ the anaylsis above is insufficient to make any conclusions.

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: $\begin{align*} \pp[F]{x} &= \answer{3x^2-3}\\ \pp[F]{y} &= \answer{4-2y}\\ \frac{\partial^2F}{\partial x^2} &= \answer{6x}\\ \frac{\partial^2F}{\partial y^2} &= \answer{-2}\\ \frac{\partial^2F}{\partial x\partial y} &= \answer{0}\\ \end{align*}$

Now check: $\begin{align*} F^{(1,0)}(-1,2) &= \answer{0}\\ F^{(0,1)}(-1,2) &= \answer{0} \end{align*}$

Since the first derivatives are zero, we can now use our new second derivative test: $D(-1,2) = \answer {12}$

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 $(1,2)$ ? First check: $\begin{align*} F^{(1,0)}(1,2) &= \answer{0}\\ F^{(0,1)}(1,2) &= \answer{0} \end{align*}$

Since the first derivatives are zero, we can now use our new second derivative test: $D(1,2) = \answer {-12}$

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: $\begin{align*} \pp[F]{x} &= \answer{2x e^{x^2+y^2}}\\ \pp[F]{y} &= \answer{2y e^{x^2+y^2}}\\ \frac{\partial^2F}{\partial x^2} &= \answer{2 e^{x^2+y^2}+4x^2 e^{x^2+y^2}}\\ \frac{\partial^2F}{\partial y^2} &= \answer{2 e^{x^2+y^2}+4y^2 e^{x^2+y^2}}\\ \frac{\partial^2F}{\partial x\partial y} &= \answer{4 x y e^{x^2+y^2}}\\ \end{align*}$

Now check: $\begin{align*} F^{(1,0)}(0,0) &= \answer{0}\\ F^{(0,1)}(0,0) &= \answer{0} \end{align*}$

Since the first derivatives are zero, we can now use our new second derivative test: $D(0,0) = \answer {4}$

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: $\begin{align*} \pp[F]{x} &= \answer{y e^{-x y}-xy^2e^{-x y}}\\ \pp[F]{y} &= \answer{x e^{-x y}-x^2ye^{-x y}}\\ \frac{\partial^2F}{\partial x^2} &= \answer{xy^3e^{-xy}-2y^2e^{-xy}}\\ \frac{\partial^2F}{\partial y^2} &= \answer{x^3ye^{-xy}-2x^2e^{-xy}}\\ \frac{\partial^2F}{\partial x\partial y} &= \answer{x^2 y^2 e^{-x y}-3 x y e^{-x y}+e^{-x y}}\\ \end{align*}$

Now check: $\begin{align*} F^{(1,0)}(0,0) &= \answer{0}\\ F^{(0,1)}(0,0) &= \answer{0} \end{align*}$

Since the first derivatives are zero, we can now use our new second derivative test: $D(0,0) = \answer {-1}$

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