Math 344: Calculus III

12.6 Cylinders and Quadratic Surfaces

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12.6.1 Implicit Functions and Curves

In two dimensional space all curves are plane curve; that is, they all lie in the same plane $\mathbb{R}^2$. We describe then algebraically using an equation. If we can describe one dependent variable as a function of an independent variable


$$ y = f(x) \def\ihat{\mathbf{\hat{\mmlToken{mi}[mathvariant="bold"]{ı}}}} \def\jhat{\mathbf{\hat{\mmlToken{mi}[mathvariant="bold"]{ȷ}}}} \def\khat{\mathbf{\hat{k}}} $$
then we say that the curve is the graph of the function $f$. In this case the curve is the locus of points

$$ \left\{ (x, y)\,:\,y=f(x)\right\} $$
We can parameterize the curve so that both $x$ and $y$ are dependent variables, and describe both variables as a function of a common independent variable. For example if

$$ \begin{align*} x &= x(t) \\ y &= y(t) \end{align*} $$
then the we have a function with $\mathbf{r}:\mathbb{R}\rightarrow\mathbb{R}^2$ with one input and two outputs that are points in $\mathbb{R}^2$

$$ \mathbf{r}(t) = \left( x(t), y(t) \right), $$
or we can think of the function $\mathbf{r}$ with one input and a vector output

$$ \mathbf{r}(t) = \langle x(t), y(t) \rangle = \begin{bmatrix} x(t) \\ y(t) \end{bmatrix} $$
We also describe a curve as a graph of an implicit function such as the unit circle

$$ x^2 + y^2 = 1 $$
Higher degree polynomials, exponential functions and transcendental functions may also be used to describe curves in the plane.

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12.6.2 Quadratic Curves

Examples of quadratic equations whose graphs are curves on the two dimensional plane

Equation	Description
$y = (x+2)^2 - 3$	parabola
$(x - 2)^2 + (y + 1)^2 = 4$	circle
$\left(\dfrac{x}{2}\right)^2 + \left(\dfrac{y}{3}\right)^2 = 1$	ellipse
$x^2 - y^2 = 4$	hyperbola
$x^2 + xy + 4y^2 = 4$	ellipse

To graph the last implicit quadratic equation, multiply both sides by 4

$$ 4x^2 + 4xy + 16y^2 = 16 $$
Rewrite the expression on the left

$$ 4x^2 + 4xy + y^2 + 15y^2 = 16 $$
Factoring this expression becomes

$$ \begin{align*} (2x + y)^2 + 15y^2 &= 16 \\ \\ 4\left(x + \frac{y}{2}\right)^2 + 15y^2 &= 16 \\ \end{align*} $$
This gives us an ellipse rotated clockwise from one with major and minor axis on the $x$ and $y$ axes.
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Recall a generic quadratic curve in a two dimensional coordinate system is given by

$$ ax^2 + bx + cy^2 + dy + exy + f = 0 $$

1. First let us consider $e=0$. To graph the quadratic curve one completes the square for $ax^2 + bx$ if $a\neq 0$, and completes the square for $cy^2 + d$ if $c\neq 0$. One obtains one of three possibilities
   
   $$ \begin{align*} a\left(x^2 + \left(\frac{b}{a}\right)x + \frac{b^2}{4a^2}\right) + dy + e &= \frac{b^2}{4a} \\ \\ bx + c\left(y^2 + \left(\frac{d}{2c}\right)y + \frac{d^2}{4c^2}\right) + e &= \frac{d^2}{4c} \\ \\ a\left(x^2 + \left(\frac{b}{a}\right)x + \frac{b^2}{4a^2}\right) + c\left(y^2 + \left(\frac{d}{2c}\right)y + \frac{d^2}{4c^2}\right) + e &= \frac{b^2}{4a} + \frac{d^2}{4c} \end{align*} $$
   After completing the square we simplify these to one of
   
   $$ \begin{align*} y = a(x - h)^2 + k \\ \\ x = c(y - k)^2 + h \\ \\ a(x - h)^2 + c(y - k)^2 &= r^2 \end{align*} $$
   Using this form of the quadratic curves one first graphs basic curve
   
   $$ \begin{align*} y &= ax^2 &\qquad &\text{a parabola} \\ \\ x &= cy^2 &\qquad &\text{a parabola} \\ \\ ax^2 + by^2 &= r^2 &\qquad &\text{a circle, ellipse or hyperbola} \end{align*} $$
   In the last case the graph depends on the value and sign of $a$ and $b$. Then we can shift the graph to the right or left $h$, and up or down $k$.
   
   
   

2. Consider the case that $e\neq 0$. Then as in our example, the graph of the quadratic equation is one with the major and minor axis rotated away from the $x$ and $y$ axes. Thus we graph an appropriate ellipse with no $exy$ term and rotate it to obtain the graph of the ellipse with the $exy$ term.

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12.6.3 Cylinders

In three dimensional space the graph of an implicit function of three variables yields a surface. We can describe a simple surface using only two of the three variables, $x$, $y$ and $z$. We start with a plane curve in one of the $xy$-plane, $yz$-plane, or $xz$-plane. For example

$$ x^2 + y^2 = 1 $$
is a plane curve in the $xy$-plane. In a two dimensional coordinate system this is the unit circle. In a three dimensional coordinate system one of variables $z$ is unspecified. Therefore $z$ is allowed to have any real value. The locus of points that satisfy this equation include the points on the unit circle in the $xy$-plane

$$ \left\{ (x,y,0)\,:\,x^2 + y^2 = 1\right\} $$
However the locus of points also includes all of the points

$$ \left\{ (x,y,z)\,:\,x^2 + y^2 = 1,\ z\in\mathbb{R}\right\} $$

The curve in the $xy$-plane is the plane curve $x^2 + y^2 = 1$. The graph is a __right circular cylinder__ since $z$ is unspecified and may thus have any real value.

Any surface that results from describing a plane curve using two variables and leaves the third variable unspecified is called a cylinder.

Example 12.6.1

Graph the equation $z = y^2$ in a three dimensional coordinate system.

The plane curve is a parabola in the $yz$-plane and the cylinder results from leaving $x$ unspecified.

Exercise 12.6.1

Graph the equation $x^2 - y^2 = 1$.

Check Your Work

The plane curve the the hyperbola $x^2 - y^2 = 1$ in the $xy$-plane and the cylinder results from leaving $z$ unspecified.

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12.6.4 Quadratic Surfaces

The general form an quadratic implicit function of three variables is given by

$$ Ax^2 + By^2 + Cz^2 + Dxy + Eyz + Fxz + Gx + Hy + Iz + J = 0 $$
As in the two dimensional case, The $Dxy$, $Eyz$, and $Fxz$ terms cause the surface to be rotated in the $xy$, $yz$, or $xz$ plane, respectively. First we graph the simpler quadratic implicit function and then rotate the graph in each plane to obtain the more general surface. To help our understanding of these surfaces let us consider the cases when $D = E = F = 0$. Thus our implicit quadratic function has the form

$$ Ax^2 + By^2 + Cz^2 + Gx + Hy + Iz + J = 0 $$

If $A$, $B$, $C$ are all nonzero we can complete the square for all three variables in the equation

$$ Ax^2 + Gx + By^2 + Hy + Cz^2 + Iz + J = 0 $$
and obtain

$$ A(x - h)^2 + B(y - k)^2 + C(z - l)^2 + J = 0 $$
Thus we can graph the quadratic surface $Ax^2 + By^2 + Cz^2 = r^2$ and then shift the surface to along each axis. We say that after rotation, and translation along the axes, the general quadratic implicit function has the form

$$ Ax^2 + By^2 + Cz^2 + J = 0,\qquad\text{or}\qquad Ax^2 + By^2 + Iz + J = 0. $$
If two of the quadratic coefficients are zero, say

$$ Gx + Hy + Cz^2 + J = 0 $$
then the graph of the function is a parabolic cylinder with the vertex of the parabola cross section along the line $Gx + Hy + J = 0$.

Notice that the edge of the parabolic cylinder when $z=0$ is the line $x + y = 1$.

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12.6.5 Ellipsoid

In the case $A$, $B$, and $C$ from the previous section are all positive the equation

$$ Ax^2 + By^2 + Cz^2 + J = 0 $$
becomes

$$ \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1 $$
Coefficient $J$ must be negative as $Ax^2 + By^2 + Cz^2 > 0$ in this case. We have $a^2 = \frac{-J}{A}$, $b^2 = \frac{-J}{B}$, and $c^2 = \frac{-J}{C}$.

We determine the graph of the equation by graphing the intersections of the surface with the $xy$, $yz$, and $xz$ planes.

• Setting $z=0$ we have the elliptic plane curve $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ in the $xy$-plane.
• Setting $x=0$ we have the elliptic plane curve $\frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$ in the $yz$-plane.
• Setting $y=0$ we have the elliptic plane curve $\frac{x^2}{a^2} + \frac{z^2}{c^2} = 1$ in the $xz$-plane.

Example 12.6.2

Graph the ellipsoid defined by

$$ x^2 + \dfrac{y^2}{4} + \dfrac{z^2}{4} = 1 $$


One graphs the plane curves in the coordinate planes first to reveal the shape of the ellipsoid.

Exercise 12.6.2

Graph the ellipsoid defined by the equation


$$ \dfrac{x^2}{9} + \dfrac{y^2}{9} + \dfrac{z^2}{16} = 1 $$

Check Your Work

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12.6.6 Paraboloid

For implicit quadratic function


$$ Ax^2 + By^2 + Iz + J = 0 $$
we simplify the equation into

$$ \dfrac{z}{c} = \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} $$
The $\frac{J}{I}$ term shifts the paraboloid up and down the $z$-axis. Here $a^2 = \frac{I}{A}$ and $b^2 = \frac{I}{B}$. In this case $I$ can be negative. There are two cases.

• $A$ and $B$ must have the same sign. In this case the quadratic surface is an elliptic paraboloid.
  
• $A$ and $B$ have opposite signs. In this case the quadratic surface is a hyperbolic paraboloid.
  
  
  

Example 12.6.4

Graph the elliptic paraboloid $z = x^2 + \dfrac{y^2}{4}$

Exercise 12.6.4

Graph the hyperbolic paraboloid $z = \frac{x^2}{16} - \frac{y^2}{9}$

Check Your Work

Notice that when $y=0$, the plane curve is a parabola that is concave up in the $xz$-plane. However, when $x=0$, the plane curve is concave down.

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12.6.7 Hyperboloids

Hyperboloids have the form


$$ \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} - \dfrac{z^2}{c^2} = d $$
Notice that there are two positive terms and one negative term on the left-hand side of the equation.

If you have one positive term and two negative terms, multiply both sides by -1 before continuing.

This means one may encounter two other forms of a hyperboloid

$$ \begin{align*} -\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} + \dfrac{z^2}{c^2} &= d \\ \\ \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} + \dfrac{z^2}{c^2} &= d \\ \end{align*} $$

In each case the axis of symmetry is the axis with the negative coefficient. The form of the hyperboloid depends on the value on the right-hand side, $d$.

1. Hyperboloid of One Sheet
   
   $$ \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} - \dfrac{z^2}{c^2} = d \gt 0 $$
   In this case, when $z=0$ we have a plane curve that is an ellipse in the $xy$-plane
   
   
   $$ \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = d $$
   For every real number $z=k$ we have a horizontal plane curve that is an ellipse in the plane $z=k$
   
   
   $$ \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = d + \dfrac{k^2}{c^2} $$
   This increases the major and minor axis as $z$ gets larger in either the positive or negative direction along the $z$-axis. This hyperboloid consists of one smooth surface
   
   
   
   
   
   

2. Cone
   $$ \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} - \dfrac{z^2}{c^2} = 0 $$
   When $d=0$, we have a plane curve of radius zero in the $xy$-plane; a point. Similar to the hyperboloid of one sheet, for every nonzero real number $z=k$ we have a horizontal plane curve that is an ellipse in the plane $z=k$
   
   
   $$ \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = \dfrac{k^2}{c^2} $$
   
   
   
   

3. Hyperboloid of Two Sheets
   $$ \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} - \dfrac{z^2}{c^2} = -d \lt 0 $$
   With $d < 0$ there is no plane curve in the $xy$-plane because when $z=0$ we have
   
   $$ \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = -d \lt 0 $$
   There is no plane curve of the surface until $z$ is large enough that $\dfrac{z^2}{c^2} - d \ge 0$. Hence there are two separate surfaces to the graph of the quadratic implicit function.
   
   
   

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