Cf. Vectors.

Cf. Vectors: Vector operations.

Cf. Vectors: Unit vectors.

A “right handed” (if you stand at the origin with your right arm along the positive $x$-axis and your left arm along the positive $y$-axis, your head will point in the direction of the positive $z$-axis) three-dimensional coordinate system has three perpendicular **coordinate planes**: the $xy$-, $xz$-, and $yz$-planes.

Equations of the coordinate planes in $\mathbb{R}^3$:

Plane | Equation |
---|---|

$yz$ | $x = 0$ |

$xy$ | $z = 0$ |

$xz$ | $y = 0$ |

In $\mathbb{R}^3$, the **distance** from the origin to $(a, b, c)$ is $d = \sqrt{a^2 + b^2 + c^2}$.

We can summarize the **distance formula** for $P_1(x_1, y_1, z_1)$ and $P_2(x_2, y_2, z_2)$

$\lVert P_1 P_2 \rVert = \sqrt{ \Delta x^2 + \Delta y^2 + \Delta x^2 }$

The **graph of an equation** in $\mathbb{R}^3$ is the collection of all points $(x, y, z)$ whose coordinates satisfy a given equation. This graph is called the **surface**.

To graph a **plane**, find some ordered triples that satisfy the equation. The best ones to use are those that fall on a coordinate axis (intercepts).

A **sphere** is defined as the collection of all points located a fixed distance (radius) from a fixed point (center).

**Equation of a sphere:** The graph of the equation

$(x - a)^2 + (y - b)^2 + (z - c)^2 = r^2$

is a sphere with center $(a, b, c)$ and radius $r$. This is the **standard form of the equation of a sphere**.

Solve given equations by rearranging to standard form and completing the square in the variables that require it.

**Warning:** Pay close attention to the signs when deriving the center coordinates from the equation since the coordinates are subtracted in the equation.

A **cylindrical surface** is a surface traced by a line moving parallel to a given fixed line and intersecting a given curve. A cylinder is defined with a **generating curve** (**directrix**) and a **generating line** (**directrix**). Lines running parallel to the directrix are called the **rulings**.

The curve is always in the plane containing the two variables in the equation. The rulings are parallel to the axis of the missing variable: e.g $y = x2$ has a generating curve in the $xy$-plane and rulings that run parallel to the $z$-axis. Any three variable equation missing one variable will be a cylindrical surface.

A vector in $\mathbb{R}^3$ is a directed line segment in space.

Representation and operations are analogous to the representation and operations defined in $\mathbb{R}^2$.

Cf. Vectors: Vector Operations, Dot (scalar) product.

Cf. Vectors: Vector Operations Cross (vector) product.

$\vec{r}(t) = \vec{r}_0 + t\vec{v} = \langle x_0, y_0, z_0 \rangle + t \langle a, b, c \rangle$

where $\vec{r}_0$ is a position vector (points to a point on the line) and $\vec{v}$ is a direction vector (a vector parallel to the vector).

A line **parallel** to the vector $v = \langle a, b, c \rangle$ that passes through the point $(x_1, y_1, z_1)$ is given by

$x = x_1 + ta
\\
y = y_1 + tb
\\
z = z_1 + tc$

for some number $t$.

The parameter $t$ can be eliminated to obtain the **symmetric form of a line**

$\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$

Converting a parametric equation to a rectangular equation is called **eliminating the parameter**. General guidelines are:

- Solve for $t$ in one of the equations
- Substitute into the second equation
- Simplify

- If the equation involve trigonometric functions, trigonometric identities may be needed
- The domain may need to be adjusted based upon the original parametric equation

- Test for parallel. If not, then
- Test for intersecting
- If intersecting, then test for perpendicular
- If not intersecting, then is skew

For example, given $\langle 3 + 2t, 4 - t, 1 + 3t \rangle$ and $\langle 1 + 4s, 3 - 2s, 4 + 5s \rangle$

**Test for parallel:** Take the ratio of each component’s direction numbers (coefficients on the parameters) and set them equal to each other. Does $\frac{2}{4} = \frac{-1}{-2} = \frac{3}{5}$? No, therefore the lines are not parallel.

**Test for intersecting:** Solve a system of simultaneous equations. If we can solve it, then the lines are intersecting. Solve

$3 + 2t = 1 + 4s
\\[16pt]
4 - t = 3 - 2s
\\[16pt]
1 + 3t = 4 + 5s$

**Test for perpendicular:** If the dot product of the two lines is $0$, then the lines are perpendicular.

A common way to specify the direction of a plane is by means of a vector $N$ (called a **normal** to the plane) that is orthogonal to every vector in the plane. The **point-normal form** comes from determining the dot product of all the points in the plane with the normal vector. The **standard form** then comes from distributing and simplifying.

A plane with normal $N = \langle A, B, C \rangle$ containing the points $(x_0, y_0, z_0)$ has the following equations

**Point-normal form**: $A(x - x_0) + B(y - y_0) + C(z - z_0) = 0$**Standard form**: $Ax + By + Cz + D= 0$

A line is **parallel** to a plane if the direction vector of the line is orthogonal to the normal vector of the plane. Remember that two vectors are orthogonal if their dot product is zero.

A line is **perpendicular** to a plane if its direction vector is a scalar multiple of the normal vector of the plane. Remember that the normal vector is already perpendicular to the plane, so both being parallel means that their directions lie along the same line in $\mathbb{R}^3$.

Use the normal $N$ of the given plane and the given point to construct the equation in point-normal form.

Given the points $P$, $Q$, and $R$, a normal $N$ to the required plane is orthogonal to the vectors $PR$ and $PQ$ and is, therefore, found by computing with the cross product.

$N = PR \times PQ$

You can now find the equation of the plane using this normal vector and any point in the plane.

Because the required line is perpendicular to the normals $N_1$ and $N_2$ of the given planes, the aligned vector is found by computing with the cross product.

$N_1 \times N_2$

You can now find the equation of the plane using this normal vector and any point in the plane.

The distance from the point $P(x_0, y_0, z_0)$ to the plane $Ax + By + Cz + D = 0$ is given by

$d = \lVert \text{proj}_n QP \rVert = \frac{| QP \cdot N|}{\lVert N \rVert} = \frac{| Ax_0 + By_0 + Cz_0 + D |}{\sqrt{A^2 + B^2 + C^2}}$

Where $Q$ is any point in the given plane and $N$ is a normal to the given plane.

Note that to determine $QP$, we will need to determine one point on the plane $Q$.

Given $C(x_0, y_0, z_0)$ and a plane, the radius $r$ is the distance from the center $C$ to the given plane (use the distance formula from above). Therefore the equation of the sphere is

$(x - x_0)^2 + (y - y_0)^2 + (z - z_0)^2 = r^2$

The distance from point %P% to the line $L$ is given by

$d = \frac{\lVert v \times QP \rVert}{\lVert v \rVert}$

where $v$ is a vector parallel to $L$ and $Q$ is any point on $L$.

Note that to determine $QP$, we will need to determine one point on the plane $Q$ (set $t = 0$).

A quadric surface is given by a degree two equation in the general form

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

where $A, \ldots , J$ are constants.

To graph a quadric surface, it is often helpful to graph the $xy$-trace, $xz$-trace, and $yz$-trace (the intersections of the surface with these three planes). To determine the $xy$-trace, sex $z = 0$. To determine the $xz$-trace, set $y = 0$. To determine the $yz$-trace, set $x = 0$.

The general equation of an ellipsoid is

$\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$

If $a = b = c$ then we will have a sphere.

The general equation of a cylinder is

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

If $a = b$ we have a cylinder whose cross section is a circle. The cylinder will be centered on the axis corresponding to the variable that does not appear in the equation.

The general equation of a hyperboloid of one sheet is

$\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1$

The variable with the negative in front of it will give the axis along which the graph is centered.

The general equation of a hyperboloid of two sheets is

$-\frac{x^2}{a^2} - \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$

The variable with the positive in front of it will give the axis along which the graph is centered.

The general equation of an elliptic cone is

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = \frac{z^2}{c^2}$

The variable that sits by itself on one side of the equal sign will determine the axis that the cone opens up along.

The general equation of an elliptic paraboloid is

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = \frac{z}{c}$

The variable that isn’t squared determines the axis upon which the paraboloid opens up. The sign of $c$ will determine the direction that the paraboloid opens. If $c$ is positive then it opens up and if $c$ is negative then it opens down.

The general equation of a hyperbolic paraboloid is

$\frac{x^2}{a^2} - \frac{y^2}{b^2} = \frac{z}{c}$

These graphs are vaguely saddle shaped and as with the elliptic paraboloid the sign of $c$ will determine the direction in which the surface opens.

A **vector-valued function** (or **vector function**) is a function where the domain is a subset of the real numbers and the range is a vector. It is given by

$r(t) = f(t)\hat{i} + g(t)\hat{j}
\\[16pt]
r(t) = \langle f(t), g(t) \rangle$

in $\mathbb{R}^2$.

$r(t) = f(t)\hat{i} + g(t)\hat{j} + h(t)\hat{k}
\\[16pt]
r(t) = \langle f(t), g(t), h(t) \rangle$

in $\mathbb{R}^3$.

- The
**domain**of a vector valued function is the intersection of the domain of its components.

Let $F$ and $G$ be vector functions of the real variable $t$, and let $f(t)$ be a scalar function.

Operation | Equation |
---|---|

Addition | $(F + G)(t) = F(t) + G(t)$ |

Subtraction | $(F - G)(t) = F(t) - G(t)$ |

Scalar multiplication | $(fF)(t) = f(t)F(t)$ |

Cross product | $(F \times G)(t) = F(t) \times G(t)$ |

Suppose the components of the vector function $F(t) = a\hat{i} + b\hat{j} + c\hat{k}$ all have finite limits as $t \rightarrow t_0$, where $t_0$ is any number or $\infin$ or $-\infin$. Then

$\lim\limits_{t \to t_0} f(t) = \left[ \lim\limits_{t \to t_0} a(t) \right]i + \left[ \lim\limits_{t \to t_0} b(t) \right]j + \left[ \lim\limits_{t \to t_0} c(t) \right]k$

Let $F$ and $G$ be vector functions of the real variable $t$, and let $h(t)$ be a scalar function so that all three functions have finite limits as $t \rightarrow t_0$, then

Limit of a… | Equation |
---|---|

Sum | $\lim\limits_{t \to t_0} [F(t) + G(t)] = \lim\limits_{t \to t_0} F(t) + \lim\limits_{t \to t_0} G(t)$ |

Difference | $\lim\limits_{t \to t_0} [F(t) - G(t)] = \lim\limits_{t \to t_0} F(t) - \lim\limits_{t \to t_0} G(t)$ |

Scalar multiple | $\lim\limits_{t \to t_0} [h(t)G(t)] = \left[ \lim\limits_{t \to t_0} h(t) \right] \left[ \lim\limits_{t \to t_0} G(t) \right]$ |

Dot product | $\lim\limits_{t \to t_0} [F(t) \cdot G(t)] = \left[ \lim\limits_{t \to t_0} F(t) \right] \cdot \left[ \lim\limits_{t \to t_0} G(t) \right]$ |

Cross product | $\lim\limits_{t \to t_0} [F(t) \times G(t)] = \left[ \lim\limits_{t \to t_0} F(t) \right] \times \left[ \lim\limits_{t \to t_0} G(t) \right]$ |

A vector function $F(t)$ is said to be continuous at $t_0$ if $t_0$ is in the domain of $F$ and $\lim\limits_{t \to t_0} F(t) = F(t_0)$

The **derivative** of the vector function $F$ is the vector function $F^{\prime}$ determined by the limit.

$F^{\prime}(t) = \lim\limits_{\Delta t \to 0} \frac{\Delta F}{\Delta t}$

wherever the limit exists. We say that the vector function $F$ is **differentiable** at $t = t_0$ if $F^{\prime}(t)$ is defined at $t_0$.

The vector function $F(t)$ is differentiable whenever the components functions are each differentiable.

$F^{\prime} = f_{1}^{\prime}(t)i + f_{2}^{\prime}(t)j +f_{3}^{\prime}(t)k$

The vector derivative can be used to find tangent vectors to curves in space.

A vector function $F$ is smooth on a given interval if $F^{\prime}(t) \neq 0$.

For example, if $F(t) = \langle a, b, c \rangle$, then $F^{\prime}(t) \neq \langle 0, 0, 0 \rangle$.

Higher-order derivatives of a vector function $F$ are obtained by successively differentiating the components of $F(t)$. For example, the second derivative of $F$ is the derivative of $F^{\prime}(t)$, and so on.

An object that moves in such a way that its position at time $t$ is given by the vector function $R(t)$ is said to have

- the
**position vector**: $R(t)$ - the
**velocity**: $V = R^{\prime}(t)$

At any time $t$,

- the
**speed**is the magnitude of the velocity: $\lVert V \rVert$ - the
**direction of motion**is the unit vector: $\frac{V}{\lVert V \rVert}$ - the
**acceleration vector**is the derivative of the velocity: $A = V^{\prime}(t) = R^{\prime\prime}(t)$

Vector integration is performed per component.

The **indefinite integral** of $F(t)$ is the vector function

$\int F(t)dt = \left[ \int f_{1}(t)dt \right]i + \left[ \int f_{2}(t)dt \right]j + \left[ \int f_{3}(t)dt \right]k + C$

where $C = C_1 i + C_2 j + C_3 k$ is an arbitrary constant vector.

The **definite integral** of $F(t)$ is the vector

$\int_{a}^{b} F(t)dt = \left[ \int_{a}^{b} f_{1}(t)dt \right]i + \left[ \int_{a}^{b} f_{2}(t)dt \right]j + \left[ \int_{a}^{b} f_{3}(t)dt \right]k$

This chapter extends the methods of single-variable differential calculus to functions of two of more independent variables.

A **function of two variables** is a rule $f$ that assigns to each ordered pair $(x, y)$ in a set $D$ a unique number $f(x, y)$. The set $D$ is called the **domain** and the corresponding values of $f(x, y)$ constitute the **range**.

When dealing with such functions, we may write $z = f(x, y)$ and refer to $x$ and $y$ as the **independent variables** and to $z$ as the **dependent** variable.

One way we can approach sketching the graph of a function of two variables, without the assistance of technology, is the same as graphing quadric surfaces in Section 9.7: Use the trace of of graph in a plane.

When the plane $z = C$ intersects the surface $z = f(x, y)$, the result is the trace, the equation $f(x, y) = C$. The set of points $(x, y)$ in the $xy$-plane that satisfy this equation is called the **level curve** (or **contour curve**) of $f$ at $C$. An entire family of level curves is generated as $C$ varies over the range of $f$. Think of a trace as a “slice” of the surface at a particular location and a level curve as its projection onto the $xy$-plane.

A more complete picture of a surface can be obtained by examining cross sections perpendicular to the other two principle directions as well.

If $f$ is a function of three variables $x$, $y$, and $z$, then the solution set of the equation $f(x, y, z) = C$ is a region of $\mathbb{R}^3$ called a **level surface** of $f$ at $C$.

Single variable functions have domains that can be expressed in terms of intervals. However, functions of two or more variables requires special terminology and notation introduced in this section followed by limits and continuity.

An **open disk** in $\mathbb{R}^2$ centered at point $C(a, b)$ with radius $r$ is the set of all points $P(x, y)$ such that $\sqrt{(x - a)^2 + (y - b)^2} < r$. If the boundary of the disk is included, it is said to be a **closed disk**. Open and closed disks are analogous to open and closed intervals on a coordinate line.

A point $P_0$ is said to be an **interior point** of a set $S$ in $\mathbb{R}^2$ if some open disk centered at $P_0$ is contained entirely within $S$. If $S$ is the empty set, or if every point of $S$ is an interior point, then $S$ is called an open set.

A point $P_0$ is said to be a **boundary point** of a set $S$ in $\mathbb{R}^2$ if every open disk centered at $P_0$ contains both points that belong to $S$ and points that do not. The collection of all boundary points of $S$ is called the **boundary** of $S$. $S$ is said to be **closed** if it contains its boundary.

The empty set $\emptyset$ is both open and closed in any topological space.

Similarly, an **open ball** centered at point $C(a, b, c)$ in $\mathbb{R}^3$ is the set of all points $P(x, y, z$) such that $\sqrt{(x - a)^2 + (y - b)^2 + (z - c)^2} < r$. (etc. as above in $\mathbb{R}^2$).

For a function of two variables, the limit statement

$\lim\limits_{(x, y) \to (x_0, y_0)} f(x, y) = L$

means that for each given number $\epsilon > 0$, there exists a number $\delta > 0$ such that for all $(x, y) \neq (a, b)$, if $(x, y)$ is in the open disk centered at $(x_0, y_0)$ in the $xy$-plane with radius $\delta$, then

$| f(x, y) - L | < \epsilon$

What this says is that the function value of $f(x, y)$ must lie in the interval $(L - \epsilon, L + \epsilon)$ whenever $(x, y)$ is a point in the domain of $f$ other than $P_0(x_0, y_0)$ that lies inside the disk of radius $\delta$ centered at $P_0$.

If the limit is not the same for *all* approaches within the domain of $f$, then *the limit does not exist*.

**Evaluate:**

- Try direct substitution. If this results in an indeterminate form, then
- Check if the equation is factorable and cancel the factored terms. If not, then
- Take the limits $(x, y) \rightarrow (0, 0)$ along the $x$-axis and $y$-axis or the lines $y = x$ and $y = -x$.

Using the definition of the limit of a function of two variables, we can define the continuity of a function of two variables analogously. The function $f(x, y)$ is **continuous** at the point $(x_0, y_0)$ if and only if

- $f(x_0, y_0)$ is defined
- $\lim\limits_{(x, y) \to (x_0, y_0)} f(x, y)$ exists
- $\lim\limits_{(x, y) \to (x_0, y_0)} f(x, y) = f(x_0, y_0)$

The function $f$ is **continuous on a set $S$** if it is continuous at each point in $S$.

The concepts introduced for functions of two variables in $\mathbb{R}^2$ extend naturally to functions of three variables in $\mathbb{R}^2$.

The limit statement

$\lim\limits_{(x, y, z) \to (x_0, y_0, z_0)} f(x, y, z) = L$

means that for each number $\epsilon > 0$, there exists a number $\delta > 0$ such that

$| f(x, y, z) - L | < \epsilon$

whenever $(x, y, z)$ is a point in the domain of $f$ such that

$0 < \sqrt{(x - x_0)^2 + (y - y_0)^2 + (z - z_0)^2}$

The function is continuous at the point $P_0(x_0, y_0, z_0)$ if

- $f(x_0, y_0, z_0)$ is defined
- $\lim\limits_{(x, y, z) \to (x_0, y_0, z_0)} f(x, y, z)$ exists
- $\lim\limits_{(x, y, z) \to (x_0, y_0, z_0)} f(x, y, z) = f(x_0, y_0, z_0)$

The process of differentiating a function of several variables with respect to one of its variables while keeping the other variable(s) fixed is called **partial differentiation**, and the resulting derivative is a **partial derivative** of the function.

If $z = f(x, y)$, then the partial derivatives of $f$ with respect to $x$ and $y$ are the functions $f_x$ and $f_y$, respectively, defined by

$f_{x}(x, y) = \lim\limits_{\Delta x \to 0} \frac{ f(x + \Delta x, y) - f(x, y) }{\Delta x}$

and

$f_{y}(x, y) = \lim\limits_{\Delta y \to 0} \frac{ f(x, y + \Delta y) - f(x, y) }{\Delta y}$

provided the limits exist.

This means that we find the partial derivative with respect to $x$ by **regarding $y$ as a constant** while differentiating the function with respect to $x$. Similarly, the partial derivative with respect to $y$ is found by **regarding $x$ as a constant** while differentiating the function with respect to $y$.

**Alternate notation for partial derivatives** for $z = f(x, y)$

$f_{x}(x, y) = \frac{\partial f}{\partial x} = \frac{\partial z}{\partial x} = \frac{\partial}{\partial x}f(x, y) = z_x = D_x(f)$

and

$f_{y}(x, y) = \frac{\partial f}{\partial y} = \frac{\partial z}{\partial y} = \frac{\partial}{\partial y}f(x, y) = z_y = D_y(f)$

Calculate the **partial derivative of an implicitly defined function**

$\frac{\partial z}{\partial x} = \frac{- f_x}{f_z}$

where $f_x \neq 0$.

The partial derivative of a function is a function, so it is possible to take the partial derivative of a partial derivative. If we take two consecutive partial derivatives with respect to the same variable, the resulting derivative is called the **second-order partial**. We can also take a second partial derivative with respect to a different variable,producing what is called a **mixed second-order partial derivative**.

$f_{xy} \equiv f_{yx}$

One independent variable $t$.

$\frac{dz}{dt} = \frac{\partial z}{\partial x} \frac{dx}{dt} + \frac{\partial z}{\partial y} \frac{dy}{dt}$

Two independent variables $u$ and $v$.

$\frac{dz}{du} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial u} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial u}
\\[16pt]
\text{and}
\\[16pt]
\frac{dz}{dv} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial v} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial v}$