Welcome to Calculus III! We begin our journey by adding a new dimension to our mathematical world. Just as we used two axes ($x$ and $y$) to locate points in a plane, we will now use three mutually perpendicular axes ($x$, $y$, and $z$) to locate points in space. This transition opens up a new universe of shapes and concepts to explore. 🌌
In two dimensions, we describe a point with an ordered pair $(a, b)$. This set of all possible ordered pairs forms the 2D Cartesian plane, which we denote as $\mathbb{R}^2$.
To move into three dimensions, we introduce a third axis, the $z$-axis. A point in space is now represented by an ordered triplet $(a, b, c)$. The set of all such triplets is called 3D space, or $\mathbb{R}^3$.
The three axes, taken two at a time, define three coordinate planes:
These three planes divide space into eight regions called octants. The first octant is where all three coordinates are positive.
Let's formally bridge the gap between two and three dimensions. In $\mathbb{R}^2$, the graph of a function of a single variable, $y = f(x)$, is the set of all ordered pairs $(x, y)$ that satisfy the equation for all $x$ in the domain of $f$. This is the set of points $(x, f(x))$. Plotting these points gives us a curve.
In $\mathbb{R}^3$, we now consider functions of two independent variables, written as $z = f(x, y)$. The domain of such a function is a set of ordered pairs $(x,y)$ in the xy-plane. For each point $(x,y)$ in the domain, the function assigns a unique real number $z = f(x,y)$. The set of all these output values constitutes the range of the function. The graph of $f$ is the set of all ordered triplets $(x, y, z)$ that satisfy the equation. To graph, we plot the set of points $(x, y, f(x,y))$. Plotting this set of points creates a surface in space.
Let's visualize the surface for the function $z = f(x, y) = x^2 + y^2$. The domain is all of $\mathbb{R}^2$. The range is $[0, \infty)$, since $x^2+y^2$ is never negative. The output $z$ is the height of the surface above the xy-plane at the point $(x,y)$.
We can create a table of values to find some points on the surface:
| x | y | $z = f(x,y) = x^2 + y^2$ | Point (x,y,z) |
|---|---|---|---|
| 0 | 0 | 0 | (0,0,0) |
| 1 | 0 | 1 | (1,0,1) |
| 0 | 1 | 1 | (0,1,1) |
| 1 | 1 | 2 | (1,1,2) |
| 2 | 0 | 4 | (2,0,4) |
| 0 | 2 | 4 | (0,2,4) |
If we plot these points and all others, we get a bowl-shaped surface called a circular paraboloid.
Now consider the equation $x^2 + y^2 = 1$. How we interpret this depends on the context.
The context—whether we are working in $\mathbb{R}^2$ or $\mathbb{R}^3$—is crucial for interpreting the graph of an equation.
To find the distance between two points $P_1(x_1, y_1, z_1)$ and $P_2(x_2, y_2, z_2)$ in space, we can use the Pythagorean theorem twice. We construct a box where the diagonal connects $P_1$ and $P_2$.
The length of the diagonal on the base of the box is given by the 2D distance formula: $d_{base}^2 = (x_2-x_1)^2 + (y_2-y_1)^2$. Now, this diagonal and the vertical edge of length $|z_2-z_1|$ form a right triangle whose hypotenuse is the distance $|P_1P_2|$ we want.
Applying the Pythagorean theorem again gives us: $$|P_1P_2|^2 = d_{base}^2 + (z_2-z_1)^2 = (x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2$$
The distance $|P_1P_2|$ between the points $P_1(x_1, y_1, z_1)$ and $P_2(x_2, y_2, z_2)$ is:
$$|P_1P_2| = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$$Find the distance between the points $P(5, -1, 6)$ and $Q(1, 7, 0)$.
We apply the distance formula directly:
We can simplify the radical: $\sqrt{116} = \sqrt{4 \cdot 29} = 2\sqrt{29}$.
Problem: Find the distance between the points $A(-2, 0, 5)$ and $B(4, 3, -1)$.
A sphere is the set of all points $P(x, y, z)$ that are equidistant from a fixed center point $C(h, k, l)$. This constant distance is the radius, $r$.
Using the distance formula, we can say that the distance between any point $P$ on the sphere and the center $C$ must be $r$:
Squaring both sides gives us the standard equation of a sphere.
An equation of a sphere with center $C(h, k, l)$ and radius $r$ is:
$$(x-h)^2 + (y-k)^2 + (z-l)^2 = r^2$$Determine whether the points $A(2,4,2)$, $B(3,7,-2)$, and $C(1,3,3)$ lie on a straight line.
Thinking process: If three points lie on a line, the sum of the distances between the two smaller segments must equal the distance of the largest segment. That is, if B is between A and C, we would have $|AB| + |BC| = |AC|$. We need to calculate all three distances and compare them.
Step 1: Calculate $|AB|$
Step 2: Calculate $|BC|$
Step 3: Calculate $|AC|$
Step 4: Compare the distances
The three distances are $\sqrt{3}$, $\sqrt{26}$, and $\sqrt{45}$. We check if the sum of the two smaller distances equals the largest distance: $\sqrt{3} + \sqrt{26} \neq \sqrt{45}$.
Since the triangle inequality $|AC| + |AB| > |BC|$ holds and they are not equal, the points are not collinear. They form a triangle.
Problem: Find the distance from the point $P(4, -2, 6)$ to the xy-plane.
Today, we extended our understanding of coordinate systems into three dimensions. We've seen how points are located with ordered triplets, how equations define surfaces, and how to measure distances between points. This led us to the equation of a sphere, our first fundamental 3D surface. These basic tools—coordinates, distance, and geometric interpretation—are the essential building blocks we will use throughout Calculus III. In our next lecture, we will introduce vectors, which provide a powerful new way to describe quantities in space. 🚀