Source: Math 126 Midterm 1 Practice Exam
Difficulty: Intermediate | Prerequisites: Distance formula in 2D, completing the square (precalculus), basic familiarity with conic sections.
Surfaces in 3D space are the backbone of multivariable calculus. Before you can do double or triple integrals, you need to recognise the shape you are integrating over. This topic covers how to identify standard surfaces from their equations, how distance formulas generate surfaces (a classic exam setup), and how to parametrise the curve where two surfaces intersect. If you are comfortable with the distance formula and completing the square, you have the tools you need.
You can build surface equations by setting two distance expressions equal. Completing the square turns messy expanded forms into standard quadric forms you can name on sight. To parametrise the intersection of two surfaces, pick a parameter for one or two coordinates and solve for the rest. These three skills cover at least two full problems on a typical midterm.
Paraboloid
A surface of the form z = a(x - h)² + b(y - k)² + c. It looks like a bowl (opens up if a, b > 0) or an inverted bowl (opens down if a, b < 0). In simple terms, it is the 3D version of a parabola, rotated around its axis.
Ellipsoid
A surface of the form (x - h)²/a² + (y - k)²/b² + (z - l)²/c² = 1. Think of it as a stretched or squished sphere.
Distance from a point to a point
For points P = (x₁, y₁, z₁) and Q = (x₂, y₂, z₂), the distance is d = √((x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²).
Distance from a point to a plane
For a point (x₀, y₀, z₀) and a plane ax + by + cz = d, the distance is |ax₀ + by₀ + cz₀ - d| / √(a² + b² + c²). For a coordinate plane like y = -2, this simplifies to |y₀ + 2|.
Completing the square
A technique for rewriting ax² + bx as a(x + b/(2a))² - b²/(4a). In simple terms, you rearrange a quadratic expression to reveal its vertex form, which makes the surface type visible.
Parametric curve (space curve)
A curve described by r(t) = ⟨x(t), y(t), z(t)⟩ where each coordinate is a function of parameter t. Think of it as a recipe for tracing out a path through 3D space.
Intersection of surfaces
The set of points that lie on both surfaces simultaneously. In 3D, two surfaces typically intersect in a curve, which can be parametrised as a vector function r(t).
The standard approach: write the distance condition as an equation, square both sides to remove the square root, expand, simplify, and identify the surface.
Find the equation of the surface of all points (x, y, z) whose distance to (1, 1, -1) equals the distance to the plane y = -2. Identify the surface.
Step 1: Write both distances
Distance to (1, 1, -1): √((x - 1)² + (y - 1)² + (z + 1)²)
Distance to y = -2: |y + 2| (since the plane y = -2 is simply a horizontal plane, the distance from any point to it is the absolute value of the difference in y-coordinates).
Step 2: Set them equal and square both sides
(x - 1)² + (y - 1)² + (z + 1)² = (y + 2)²
Step 3: Expand and simplify
(x - 1)² + y² - 2y + 1 + (z + 1)² = y² + 4y + 4
(x - 1)² + (z + 1)² = 6y + 3
(x - 1)² + (z + 1)² = 6(y + 1/2)
Step 4: Identify the surface
This is an elliptic paraboloid (in fact a circular paraboloid, since the x and z coefficients are equal) opening in the positive y-direction, with vertex shifted to (1, -1/2, -1).
The key recognition: the locus of points equidistant from a point and a plane is a paraboloid. This is the 3D analogue of the 2D definition of a parabola (equidistant from a focus and a directrix).
Strategy: let one or two coordinates be the parameter, then use both surface equations to solve for the remaining coordinate(s).
Find a vector function r(t) whose space curve is the intersection of z = 2(x - 1/2)² + 2(y - 1)² and the plane 2x + 4y + z = 5.
Step 1: Substitute the paraboloid into the plane equation
2x + 4y + 2(x - 1/2)² + 2(y - 1)² = 5
Step 2: Expand and complete the square
2x + 4y + 2(x² - x + 1/4) + 2(y² - 2y + 1) = 5
2x + 4y + 2x² - 2x + 1/2 + 2y² - 4y + 2 = 5
2x² + 2y² + 5/2 = 5
2x² + 2y² = 5/2
x² + y² = 5/4
This is a circle of radius √(5)/2 in the xy-plane.
Step 3: Parametrise the circle
Let x = (√5/2) cos t, y = (√5/2) sin t.
Step 4: Find z from either equation
z = 5 - 2x - 4y = 5 - √5 cos t - 2√5 sin t
Result:
r(t) = ⟨(√5/2) cos t, (√5/2) sin t, 5 - √5 cos t - 2√5 sin t⟩
The key insight: substituting one surface equation into the other often reveals a circle or ellipse in two of the variables, which you parametrise with cos t and sin t.
d(P, Q) = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}d(P, \text{plane}) = \frac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2 + b^2 + c^2}}Standard paraboloid: z = a(x - h)² + b(y - k)² + c
Standard ellipsoid: (x - h)²/a² + (y - k)²/b² + (z - l)²/c² = 1
Circle parametrisation: x = r cos t, y = r sin t, for t in [0, 2π)
Satellite dishes are paraboloids: they focus incoming parallel signals to a single point (the receiver at the focus). The distance-to-point-equals-distance-to-plane property is what makes this focusing work. Parametric intersection curves appear in CAD software whenever two surfaces meet, for instance where the body of a car meets the windscreen.
Students often forget to square both sides when setting distances equal. You cannot simplify a square root equation without squaring. And once you do, the absolute value signs vanish.
Students sometimes mix up the distance-to-a-plane formula with the distance-to-a-point formula. For a simple coordinate plane like y = -2, the distance is just |y - (-2)| = |y + 2|. You do not need the full formula.
Students frequently forget to complete the square after expanding. The expanded form looks like a mess of terms; the completed-square form reveals the surface. Skipping this step means you cannot identify the surface.
Students sometimes try to parametrise an intersection without first reducing to a recognisable curve. Always substitute one equation into the other first.
⚠️ "Find the equation of the surface and identify it by name" is a two-part question. You lose marks if you give the equation but forget to write the name.
⚠️ When a problem gives you a point and a plane and says "distance to the point equals distance to the plane," the answer is always a paraboloid. Recognise this pattern instantly.
⚠️ Intersection problems on the midterm almost always reduce to a circle or ellipse in two variables. If your algebra does not lead to x² + y² = constant or similar, re-check your expansion.
⚠️ Give answers in exact form. Leave √5/2, not 1.118.
True or False: The set of points equidistant from a point and a plane is a sphere. (False, it is a paraboloid.)
Fill in the blank: To parametrise a circle of radius r, use x = ___ cos t, y = ___ sin t. (r, r)
True or False: When you square both sides of a distance equation, the absolute value signs disappear. (True.)
Fill in the blank: The surface z = 3x² + 3y² is called a ___. (circular paraboloid)
Q: Find the equation of the surface of all points (x, y, z) equidistant from (0, 0, 3) and the plane z = -3. Identify the surface.
A: √(x² + y² + (z - 3)²) = |z + 3|. Squaring: x² + y² + z² - 6z + 9 = z² + 6z + 9. Simplify: x² + y² = 12z. This is a circular paraboloid opening upward.
Q: Parametrise the intersection of the cylinder x² + y² = 4 and the plane x + z = 3.
A: Let x = 2 cos t, y = 2 sin t. Then z = 3 - x = 3 - 2 cos t. So r(t) = ⟨2 cos t, 2 sin t, 3 - 2 cos t⟩.
Q: What type of surface is (x - 2)² + 4(y + 1)² = 8(z - 5)?
A: Elliptic paraboloid opening in the positive z-direction, vertex at (2, -1, 5).
This topic connects to vector-valued functions: the parametric curves you find as intersections become the r(t) you differentiate to find tangent vectors and arc length. It also connects to planes and lines (the next set of notes), because the plane equation is one of the two surfaces you intersect. Later in Math 126, double integrals will require you to identify the region of integration, which often involves these same surface equations.
Tags: paraboloid, elliptic paraboloid, circular paraboloid, quadric surface, distance formula 3D, distance point to plane, completing the square, parametric curve, space curve, intersection of surfaces, cylinder, ellipsoid, vector function, Math 126, Calculus III, UW, midterm 1, focus and directrix 3D