Difficulty: Intermediate to Advanced | Prerequisites: §12.1–12.5 (3D coordinates, vectors, dot/cross product, planes, quadric surfaces)
This set of topics ties together the geometry of surfaces in three dimensions with the parametric description of curves that live on those surfaces. You will need to recognise standard quadric surfaces from their equations, parametrise curves formed by intersecting two surfaces, and sketch bounded 3D regions defined by inequalities. The hyperboloid modelling problem connects the algebra to physical engineering design. If you are not comfortable identifying quadric surfaces from their equations, review §12.6 before proceeding.
Parametric curves in 3D are found by choosing a parameter (often an angle) that satisfies both surface equations simultaneously. Quadric surfaces are classified by which squared terms appear and their signs. Sketching regions defined by linear inequalities means finding intercepts and shading the correct side of each bounding plane.
Parametric equations (space curves)
A set of equations x = f(t), y = g(t), z = h(t) that trace out a curve in 3D as the parameter t varies. Think of t as time: at each moment the point is at a specific location in space.
Cylinder (in multivariable calculus)
A surface formed by translating a 2D curve along an axis. For example, x² + y² = 64 is a circular cylinder of radius 8, extending infinitely along the z-axis. In simple terms, it is a tube.
Paraboloid of revolution
A surface generated by rotating a parabola around its axis of symmetry. The equation x² + y² + 18z = 0 describes one opening downward. Think of it as an upside-down bowl whose cross-sections are circles.
Hyperboloid of one sheet
A quadric surface shaped like an hourglass or cooling tower, given by x²/a² + y²/b² − z²/c² = 1. It is connected (one piece) and has a minimum cross-section at z = 0. In simple terms, it is a surface that curves inward in the middle and flares out at top and bottom.
First octant
The region in 3D space where x ≥ 0, y ≥ 0, and z ≥ 0. The 3D equivalent of the first quadrant.
Quadric surface
Any surface whose equation is a second-degree polynomial in x, y, z. The six standard types are ellipsoids, hyperboloids (one and two sheets), elliptic paraboloids, hyperbolic paraboloids, and cones.
Vector-valued function
r(t) = ⟨f(t), g(t), h(t)⟩ gives a position vector in 3D for each value of t. The image of r is a space curve.
When two surfaces intersect, the intersection is typically a curve. To parametrise it, choose a parameter that naturally satisfies both equations.
For a cylinder x² + y² = 64 and a paraboloid x² + y² + 18z = 0, set x = 8cos(t), y = 8sin(t). Then x² + y² = 64, and from the paraboloid, 64 + 18z = 0, so z = −64/18 = −32/9.
The parametric equations are: x = 8cos(t), y = 8sin(t), z = −32/9. This is a circle at height z = −32/9.
Given r(t) = ⟨5t sin(3t), 6t cos(3t), t²⟩, you can identify a quadric surface by eliminating the parameter.
Compute x² + y²: (5t sin(3t))² + (6t cos(3t))² = 25t² sin²(3t) + 36t² cos²(3t).
This does not simplify to a clean form unless the coefficients match. Look for relationships involving z = t² and try x²/25 + y²/36 = t² = z. This gives x²/25 + y²/36 = z, an elliptic paraboloid.
For r(t) = ⟨3 + 5sin(t⁴), 2 − 6sin(t⁴), 3 + 7sin(t⁴)⟩, all three components are affine functions of one variable u = sin(t⁴).
This means the curve is a straight line segment (as u ranges over [−1, 1]) in the direction ⟨5, −6, 7⟩ through (3, 2, 3).
Any plane containing this line is a valid 2D surface. For example, any plane through the point (3, 2, 3) whose normal is perpendicular to ⟨5, −6, 7⟩.
For x + 2y + 3z ≤ 4 in the first octant (x, y, z ≥ 0):
Find the intercepts of the plane x + 2y + 3z = 4: (4, 0, 0), (0, 2, 0), (0, 0, 4/3).
The region is a tetrahedron with vertices at the origin and these three intercepts.
Sketch: draw the triangular face connecting the three intercepts, then connect each to the origin.
For |x| + 2|y| + 3|z| ≤ 4:
In the first octant this reduces to x + 2y + 3z ≤ 4 (the same tetrahedron).
The absolute values create mirror symmetry across each coordinate plane, so the full region is the union of eight reflected copies of that tetrahedron. The result is an octahedron-like shape, stretched differently along each axis.
The hyperboloid x²/a² + y²/b² − z²/c² = 1 has circular cross-sections when a = b.
At z = 0, the cross-section is a circle of radius a (the minimum, sometimes called the "waist").
At other heights, the radius grows: at height z, the radius is a√(1 + z²/c²).
For the cooling tower with base diameter 80m and minimum diameter 40m at 30m above the ground:
Place the origin at the waist (minimum diameter), so the waist is at z = 0 with radius 20. Thus a = b = 20.
The base is 30m below the waist (z = −30), with radius 40. Substituting: 40²/20² = 1 + 30²/c², so 4 = 1 + 900/c², giving c² = 300, c = 10√3.
The equation is x²/400 + y²/400 − z²/300 = 1.
A line on the hyperboloid through (0, b, 0) = (0, 20, 0): ruling lines on a hyperboloid of one sheet are found by factoring the equation. Each ruling line is straight and lies entirely on the surface.
\text{Circular cylinder: } x^2 + y^2 = R^2 \quad\Rightarrow\quad x = R\cos t,\; y = R\sin t\text{Elliptic paraboloid: } \frac{x^2}{a^2} + \frac{y^2}{b^2} = z\text{Hyperboloid of one sheet: } \frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1\text{Cross-sectional radius at height } z\colon \quad r(z) = a\sqrt{1 + z^2/c^2} \quad (\text{when } a = b)Cooling towers at power stations are built as hyperboloids of one sheet because the shape is structurally strong (doubly ruled, so it can be constructed from straight steel beams) and promotes natural air circulation. Parametric curves on surfaces appear in CNC machining, where a cutting tool follows a space curve defined by two intersecting surfaces.
Students often confuse a cylinder in multivariable calculus with a finite tube. In this course, x² + y² = 64 extends infinitely along z. The word "cylinder" here means any surface generated by translating a curve along a line.
When parametrising a curve of intersection, students sometimes introduce two separate parameters. You only need one: the trick is choosing a single parameter that satisfies both surface equations.
The hyperboloid of one sheet (one minus sign) is often confused with the hyperboloid of two sheets (two minus signs, or equivalently, = −1 on the right side). One sheet is connected; two sheets has two separate pieces.
For the absolute-value region |x| + 2|y| + 3|z| ≤ 4, students sometimes forget that removing the absolute values creates eight cases (one per octant). The full shape is not a tetrahedron but an octahedron-like solid.
⚠️ Identifying and sketching quadric surfaces from their equations is explicitly flagged as exam material (the homework says "hint hint"). Practise completing the square and matching to the standard forms.
⚠️ The technique of parametrising a curve by intersecting two surfaces is a core skill for §13.1 and beyond. Expect an exam question that gives two surfaces and asks for the parametric curve.
⚠️ Sketching 3D regions from inequalities (first-octant constraints, absolute-value regions) tests your ability to visualise. Drawing the intercepts and bounding planes is the reliable method.
True or false: the cylinder x² + y² = 9 in 3D extends infinitely along the z-axis.
Fill in the blank: the standard parametrisation of x² + y² = R² is x = ____, y = ____.
True or false: a hyperboloid of one sheet has two separate pieces.
Fill in the blank: the intercepts of x + 2y + 3z = 4 are (____, 0, 0), (0, ____, 0), and (0, 0, ____).
True or false: r(t) = ⟨3 + 5sin(t⁴), 2 − 6sin(t⁴), 3 + 7sin(t⁴)⟩ traces a straight line segment.
Answers: 1. True. 2. R cos(t), R sin(t). 3. False (one sheet is connected; two sheets has two pieces). 4. (4, 0, 0), (0, 2, 0), (0, 0, 4/3). 5. True (all components are affine in sin(t⁴)).
Q: Find parametric equations for the curve where x² + y² = 64 meets x² + y² + 18z = 0.
A: Let x = 8cos(t), y = 8sin(t). Then 64 + 18z = 0 gives z = −32/9. The curve is r(t) = ⟨8cos(t), 8sin(t), −32/9⟩, a circle of radius 8 at constant height.
Q: What quadric surface does r(t) = ⟨5t sin(3t), 6t cos(3t), t²⟩ lie on?
A: Compute x²/25 + y²/36 = t² sin²(3t) + t² cos²(3t) = t² = z. The surface is the elliptic paraboloid x²/25 + y²/36 = z.
Q: A hyperboloid x²/a² + y²/a² − z²/c² = 1 has a waist radius of 20 and a radius of 40 at z = 30. Find a and c.
A: At z = 0, radius = a = 20. At z = 30, radius = a√(1 + 900/c²) = 40, so √(1 + 900/c²) = 2, giving 1 + 900/c² = 4, so c² = 300 and c = 10√3.
Q: Describe the region |x| + 2|y| + 3|z| ≤ 4 in words.
A: It is a centrally symmetric solid (an octahedron stretched along the coordinate axes) with semi-axis lengths 4 along x, 2 along y, and 4/3 along z. In any single octant, the region is a tetrahedron; the eight tetrahedra fit together to form the full solid.
Parametric curves connect directly to §13.1 (vector functions and space curves), where you will compute derivatives, arc length, and curvature of these curves. The quadric surface classification from §12.6 is the toolkit you use to identify what surface a curve lives on. The hyperboloid modelling problem bridges into engineering applications and foreshadows the surface integrals in §16.
Tags: parametric equations, space curves, cylinder, paraboloid, hyperboloid of one sheet, quadric surfaces, first octant, region sketching, absolute value inequalities, vector-valued functions, ruling lines, cooling tower, MATH 215, Calculus III, §12.6, §13.1, University of Michigan