Source: Homework Set 3, §13.1–13.4
Tags: parametric curves, vector-valued functions, tangent line, tangent vector, unit tangent vector, singular point, Calculus III, MATH 215
Difficulty: Intermediate | Prerequisites: Single-variable differentiation, basic vectors in R³, parametric equations from Calculus II.
This material is the foundation for everything else in multivariable calculus that involves curves. You are moving from single-variable parametric curves (Calc II) to vector-valued functions in three dimensions. The core skill here is computing tangent vectors and tangent lines to space curves, which you will need again for arc length, curvature, and eventually line integrals. If you are comfortable differentiating trig functions and working with vectors, you are ready.
A vector-valued function r(t) traces a curve in space. Its derivative r'(t) gives the tangent vector at each point. Where r'(t) = 0, the curve has a singular point and the tangent line is undefined. The unit tangent vector T(t) = r'(t)/|r'(t)| gives direction only.
Vector-valued function
A function r(t) = ⟨f(t), g(t), h(t)⟩ that assigns a vector in R³ to each value of the parameter t. Think of it as a set of instructions that traces out a path in space as t changes.
Parametric curve (space curve)
The set of points traced by a vector-valued function as t varies over its domain. In simple terms, it is the path itself, the trail left behind as t runs through its interval.
Tangent vector
The derivative r'(t) = ⟨f'(t), g'(t), h'(t)⟩, computed component by component. It points in the direction the curve is moving at time t and its magnitude reflects the speed of traversal.
Unit tangent vector T(t)
Defined as T(t) = r'(t) / |r'(t)|. This normalises the tangent vector to length 1, stripping out speed information and leaving only direction. Only defined where r'(t) ≠ 0.
Singular point (cusp)
A point on the curve where r'(t) = 0, so the tangent vector vanishes and the unit tangent vector is undefined. The curve may have a cusp, a corner, or reverse direction at such a point. Think of it as a moment where the curve "stalls" before changing behaviour.
Tangent line
The line through a point on the curve in the direction of the tangent vector. Given a point r(t₀) and tangent vector r'(t₀), the tangent line is L(s) = r(t₀) + s · r'(t₀), where s is a scalar parameter.
Differentiate each component of r(t) separately: if r(t) = ⟨f(t), g(t), h(t)⟩, then r'(t) = ⟨f'(t), g'(t), h'(t)⟩.
The tangent vector r'(t₀) at a specific point tells you the instantaneous direction and speed of the curve at that parameter value.
First, identify the parameter value t₀ that produces the given point. Solve r(t₀) = (x₀, y₀, z₀) for t₀.
Compute r'(t₀).
Write the tangent line as: L(s) = r(t₀) + s · r'(t₀), or equivalently in parametric form:
x = x₀ + f'(t₀) · s
y = y₀ + g'(t₀) · s
z = z₀ + h'(t₀) · s
r'(t) = ⟨3sin²(t)cos(t), −3cos²(t)sin(t), 2sin(t)cos(t)⟩.
At the point (1/(2√2), 1/(2√2), 1/2): solving gives t = π/4.
r'(π/4) = ⟨3/4, −3/4, 1⟩ (after substitution).
Tangent line: L(s) = (1/(2√2), 1/(2√2), 1/2) + s · (3/4, −3/4, 1).
At the point (1, 0, 1): solving r(t) = (1, 0, 1) gives t = π/2.
r'(π/2) = ⟨0, 0, 0⟩. The tangent vector is the zero vector.
The unit tangent vector T(t) = r'(t)/|r'(t)| is undefined here because you would divide by zero.
Geometrically, the curve has a cusp at this point. The curve arrives, stalls (zero velocity), and changes direction. There is no single well-defined tangent direction.
This is a key concept: a parametrisation can fail to produce a tangent line at certain points, even though the curve itself exists there. The issue is with the parametrisation, or genuinely with the geometry of the curve (a cusp or corner).
This technique appears in Question 4 and previews Chapter 14. It uses parametric curves on a surface to find a tangent plane.
Given a surface z = f(x, y), you can create parametric curves that lie on the surface by holding one variable constant.
For the paraboloid z = 2x² + y²:
Fixing x = 1 gives r₁(t) = ⟨1, t, 2 + t²⟩. This traces the cross-section of the paraboloid in the plane x = 1.
Fixing y = 3 gives r₂(s) = ⟨s, 3, 2s² + 9⟩. This traces the cross-section in the plane y = 3.
r₁'(t) and r₂'(s) are tangent to these cross-sectional curves, which lie on the surface.
Therefore these velocity vectors are also tangent to the surface itself.
Two tangent vectors at a point determine a tangent plane.
At the point (1, 3, 11): use t = 3 in r₁ and s = 1 in r₂.
r₁'(3) = ⟨0, 1, 6⟩ and r₂'(1) = ⟨1, 0, 4⟩.
The normal vector to the tangent plane is the cross product: n = r₁'(3) × r₂'(1) = ⟨4, 6, −1⟩ (compute via determinant).
The tangent plane equation: 4(x − 1) + 6(y − 3) − 1(z − 11) = 0, which simplifies to 4x + 6y − z = 11.
\mathbf{r}'(t) = \langle f'(t),\, g'(t),\, h'(t) \rangle\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|}L(s) = \mathbf{r}(t_0) + s \cdot \mathbf{r}'(t_0)\mathbf{n} = \mathbf{r}_1'(t_0) \times \mathbf{r}_2'(s_0)The tangent plane through point (x₀, y₀, z₀) with normal n = ⟨a, b, c⟩ is: a(x − x₀) + b(y − y₀) + c(z − z₀) = 0.
Tangent vectors describe instantaneous velocity of objects moving through 3D space. Every time an engineer models a particle path, a robot arm trajectory, or a satellite orbit, they are working with vector-valued functions and their derivatives. Tangent planes are essential in computer graphics for shading and in engineering for approximating curved surfaces locally as flat ones.
Students often assume that if r'(t₀) = 0, the curve itself does not exist at that point. The curve is still there; the parametrisation simply fails to produce a tangent direction. The point may be a cusp or a place where the curve reverses.
Confusing the tangent vector with the unit tangent vector. The tangent vector r'(t) carries speed information (its magnitude tells you how fast the curve is traversed). The unit tangent vector T(t) gives direction only.
Forgetting that multiple t values can map to the same point on the curve. Always check whether the parameter value you found actually gives the correct point.
When finding tangent planes, students sometimes forget to take the cross product. Two tangent vectors alone define a plane, but you need their cross product to get the normal vector for the plane equation.
⚠️ "Find the tangent line at a point" is a standard exam question. Know the full procedure: find t₀, compute r'(t₀), write L(s).
⚠️ Singular points where r'(t) = 0 are a favourite for exams. Be prepared to explain geometrically what is happening, not just state that the derivative is zero.
⚠️ The tangent-plane-via-cross-sections technique is a preview of Chapter 14 methods. Understanding it now gives you a head start on partial derivatives and tangent planes to surfaces.
True or false: If r'(t₀) = 0, the curve does not pass through r(t₀). ___
The unit tangent vector is computed by dividing r'(t) by ___.
True or false: The tangent line requires both a point and a direction vector. ___
To find a tangent plane to a surface at a point, you need ___ tangent vectors and you take their ___.
True or false: A singular point means the curve does not exist at that location. ___
Answers: 1. False (the curve still passes through the point). 2. Its magnitude |r'(t)|. 3. True. 4. Two; cross product. 5. False (the curve exists, but the parametrisation gives a zero derivative there).
Q: Given r(t) = ⟨t², t³, t⟩, find the tangent line at t = 1.
A: r(1) = (1, 1, 1), r'(t) = ⟨2t, 3t², 1⟩, so r'(1) = ⟨2, 3, 1⟩. Tangent line: L(s) = (1, 1, 1) + s(2, 3, 1).
Q: At what parameter value does r(t) = ⟨sin³(t), cos³(t), sin²(t)⟩ have a singular point on [0, π]?
A: r'(t) = 0 when sin(t)cos(t) = 0 and sin(t) = 0 simultaneously, or when all components vanish. At t = π/2, r'(π/2) = ⟨0, 0, 0⟩. The singular point is at t = π/2, corresponding to the point (1, 0, 1).
Q: Why can you use two parametric curves on a surface to determine a tangent plane?
A: Each curve lies on the surface, so its velocity vector is tangent to the surface. Two independent tangent vectors at a point span the tangent plane. Their cross product gives the normal to that plane.
Q: For the paraboloid z = 2x² + y², write a parametrisation of the trace in the plane y = 3.
A: r₂(s) = ⟨s, 3, 2s² + 9⟩. You fix y = 3 and let x = s vary, then z = 2s² + 9 follows from the surface equation.
This connects directly to arc length (§13.3), which uses |r'(t)| as the integrand. Curvature (§13.4) builds on the unit tangent vector by measuring how fast T(t) changes direction. The tangent plane technique previews the gradient and partial derivatives in Chapter 14, where you will find tangent planes using a general formula rather than constructing cross-sectional curves by hand.
Vector-valued function, parametric curve, space curve, tangent vector, velocity vector, unit tangent vector, singular point, cusp, tangent line to a curve, tangent plane, cross product, normal vector, paraboloid, cross-section, trace of a surface, parametrisation, MATH 215, Calculus III, §13.1, §13.2, §14.1