Source: Homework Set 2, §13.1–13.2
Tags: space curves, parametric equations, tangent vector, tangent line, curve of intersection, vector-valued functions
Difficulty: Intermediate Prerequisites: Equations of planes (§12.5), dot product, cross product, derivatives of basic functions.
Space curves extend the idea of parametric curves in 2D into three dimensions. Instead of y = f(x), you now describe a path through 3D space with a vector-valued function r(t). The derivative r’(t) gives you the tangent vector at each point, which tells you the direction the curve is heading. This section is the gateway to arc length, curvature and the TNB frame that appear in the next few weeks. If you can differentiate component functions and write the equation of a line in 3D, you are ready.
A space curve r(t) = ⟨x(t), y(t), z(t)⟩ traces a path in 3D. Its derivative r’(t) is the tangent vector. To find where the tangent is parallel to a plane, set the dot product of r’(t) with the plane’s normal to zero. To show a curve lies in a plane, verify that every point satisfies a single plane equation.
Space curve / vector-valued function
A function r(t) = ⟨x(t), y(t), z(t)⟩ that maps a parameter t to a point in 3D space. Think of it as a set of instructions telling a particle where to be at each moment.
Tangent vector
The derivative r’(t) = ⟨x’(t), y’(t), z’(t)⟩. It points in the direction the curve is moving at time t. In simple terms, it is the velocity vector of a particle travelling along the curve.
Tangent line
The line through the point r(t₀) in the direction of r’(t₀). Written as L(s) = r(t₀) + s r’(t₀). This is the best straight-line approximation to the curve at that point.
Curve of intersection
The space curve formed where two surfaces meet. You find it by solving the surface equations simultaneously, then parameterising the result.
Parallel to a plane
A vector v is parallel to a plane when v is perpendicular to the plane’s normal, i.e. v · n = 0. A tangent line is parallel to a plane when the tangent vector satisfies this condition.
Given r(t) = ⟨x(t), y(t), z(t)⟩, the tangent vector at t = t₀ is r’(t₀) = ⟨x’(t₀), y’(t₀), z’(t₀)⟩.
The tangent line at t₀ is L(s) = r(t₀) + s r’(t₀).
The unit tangent vector is T(t) = r’(t) / |r’(t)|.
A tangent vector r’(t) is parallel to the plane ax + by + cz = d when r’(t) · ⟨a, b, c⟩ = 0.
Compute r’(t), form the dot product with the plane’s normal, set it equal to zero, and solve for t.
Example: for r(t) = ⟨2cos t, 2sin t, eᵗ⟩, we get r’(t) = ⟨–2sin t, 2cos t, eᵗ⟩. For the plane √3 x + y = 1, the normal is ⟨√3, 1, 0⟩. Set –2√3 sin t + 2cos t = 0, which gives tan t = 1/√3, so t = π/6. The point is r(π/6) = ⟨√3, 1, e^(π/6)⟩.
When two surfaces F(x,y,z) = c₁ and G(x,y,z) = c₂ intersect, the tangent to the curve of intersection at a point P is perpendicular to both surface gradients at P.
The tangent direction is ∇F × ∇G evaluated at P.
Example: cylinders x² + y² = 25 and y² + z² = 20 at (3, 4, 2). Gradients: ∇F = ⟨2x, 2y, 0⟩ = ⟨6, 8, 0⟩ and ∇G = ⟨0, 2y, 2z⟩ = ⟨0, 8, 4⟩. Cross product: ⟨6, 8, 0⟩ × ⟨0, 8, 4⟩ = ⟨32, –24, 48⟩ = 8⟨4, –3, 6⟩. The tangent line is r(s) = ⟨3, 4, 2⟩ + s⟨4, –3, 6⟩.
To prove r(t) = ⟨x(t), y(t), z(t)⟩ lies in a plane, find constants a, b, c, d such that ax(t) + by(t) + cz(t) = d for all t.
One approach: substitute the component functions and check whether the equation holds identically.
Another approach: show that the vector from one point on the curve to any other point is always perpendicular to some fixed normal.
Example: r(t) = ⟨4t + 1, 2 + 6t – 2t², t² + t⟩. Try ax(t) + by(t) + cz(t) = d. Substituting and collecting powers of t: the t² terms give –2b + c = 0, the t terms give 4a + 6b + c = 0, and the constant terms give a + 2b = d. Setting c = 2b and substituting into the second equation gives a = –2b. With b = 1: a = –2, c = 2, d = 0. Check: –2(4t+1) + (2+6t–2t²) + 2(t²+t) = –8t–2+2+6t–2t²+2t²+2t = 0. The plane is –2x + y + 2z = 0.
\mathbf{r}'(t) = \langle x'(t),\, y'(t),\, z'(t) \rangle\text{Tangent line: } \mathbf{L}(s) = \mathbf{r}(t_0) + s\,\mathbf{r}'(t_0)\text{Tangent parallel to plane } ax+by+cz=d \text{: } \mathbf{r}'(t) \cdot \langle a,b,c \rangle = 0\text{Tangent to curve of intersection: } \nabla F \times \nabla G \text{ at the point}Tangent vectors describe instantaneous velocity. In robotics, the tangent to a pre-programmed path tells the controller which direction the arm should be moving at each moment. In road design, the tangent line at any point on a highway’s centreline determines the direction a car faces, which matters for banking and sight-line calculations.
Students sometimes confuse “tangent line parallel to a plane” with “tangent line lies in the plane.” Parallel to a plane only means the direction vector is perpendicular to the normal. The line need not touch the plane at all.
When finding the tangent to a curve of intersection, students often try to parameterise the whole curve first. You do not need to. The gradient cross-product method gives the tangent direction directly at the point.
Forgetting that r’(t) is a direction, not a point. The tangent line passes through the point r(t₀), and r’(t₀) only tells you which way it goes.
Mixing up “the curve lies in a plane” with “the curve touches a plane.” A curve lies in a plane when every single point satisfies the plane equation, not just one or two.
⚠️ The gradient cross-product technique for curves of intersection is a high-value exam skill. It avoids messy parameterisation.
⚠️ “Show the curve lies in a plane” problems test your ability to set up and solve a system of linear equations for the plane coefficients. Practise the method of undetermined coefficients (grouping by powers of t).
⚠️ Know how to interpret “parallel to a plane” as a dot-product condition, not a geometric coincidence.
True or False: r’(t) gives the position of the curve at time t. (False, it gives the tangent/velocity vector.)
Fill in the blank: The tangent to the curve of intersection of two surfaces has direction ______ evaluated at the point. (∇F × ∇G)
True or False: If r’(t) · n = 0 for a specific t₀, the tangent line at t₀ is parallel to the plane with normal n. (True)
Fill in the blank: To show a curve lies in a plane, you must verify that every point satisfies ______. (the plane equation)
True or False: The unit tangent vector T(t) always has magnitude 1. (True)
Q: For r(t) = ⟨t², t, 3t⟩, find the tangent line at t = 1.
A: r(1) = ⟨1, 1, 3⟩, r’(t) = ⟨2t, 1, 3⟩, so r’(1) = ⟨2, 1, 3⟩. Tangent line: L(s) = ⟨1 + 2s, 1 + s, 3 + 3s⟩.
Q: At what value of t is the tangent to r(t) = ⟨sin t, cos t, t⟩ perpendicular to the z-axis?
A: The z-axis direction is ⟨0, 0, 1⟩. We need r’(t) · ⟨0, 0, 1⟩ = 0. Since r’(t) = ⟨cos t, –sin t, 1⟩, the z-component is always 1, so the dot product is always 1, which is never zero. The tangent is never perpendicular to the z-axis.
Q: Two surfaces x² + z² = 10 and y² + z² = 10 intersect. Find the tangent direction at (1, 1, 3).
A: ∇F = ⟨2, 0, 6⟩, ∇G = ⟨0, 2, 6⟩. Cross product: ⟨(0)(6) – (6)(2), (6)(0) – (2)(6), (2)(2) – (0)(0)⟩ = ⟨–12, –12, 4⟩, or simplified ⟨–3, –3, 1⟩.
Q: Show that r(t) = ⟨t, 2t, 3t⟩ lies in a plane and find the plane.
A: This is a straight line through the origin, so it lies in infinitely many planes. One is 2x – y = 0 (check: 2t – 2t = 0 for all t). Another is 3x – z = 0.
Tangent vectors lead directly into arc length (§13.3) and curvature (§13.3), where you use |r’(t)| as the speed and the rate of change of the unit tangent to measure how sharply the curve bends. The gradient cross-product method here reappears when computing normal vectors to surfaces in §14.6.
Related Terms / Search Tags
space curve, vector-valued function, parametric curve 3D, tangent vector derivative, tangent line 3D, curve of intersection, gradient cross product, parallel to plane tangent, curve in a plane, MATH 215 tangent, Calc III curves, unit tangent vector