Difficulty: Intermediate | Prerequisites: Part 1 study notes (vector functions, space curves), derivatives of trig and exponential functions, dot product formula for angles.
Once you can write down a vector function for a curve, the next questions are geometric: which way is the curve heading at a given point, how fast is a particle accelerating along it, how long is the curve, and at what angle do two curves cross? This section covers the derivative and integral machinery that answers those questions. Everything here rests on the tangent vector r′(t), so make sure you are comfortable differentiating vector functions component-by-component before diving in.
The tangent vector r′(t) tells you the direction and speed of motion along a curve. The acceleration vector r″(t) is its derivative. Arc length is the integral of |r′(t)|, and two curves' angle of intersection is the angle between their tangent vectors at the crossing point.
Tangent vector
The derivative r′(t) = ⟨f′(t), g′(t), h′(t)⟩ at a point on the curve. It points in the instantaneous direction of travel. Think of it as the velocity of a particle moving along the curve.
Unit tangent vector
The normalised tangent vector: T(t) = r′(t) / |r′(t)|. In simple terms, it keeps the direction but strips away the speed, giving a pure direction of length 1.
Tangent line
The line through r(t₀) in the direction of r′(t₀). Its parametric equation is L(s) = r(t₀) + s r′(t₀). Think of it as the straight path the particle would follow if it stopped turning at that instant.
Acceleration vector
The second derivative r″(t) = ⟨f″(t), g″(t), h″(t)⟩. It captures how the velocity is changing, both in speed and in direction.
Angle of intersection
When two curves cross at a point, the angle between them is the angle between their tangent vectors at that point, computed via the dot product formula. In simple terms, it is how sharply the two paths diverge where they meet.
Parallel to a plane
A line (or tangent vector) is parallel to a plane when it is perpendicular to the plane's normal vector. The condition is r′(t) · n = 0, where n is the normal to the plane.
Differentiate each component of r(t) separately: r′(t) = ⟨f′(t), g′(t), h′(t)⟩.
Example: if r(t) = (1 + cos t)i + (3 + sin t)j, then r′(t) = (-sin t)i + (cos t)j.
The tangent vector has both magnitude and direction. Its magnitude |r′(t)| is the speed of the particle.
At each point r(t₀), draw r′(t₀) as an arrow with its tail at that point.
For the circle r(t) = (1 + cos t, 3 + sin t), the tangent vector at t = 0 is (0, 1), pointing straight up: the particle is at the rightmost point of the circle and heading upward.
The tangent line at t = t₀ is L(s) = r(t₀) + s r′(t₀).
This is a line in the direction of the tangent vector, passing through the point of tangency.
"Tangent line parallel to a plane" means the tangent vector is perpendicular to the plane's normal.
Differentiate r′(t) to get r″(t) = ⟨f″(t), g″(t), h″(t)⟩.
Example: for r(t) = (2 cos t, 2 sin t, eᵗ), r′(t) = (-2 sin t, 2 cos t, eᵗ) and r″(t) = (-2 cos t, -2 sin t, eᵗ).
At t = π/6 this gives r″(π/6) = (-√3, -1, e^(π/6)).
The tangent line is parallel to a plane ax + by + cz = d when r′(t) · n = 0, where n = ⟨a, b, c⟩.
Write out the dot product as a single equation in t.
Solve for t within the given domain.
Substitute back into r(t) to get the point.
Example from the homework: r(t) = (2 cos t, 2 sin t, eᵗ) and the plane √3 x + y = 7 (normal n = ⟨√3, 1, 0⟩).
r′(t) · n = -2√3 sin t + 2 cos t = 0.
This gives tan t = 1/√3, so t = π/6 (within [0, π]).
The point is r(π/6) = (√3, 1, e^(π/6)).
For a curve r(t) from t = a to t = b:
L = \int_a^b |\mathbf{r}'(t)|\, dt = \int_a^b \sqrt{[f'(t)]^2 + [g'(t)]^2 + [h'(t)]^2}\, dtExam problems are often designed so that the expression under the square root factors as a perfect square. Always check before integrating.
Example: r(t) = 12t i + 8t^(3/2) j + 3t² k, for 0 ≤ t ≤ 1.
r′(t) = ⟨12, 12t^(1/2), 6t⟩.
|r′(t)|² = 144 + 144t + 36t² = 4(36 + 36t + 9t²) = 4(6 + 3t)² = (12 + 6t)².
So |r′(t)| = 12 + 6t, and the integral is straightforward: L = [12t + 3t²] from 0 to 1 = 15.
When a cylinder (e.g. 4x² + y² = 4) meets a plane (e.g. x + y + z = 2):
Parametrise the cylinder: x = cos t, y = 2 sin t (so that 4cos²t + 4sin²t = 4).
Use the plane to get z: z = 2 - cos t - 2 sin t.
Write r(t) = ⟨cos t, 2 sin t, 2 - cos t - 2 sin t⟩ for t ∈ [0, 2π].
Compute |r′(t)| and integrate (this one typically requires a calculator for the final decimal).
When two curves r₁(t) and r₂(t) cross at a point, first find the parameter values where the intersection occurs. Then compute each curve's tangent vector at that point. The angle θ between the curves is:
\cos\theta = \frac{|\mathbf{r}_1'(t_1) \cdot \mathbf{r}_2'(t_2)|}{|\mathbf{r}_1'(t_1)|\,|\mathbf{r}_2'(t_2)|}Use the absolute value of the dot product because two curves meeting at a point have two supplementary angles, and the convention is to report the acute one.
Given r₁(t) = (t, 1 - t, 3 + t²) and r₂(t) = (3 - t, t - 2, t²):
Set components equal (using different parameters t and s if needed, though here both use t): t = 3 - s, 1 - t = s - 2, 3 + t² = s². Solving gives the intersection point and the parameter values.
Compute r₁′ and r₂′ at those values.
Apply the dot product formula and take arccos to find the angle. Round to the nearest degree as required.
\mathbf{r}'(t) = \langle f'(t),\, g'(t),\, h'(t) \rangle\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|}\text{Tangent line: } \mathbf{L}(s) = \mathbf{r}(t_0) + s\,\mathbf{r}'(t_0)\text{Parallel to plane: } \mathbf{r}'(t) \cdot \mathbf{n} = 0L = \int_a^b |\mathbf{r}'(t)|\, dt\cos\theta = \frac{|\mathbf{r}_1' \cdot \mathbf{r}_2'|}{|\mathbf{r}_1'|\,|\mathbf{r}_2'|}Tangent vectors are how navigation systems compute the heading of a moving vehicle at each instant. Arc length calculations arise whenever engineers need the total length of a cable, pipeline, or road that follows a curved path. The angle-of-intersection concept is used in road design (the angle at which two roads meet determines sight lines and turning radii) and in robotic arm path planning.
Students often forget to use the absolute value of the dot product when finding the angle of intersection. Without it you may get an obtuse angle when the convention calls for the acute one.
Students often think the tangent vector must be a unit vector. It does not. r′(t) gives the tangent direction with magnitude equal to the speed. You normalise it to get T(t) only when you specifically need a unit vector.
Students often try to integrate the components of r′(t) separately to find arc length. That does not work. You integrate the magnitude of r′(t), which involves a square root of the sum of squares.
Students often miss that "parallel to a plane" means perpendicular to the plane's normal, not parallel to any vector in the plane.
⚠️ "Find the point where the tangent line is parallel to a given plane" is a high-frequency exam question. The method is always: set r′(t) · n = 0 and solve.
⚠️ Arc length problems on exams almost always simplify to a perfect square under the radical. If your integrand looks horrible, re-check your algebra.
⚠️ The angle-of-intersection problem requires you to find the point of intersection first, then compute tangent vectors. Missing either step is a common source of lost marks.
⚠️ The acceleration vector at a specific point is a standard follow-up after finding the tangent line. Do not forget to evaluate r″(t) at the correct t-value.
True or False: The tangent vector r′(t) always has magnitude 1. (False, only the unit tangent vector T(t) does.)
Fill in the blank: A tangent line is parallel to a plane when the tangent vector is ______ to the plane's normal. (perpendicular)
True or False: To find arc length, you can integrate each component of r′(t) separately and add the results. (False, you integrate the magnitude of the whole vector.)
Fill in the blank: The angle between two intersecting curves is found using the ______ formula applied to their tangent vectors. (dot product)
True or False: If |r′(t)|² simplifies to (12 + 6t)², then |r′(t)| = 12 + 6t (assuming 12 + 6t > 0 on the interval). (True)
Q: For r(t) = (1 + cos t) i + (3 + sin t) j, find the tangent vector r′(t) and evaluate it at t = π/2.
A: r′(t) = (-sin t, cos t). At t = π/2: r′(π/2) = (-1, 0). The tangent points directly to the left.
Q: Find the arc length of r(t) = 12t i + 8t^(3/2) j + 3t² k for 0 ≤ t ≤ 1.
A: r′(t) = (12, 12√t, 6t). |r′(t)|² = 144 + 144t + 36t² = (12 + 6t)². So L = ∫₀¹ (12 + 6t) dt = [12t + 3t²]₀¹ = 15.
Q: At what point do r₁(t) = (t, 1 - t, 3 + t²) and r₂(t) = (3 - t, t - 2, t²) intersect? Find the angle of intersection.
A: Setting components equal: t₁ = 3 - t₂ and 1 - t₁ = t₂ - 2. From the second: t₁ + t₂ = 3, combined with t₁ = 3 - t₂ gives t₂ = ... (solve the system). Then compute r₁′ and r₂′ at those values, take the dot product, and use arccos. The answer is approximately 73°.
Q: Find the acceleration of r(t) = (2 cos t, 2 sin t, eᵗ) at t = π/6.
A: r″(t) = (-2 cos t, -2 sin t, eᵗ). At t = π/6: r″(π/6) = (-√3, -1, e^(π/6)).
Q: Parametrise the intersection of the cylinder 4x² + y² = 4 and the plane x + y + z = 2.
A: Let x = cos t, y = 2 sin t (satisfies 4cos²t + 4sin²t = 4). Then z = 2 - cos t - 2 sin t. So r(t) = (cos t, 2 sin t, 2 - cos t - 2 sin t), t ∈ [0, 2π].
Tangent vectors feed directly into the TNB (tangent-normal-binormal) frame, which is the next major topic in most Calculus III courses. Arc length parametrisation, where t is replaced by s (arc length itself), uses the integral from this section and simplifies many later formulas. The condition for tangent-line parallelism to a plane foreshadows the gradient and normal-vector work that comes with surfaces and partial derivatives.
tangent vector, unit tangent vector, tangent line, acceleration vector, velocity vector, arc length, arc length formula, speed, angle of intersection, dot product angle, parallel to plane, normal vector, perpendicular condition, TNB frame, curvature, Math 215, Calculus III, space curves, parametric derivatives, multivariable calculus