This material covers how to describe curves in 3D using vector functions, how to measure how sharply a curve bends (curvature), and how to recover velocity from acceleration using integration. These problems tie together differentiation, integration, and vector algebra in one place.
Difficulty: Intermediate | Prerequisites: Vectors, dot and cross products (see Part 1 notes), single-variable differentiation and integration.
Once you know how to describe lines and planes, the next step in MATH 126 is to describe curves that twist through 3D space. A vector function r(t) = ⟨x(t), y(t), z(t)⟩ traces out a space curve as t varies. Differentiation gives you velocity and acceleration; the cross product of these gives curvature, which measures bending. These ideas set up the machinery for later topics such as line integrals, where you integrate a function along a curve. If you are comfortable with single-variable derivatives and integrals component-by-component, the mechanics here are approachable.
Vector function / vector-valued function
r(t) = ⟨x(t), y(t), z(t)⟩. A function whose output is a vector in 3D, tracing a curve as t varies. Think of it as: the position of a particle at time t.
Space curve
The path traced out by a vector function as the parameter changes. In simple terms, a curve that lives in three dimensions rather than on a flat plane.
Derivative of a vector function
r'(t) = ⟨x'(t), y'(t), z'(t)⟩. Differentiate each component separately. The result is the velocity vector, tangent to the curve.
Velocity vector
v(t) = r'(t). The direction and speed of a particle at time t. Its magnitude |v(t)| is the speed.
Acceleration vector
a(t) = v'(t) = r''(t). The rate of change of velocity.
Curvature
A scalar κ measuring how sharply a curve bends at a point. Defined as κ = |r'(t) × r''(t)| / |r'(t)|³. Think of it as: the reciprocal of the radius of the best-fitting circle at that point. A straight line has curvature zero; a tight turn has high curvature.
Speeding up vs slowing down
A particle is speeding up when v(t) · a(t) > 0 and slowing down when v(t) · a(t) < 0. Equivalently, it speeds up when the angle between velocity and acceleration is acute (less than 90°), and slows down when that angle is obtuse.
A vector function r(t) = ⟨x(t), y(t), z(t)⟩ describes a curve in 3D. As t changes, the tip of the position vector traces out the curve.
Differentiation: differentiate each component: r'(t) = ⟨x'(t), y'(t), z'(t)⟩
Integration: integrate each component, adding a constant vector C: ∫r(t) dt = ⟨∫x(t) dt, ∫y(t) dt, ∫z(t) dt⟩ + C
Initial conditions determine the constant vector. For example, if you know v(0) = ⟨1, -1, 1⟩, integrate a(t) to get v(t) with unknown constants, then plug in t = 0 to solve for them.
Given r(t) = ⟨x(t), y(t), z(t)⟩, find an equation relating x, y, z (with no t) that the curve satisfies.
The strategy is to use trigonometric identities or algebraic manipulation to eliminate t from the parametric equations.
Example from the exam: r(t) = ⟨1 + 2 sin t, t², 3 cos t⟩.
x = 1 + 2 sin t, so sin t = (x - 1)/2
z = 3 cos t, so cos t = z/3
Use sin²t + cos²t = 1: ((x - 1)/2)² + (z/3)² = 1
This gives (x - 1)²/4 + z²/9 = 1, an elliptic cylinder (no y term, so it extends along the y-axis)
The key insight is that x and z both depend on trig functions of t, so the Pythagorean identity eliminates t. The y-component (t²) is independent of the trig part, so it does not appear in the surface equation.
Curvature κ measures the rate at which the curve changes direction per unit arc length.
Formula: κ = |r'(t) × r''(t)| / |r'(t)|³
Step-by-step for the exam problem:
Given r(t) = ⟨1 + 2 sin t, t², 3 cos t⟩, find curvature at (1, 0, 3).
Find the value of t: x = 1 requires sin t = 0, so t = 0. Check: y = 0² = 0, z = 3 cos 0 = 3. Confirmed, t = 0.
Compute r'(t) = ⟨2 cos t, 2t, -3 sin t⟩. At t = 0: r'(0) = ⟨2, 0, 0⟩.
Compute r''(t) = ⟨-2 sin t, 2, -3 cos t⟩. At t = 0: r''(0) = ⟨0, 2, -3⟩.
Cross product r'(0) × r''(0) = |i j k; 2 0 0; 0 2 -3| = ⟨(0)(-3) - (0)(2), (0)(0) - (2)(-3), (2)(2) - (0)(0)⟩ = ⟨0, 6, 4⟩.
|r'(0) × r''(0)| = √(0 + 36 + 16) = √52 = 2√13.
|r'(0)| = √(4 + 0 + 0) = 2, so |r'(0)|³ = 8.
κ = 2√13 / 8 = √13 / 4.
Recovering velocity from acceleration:
Given a(t) = ⟨2, 6e^(3t), cos t⟩ and v(0) = ⟨1, -1, 1⟩:
Integrate component by component: v(t) = ⟨2t + C₁, 2e^(3t) + C₂, sin t + C₃⟩
Apply v(0) = ⟨1, -1, 1⟩: C₁ = 1, 2 + C₂ = -1 so C₂ = -3, C₃ = 1
Result: v(t) = ⟨2t + 1, 2e^(3t) - 3, sin t + 1⟩
Angle between velocity and acceleration:
At t = 0: v(0) = ⟨1, -1, 1⟩ and a(0) = ⟨2, 6, 1⟩.
v · a = (1)(2) + (-1)(6) + (1)(1) = 2 - 6 + 1 = -3
|v| = √(1 + 1 + 1) = √3
|a| = √(4 + 36 + 1) = √41
cos θ = -3 / (√3 · √41) = -3 / √123
θ = arccos(-3/√123)
Speeding up or slowing down:
The particle speeds up when v · a > 0 and slows down when v · a < 0. At t = 0, v · a = -3 < 0, so the particle is slowing down.
The logic: if the acceleration has a component in the direction of motion (dot product positive), it pushes the particle faster. If the component opposes the motion (dot product negative), it brakes the particle.
Velocity from acceleration: v(t) = ∫a(t) dt + C, where C is found from the initial velocity
Speed: |v(t)| = √(v₁² + v₂² + v₃²)
Curvature: κ = |r'(t) × r''(t)| / |r'(t)|³
Angle between vectors: θ = arccos((v · a) / (|v| |a|))
Speed test: if v · a > 0 then speeding up; if v · a < 0 then slowing down; if v · a = 0 then speed is momentarily constant
Students often confuse speed (a scalar, |v|) with velocity (a vector, v). Speed is always non-negative.
Curvature is not the same as the second derivative. Curvature uses the cross product of r' and r'', divided by |r'|³. Forgetting the denominator is a common error.
When integrating a(t) to find v(t), students frequently forget to add the constant vector C and then solve for it using the initial condition. Each component gets its own constant.
"The particle is slowing down" does not mean the acceleration is zero. It means the acceleration has a component opposing the velocity. The dot product v · a captures this.
Students sometimes try to find the angle between velocity and acceleration using the cross product alone. You need the dot product for the angle formula (arccos of the normalised dot product).
⚠️ "Find a surface containing the curve" is a standard question. Look for trig components you can combine via sin²t + cos²t = 1, or polynomial components you can relate algebraically.
⚠️ The curvature formula κ = |r' × r''| / |r'|³ appears on nearly every MATH 126 midterm. Memorise it and practise the cross-product computation by hand.
⚠️ "Speeding up or slowing down" is a one-mark question that students lose by forgetting the method. Compute v · a and check the sign.
⚠️ When recovering v(t) from a(t), show every constant of integration and every substitution of the initial condition. Partial credit depends on visible work.
True or false: the velocity vector v(t) is always tangent to the curve at the point r(t). (True)
Fill in the blank: curvature κ = |______ × | / ||³. (r'(t); r''(t); r'(t))
True or false: if v · a = 0, the particle is speeding up. (False, speed is momentarily constant)
Fill in the blank: to find v(t) from a(t), you ______ each component of a(t) and use the ______ to find the constants. (integrate; initial condition)
True or false: a curve with curvature zero at a point is curving sharply at that point. (False, curvature zero means a straight-line segment)
Q: Given r(t) = ⟨1 + 2 sin t, t², 3 cos t⟩, find the equation of a surface containing the curve.
A: sin t = (x - 1)/2 and cos t = z/3. Then sin²t + cos²t = 1 gives (x - 1)²/4 + z²/9 = 1.
Q: For the same curve, find the curvature at the point (1, 0, 3).
A: t = 0 at that point. r'(0) = ⟨2, 0, 0⟩, r''(0) = ⟨0, 2, -3⟩. Cross product: ⟨0, 6, 4⟩. Magnitude: 2√13. |r'(0)|³ = 8. κ = 2√13/8 = √13/4.
Q: The acceleration of a particle is a(t) = 2i** + 6e^(3t)j + cos t k, and v(0) = i - j + k. Find v(t).**
A: Integrate: v(t) = ⟨2t + C₁, 2e^(3t) + C₂, sin t + C₃⟩. From v(0) = ⟨1, -1, 1⟩: C₁ = 1, C₂ = -3, C₃ = 1. So v(t) = ⟨2t + 1, 2e^(3t) - 3, sin t + 1⟩.
Q: Is the particle from the previous question speeding up or slowing down at t = 0?
A: v(0) · a(0) = ⟨1, -1, 1⟩ · ⟨2, 6, 1⟩ = 2 - 6 + 1 = -3 < 0. The particle is slowing down.
Q: Find the angle between the velocity and acceleration at t = 0.
A: cos θ = -3 / (√3 · √41) = -3/√123. θ = arccos(-3/√123), which is an obtuse angle (greater than 90°), consistent with the particle slowing down.
Curvature comes back when studying motion along curves (the TNB frame, normal and tangential components of acceleration). The integration technique used to recover velocity from acceleration is the same pattern you will use to solve separable differential equations later. Finding surfaces containing curves is directly related to understanding level surfaces and the classification of quadric surfaces, which appear throughout the rest of MATH 126.
Vector function, vector-valued function, space curve, parametric curve 3D, velocity vector, acceleration vector, speed, curvature formula, cross product curvature, speeding up slowing down dot product, integrating acceleration, initial velocity condition, surface containing a curve, eliminating the parameter, Pythagorean identity, elliptic cylinder, MATH 126, Calculus III, UW, University of Washington, multivariable calculus, TNB frame, unit tangent vector