Difficulty: Intermediate to Advanced | Prerequisites: Ch. 12 (3D space, vectors, dot and cross products, lines and planes), single-variable calculus (derivatives, integrals, chain rule, Calc I max/min).
This document covers the second half of the course: moving from static vectors to curves in motion (Chapter 13) and then to functions of two variables and optimisation (Chapter 14). You will learn to differentiate and integrate vector functions component-wise, measure how curves bend (curvature), decompose acceleration, compute partial derivatives, build tangent planes, find critical points of multivariable functions, and locate absolute maxima and minima over closed regions.
If you are catching up, make sure you are solid on Chapter 12 material (vectors, dot/cross products, lines and planes) and single-variable calculus (derivatives, integrals, the chain rule, and Calc I optimisation). Everything here generalises those tools to higher dimensions.
Vector-valued function / parametric curve, r(t)
A function r(t) = ⟨f(t), g(t), h(t)⟩ that traces out a curve in 3D as t varies. Think of t as time and r(t) as the position of a moving particle.
Tangent vector, r'(t)
The derivative of r(t), computed component-wise: r'(t) = ⟨x'(t), y'(t), z'(t)⟩. It points in the direction the curve is heading at time t.
Unit tangent vector, T(t)
r'(t) divided by its own magnitude: T(t) = r'(t) / |r'(t)|. Pure direction, no speed information.
Principal unit normal, N(t)
T'(t) divided by its magnitude. It points "inward" toward the centre of the curve's turning circle.
Arc length
The total distance travelled along the curve from t = a to t = b: ∫ |r'(t)| dt. Not the same as displacement.
Curvature, κ
A measure of how sharply a curve bends. Computed as κ = |r' × r''| / |r'|³ in 3D or κ = |f''(x)| / (1 + (f'(x))²)^(3/2) in 2D.
Partial derivative
The derivative of a multivariable function with respect to one variable while treating all others as constants. f_x means "differentiate with respect to x, hold y fixed."
Tangent plane
The plane that best approximates the surface z = f(x, y) near a point. Equation: z = f(a, b) + f_x(a, b)(x − a) + f_y(a, b)(y − b).
Critical point (multivariable)
A point (a, b) where both f_x(a, b) = 0 and f_y(a, b) = 0 (or where either partial does not exist).
Second derivative test (multivariable)
Uses D = f_xx · f_yy − (f_xy)² at a critical point to classify it as a local min, local max, or saddle point.
Parametric form: x = f(t), y = g(t), z = h(t). Vector form: r(t) = ⟨f(t), g(t), h(t)⟩.
Eliminate the parameter to find relationships between x, y and z. This tells you what surface the curve lives on.
Plot a few points (try t = 0, 1, 2 or t = 0, π/2, π) and connect with arrows in the direction of increasing t.
x = r cos(t), y = r sin(t) traces a circle of radius r. Use sin²(t) + cos²(t) = 1 to eliminate t and get x² + y² = r².
For an ellipse: x = a cos(t), y = b sin(t) gives x²/a² + y²/b² = 1.
Adding z = t on top of circular motion gives a helix spiralling upward.
Two curves intersecting: use different parameters (t and u), set coordinates equal, solve the system. Check all three coordinate equations.
Two curves colliding: same as intersection, but both parameters must be equal (same time).
Parameterising the curve of intersection of two surfaces: pick one variable equal to t, solve for the others. If there is a circle or ellipse involved, use cos(t) and sin(t).
Differentiate and integrate component-wise:
r'(t) = ⟨x'(t), y'(t), z'(t)⟩ (tangent vector)
r''(t) = ⟨x''(t), y''(t), z''(t)⟩ (acceleration vector)
∫ r(t) dt = ⟨∫ x(t) dt, ∫ y(t) dt, ∫ z(t) dt⟩, each with its own constant of integration
The tangent line to r(t) at t = a uses the point r(a) and the direction r'(a):
x = x(a) + x'(a) u, y = y(a) + y'(a) u, z = z(a) + z'(a) u.
T(t) = r'(t) / |r'(t)| (unit tangent, the direction of travel)
N(t) = T'(t) / |T'(t)| (principal unit normal, points inward)
B(t) = T(t) × N(t) (binormal, perpendicular to both T and N)
The normal plane at a point is the plane with normal vector T(t), passing through that point.
Distance from t = a to t = b:
Arc length = ∫_a^b |r'(t)| dt = ∫_a^b √(x'² + y'² + z'²) dt.
The arc length function from starting time a: s(t) = ∫_a^t |r'(u)| du.
Compute s(t) and simplify.
Solve for t in terms of s.
Replace t with that expression in r(t) to get r(s).
This is useful when you want to describe position in terms of distance travelled rather than time.
Curvature κ measures how sharply a curve bends, independent of the speed you traverse it.
Three equivalent formulas:
κ(t) = |T'(t)| / |r'(t)|
κ(t) = |r'(t) × r''(t)| / |r'(t)|³ (the most commonly used in 3D)
κ(x) = |f''(x)| / (1 + (f'(x))²)^(3/2) (for a 2D curve y = f(x))
The radius of curvature is 1/κ: it is the radius of the circle that best fits the curve at that point.
When t represents time:
Velocity: v(t) = r'(t)
Speed: |v(t)| = |r'(t)|
Acceleration: a(t) = r''(t)
Acceleration splits into a component along the direction of travel and a component toward the centre of curvature:
Tangential component: a_T = (r'(t) · r''(t)) / |r'(t)|
Normal component: a_N = |r'(t) × r''(t)| / |r'(t)|
a_T measures how the speed is changing. a_N measures how the direction is changing.
F(t) = m a(t). To model motion: find all the forces, divide by mass to get acceleration, then integrate to find velocity and position (using initial conditions to determine constants).
z = f(x, y) gives the height of a surface above the xy-plane. The domain is the set of (x, y) pairs for which the function is defined. Recall the standard restrictions: square roots need non-negative arguments, denominators cannot be zero, logarithms need positive arguments.
Level curves are the 2D curves you get by fixing z at different constants and graphing the resulting equation in x and y. They are the contour lines of the surface, like a topographic map.
f_x(x, y) = ∂z/∂x: differentiate with respect to x, treat y as a constant.
f_y(x, y) = ∂z/∂y: differentiate with respect to y, treat x as a constant.
Geometric meaning: f_x(a, b) is the slope of the surface in the x-direction at (a, b). f_y(a, b) is the slope in the y-direction. Positive means uphill, negative means downhill.
Higher-order partials:
f_xx = ∂²z/∂x² (differentiate f_x with respect to x again)
f_yy = ∂²z/∂y²
f_xy = ∂²z/∂y∂x (differentiate f_x with respect to y)
f_yx = ∂²z/∂x∂y
By Clairaut's theorem, f_xy = f_yx when both are continuous.
The tangent plane to z = f(x, y) at the point (x₀, y₀, z₀) is:
z − z₀ = f_x(x₀, y₀)(x − x₀) + f_y(x₀, y₀)(y − y₀)
This doubles as the linear approximation: for (x, y) near (x₀, y₀),
f(x, y) ≈ L(x, y) = f(x₀, y₀) + f_x(x₀, y₀)(x − x₀) + f_y(x₀, y₀)(y − y₀).
Finding critical points: set f_x = 0 and f_y = 0 simultaneously and solve for x and y. Also include any points where a partial derivative does not exist.
Classifying critical points: at each critical point (a, b), compute:
D = f_xx(a, b) · f_yy(a, b) − [f_xy(a, b)]²
D > 0 and f_xx > 0: local minimum.
D > 0 and f_xx < 0: local maximum.
D < 0: saddle point.
D = 0: inconclusive (try other methods).
By the extreme value theorem, a continuous function on a closed, bounded region attains its absolute maximum and minimum. To find them:
Find all critical points inside the region and evaluate f at each.
For each boundary curve, substitute to reduce f to a single-variable function, then find its critical numbers and endpoints. Evaluate f at all of them.
The largest value among all evaluations is the absolute max; the smallest is the absolute min.
This mirrors the Calc I method (find critical numbers, check endpoints) but with more boundary work.
\mathbf{r}'(t) = \langle x'(t),\, y'(t),\, z'(t) \rangle\text{Arc length} = \int_a^b |\mathbf{r}'(t)|\, dt\kappa = \frac{|\mathbf{r}' \times \mathbf{r}''|}{|\mathbf{r}'|^3}a_T = \frac{\mathbf{r}' \cdot \mathbf{r}''}{|\mathbf{r}'|}, \quad a_N = \frac{|\mathbf{r}' \times \mathbf{r}''|}{|\mathbf{r}'|}z - z_0 = f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)D = f_{xx} f_{yy} - (f_{xy})^2Students often confuse speed with velocity. Speed is |r'(t)|, a scalar. Velocity is r'(t), a vector.
When computing curvature, a common mistake is using |r'|² in the denominator instead of |r'|³. The cube is correct.
For partial derivatives, students sometimes differentiate with respect to both variables at once instead of holding one constant. Each partial derivative treats the other variable as a fixed number.
In absolute max/min problems, students frequently check the interior critical points but forget to check the boundary, or they check only some boundary segments. Every piece of the boundary must be examined.
When using the second derivative test, students sometimes forget that D = 0 is inconclusive, not a saddle point.
⚠️ Arc length and curvature computations are standard exam fare. Practise until you can compute |r'(t)|, the cross product r' × r'', and their magnitudes quickly and cleanly.
⚠️ Tangent line problems are straightforward if you remember the recipe: evaluate r(a) for the point and r'(a) for the direction.
⚠️ For partial derivatives, expect to differentiate functions involving products, quotients, trig, exponentials, and logarithms. Being fluent with these is non-negotiable.
⚠️ The absolute max/min procedure is lengthy but mechanical. Draw the region, label the boundary pieces, and work through each one systematically. Show all your work.
⚠️ The second derivative test (computing D) is heavily tested. Memorise the classification rules.
Fill in the blank: If r(t) = ⟨t², sin(t), e^t⟩, then r'(t) = ___. (⟨2t, cos(t), e^t⟩)
True or false: Arc length can be negative. (False, it is always non-negative.)
Fill in the blank: The curvature of a straight line is ___. (0)
True or false: f_xy and f_yx are always equal. (True when both are continuous, by Clairaut's theorem.)
Fill in the blank: If D > 0 and f_xx < 0, the critical point is a ___. (local maximum)
Q: Find the unit tangent vector T(t) for r(t) = ⟨3t, 4cos(t), 4sin(t)⟩.
A: r'(t) = ⟨3, −4sin(t), 4cos(t)⟩. |r'(t)| = √(9 + 16sin²(t) + 16cos²(t)) = √25 = 5. T(t) = ⟨3/5, −4sin(t)/5, 4cos(t)/5⟩.
Q: Find the curvature of f(x) = 5x² and determine where it is maximised.
A: f'(x) = 10x, f''(x) = 10. κ(x) = 10 / (1 + 100x²)^(3/2). This is maximised when x = 0 (denominator is smallest), giving κ(0) = 10.
Q: Find the critical points of f(x, y) = x² − 8y³ + 4xy + 1 and classify them.
A: f_x = 2x + 4y = 0, so x = −2y. f_y = −24y² + 4x = 0, so x = 6y². Combining: −2y = 6y², giving y(3y + 1) = 0, so y = 0 or y = −1/3. Critical points: (0, 0) and (2/3, −1/3). f_xx = 2, f_yy = −48y, f_xy = 4. D = 2(−48y) − 16 = −96y − 16. At (0, 0): D = −16 < 0, so saddle point. At (2/3, −1/3): D = 32 − 16 = 16 > 0 and f_xx = 2 > 0, so local minimum.
Q: Find the tangent plane to z = x cos(y + 2x) + y² at (−1, 2).
A: f(−1, 2) = −cos(0) + 4 = 3. f_x = cos(y + 2x) − 2x sin(y + 2x); f_x(−1, 2) = 1. f_y = −x sin(y + 2x) + 2y; f_y(−1, 2) = 4. Tangent plane: z = 3 + 1(x + 1) + 4(y − 2) = x + 4y − 4.
The tangent vector r'(t) and curvature κ reappear when studying line integrals and flux in Chapter 16. Partial derivatives are the foundation of the gradient vector (14.6), which in turn drives directional derivatives, the chain rule for multivariable functions, and Lagrange multipliers (14.8). The second derivative test generalises to the bordered Hessian in constrained optimisation.
vector-valued function, parametric curve, space curve, helix, tangent vector, unit tangent, unit normal, binormal, TNB frame, arc length, arc length reparameterisation, curvature, radius of curvature, velocity, speed, acceleration, tangential acceleration, normal acceleration, Newton's second law, projectile motion, partial derivative, mixed partial, Clairaut's theorem, tangent plane, linear approximation, linearisation, critical point, second derivative test, discriminant D, saddle point, local max, local min, absolute max, absolute min, boundary, closed region, extreme value theorem, optimisation, Math 126, Calculus III, UW Calculus