Difficulty: Intermediate to Advanced | Prerequisites: Lines and planes in 3D, basic differentiation rules, vector-valued functions. You should be comfortable with the chain rule and product rule from single-variable calculus.
This covers the core differential calculus of MATH 126: computing partial derivatives, using them to find tangent planes to surfaces, measuring curvature of space curves, and solving optimisation problems for functions of two variables. These skills form the analytical heart of the course and are tested heavily on every final exam. If you can handle these topics fluently, you are well placed for the rest of the course.
Partial derivatives let you find the slope of a surface in one direction at a time, and plugging them into the tangent plane formula gives you the best linear approximation to a surface at a point. Curvature measures how sharply a space curve bends, computed from cross products of velocity and acceleration. For optimisation, set both partial derivatives to zero, solve the system, then verify that your critical point is actually a maximum (or minimum) by checking the boundary or using the second derivative test.
Partial derivative
The derivative of a multivariable function with respect to one variable, treating all others as constants. Think of it as asking "how does the output change if I nudge just this one input?"
Tangent plane
The plane that best approximates a surface at a given point. It is the 3D analogue of the tangent line in single-variable calculus.
Curvature (κ)
A scalar that measures how sharply a curve bends at a particular point. A straight line has curvature zero; a tight circle has high curvature. In simple terms, it tells you how quickly the curve is changing direction.
Critical point
A point where all first partial derivatives equal zero (or do not exist). These are the candidates for local maxima, minima, and saddle points.
Vector-valued function r(t)
A function that outputs a vector for each input t, tracing out a curve in space. Think of it as a position function: at time t, the particle is at position r(t).
To find ∂z/∂x, differentiate z with respect to x, treating y as a constant.
To find ∂z/∂y, differentiate z with respect to y, treating x as a constant.
Apply the standard rules (chain rule, product rule, quotient rule) exactly as in single-variable calculus.
Example: z = x²y − √(3x − y) − x/y. Then ∂z/∂x = 2xy − 3/(2√(3x − y)) − 1/y.
The tangent plane at (a, b, f(a, b)) is: z = fₓ(a, b)(x − a) + f_y(a, b)(y − b) + f(a, b).
In the form Ax + By + Cz = D, rearrange after computing the partial derivatives at the point.
Example: z = x²y − √(3x − y) − x/y at (6, 2, 65). Compute ∂z/∂x|(6,2) = 185/8 and ∂z/∂y|(6,2) = 301/8. The tangent plane becomes 185x + 301y − 8z = 1192.
Always compute the general partial derivative first, then substitute the specific coordinates.
Watch for terms like √(3x − y) that need the chain rule: ∂/∂x[√(3x − y)] = 3/(2√(3x − y)).
Fractions are common in the answer. Do not panic if you get 185/8; leave it or clear denominators as the problem directs.
For a vector-valued function r(t), the curvature at a point is: κ = |r′(t) × r″(t)| / |r′(t)|³.
This requires computing the first and second derivatives of r(t), their cross product, and the magnitudes.
Find the value of t where the curve meets the given condition (e.g. intersects a plane).
Example: r(t) = ⟨t² + t, t³, −3t²/2 − 2t⟩ intersects y = −8 when t³ = −8, so t = −2.
Compute r′(t) and r″(t) in general, then evaluate at the specific t value.
r′(t) = ⟨2t + 1, 3t², −3t − 2⟩, so r′(−2) = ⟨−3, 12, 4⟩.
r″(t) = ⟨2, 6t, −3⟩, so r″(−2) = ⟨2, −12, −3⟩.
Compute the cross product r′ × r″.
r′(−2) × r″(−2) = ⟨12, −1, 12⟩.
Compute the magnitudes: |r′(−2)| = √(9 + 144 + 16) = 13. |r′(−2) × r″(−2)| = √(144 + 1 + 144) = 17.
Final answer: κ = 17/13³ = 17/2197.
Set fₓ(x, y) = 0 and f_y(x, y) = 0 simultaneously.
Solve the resulting system of equations for x and y.
Example: Maximise V = xyz subject to z = 20 − x² − y in the first octant. Write V = f(x, y) = xy(20 − x² − y) = 20xy − x³y − xy².
fₓ = 20y − 3x²y − y² = 0 → y(20 − 3x² − y) = 0.
f_y = 20x − x³ − 2xy = 0 → x(20 − x² − 2y) = 0.
In many optimisation problems (especially volume), x = 0 or y = 0 gives a minimum (zero volume), not a maximum. Discard those and solve the non-trivial system.
From the example: 20 − 3x² − y = 0 and 20 − x² − 2y = 0. Subtract to eliminate: solving gives y = 8, x = 2, z = 8.
Check the boundary of the domain. If the function is zero on the boundary (e.g. where x, y, or z = 0), and positive at the critical point, then the critical point must be the maximum.
Alternatively, use the second derivative test: compute D = fₓₓf_yy − (fₓᵧ)². If D > 0 and fₓₓ < 0, you have a local maximum.
In the example: the only critical point with positive volume is (2, 8, 8), giving V = 128. On the boundary, V = 0. So the maximum volume is 128.
Tangent plane to z = f(x, y) at (a, b)
z − f(a, b) = fₓ(a, b)(x − a) + f_y(a, b)(y − b)
Curvature of r(t)
κ = |r′(t) × r″(t)| / |r′(t)|³
Critical point condition
fₓ(x, y) = 0 and f_y(x, y) = 0 simultaneously
Second derivative test
D = fₓₓf_yy − (fₓᵧ)². If D > 0 and fₓₓ < 0: local max. If D > 0 and fₓₓ > 0: local min. If D < 0: saddle point.
Tangent planes are used in computer graphics to approximate curved surfaces with flat patches for rendering. Curvature matters in road and railway design: sharper curves require lower speed limits. Optimisation of box volume under surface constraints is directly relevant to packaging design, where manufacturers maximise internal volume while minimising material cost.
Students often compute partial derivatives correctly but forget to evaluate them at the given point before writing the tangent plane equation.
When computing curvature, a common error is dividing by |r′(t)|² instead of |r′(t)|³. The cube is correct.
Students sometimes assume a critical point is automatically a maximum without checking the boundary or applying the second derivative test. Always verify.
When the problem says "write in the form Ax + By + Cz = D," students sometimes leave the answer as z = ... and lose marks. Read the instructions.
⚠️ Tangent plane problems are worth 12 points on recent MATH 126 finals. Practise computing both partial derivatives and evaluating at the point until the process is automatic.
⚠️ Curvature problems require careful cross-product arithmetic. Triple-check your determinant expansion.
⚠️ Optimisation questions often require you to prove that your answer is the maximum (or minimum), not just a critical point. State why the boundary gives a smaller value.
⚠️ Read the output format instruction carefully. "Write in the form Ax + By + Cz = D" means clear all fractions and have integer coefficients if possible.
True or False: The tangent plane at a point uses the values of the partial derivatives at that point, not the general formulas. (True)
Fill in the blank: Curvature κ = |r′ × r″| / |r′|^______. (3)
True or False: If fₓ = 0 and f_y = 0 at a point, that point must be a local maximum. (False, it could be a minimum or saddle point)
Fill in the blank: In the second derivative test, D = fₓₓf_yy − (______)². (fₓᵧ)
True or False: A curvature of zero means the curve is a straight line at that point. (True)
Q: Find the tangent plane to z = x²y − √(3x − y) − x/y at the point (6, 2, 65). Write your answer in the form Ax + By + Cz = D.
A: Compute ∂z/∂x = 2xy − 3/(2√(3x − y)) − 1/y. At (6, 2): 24 − 3/8 − 1/2 = 185/8. Compute ∂z/∂y = x² + 1/(2√(3x − y)) + x/y². At (6, 2): 36 + 1/8 + 3/2 = 301/8. Tangent plane: z = (185/8)(x − 6) + (301/8)(y − 2) + 65. Multiply through by 8: 8z = 185(x − 6) + 301(y − 2) + 520. Final answer: 185x + 301y − 8z = 1192.
Q: Find the curvature of r(t) = ⟨t² + t, t³, −3t²/2 − 2t⟩ at the point where the curve intersects y = −8.
A: t³ = −8, so t = −2. r′(−2) = ⟨−3, 12, 4⟩, r″(−2) = ⟨2, −12, −3⟩. Cross product: ⟨12, −1, 12⟩. |r′| = 13, |r′ × r″| = 17. κ = 17/13³ = 17/2197.
Q: Find the maximum volume of a rectangular box in the first octant with one vertex at the origin and the opposite vertex on z = 20 − x² − y.
A: V = xy(20 − x² − y). Set Vₓ = 0 and V_y = 0. Solving: x = 2, y = 8, z = 8. V = 2 × 8 × 8 = 128. Boundary gives V = 0, so 128 is the maximum.
Partial derivatives lead directly into the gradient vector and directional derivatives, which underpin Lagrange multipliers and the chain rule for multivariable functions. Curvature connects to the TNB (tangent-normal-binormal) frame used in physics for describing motion along curves. Optimisation techniques extend to double and triple integrals when you need to find volumes or centres of mass over constrained regions.
Partial derivatives, tangent plane equation, curvature formula, cross product, vector-valued function, critical points, second derivative test, optimisation, maximum volume, first octant, gradient, directional derivative, Lagrange multipliers, chain rule multivariable, MATH 126, Calculus III, UW