Difficulty: Intermediate | Prerequisites: Chapters 12–13 (vectors, 3D geometry), single-variable differentiation
Chapter 14 is where calculus meets surfaces. Instead of a single derivative for a curve, you now take partial derivatives of functions of two (or more) variables. These let you build tangent planes, approximate surfaces locally, and find maxima and minima on surfaces, which is the multivariable version of the optimisation you did in MATH 124.
Partial derivatives let you measure how a function of several variables changes when you nudge one variable at a time. With them, you can build tangent planes (the best flat approximation to a surface at a point) and find local and global maxima and minima using the second derivative test and the extreme value theorem.
Partial derivative (∂z/∂x or fₓ)
The derivative of f(x, y) with respect to x, treating y as a constant. Think of it as the slope of the surface in the x-direction at a given point.
Second-order partial derivatives (fₓₓ, f_yy)
The partial derivative taken twice. fₓₓ measures concavity in the x-direction; f_yy measures concavity in the y-direction.
Mixed partials (fₓᵧ, fᵧₓ)
Partial derivatives taken with respect to two different variables. By Clairaut's theorem, fₓᵧ = fᵧₓ provided both are continuous.
Tangent plane
The plane that best approximates the surface z = f(x, y) at a point. Its equation uses the two partial derivatives at that point.
Linear approximation L(x, y)
The tangent plane used as an approximation of f near (x₀, y₀). Think of it as the multivariable version of the tangent line approximation from MATH 124.
Critical point
A point (a, b) where fₓ(a, b) = 0 and fᵧ(a, b) = 0. These are the candidates for local maxima, local minima, or saddle points.
Saddle point
A critical point that is neither a local max nor a local min. The surface curves upward in one direction and downward in another, like a horse saddle.
Second derivative test (discriminant D)
D = fₓₓ(a, b) fᵧᵧ(a, b) - [fₓᵧ(a, b)]². The sign of D, combined with the sign of fₓₓ, tells you whether a critical point is a max, min, or saddle.
Extreme value theorem (for two variables)
If f is continuous on a closed, bounded region, then f attains an absolute maximum and an absolute minimum somewhere in that region, either at a critical point, on the boundary, or at a corner.
fₓ(x, y) = lim(h→0) [f(x+h, y) - f(x, y)] / h
fᵧ(x, y) = lim(h→0) [f(x, y+h) - f(x, y)] / h
In practice, to compute fₓ, differentiate with respect to x while treating y as a constant (and vice versa)
fₓₓ = second partial with respect to x (concavity in x-direction)
fᵧᵧ = second partial with respect to y (concavity in y-direction)
Mixed partials: fₓᵧ = fᵧₓ (Clairaut's theorem, provided continuity)
Domain restrictions to watch for:
Expression | Restriction |
|---|---|
√(BLAH) | BLAH ≥ 0 |
STUFF/BLAH | BLAH ≠ 0 |
ln(BLAH) | BLAH > 0 |
sin⁻¹(BLAH) | -1 ≤ BLAH ≤ 1 |
Tangent plane at (x₀, y₀): z - z₀ = fₓ(x₀, y₀)(x - x₀) + fᵧ(x₀, y₀)(y - y₀)
where z₀ = f(x₀, y₀)
Tangent in the x-direction: the vector ⟨1, 0, fₓ(x₀, y₀)⟩
Tangent in the y-direction: the vector ⟨0, 1, fᵧ(x₀, y₀)⟩
Normal to the surface: cross product of the two tangent vectors = ⟨-fₓ, -fᵧ, 1⟩
Linear approximation: L(x, y) = z₀ + fₓ(x₀, y₀)(x - x₀) + fᵧ(x₀, y₀)(y - y₀)
This is the same equation as the tangent plane, used to estimate f(x, y) near (x₀, y₀)
Set fₓ(a, b) = 0 and fᵧ(a, b) = 0 simultaneously
Solve the resulting system of equations
Compute D = fₓₓ(a, b) fᵧᵧ(a, b) - [fₓᵧ(a, b)]²
If D > 0 and fₓₓ > 0: local minimum
If D > 0 and fₓₓ < 0: local maximum
If D < 0: saddle point
If D = 0: the test is inconclusive
Find all critical points inside the region
Find critical numbers and corners on each boundary:
For each boundary, write an equation relating x and y
Substitute into f to get a single-variable function
Find that function's critical numbers and note the endpoints
Evaluate f at every point found in steps 1 and 2
The largest value is the global max; the smallest is the global min
Topic | Formula |
|---|---|
Partial derivative (x) | fₓ = lim(h→0) [f(x+h, y) - f(x, y)] / h |
Tangent plane | z - z₀ = fₓ(x₀, y₀)(x - x₀) + fᵧ(x₀, y₀)(y - y₀) |
Normal to surface | ⟨-fₓ(x₀, y₀), -fᵧ(x₀, y₀), 1⟩ |
Linear approximation | L(x, y) = f(x₀, y₀) + fₓ(x - x₀) + fᵧ(y - y₀) |
Discriminant (2nd deriv test) | D = fₓₓ fᵧᵧ - (fₓᵧ)² |
Students sometimes think fₓ = 0 alone is enough to identify a critical point. You need both fₓ = 0 and fᵧ = 0 simultaneously.
Students confuse the discriminant D with the second derivative test from single-variable calculus. In multivariable, D = fₓₓ fᵧᵧ - (fₓᵧ)². The sign of fₓₓ only matters after you have confirmed D > 0.
Students forget to check the boundary when finding absolute extrema on a closed region. The extreme value theorem guarantees an absolute max and min exist, but they can occur on the boundary, not just at interior critical points.
⚠️ The second derivative test (computing D and classifying the critical point) is tested on virtually every exam.
⚠️ Finding the absolute max/min on a closed bounded region is a multi-step process. Practise the full procedure: interior critical points, then boundary analysis, then compare all values.
⚠️ Tangent plane equations appear regularly. Know how to find z₀, fₓ, and fᵧ at the given point.
⚠️ Domain restrictions (the table above) affect where partial derivatives exist. Exam problems sometimes ask for the domain of f before computing derivatives.
True or false: fₓᵧ = fᵧₓ always.
True (assuming the mixed partials are continuous, which they are in this course).
Fill in the blank: at a critical point, fₓ = ______ and fᵧ = ______.
0 and 0.
True or false: if D < 0 at a critical point, the point is a local minimum.
False. D < 0 means it is a saddle point.
Fill in the blank: the normal vector to the surface z = f(x, y) at (x₀, y₀) is ⟨______, ______, 1⟩.
-fₓ(x₀, y₀), -fᵧ(x₀, y₀).
Q: Find the critical points of f(x, y) = x² + y² - 4x + 6y.
A: fₓ = 2x - 4 = 0 gives x = 2. fᵧ = 2y + 6 = 0 gives y = -3. The only critical point is (2, -3).
Q: Classify the critical point (2, -3) of f(x, y) = x² + y² - 4x + 6y using the second derivative test.
A: fₓₓ = 2, fᵧᵧ = 2, fₓᵧ = 0. D = (2)(2) - 0² = 4 > 0, and fₓₓ = 2 > 0. So (2, -3) is a local minimum.
Q: Find the equation of the tangent plane to z = x² + y² at the point (1, 1, 2).
A: fₓ = 2x, so fₓ(1,1) = 2. fᵧ = 2y, so fᵧ(1,1) = 2. Tangent plane: z - 2 = 2(x - 1) + 2(y - 1), which simplifies to z = 2x + 2y - 2.
Partial derivatives generalise the single-variable derivative from MATH 124. The tangent plane is the 2D analogue of the tangent line. The second derivative test here extends the concavity test from single-variable calculus, with the added complication of the mixed partial fₓᵧ.
This chapter sets up Chapter 15: double integrals compute volumes under surfaces, and the regions you integrate over are often defined by the boundaries you would analyse here.
Related Terms / Search Tags
partial derivative, partial differentiation, tangent plane, linear approximation, linearisation, critical point, saddle point, second derivative test, discriminant, absolute maximum, absolute minimum, extreme value theorem, closed bounded region, mixed partials, Clairaut's theorem, MATH 126, Calculus III, UW, multivariable optimisation