Difficulty: Intermediate | Prerequisites: Partial derivatives (MATH 126, Ch. 14 basics)
This material sits at the heart of multivariable optimisation. If single-variable calculus gave you local extrema via the first and second derivative tests, this extends those ideas to functions of two variables, where you now have partial derivatives, a discriminant, and the possibility of saddle points. You need to be comfortable with partial differentiation before diving in. If tangent lines made sense in Calc I, tangent planes are the natural next step, and linearization is how you use them to approximate.
A tangent plane is the flat surface that best approximates a function z = f(x, y) at a given point, and linearization uses that plane to estimate nearby values. Critical points of multivariable functions are found by setting both partial derivatives to zero, then classified as local min, local max, or saddle point using the second derivative test (the discriminant D).
Tangent plane
The plane that touches the surface z = f(x, y) at a point (x₀, y₀, z₀) and matches the surface's slope in both the x- and y-directions at that point. Think of it as the flat surface you would feel if you pressed a sheet of glass against a hill at one spot.
Linearization (linear approximation)
The function L(x, y) that uses the tangent plane to approximate f(x, y) near the point of tangency. In simple terms, this means replacing a complicated surface with a flat one for values close to a known point.
Critical point
A point (a, b) where both partial derivatives fₓ(a, b) = 0 and f_y(a, b) = 0, meaning the surface is momentarily flat in every direction. Think of it as a hilltop, a valley floor, or a mountain pass, depending on what happens around it.
Second derivative test (discriminant D)
A test using the second-order partial derivatives to classify a critical point. D(a, b) = fₓₓ(a, b) · f_yy(a, b) - [f_xy(a, b)]². In simple terms, D tells you whether a critical point is a peak, a valley, or a saddle.
Saddle point
A critical point where D < 0. The surface curves upward in one direction and downward in another, like the middle of a horse saddle or a Pringles crisp. It is neither a local max nor a local min.
Local minimum
A critical point where D > 0 and fₓₓ > 0. The surface curves upward in all directions from this point, like the bottom of a bowl.
Local maximum
A critical point where D > 0 and fₓₓ < 0 (and f_yy < 0). The surface curves downward in all directions from this point, like the top of a dome.
For a surface z = f(x, y) at the point (x₀, y₀, z₀):
z = fₓ(x₀, y₀)(x - x₀) + f_y(x₀, y₀)(y - y₀) + z₀
The normal vector to the surface at that point is ⟨fₓ, f_y, -1⟩
If an exam asks you to find the tangent plane, this is the normal vector you plug into the plane equation along with the given point
To use: evaluate fₓ and f_y at (x₀, y₀), find z₀ = f(x₀, y₀), and substitute everything in
L(x, y) = fₓ(x₀, y₀)(x - x₀) + f_y(x₀, y₀)(y - y₀) + f(x₀, y₀)
This is the same equation as the tangent plane, just written as an approximation function
Use it to estimate f at points near (x₀, y₀) without computing f itself
Works well when (x, y) is close to (x₀, y₀); accuracy drops as you move further away
Step 1: Find partial derivatives fₓ and f_y
Step 2: Set both equal to zero and solve for the variables to find all critical points
Step 3: Apply the second derivative test at each critical point
Compute D(a, b) = fₓₓ(a, b) · f_yy(a, b) - [f_xy(a, b)]²
If D > 0 and fₓₓ > 0, the point is a local minimum
If D > 0 and fₓₓ < 0, the point is a local maximum
If D < 0, the point is a saddle point
If D = 0, the test is inconclusive (you need other methods)
fₓ: the slope of the surface as you move in the positive x-direction (holding y constant)
f_y: the slope of the surface as you move in the positive y-direction (holding x constant)
fₓₓ: concavity in the x-direction. Positive means the surface curves upward (bowl shape), negative means it curves downward (dome shape)
f_yy: concavity in the y-direction, same interpretation
f_xy: the rate at which the x-slope changes as you move in the y-direction (equivalently, f_yx, since f_xy = f_yx for well-behaved functions)
z = f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0) + f(x_0, y_0)D(a, b) = f_{xx}(a, b) \cdot f_{yy}(a, b) - \bigl[f_{xy}(a, b)\bigr]^2\vec{n} = \langle f_x,\; f_y,\; -1 \rangleTangent planes are how engineers approximate curved surfaces locally. If you are machining a curved part, the tangent plane at any point tells you the angle of the cutting tool. Linearization is the principle behind every "small perturbation" analysis in physics and engineering, where you approximate a complicated function near a known operating point.
Students often think D > 0 alone is enough to declare a local min. It is not. You also need fₓₓ > 0 for a min (or fₓₓ < 0 for a max). D > 0 only tells you the point is an extremum of some kind.
Students sometimes confuse the tangent plane equation with linearization. They are the same formula. Linearization is just the name for using the tangent plane as an approximation.
A common error is forgetting to evaluate all partial derivatives at the specific point (x₀, y₀) before substituting. The derivatives must be numbers, not expressions.
Students sometimes assume D = 0 means saddle point. It does not. D = 0 means the test gives no answer, and you need further analysis.
⚠️ The second derivative test is tested heavily. Know the decision tree cold: D > 0 with fₓₓ > 0 = min, D > 0 with fₓₓ < 0 = max, D < 0 = saddle.
⚠️ Tangent plane problems often ask for the equation of the plane and the normal vector. Write both forms.
⚠️ Linearization questions typically give you a point and ask you to approximate f at a nearby point. Show the full substitution.
⚠️ When classifying critical points, show all your working: partial derivatives set to zero, the system of equations solved, D computed at each critical point. Marks go to the process, not just the answer.
True or false: If D(a, b) > 0 and f_yy(a, b) > 0, the critical point is a local minimum. (True, because D > 0 with positive concavity in either direction confirms a minimum.)
Fill in the blank: The normal vector to the surface z = f(x, y) is ⟨fₓ, f_y, ____⟩. (-1)
True or false: A saddle point is where D = 0. (False. A saddle point is where D < 0. D = 0 is inconclusive.)
Fill in the blank: Linearization approximates f(x, y) using the ____ at the point. (tangent plane)
Q: Given f(x, y) = x² + y² - 2x - 4y + 5, find and classify all critical points.
A: fₓ = 2x - 2 = 0 gives x = 1. f_y = 2y - 4 = 0 gives y = 2. Critical point: (1, 2). fₓₓ = 2, f_yy = 2, f_xy = 0. D = (2)(2) - 0² = 4 > 0 and fₓₓ = 2 > 0, so (1, 2) 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. f_y = 2y, so f_y(1,1) = 2. Tangent plane: z = 2(x - 1) + 2(y - 1) + 2, which simplifies to z = 2x + 2y - 2.
Q: If D(a, b) < 0, what type of critical point is (a, b)?
A: A saddle point. The surface curves upward in one direction and downward in another.
Q: What is the normal vector to z = 3x - y² at the point (1, 2, -1)?
A: fₓ = 3, f_y = -2y, so f_y(1, 2) = -4. Normal vector: ⟨3, -4, -1⟩.
This material connects directly to absolute extrema on bounded domains, which extends the critical-point analysis by adding boundary checks. The gradient vector ⟨fₓ, f_y⟩, covered in the next set of notes, points in the direction of steepest ascent and is perpendicular to level curves, tying together slopes, tangent planes, and directional derivatives. The normal vector ⟨fₓ, f_y, -1⟩ also appears when setting up surface integrals later in the course.
Tangent plane, linearization, linear approximation, second derivative test, discriminant, critical point classification, saddle point, local extrema, local minimum, local maximum, normal vector to surface, fₓₓ, f_yy, f_xy, Hessian, multivariable optimization, MATH 126, Calculus III, UW