Difficulty: Intermediate to Advanced
Prerequisites: Sections 14.1 and 14.3 (functions of two variables, partial derivatives). Familiarity with single-variable limits, the chain rule, and polar coordinates.
Limits in two variables are harder than in one variable because you can approach a point from infinitely many directions, not just from the left and right. This means showing a limit exists requires more work, and showing it does not exist is often easier (find two paths that give different answers). Continuity builds on limits, and several homework problems explore functions that have partial derivatives at a point yet are not continuous there, which is a genuinely surprising result. The multivariable chain rule extends the single-variable chain rule to compositions where the intermediate variables depend on other parameters. Clairaut's theorem on the equality of mixed partial derivatives rounds out the picture: it usually holds, but the homework shows a careful counterexample when the hypotheses fail.
To show a two-variable limit does not exist, find two paths to the point that give different limiting values. The multivariable chain rule tracks how a function changes through intermediate variables. Clairaut's theorem says f_xy = f_yx whenever the mixed partials are continuous, but the homework gives an example where they are not continuous and the theorem fails.
Limit of f(x, y) as (x, y) approaches (a, b)
The value L that f(x, y) gets arbitrarily close to as (x, y) approaches (a, b) along every possible path. Unlike one-variable limits, "every possible path" includes straight lines, curves, spirals, and anything else.
Path test for non-existence of a limit
If two different paths to (a, b) give different limiting values, the limit does not exist. Think of it as a quick disproof: you only need two paths that disagree.
Continuity at a point
f is continuous at (a, b) if: (1) f(a, b) is defined, (2) the limit of f as (x, y) approaches (a, b) exists, and (3) the limit equals f(a, b). All three conditions must hold.
Multivariable chain rule
If z = f(x, y) and x and y each depend on one or more other variables (say t, or r and s), then the chain rule expresses dz/dt (or the partials of z with respect to r and s) in terms of the partials of f and the derivatives of x and y.
Clairaut's theorem (equality of mixed partials)
If f_xy and f_yx are both continuous on a region containing (a, b), then f_xy(a, b) = f_yx(a, b). In simple terms, the order of differentiation does not matter, provided the mixed partials behave nicely.
Second-order partial derivatives
The four second partials of f(x, y): f_xx, f_yy, f_xy, and f_yx. The first two are the "pure" second partials; the last two are the "mixed" partials.
Coordinate transformation (polar)
Writing x = r cos theta, y = r sin theta. The chain rule then rewrites partial derivatives in terms of r and theta. The Laplacian in polar coordinates is a standard result: u_xx + u_yy = u_rr + (1/r) u_r + (1/r^2) u_{theta theta}.
The standard technique: approach (0, 0) along two different paths and show the limits disagree.
Straight-line paths: Set y = ax (or x = ay) and take the limit as x approaches 0. If the result depends on a, the limit does not exist.
Curved paths: Sometimes all straight-line paths give the same limit, but a curved path (like y = x^2) gives a different answer. This is the key lesson of Problem 8.
Along y = ax (straight lines): f = x(ax)^3 / (x^2 + (ax)^6) = a^3 x^4 / (x^2 + a^6 x^6). As x approaches 0, this goes to 0 for any fixed a. All straight-line limits are 0.
Along y = x^(1/3) (the curve where y^3 = x): f = x * x / (x^2 + x^2) = x^2 / (2x^2) = 1/2.
Since 0 is not equal to 1/2, the limit as (x, y) approaches (0, 0) does not exist, so f is not continuous at the origin.
If z = f(x, y), x = x(t), y = y(t), then dz/dt = f_x * dx/dt + f_y * dy/dt.
More generally, if x = x(r, s) and y = y(r, s), then the partial of z with respect to r is f_x * (partial x / partial r) + f_y * (partial y / partial r), and similarly for s.
Temperature T(x, y) with x = sqrt(1 + t), y = 2 + t/3. At t = 3: x = 2, y = 3.
dx/dt = 1 / (2 sqrt(1 + t)), so at t = 3: dx/dt = 1/4.
dy/dt = 1/3.
dT/dt = T_x * dx/dt + T_y * dy/dt = 4 * (1/4) + 3 * (1/3) = 1 + 1 = 2 degrees per second.
For f(x, y) = (x^3 y - x y^3) / (x^2 + y^2) with f(0, 0) = 0:
Using the limit definition, f_x(0, 0) = 0 and f_y(0, 0) = 0.
Computing f_xy(0, 0) and f_yx(0, 0) from the definition gives f_xy(0, 0) = -1 and f_yx(0, 0) = 1.
These are not equal. This does not contradict Clairaut's theorem because f_xy and f_yx are not continuous at (0, 0). The hypothesis of Clairaut's theorem fails.
When x = r cos theta, y = r sin theta, the chain rule gives:
partial z / partial r = f_x cos theta + f_y sin theta.
partial z / partial theta = -f_x r sin theta + f_y r cos theta.
The Laplacian transforms as: f_xx + f_yy = f_rr + (1/r) f_r + (1/r^2) f_{theta theta}.
The gradient magnitude transforms as: (f_x)^2 + (f_y)^2 = (f_r)^2 + (1/r^2)(f_theta)^2.
These identities appear repeatedly in physics and engineering whenever you switch between Cartesian and polar coordinates.
\frac{dz}{dt} = \frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt}\frac{\partial z}{\partial r} = \frac{\partial f}{\partial x}\cos\theta + \frac{\partial f}{\partial y}\sin\theta\frac{\partial z}{\partial \theta} = -\frac{\partial f}{\partial x}\,r\sin\theta + \frac{\partial f}{\partial y}\,r\cos\theta\frac{\partial^2 z}{\partial x^2} + \frac{\partial^2 z}{\partial y^2} = \frac{\partial^2 z}{\partial r^2} + \frac{1}{r}\frac{\partial z}{\partial r} + \frac{1}{r^2}\frac{\partial^2 z}{\partial \theta^2}\left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2 = \left(\frac{\partial z}{\partial r}\right)^2 + \frac{1}{r^2}\left(\frac{\partial z}{\partial \theta}\right)^2Students often think that if the limit along every straight line through (a, b) is the same, the overall limit must exist. It does not: the limit along a curved path can differ. Problem 8 is a perfect example.
Students sometimes believe that having partial derivatives at a point guarantees continuity. This is false in two or more variables. A function can have well-defined partial derivatives at (0, 0) without being continuous there.
Students often assume f_xy = f_yx always. Clairaut's theorem requires the mixed partials to be continuous. Problem 9 shows a case where f_xy(0, 0) is not equal to f_yx(0, 0) because continuity of the mixed partials fails at the origin.
When applying the chain rule, students sometimes forget a term. If z = f(x, y) and both x and y depend on t, then dz/dt has two terms (one from each variable), not one.
"Show the limit does not exist" questions are standard. The technique: try y = ax first, and if that gives 0 for all a, try a curved path like y = x^2 or y = x^(1/3).
The chain rule for multivariable functions is tested both with single-parameter paths (z depends on t through x(t) and y(t)) and with multi-parameter substitutions (polar coordinates). Know the tree-diagram approach for setting up the chain rule.
Clairaut's theorem is usually a true/false or short-answer question. Know the precise statement (both mixed partials must be continuous) and be ready to cite Problem 9 as a counterexample when the hypothesis fails.
The Laplacian in polar coordinates is a result worth memorising; it appears in physics and engineering courses and sometimes on Calc III finals.
True or false: If the limit of f(x, y) along every straight line through (0, 0) is 0, then the limit as (x, y) approaches (0, 0) is 0.
False. A curved path might give a different limit.
Fill in the blank: Clairaut's theorem requires that both ______ are continuous near the point.
Mixed partial derivatives (f_xy and f_yx).
True or false: If z = f(x, y), x = t^2, y = e^t, then dz/dt = f_x * 2t + f_y * e^t.
True.
Fill in the blank: The Laplacian in polar coordinates is u_rr + (1/r) u_r + ______.
(1/r^2) u_{theta theta}.
True or false: A function can have partial derivatives at a point without being continuous there.
True.
Q: Show that the limit of f(x, y) = xy^3 / (x^2 + y^6) as (x, y) approaches (0, 0) does not exist.
A: Along y = ax (straight lines), the limit is 0 for all a. Along y = x^(1/3) (so y^3 = x), f = x * x / (x^2 + x^2) = 1/2. Since 0 is not equal to 1/2, the limit does not exist.
Q: A bug crawls on a heated surface T(x, y) along the path x = sqrt(1 + t), y = 2 + t/3. Given T_x(2, 3) = 4 and T_y(2, 3) = 3, find dT/dt at t = 3.
A: At t = 3, x = 2, y = 3. dx/dt = 1/(2 sqrt(4)) = 1/4. dy/dt = 1/3. By the chain rule: dT/dt = 4(1/4) + 3(1/3) = 1 + 1 = 2 degrees Celsius per second.
Q: For the function f(x, y) = (x^3 y - xy^3)/(x^2 + y^2) with f(0,0) = 0, use the limit definition to find f_x(0, 0).
A: f_x(0, 0) = lim_{h to 0} (f(h, 0) - f(0, 0))/h = lim_{h to 0} (0 - 0)/h = 0.
Q: State Clairaut's theorem precisely and explain why f_xy(0, 0) can differ from f_yx(0, 0) in Problem 9.
A: If f_xy and f_yx are both continuous on an open region containing (a, b), then f_xy(a, b) = f_yx(a, b). In Problem 9, the mixed partials exist at (0, 0) but are not continuous there, so the hypothesis is not satisfied and the conclusion need not hold.
The path test for limits connects to the epsilon-delta definition used in real analysis. The chain rule in polar coordinates feeds directly into double integrals in polar form (Chapter 15). Clairaut's theorem underpins the symmetry of the Hessian matrix used in optimization (Section 14.7). The Laplacian in polar coordinates is essential in physics for problems with circular or spherical symmetry (heat equation, wave equation, Schrodinger equation).
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