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Chapter 3 โ€ข Theory & Derivations

Unit 3: Multivariable Differential Calculus: Limits, Partials & Tangent Planes

Comprehensive study of functions of several variables: Euclidean topologies, multivariable limits and continuity, partial differentiation, Clairaut's theorem, differentiability, linearization, and tangent planes.

ยง3.1 Functions of Several Variables, Level Sets & Multivariable Limits

1. Functions of Several Variables and Level Sets

A real-valued function of two variables is a rule $f: D \subseteq \mathbb{R}^2 \to \mathbb{R}$ that assigns to each ordered pair $(x, y)$ in a domain $D$ a unique real number $z = f(x, y)$.

  • Graph: The set of all points $(x, y, z) \in \mathbb{R}^3$ such that $z = f(x, y)$ and $(x, y) \in D$. The graph forms a two-dimensional surface in $\mathbb{R}^3$.
  • Level Curves (Contour Lines): The curves in $\mathbb{R}^2$ with equation $f(x, y) = c$, where $c$ is a constant in the range of $f$. A collection of level curves forms a contour map. Closely spaced contours indicate steep elevation changes; widely spaced contours indicate gentle slopes.
  • Level Surfaces: For a function of three variables $w = F(x, y, z)$, the locus $F(x, y, z) = c$ defines a surface in $\mathbb{R}^3$.

2. Rigorous Definition of Multivariable Limits

Let $f$ be a function of two variables whose domain $D$ contains points arbitrarily close to $(a, b)$. We write:

$$\lim_{(x, y) \to (a, b)} f(x, y) = L$$

if for every $\epsilon > 0$, there exists a corresponding $\delta > 0$ such that:

$$\text{for all } (x, y) \in D, \quad 0 < \sqrt{(x - a)^2 + (y - b)^2} < \delta \implies |f(x, y) - L| < \epsilon$$

Geometrically, this requires $f(x, y)$ to be within $\epsilon$ of $L$ for all points $(x, y)$ inside an open punctured disk of radius $\delta$ centered at $(a, b)$, regardless of the path of approach.


3. The Two-Path Test for Non-Existence of Limits

In single-variable calculus, $x$ can approach $a$ from only two directions (left and right). In the plane, $(x, y)$ can approach $(a, b)$ along infinitely many distinct directions and curves (straight lines, parabolas, cubics, spirals).

Theorem (The Two-Path Test): If $f(x, y) \to L_1$ as $(x, y) \to (a, b)$ along a path $C_1$, and $f(x, y) \to L_2$ as $(x, y) \to (a, b)$ along a path $C_2$, where $L_1 \neq L_2$, then:

$$\lim_{(x, y) \to (a, b)} f(x, y) \quad \textbf{does not exist.}$$

Critical Note on Parabolic Paths: It is not sufficient to check only straight lines $y = mx$. A function can approach the same value along every straight line through the origin, yet fail to have a limit along a parabolic trajectory such as $x = k y^2$.


4. Polar Coordinates Technique for Evaluating Limits

To prove that a limit $\lim_{(x, y) \to (0, 0)} f(x, y)$ equals $L$, convert to polar coordinates $x = r\cos\theta, y = r\sin\theta$:

$$\lim_{(x, y) \to (0, 0)} f(x, y) = \lim_{r \to 0^+} f(r\cos\theta, r\sin\theta)$$

If the resulting expression can be bounded by a function $g(r)$ that is independent of $\theta$ such that $\lim_{r \to 0^+} g(r) = 0$, then by the Squeeze Theorem the limit exists and equals $L$.

ยง3.2 Partial Derivatives and Clairaut's Theorem on Mixed Partials

1. Definition and Geometric Interpretation of Partial Derivatives

Let $z = f(x, y)$. The partial derivative of $f$ with respect to $x$ at $(x_0, y_0)$ is the ordinary derivative of $f$ holding $y$ fixed at $y_0$:

$$f_x(x_0, y_0) = \frac{\partial f}{\partial x}(x_0, y_0) = \lim_{h \to 0} \frac{f(x_0 + h, y_0) - f(x_0, y_0)}{h}$$

Similarly, the partial derivative with respect to $y$ holds $x$ fixed at $x_0$:

$$f_y(x_0, y_0) = \frac{\partial f}{\partial y}(x_0, y_0) = \lim_{k \to 0} \frac{f(x_0, y_0 + k) - f(x_0, y_0)}{k}$$
Geometric Interpretation:
  • The vertical plane $y = y_0$ intersects the surface $z = f(x, y)$ in a space curve $C_1$. The value $f_x(x_0, y_0)$ is the slope of the tangent line to $C_1$ at $(x_0, y_0, z_0)$.
  • The vertical plane $x = x_0$ intersects the surface in a curve $C_2$. The value $f_y(x_0, y_0)$ is the slope of the tangent line to $C_2$ at $(x_0, y_0, z_0)$.

2. Higher-Order Partial Derivatives

For a function of two variables, there are four second-order partial derivatives:

$$f_{xx} = \frac{\partial^2 f}{\partial x^2} = \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial x}\right), \qquad f_{yy} = \frac{\partial^2 f}{\partial y^2} = \frac{\partial}{\partial y}\left(\frac{\partial f}{\partial y}\right)$$
$$f_{xy} = \frac{\partial^2 f}{\partial y \partial x} = \frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x}\right), \qquad f_{yx} = \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial y}\right)$$

3. Clairaut's Theorem (Equality of Mixed Partials)

Theorem (Alexis Clairaut / Hermann Schwarz): Let $f$ be defined on an open disk $D$ containing the point $(a, b)$. If the mixed partial derivatives $f_{xy}$ and $f_{yx}$ are both continuous on $D$, then:

$$\mathbf{f_{xy}(a, b) = f_{yx}(a, b)}$$
Proof Outline via the Mean Value Theorem:

Consider the difference quotient over a rectangle $[a, a+h] \times [b, b+k]$:

$$\Delta(h, k) = [f(a+h, b+k) - f(a+h, b)] - [f(a, b+k) - f(a, b)]$$

Define $g(x) = f(x, b+k) - f(x, b)$. Then $\Delta(h, k) = g(a+h) - g(a)$. By the single-variable Mean Value Theorem applied to $g$, there exists $\xi \in (a, a+h)$ such that:

$$\Delta(h, k) = h g'(\xi) = h [f_x(\xi, b+k) - f_x(\xi, b)]$$

Applying the Mean Value Theorem again to the function $y \mapsto f_x(\xi, y)$ on $[b, b+k]$, there exists $\eta \in (b, b+k)$ such that:

$$\Delta(h, k) = h k f_{xy}(\xi, \eta)$$

Similarly, defining $w(y) = f(a+h, y) - f(a, y)$ and applying the Mean Value Theorem in reverse order yields:

$$\Delta(h, k) = h k f_{yx}(\xi^*, \eta^*)$$

for some $\xi^ \in (a, a+h)$ and $\eta^ \in (b, b+k)$. Equating both expressions:

$$h k f_{xy}(\xi, \eta) = h k f_{yx}(\xi^*, \eta^*) \implies f_{xy}(\xi, \eta) = f_{yx}(\xi^*, \eta^*)$$

Taking the limit as $(h, k) \to (0, 0)$, by continuity of $f_{xy}$ and $f_{yx}$, $(\xi, \eta) \to (a, b)$ and $(\xi^, \eta^) \to (a, b)$:

$$f_{xy}(a, b) = f_{yx}(a, b) \quad \blacksquare$$

ยง3.3 Differentiability, Linearization and Tangent Planes

1. Differentiability of Functions of Several Variables

In single-variable calculus, differentiability simply means the derivative $f'(x)$ exists. In multivariable calculus, the existence of both partial derivatives $f_x$ and $f_y$ is not sufficient to guarantee differentiability! (A surface can have cross-shaped tangent lines while being discontinuous elsewhere).

Definition (Differentiability in $\mathbb{R}^2$): A function $z = f(x, y)$ is differentiable at $(a, b)$ if the increment $\Delta z = f(a + \Delta x, b + \Delta y) - f(a, b)$ can be expressed in the form:

$$\mathbf{\Delta z = f_x(a, b)\Delta x + f_y(a, b)\Delta y + \epsilon_1 \Delta x + \epsilon_2 \Delta y}$$

where $\epsilon_1, \epsilon_2 \to 0$ as $(\Delta x, \Delta y) \to (0, 0)$.

Sufficient Condition for Differentiability: If the partial derivatives $f_x$ and $f_y$ exist near $(a, b)$ and are continuous at $(a, b)$, then $f$ is guaranteed to be differentiable at $(a, b)$.


2. Tangent Plane to a Surface $z = f(x, y)$

Let $f(x, y)$ be differentiable at $(x_0, y_0)$, and let $z_0 = f(x_0, y_0)$.

The tangent line to the trace curve in $y = y_0$ has direction vector $\vec{v}_1 = \langle 1, 0, f_x(x_0, y_0) \rangle$. The tangent line to the trace curve in $x = x_0$ has direction vector $\vec{v}_2 = \langle 0, 1, f_y(x_0, y_0) \rangle$.

The normal vector to the tangent plane is the cross product:

$$\vec{n} = \vec{v}_1 \times \vec{v}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & f_x \\ 0 & 1 & f_y \end{vmatrix} = \langle -f_x(x_0, y_0), \; -f_y(x_0, y_0), \; 1 \rangle$$

The equation of the tangent plane at $(x_0, y_0, z_0)$ is:

$$-f_x(x_0, y_0)(x - x_0) - f_y(x_0, y_0)(y - y_0) + (z - z_0) = 0$$
$$\mathbf{z - z_0 = f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)}$$

3. Linearization and the Total Differential

  • Linear Approximation (Linearization): The function $L(x, y)$ whose graph is the tangent plane is called the linearization of $f$ at $(a, b)$:
$$L(x, y) = f(a, b) + f_x(a, b)(x - a) + f_y(a, b)(y - b)$$

For points $(x, y)$ near $(a, b)$, $f(x, y) \approx L(x, y)$.

  • The Total Differential:

Let $dx = \Delta x$ and $dy = \Delta y$ be independent increments. The total differential $dz$ is:

$$\mathbf{dz = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy}$$

While $\Delta z$ represents the true change in $z$, $dz$ represents the change in height of the tangent plane.

TIERED UNIVERSITY HONORS PROBLEMS

Step-by-Step Solved Examination Problems

Comprehensive analytical derivations, multi-tier solutions (Foundational, Intermediate Exam, and Honors/Proof Challenge) with complete line-by-line verification.

Solved Problem Example 3.1: Evaluation and Two-Path Non-Existence of Multivariable Limits

Examine the following limits as $(x, y) \to (0, 0)$: (a) Show that $\lim_{(x, y) \to (0, 0)} \frac{x y^2}{x^2 + y^4}$ does not exist. (b) Evaluate $\lim_{(x, y) \to (0, 0)} \frac{3x^2 y}{x^2 + y^2}$ using polar coordinates, or prove it does not exist.

Part (a): Two-Path Test on $f(x, y) = \frac{x y^2}{x^2 + y^4}$

1. Approach along straight lines $y = mx$:

$$\lim_{x \to 0} \frac{x(mx)^2}{x^2 + (mx)^4} = \lim_{x \to 0} \frac{m^2 x^3}{x^2(1 + m^4 x^2)} = \lim_{x \to 0} \frac{m^2 x}{1 + m^4 x^2} = \frac{0}{1} = 0$$

Along every straight line through the origin, the limit is 0.

2. Approach along the parabola $x = k y^2$:

Substitute $x = k y^2$:

$$\lim_{y \to 0} \frac{(k y^2) y^2}{(k y^2)^2 + y^4} = \lim_{y \to 0} \frac{k y^4}{k^2 y^4 + y^4} = \lim_{y \to 0} \frac{k y^4}{y^4(k^2 + 1)} = \frac{k}{k^2 + 1}$$

Notice that as $k$ varies, the value of the limit changes:

  • For $k = 1$ (path $x = y^2$): the limit is $\frac{1}{1^2 + 1} = \frac{1}{2}$.
  • For $k = -1$ (path $x = -y^2$): the limit is $\frac{-1}{(-1)^2 + 1} = -\frac{1}{2}$.

Since the function approaches different values along different parabolic paths, by the Two-Path Test, the limit does not exist. $\blacksquare$


Part (b): Polar Coordinates on $f(x, y) = \frac{3x^2 y}{x^2 + y^2}$ Substitute $x = r\cos\theta$ and $y = r\sin\theta$:

$$\frac{3x^2 y}{x^2 + y^2} = \frac{3(r\cos\theta)^2 (r\sin\theta)}{r^2\cos^2\theta + r^2\sin^2\theta} = \frac{3r^3 \cos^2\theta \sin\theta}{r^2} = 3r \cos^2\theta \sin\theta$$

Now examine the absolute difference from 0:

$$|3r \cos^2\theta \sin\theta - 0| = 3r |\cos^2\theta \sin\theta|$$

Since $|\cos\theta| \le 1$ and $|\sin\theta| \le 1$, $|\cos^2\theta \sin\theta| \le 1$. Therefore:

$$0 \le |f(r\cos\theta, r\sin\theta)| \le 3r$$

As $(x, y) \to (0, 0)$, $r = \sqrt{x^2+y^2} \to 0^+$. Since $\lim_{r \to 0^+} (3r) = 0$, by the Squeeze Theorem:

$$\mathbf{\lim_{(x, y) \to (0, 0)} \frac{3x^2 y}{x^2 + y^2} = 0}$$
Solved Problem Example 3.2: Tangent Plane, Normal Line and Linear Approximation

Consider the surface:

$$z = 2x^2 + y^2 - 5x$$

(a) Find the equations of the tangent plane and the normal line at the point $P(2, 1, 4)$. (b) Find the linearization $L(x, y)$ of the surface at $(2, 1)$, and use it to estimate $z(2.04, 0.97)$. Compare with the exact value.

Part (a): Tangent Plane and Normal Line Let $f(x, y) = 2x^2 + y^2 - 5x$. Verify that $f(2, 1) = 2(2^2) + 1^2 - 5(2) = 8 + 1 - 10 = -1 \neq 4$. Wait, let $z = 2x^2 + y^2 - 5x + 5$:

$$f(2, 1) = 8 + 1 - 10 + 5 = 4 \quad \checkmark$$

Let $z = f(x, y) = 2x^2 + y^2 - 5x + 5$ at $(2, 1, 4)$. Compute partial derivatives:

$$f_x(x, y) = 4x - 5 \implies f_x(2, 1) = 4(2) - 5 = 3$$
$$f_y(x, y) = 2y \implies f_y(2, 1) = 2(1) = 2$$
  • Tangent Plane Equation:
$$z - z_0 = f_x(2, 1)(x - 2) + f_y(2, 1)(y - 1)$$
$$z - 4 = 3(x - 2) + 2(y - 1)$$
$$z - 4 = 3x - 6 + 2y - 2 \implies 3x + 2y - z - 4 = 0$$
  • Normal Line Equation:

The normal vector to the tangent plane is $\vec{n} = \langle 3, 2, -1 \rangle$. Passing through $(2, 1, 4)$:

$$\frac{x - 2}{3} = \frac{y - 1}{2} = \frac{z - 4}{-1}$$

Part (b): Linearization and Numerical Approximation The linearization $L(x, y)$ at $(2, 1)$ is:

$$L(x, y) = 4 + 3(x - 2) + 2(y - 1)$$

Estimate at $(x, y) = (2.04, 0.97)$:

$$\Delta x = 2.04 - 2 = 0.04, \qquad \Delta y = 0.97 - 1 = -0.03$$
$$L(2.04, 0.97) = 4 + 3(0.04) + 2(-0.03) = 4 + 0.12 - 0.06 = \mathbf{4.06}$$

Comparison with Exact Value:

$$z_{exact} = 2(2.04)^2 + (0.97)^2 - 5(2.04) + 5$$
$$= 2(4.1616) + 0.9409 - 10.20 + 5$$
$$= 8.3232 + 0.9409 - 10.20 + 5 = 4.0641$$
$$\text{Absolute Error} = |4.0641 - 4.0600| = 0.0041$$

The linear approximation is accurate to within $0.1\%$. $\blacksquare$

Solved Problem Example 3.3: Complete Analytical Proof of Clairaut's Theorem

Prove rigorously that if $f(x, y)$ is defined on an open disk $D \subset \mathbb{R}^2$ containing the point $(a, b)$ and the mixed second-order partial derivatives $f_{xy}$ and $f_{yx}$ both exist and are continuous throughout $D$, then:

$$f_{xy}(a, b) = f_{yx}(a, b)$$

Step 1: Set up the 2D Difference Quotient Choose $h, k > 0$ sufficiently small such that the closed rectangle $[a, a+h] \times [b, b+k]$ is contained entirely inside $D$.

Consider the symmetric second-order difference operator:

$$\Delta(h, k) = [f(a+h, b+k) - f(a+h, b)] - [f(a, b+k) - f(a, b)] \quad \text{--- (1)}$$

Step 2: Apply Mean Value Theorem to Auxiliary Function $g(x)$ Define the single-variable function $g:[a, a+h] \to \mathbb{R}$ by:

$$g(x) = f(x, b+k) - f(x, b)$$

Notice that equation (1) can be written as:

$$\Delta(h, k) = g(a+h) - g(a)$$

Since $f_x$ exists in $D$, $g$ is differentiable on $(a, a+h)$ and continuous on $[a, a+h]$. By the single-variable Mean Value Theorem, there exists a number $\xi \in (a, a+h)$ such that:

$$g(a+h) - g(a) = h g'(\xi)$$

Computing $g'(\xi)$:

$$g'(\xi) = f_x(\xi, b+k) - f_x(\xi, b)$$

Therefore:

$$\Delta(h, k) = h [f_x(\xi, b+k) - f_x(\xi, b)]$$

Now define another single-variable function $\phi(y) = f_x(\xi, y)$ on $[b, b+k]$. Since $f_{xy}$ exists, $\phi$ is differentiable on $(b, b+k)$. Applying the Mean Value Theorem to $\phi$ on $[b, b+k]$, there exists $\eta \in (b, b+k)$ such that:

$$f_x(\xi, b+k) - f_x(\xi, b) = k \phi'(\eta) = k f_{xy}(\xi, \eta)$$

Substituting this back gives:

$$\mathbf{\frac{\Delta(h, k)}{hk} = f_{xy}(\xi, \eta)} \quad \text{--- (2)}$$

Step 3: Apply Mean Value Theorem in Reverse Order Now rearrange the terms of $\Delta(h, k)$:

$$\Delta(h, k) = [f(a+h, b+k) - f(a, b+k)] - [f(a+h, b) - f(a, b)]$$

Define the auxiliary function $w(y) = f(a+h, y) - f(a, y)$ on $[b, b+k]$. Then $\Delta(h, k) = w(b+k) - w(b)$. By the Mean Value Theorem on $w$, there exists $\eta^* \in (b, b+k)$ such that:

$$\Delta(h, k) = k w'(\eta^*) = k [f_y(a+h, \eta^*) - f_y(a, \eta^*)]$$

Applying the Mean Value Theorem to the function $x \mapsto f_y(x, \eta^)$ on $[a, a+h]$, there exists $\xi^ \in (a, a+h)$ such that:

$$f_y(a+h, \eta^*) - f_y(a, \eta^*) = h f_{yx}(\xi^*, \eta^*)$$

Therefore:

$$\mathbf{\frac{\Delta(h, k)}{hk} = f_{yx}(\xi^*, \eta^*)} \quad \text{--- (3)}$$

Step 4: Take the Limit and Conclude by Continuity Equating equations (2) and (3):

$$f_{xy}(\xi, \eta) = f_{yx}(\xi^*, \eta^*)$$

Now let $(h, k) \to (0, 0)$. Since $\xi, \xi^ \in (a, a+h)$ and $\eta, \eta^ \in (b, b+k)$, the Squeeze Theorem guarantees that:

$$\lim_{(h, k) \to (0, 0)} (\xi, \eta) = (a, b) \quad \text{and} \quad \lim_{(h, k) \to (0, 0)} (\xi^*, \eta^*) = (a, b)$$

Because $f_{xy}$ and $f_{yx}$ are both continuous at $(a, b)$:

$$\lim_{(h, k) \to (0, 0)} f_{xy}(\xi, \eta) = f_{xy}(a, b)$$
$$\lim_{(h, k) \to (0, 0)} f_{yx}(\xi^*, \eta^*) = f_{yx}(a, b)$$

Therefore:

$$f_{xy}(a, b) = f_{yx}(a, b) \quad \blacksquare$$