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:
if for every $\epsilon > 0$, there exists a corresponding $\delta > 0$ such that:
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:
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 $x = r\cos\theta, y = r\sin\theta$:
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 $y_0$:
Similarly, the partial derivative with respect to $y$ holds $x$ fixed at $x_0$:
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:
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:
Proof Outline via the Mean Value Theorem:
Consider the difference quotient over a rectangle $[a, a+h] \times [b, b+k]$:
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:
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:
Similarly, defining $w(y) = f(a+h, y) - f(a, y)$ and applying the Mean Value Theorem in reverse order yields:
for some $\xi^ \in (a, a+h)$ and $\eta^ \in (b, b+k)$. Equating both expressions:
Taking the limit as $(h, k) \to (0, 0)$, by continuity of $f_{xy}$ and $f_{yx}$, $(\xi, \eta) \to (a, b)$ and $(\xi^, \eta^) \to (a, b)$:
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:
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:
The equation of the tangent plane at $(x_0, y_0, z_0)$ is:
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 $(a, b)$:
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:
While $\Delta z$ represents the true change in $z$, $dz$ represents the change in height of the tangent plane.
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.
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 $y = mx$:
Along every straight line through the origin, the limit is 0.
2. Approach along the parabola $x = k y^2$:
Substitute $x = k y^2$:
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 $y = r\sin\theta$:
Now examine the absolute difference from 0:
Since $|\cos\theta| \le 1$ and $|\sin\theta| \le 1$, $|\cos^2\theta \sin\theta| \le 1$. Therefore:
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:
Consider the surface:
(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 $z = 2x^2 + y^2 - 5x + 5$:
Let $z = f(x, y) = 2x^2 + y^2 - 5x + 5$ at $(2, 1, 4)$. Compute partial derivatives:
- Tangent Plane Equation:
- Normal Line Equation:
The normal vector to the tangent plane is $\vec{n} = \langle 3, 2, -1 \rangle$. Passing through $(2, 1, 4)$:
Part (b): Linearization and Numerical Approximation The linearization $L(x, y)$ at $(2, 1)$ is:
Estimate at $(x, y) = (2.04, 0.97)$:
Comparison with Exact Value:
The linear approximation is accurate to within $0.1\%$. $\blacksquare$
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:
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:
Step 2: Apply Mean Value Theorem to Auxiliary Function $g(x)$ Define the single-variable function $g:[a, a+h] \to \mathbb{R}$ by:
Notice that equation (1) can be written as:
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:
Computing $g'(\xi)$:
Therefore:
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:
Substituting this back gives:
Step 3: Apply Mean Value Theorem in Reverse Order Now rearrange the terms of $\Delta(h, k)$:
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:
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:
Therefore:
Step 4: Take the Limit and Conclude by Continuity Equating equations (2) and (3):
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:
Because $f_{xy}$ and $f_{yx}$ are both continuous at $(a, b)$:
Therefore: