Multivariable Functions

Multivariable Functions

Multivariable functions take multiple inputs and produce one or more outputs. Here's what you need to know:

Basic Concepts

Definition

A function of several variables assigns a single value to an ordered collection of inputs.

Example: $$f(x, y) = x^2 + y^2$$ is a function of two variables.

Domain and Range

• Domain: The set of all input values for which the function is defined
• Range: The set of all possible output values

Example: For $$f(x, y) = \sqrt{1 - x^2 - y^2}$$, the domain is $$\{(x, y) : x^2 + y^2 \leq 1\}$$ (the unit disk).

Visualization and Level Curves/Surfaces

Graphs of Functions of Two Variables

The graph of $$z = f(x, y)$$ is a surface in three-dimensional space.

Level Curves

For a function $$f(x, y)$$, a level curve is the set of points where $$f(x, y) = c$$ for some constant c.

Example: The level curves of $$f(x, y) = x^2 + y^2$$ are circles centered at the origin.

Level Surfaces

For a function $$f(x, y, z)$$, a level surface is the set of points where $$f(x, y, z) = c$$ for some constant c.

Example: The level surfaces of $$f(x, y, z) = x^2 + y^2 + z^2$$ are spheres centered at the origin.

Limits and Continuity

Limit Definition

We say $$\lim_{(x,y) \to (a,b)} f(x,y) = L$$ if f(x,y) can be made arbitrarily close to L by taking (x,y) sufficiently close to (a,b).

Properties of Limits

• Limits of sums, products, and quotients behave as in single-variable calculus
• A limit may not exist if the function approaches different values along different paths

Continuity

A function f is continuous at a point (a,b) if:

1. f(a,b) is defined
2. $$\lim_{(x,y) \to (a,b)} f(x,y)$$ exists
3. $$\lim_{(x,y) \to (a,b)} f(x,y) = f(a,b)$$

Partial Derivatives

Definition

The partial derivative of f with respect to x, denoted $$\frac{\partial f}{\partial x}$$ or $$f_x$$, is the derivative of f with respect to x while holding all other variables constant.

Computation

To compute $$\frac{\partial f}{\partial x}(a, b)$$:

1. Treat y as a constant
2. Differentiate with respect to x using single-variable calculus rules
3. Evaluate at the point (a, b)

Example

For $$f(x, y) = x^2y + y^3$$:

$$\frac{\partial f}{\partial x} = 2xy$$ (treating y as a constant)

$$\frac{\partial f}{\partial y} = x^2 + 3y^2$$ (treating x as a constant)

Geometric Interpretation

$$\frac{\partial f}{\partial x}(a, b)$$ is the slope of the tangent line to the curve formed by intersecting the surface z = f(x, y) with the plane y = b, at the point (a, b, f(a, b)).

Higher-Order Partial Derivatives

Second-Order Partial Derivatives

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

Mixed Partial Derivatives

If f and its partial derivatives are continuous, then the mixed partial derivatives are equal: $$f_{xy} = f_{yx}$$.

Example

For $$f(x, y) = x^3y^2 + xy$$:

$$f_x = 3x^2y^2 + y$$

$$f_y = 2x^3y + x$$

$$f_{xy} = 6x^2y + 1$$

$$f_{yx} = 6x^2y + 1$$

The Gradient

Definition

The gradient of a function f(x, y, z) is the vector:

$$\nabla f = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\right)$$

Properties

• The gradient points in the direction of steepest increase of the function
• The gradient is perpendicular to the level curves/surfaces
• The magnitude of the gradient gives the rate of change in the direction of steepest increase

Example

For $$f(x, y, z) = x^2 + y^2 + z^2$$:

$$\nabla f = (2x, 2y, 2z)$$

At the point (1, 2, 3), $$\nabla f = (2, 4, 6)$$

Directional Derivatives

Definition

The directional derivative of f at a point in the direction of a unit vector $$\vec{u}$$ is:

$$D_{\vec{u}}f = \nabla f \cdot \vec{u}$$

Interpretation

The directional derivative gives the rate of change of f in the direction of $$\vec{u}$$.

Example

For $$f(x, y) = x^2 + xy$$ at the point (2, 1) in the direction $$\vec{u} = (\frac{3}{5}, \frac{4}{5})$$:

$$\nabla f = (2x + y, x) = (5, 2)$$ at (2, 1)

$$D_{\vec{u}}f = \nabla f \cdot \vec{u} = (5, 2) \cdot (\frac{3}{5}, \frac{4}{5}) = 3 + \frac{8}{5} = \frac{23}{5}$$

Tangent Planes and Normal Lines

Tangent Plane

The equation of the tangent plane to the surface z = f(x, y) at the point (a, b, f(a, b)) is:

$$z - f(a, b) = f_x(a, b)(x - a) + f_y(a, b)(y - b)$$

Normal Line

The normal line to the surface z = f(x, y) at the point (a, b, f(a, b)) has direction vector:

$$\vec{n} = (-f_x(a, b), -f_y(a, b), 1)$$

Parametric equations: $$x = a + t(-f_x(a, b))$$, $$y = b + t(-f_y(a, b))$$, $$z = f(a, b) + t$$

Extrema of Multivariable Functions

Critical Points

A critical point of f(x, y) is a point where:

• $$\nabla f = (0, 0)$$ (both partial derivatives are zero), or
• At least one partial derivative does not exist

Second Derivative Test

At a critical point (a, b) where both partial derivatives are zero, compute:

$$D = f_{xx}(a, b)f_{yy}(a, b) - [f_{xy}(a, b)]^2$$

• If D > 0 and $$f_{xx}(a, b) > 0$$, then f has a local minimum at (a, b)
• If D > 0 and $$f_{xx}(a, b) < 0$$, then f has a local maximum at (a, b)
• If D < 0, then f has a saddle point at (a, b)
• If D = 0, the test is inconclusive

Example

For $$f(x, y) = x^2 + y^2 - 2x - 4y + 5$$:

$$f_x = 2x - 2$$ and $$f_y = 2y - 4$$

Setting both equal to zero: $$2x - 2 = 0 \Rightarrow x = 1$$ and $$2y - 4 = 0 \Rightarrow y = 2$$

So the critical point is (1, 2)

$$f_{xx} = 2$$, $$f_{yy} = 2$$, $$f_{xy} = 0$$

$$D = f_{xx}f_{yy} - (f_{xy})^2 = 2 \cdot 2 - 0^2 = 4 > 0$$

Since D > 0 and $$f_{xx} > 0$$, f has a local minimum at (1, 2)

Constrained Optimization and Lagrange Multipliers

Method of Lagrange Multipliers

To find extrema of f(x, y, z) subject to the constraint g(x, y, z) = 0:

1. Form the Lagrangian: $$L(x, y, z, \lambda) = f(x, y, z) - \lambda g(x, y, z)$$
2. Find critical points by solving the system:
   • $$\frac{\partial L}{\partial x} = 0$$
   • $$\frac{\partial L}{\partial y} = 0$$
   • $$\frac{\partial L}{\partial z} = 0$$
   • $$\frac{\partial L}{\partial \lambda} = 0$$ (which gives the constraint g(x, y, z) = 0)
3. Evaluate f at each critical point to find extrema

Example

Find the maximum value of $$f(x, y) = xy$$ subject to the constraint $$x^2 + y^2 = 1$$.

Lagrangian: $$L(x, y, \lambda) = xy - \lambda(x^2 + y^2 - 1)$$

$$\frac{\partial L}{\partial x} = y - 2\lambda x = 0 \Rightarrow y = 2\lambda x$$

$$\frac{\partial L}{\partial y} = x - 2\lambda y = 0 \Rightarrow x = 2\lambda y$$

$$\frac{\partial L}{\partial \lambda} = -(x^2 + y^2 - 1) = 0 \Rightarrow x^2 + y^2 = 1$$

From the first two equations: $$y = 2\lambda x$$ and $$x = 2\lambda y$$

Substituting: $$y = 2\lambda(2\lambda y) = 4\lambda^2 y$$

If y ≠ 0, then $$1 = 4\lambda^2 \Rightarrow \lambda = \pm \frac{1}{2}$$

Using $$\lambda = \frac{1}{2}$$: $$y = 2\lambda x = x$$

With $$x^2 + y^2 = 1$$ and $$y = x$$: $$x^2 + x^2 = 1 \Rightarrow 2x^2 = 1 \Rightarrow x = \pm \frac{1}{\sqrt{2}}$$

So we have critical points $$(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$$ and $$(-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}})$$

Evaluating f: $$f(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}) = \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = \frac{1}{2}$$

$$f(-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}) = (-\frac{1}{\sqrt{2}}) \cdot (-\frac{1}{\sqrt{2}}) = \frac{1}{2}$$

Using $$\lambda = -\frac{1}{2}$$: $$y = 2\lambda x = -x$$

With $$x^2 + y^2 = 1$$ and $$y = -x$$: $$x^2 + (-x)^2 = 1 \Rightarrow 2x^2 = 1 \Rightarrow x = \pm \frac{1}{\sqrt{2}}$$

So we have critical points $$(\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}})$$ and $$(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$$

Evaluating f: $$f(\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}) = \frac{1}{\sqrt{2}} \cdot (-\frac{1}{\sqrt{2}}) = -\frac{1}{2}$$

$$f(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}) = (-\frac{1}{\sqrt{2}}) \cdot \frac{1}{\sqrt{2}} = -\frac{1}{2}$$

Therefore, the maximum value is $$\frac{1}{2}$$ at $$(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$$ and $$(-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}})$$.

Common Pitfalls and Tips

• When finding limits, check the limit along different paths to ensure it exists
• Remember that the order of differentiation doesn't matter for mixed partial derivatives of continuous functions
• The gradient vector is perpendicular to level curves/surfaces
• For constrained optimization, make sure to check all critical points found using Lagrange multipliers
• Be careful with the second derivative test—it only applies at critical points where both partial derivatives are zero
• When working with functions of three or more variables, the same principles apply but with additional variables