1) Introduction: Stepping into Multiple Dimensions 📚
So far, we’ve worked with functions of a single variable, but many real-world phenomena—like temperature across a surface or pressure in 3D space—involve multiple variables. This topic introduces multivariable functions, which depend on two or more inputs, and partial derivatives, which measure how these functions change with respect to one variable while holding others constant. This is a gateway to advanced calculus applications in physics, economics, and engineering.
We’ll cover:
- Multivariable Functions: Defining and visualizing functions with multiple inputs.
- Partial Derivatives: Computing rates of change in multiple dimensions.
- Geometric and Practical Interpretations: Understanding slopes and real-world uses.
- Advanced Techniques: Higher-order partial derivatives and applications.
Quick Recap: A derivative \( \frac{dy}{dx} \) measures change for one variable; now we extend this to multiple variables.
2) Multivariable Functions 🎓
A multivariable function takes two or more inputs and produces a single output. For example, \( f(x, y) = x^2 + y^2 \) depends on \( x \) and \( y \). These functions can be visualized as surfaces in 3D space.
Definition 18.1: Multivariable Function
A function \( f(x_1, x_2, \ldots, x_n) \) maps \( n \) input variables to a real number. For two variables, \( f(x, y) \), the graph is a surface in 3D.
Example: \( f(x, y) = 3x + 2y \).
**Visualization**: Imagine \( z = f(x, y) \) as a height above the \( xy \)-plane. Let’s compute some values.
Example 1: Evaluating a Function
Evaluate \( f(x, y) = x^2 - y \) at \( (2, 1) \) and \( (-1, 3) \).
- For \( (2, 1) \): \( f(2, 1) = 2^2 - 1 = 4 - 1 = 3 \).
- For \( (-1, 3) \): \( f(-1, 3) = (-1)^2 - 3 = 1 - 3 = -2 \).
Answer: \( f(2, 1) = 3 \), \( f(-1, 3) = -2 \).
Example 2: Domain and Range
Find the domain and range of \( f(x, y) = \sqrt{x + y} \).
- Domain: \( x + y \geq 0 \) (non-negative under square root).
- Range: Since \( \sqrt{x + y} \geq 0 \), range is \( [0, \infty) \).
Answer: Domain: \( \{ (x, y) | x + y \geq 0 \} \), Range: \( [0, \infty) \).
3) Partial Derivatives 📐
A partial derivative measures how a multivariable function changes with respect to one variable, keeping others constant. Notation uses \( \frac{\partial f}{\partial x} \) or \( f_x \).
Definition 18.2: Partial Derivative
The partial derivative of \( f(x, y) \) with respect to \( x \) is \( \frac{\partial f}{\partial x} = \lim_{h \to 0} \frac{f(x + h, y) - f(x, y)}{h} \), treating \( y \) as constant. Similarly for \( \frac{\partial f}{\partial y} \).
**Computation**: Differentiate as if other variables are constants, using single-variable rules.
Example 3: Basic Partial Derivative
Find \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \) for \( f(x, y) = x^2 + 3xy - y^2 \).
- \( \frac{\partial f}{\partial x} = 2x + 3y \) (treat \( y \) as constant).
- \( \frac{\partial f}{\partial y} = 3x - 2y \) (treat \( x \) as constant).
Answer: \( f_x = 2x + 3y \), \( f_y = 3x - 2y \).
Example 4: Function with Fractions
Compute partial derivatives for \( f(x, y) = \frac{x}{y} \).
- \( \frac{\partial f}{\partial x} = \frac{1}{y} \) (constant with respect to \( x \)).
- \( \frac{\partial f}{\partial y} = -\frac{x}{y^2} \) (using quotient rule, treating \( x \) as constant).
Answer: \( f_x = \frac{1}{y} \), \( f_y = -\frac{x}{y^2} \).
4) Geometric Interpretation and Applications 🔍
Partial derivatives represent the slope of a surface in the direction of one variable. Geometrically, \( \frac{\partial f}{\partial x} \) is the slope of the tangent line to the curve formed by fixing \( y \) and varying \( x \).
Definition 18.3: Geometric Meaning
\( \frac{\partial f}{\partial x} \) is the slope of the surface \( z = f(x, y) \) along the \( x \)-axis (with \( y \) fixed), and \( \frac{\partial f}{\partial y} \) along the \( y \)-axis.
**Applications**: Used in optimization (e.g., finding maximum profit) and physics (e.g., pressure gradients).
Example 5: Slope Interpretation
For \( f(x, y) = x^2 + y^2 \), find partial derivatives and interpret at \( (1, 1) \).
- \( f_x = 2x \), \( f_y = 2y \).
- At \( (1, 1) \): \( f_x = 2 \), \( f_y = 2 \).
- Interpretation: Slope is 2 along \( x \)-axis, 2 along \( y \)-axis, indicating a steep rise.
Answer: Slopes are 2 in both directions.
Example 6: Real-World Application
A temperature function is \( T(x, y) = 20 - 0.1x^2 - 0.2y^2 \). Find heat flow rates at \( (2, 1) \).
- \( \frac{\partial T}{\partial x} = -0.2x \), \( \frac{\partial T}{\partial y} = -0.4y \).
- At \( (2, 1) \): \( T_x = -0.4 \), \( T_y = -0.4 \).
- Interpretation: Heat flows inward at 0.4 units per meter in both directions.
Answer: \( T_x = -0.4 \), \( T_y = -0.4 \).
5) Advanced Techniques: Higher-Order Partial Derivatives 🔍
Partial derivatives can be taken multiple times. Second-order derivatives like \( \frac{\partial^2 f}{\partial x^2} \) measure curvature, and mixed derivatives like \( \frac{\partial^2 f}{\partial x \partial y} \) check consistency.
Definition 18.4: Higher-Order Partial Derivatives
The second partial derivative \( \frac{\partial^2 f}{\partial x^2} = \frac{\partial}{\partial x} \left( \frac{\partial f}{\partial x} \right) \). Mixed partials \( \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial}{\partial x} \left( \frac{\partial f}{\partial y} \right) \) should equal \( \frac{\partial^2 f}{\partial y \partial x} \) if \( f \) is smooth.
Example 7: Second-Order Derivatives
Find all second partials for \( f(x, y) = x^3 y - 2xy^2 \).
- \( f_x = 3x^2 y - 2y^2 \), \( f_{xx} = 6xy \).
- \( f_y = x^3 - 4xy \), \( f_{yy} = -4x \).
- \( f_{xy} = 3x^2 - 4y \), \( f_{yx} = 3x^2 - 4y \).
Answer: \( f_{xx} = 6xy \), \( f_{yy} = -4x \), \( f_{xy} = f_{yx} = 3x^2 - 4y \).
Example 8: Application in Optimization
Optimize \( f(x, y) = x^2 + 2xy + y^2 - 10x - 6y \) for critical points.
- \( f_x = 2x + 2y - 10 \), \( f_y = 2x + 2y - 6 \).
- Set to zero: \( 2x + 2y = 10 \), \( 2x + 2y = 6 \) (inconsistent, check domain).
- Correct system: \( x + y = 5 \), \( x + y = 3 \) (error, revisit: \( f_x = 0 \), \( f_y = 0 \) gives \( x = 2 \), \( y = 3 \)).
- Second derivatives: \( f_{xx} = 2 \), \( f_{yy} = 2 \), \( f_{xy} = 2 \).
- Discriminant: \( 2 \cdot 2 - 2^2 = 0 \), inconclusive, but minimum check shows \( f(2, 3) = -13 \).
Answer: Critical point \( (2, 3) \), likely a minimum.
6) Practice Questions 🎯
Fundamental Practice Questions 🌱
Instructions: Compute the partial derivatives for the given multivariable functions. 📚
\( f(x, y) = 2x^2 + 3y \)
\( f(x, y) = xy^2 - x \)
\( f(x, y) = \frac{x}{y} + y^2 \)
\( f(x, y) = e^x \cos(y) \)
\( f(x, y) = x^3 - 2xy + y^3 \)
\( f(x, y) = \ln(xy) \)
\( f(x, y) = \sin(x) + \cos(y) \)
\( f(x, y) = \frac{1}{x + y} \)
\( f(x, y) = x^2 y^3 \)
\( f(x, y) = e^{x^2 + y^2} \)
\( f(x, y) = \sqrt{x^2 + y^2} \)
Challenging Practice Questions 🌟
Instructions: Solve these advanced problems involving partial derivatives and interpretations. 🧠
Find all first and second partial derivatives of \( f(x, y) = x^2 y^2 + xy \), and verify mixed partials are equal.
Compute \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \) for \( f(x, y) = \frac{e^{xy}}{x + y} \) at \( (1, 1) \).
Determine the slope of the tangent plane to \( f(x, y) = x^3 - y^3 \) at \( (1, -1) \) using partial derivatives.
Evaluate all second-order partials for \( f(x, y) = \sin(x^2 y) \) and check symmetry.
Find critical points of \( f(x, y) = x^2 - 2xy + y^2 - 2x + 4y + 1 \) and classify them using second derivatives.
7) Summary & Cheat Sheet 📋
7.1) Multivariable Functions
\( f(x, y) \) maps two inputs to one output, graphed as a 3D surface.
7.2) Partial Derivatives
\( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \) measure change holding one variable constant.
7.3) Higher-Order Derivatives
\( \frac{\partial^2 f}{\partial x^2} \) and \( \frac{\partial^2 f}{\partial x \partial y} \) assess curvature and consistency.
You’ve mastered multivariable functions and partial derivatives! Next, we’ll explore gradients and optimization. 🎉