Section 15.3 Partial Derivatives

Similar documents
Calculus II Fall 2014

Lecture 19 - Partial Derivatives and Extrema of Functions of Two Variables

Section 14.3 Partial Derivatives

Exam 2 Review Sheet. r(t) = x(t), y(t), z(t)

MATH 105: Midterm #1 Practice Problems

Functions of several variables

Level Curves, Partial Derivatives

Unit 7 Partial Derivatives and Optimization

Math 2411 Calc III Practice Exam 2

Similarly, the point marked in red below is a local minimum for the function, since there are no points nearby that are lower than it:

The Chain Rule, Higher Partial Derivatives & Opti- mization

CHAPTER 11 PARTIAL DERIVATIVES

MATH 12 CLASS 9 NOTES, OCT Contents 1. Tangent planes 1 2. Definition of differentiability 3 3. Differentials 4

[f(t)] 2 + [g(t)] 2 + [h(t)] 2 dt. [f(u)] 2 + [g(u)] 2 + [h(u)] 2 du. The Fundamental Theorem of Calculus implies that s(t) is differentiable and

LECTURE 19 - LAGRANGE MULTIPLIERS

Math 5BI: Problem Set 1 Linearizing functions of several variables

Differentiable functions (Sec. 14.4)

Exam 2 Summary. 1. The domain of a function is the set of all possible inputes of the function and the range is the set of all outputs.

Math 32, October 22 & 27: Maxima & Minima

Test Yourself. 11. The angle in degrees between u and w. 12. A vector parallel to v, but of length 2.

ANSWER KEY. (a) For each of the following partials derivatives, use the contour plot to decide whether they are positive, negative, or zero.

11.2 LIMITS AND CONTINUITY

Review Sheet for Math 230, Midterm exam 2. Fall 2006

14.4. Tangent Planes. Tangent Planes. Tangent Planes. Tangent Planes. Partial Derivatives. Tangent Planes and Linear Approximations

i + u 2 j be the unit vector that has its initial point at (a, b) and points in the desired direction. It determines a line in the xy-plane:

Maxima and Minima. Terminology note: Do not confuse the maximum f(a, b) (a number) with the point (a, b) where the maximum occurs.

Calculus 3 Exam 2 31 October 2017

ES 111 Mathematical Methods in the Earth Sciences Lecture Outline 6 - Tues 17th Oct 2017 Functions of Several Variables and Partial Derivatives

Lecture 4 : Monday April 6th

Solutions to the problems from Written assignment 2 Math 222 Winter 2015

2.1 Partial Derivatives

FUNCTIONS OF SEVERAL VARIABLES AND PARTIAL DIFFERENTIATION

Review guide for midterm 2 in Math 233 March 30, 2009

11.7 Maximum and Minimum Values

Practice problems from old exams for math 233

Partial Differentiation 1 Introduction

Definitions and claims functions of several variables

Multivariate Calculus

B) 0 C) 1 D) No limit. x2 + y2 4) A) 2 B) 0 C) 1 D) No limit. A) 1 B) 2 C) 0 D) No limit. 8xy 6) A) 1 B) 0 C) π D) -1

MATH 8 FALL 2010 CLASS 27, 11/19/ Directional derivatives Recall that the definitions of partial derivatives of f(x, y) involved limits

Chapter 9 Linear equations/graphing. 1) Be able to graph points on coordinate plane 2) Determine the quadrant for a point on coordinate plane

18.3. Stationary Points. Introduction. Prerequisites. Learning Outcomes

Review Problems. Calculus IIIA: page 1 of??

MATH 234 THIRD SEMESTER CALCULUS

REVIEW SHEET FOR MIDTERM 2: ADVANCED

Math 148 Exam III Practice Problems

MULTI-VARIABLE OPTIMIZATION NOTES. 1. Identifying Critical Points

Partial derivatives and their application.

14.1 Functions of Several Variables

Math for Economics 1 New York University FINAL EXAM, Fall 2013 VERSION A

MATH 20C: FUNDAMENTALS OF CALCULUS II FINAL EXAM

MATH Review Exam II 03/06/11

Section 1.3. Slope formula: If the coordinates of two points on the line are known then we can use the slope formula to find the slope of the line.

4.4 Equations of Parallel and Perpendicular

y-intercept remains constant?

Practice problems from old exams for math 233

Math 233. Extrema of Functions of Two Variables Basics

The Picture Tells the Linear Story

10.1 Curves defined by parametric equations

4 to find the dimensions of the rectangle that have the maximum area. 2y A =?? f(x, y) = (2x)(2y) = 4xy

Section 3: Functions of several variables.

Math Final Exam - 6/11/2015

Math 206 First Midterm February 1, 2012

Discussion 8 Solution Thursday, February 10th. Consider the function f(x, y) := y 2 x 2.

MATH Exam 2 Solutions November 16, 2015

Exam 1 Study Guide. Math 223 Section 12 Fall Student s Name

Directional Derivative, Gradient and Level Set

Now we are going to introduce a new horizontal axis that we will call y, so that we have a 3-dimensional coordinate system (x, y, z).

Determine the intercepts of the line and ellipse below: Definition: An intercept is a point of a graph on an axis. Line: x intercept(s)

Plotting Points in 2-dimensions. Graphing 2 variable equations. Stuff About Lines

Math 122: Final Exam Review Sheet

Chapter 16. Partial Derivatives

Name: ID: Section: Math 233 Exam 2. Page 1. This exam has 17 questions:

Calculus I Handout: Curves and Surfaces in R 3. 1 Curves in R Curves in R 2 1 of 21

DIFFERENTIAL EQUATIONS. A principal model of physical phenomena.

SYDE 112, LECTURE 34 & 35: Optimization on Restricted Domains and Lagrange Multipliers

Section 5.2 Graphs of the Sine and Cosine Functions

Calculus of Several Variables

Limits and Continuity

Lecture 19. Vector fields. Dan Nichols MATH 233, Spring 2018 University of Massachusetts. April 10, 2018.

Examples: Find the domain and range of the function f(x, y) = 1 x y 2.

14.2 Limits and Continuity

4 The Cartesian Coordinate System- Pictures of Equations

33. Riemann Summation over Rectangular Regions

Review #Final Exam MATH 142-Drost

LEIBNIZ INDIFFERENCE CURVES AND THE MARGINAL RATE OF SUBSTITUTION

MATH 261 EXAM II PRACTICE PROBLEMS

WESI 205 Workbook. 1 Review. 2 Graphing in 3D

11/1/2017 Second Hourly Practice 2 Math 21a, Fall Name:

Review for Mastery. Identifying Linear Functions

Lecture 15. Global extrema and Lagrange multipliers. Dan Nichols MATH 233, Spring 2018 University of Massachusetts

Section 7.2 Logarithmic Functions

DIFFERENTIAL EQUATIONS. A principal model of physical phenomena.

1. Let f(x, y) = 4x 2 4xy + 4y 2, and suppose x = cos t and y = sin t. Find df dt using the chain rule.

Level Curves in Matlab

6.1 - Introduction to Periodic Functions

Study Guide and Review - Chapter 3. Find the x-intercept and y-intercept of the graph of each linear function.

Mathematics 205 HWK 2 Solutions Section 12.4 p588. x\y 0 1 2

LEVEL 9 Mathematics Observation

Math 259 Winter Recitation Handout 9: Lagrange Multipliers

Transcription:

Section 5.3 Partial Derivatives Differentiating Functions of more than one Variable. Basic Definitions In single variable calculus, the derivative is defined to be the instantaneous rate of change of a function f(x). In multivariable, this definition no longer makes sense because there are many different directions in which one could move, so the rate of change will depend not only upon the point we are at, but also the direction we choose to move. We illustrate with an example. Example.. A sheet of unevenly heated metal lies in the xy-plane with the lower left corner at the origin. The temperature at any point of the sheet is a function of x and y, T(x, y). After taking some measurements, you gather the following information: 3 85 90 0 35 55 80 2 00 0 20 45 90 70 25 28 35 60 75 60 0 20 35 55 60 60 50 y/x 0 2 3 4 5 If we are stood at the point (2, ) in the xy-plane, the temperature changes depending upon the direction we choose to mover. If we fix the y-value at and move in the positive x-direction, then the function becomes a function of a single variable, so we can consider its rate of change in the x-direction. Specifically, we have T T(3, ) T(2, ) 60 35 (x, ) = = 25, so the rate of change of temperature in the x-direction is approximately 25 degrees per unit moved. This means that the temperature is increasing in the x-direction quite quickly. Likewise, in the y-direction we can consider the rate of change like with a single variable function. Specifically, fixing x = 2, we have T T(2, 2) T(2, ) 20 35 (2, y) = = 5, so the rate of change is negative in the y-direction (meaning the temperature is dropping in the y-direction). In order to define derivatives for functions of more than one variable, we can generalize the ideas from this example. That is, we can consider rates of change of a function in the directions of the different coordinate

2 axis by keeping all other variables fixed and treating the function like a function of a single variable. We formalize the definition: Definition.2. Suppose f(x, y) is a function of two variables. Then we define its partial derivative to be the functions f x and f y defined by f(x + h, y) f(x, y) f x = lim h 0 h f(x, y + h) f(x, y) f y = lim h 0 h There are many different notations used to denote partial derivatives. Some of the more common are: f x (x, y) = f x = x = x f(x, y) = f = D f = D x f and likewise for y. Since the definition for partial derivatives is a direction generalization of regular single variable derivatives, calculation is almost identical. Specifically, we have: Result.3. To find the partial derivatives of f(x, y), we do the following: (i) To find f x, treat y like a constant and differentiate with respect to x. (ii) To find f y, treat x like a constant and differentiate with respect to y. Example.4. Let f(x, y) = sin (x) + y 2 xy. (i) Find the partial derivatives of f(x, y). This is straight forward - f x = cos (x) y and f y = 2y x. (ii) Find f x (0, 2). We know f x = cos (x) y, so f x (0, 2) = 2 = 2. 2. Interpreting Partial Derivatives As with single variable calculus, partial derivatives have a meaningful geometric interpretation. Specifically, the value of f x (a, b) is the slope of the function f(x, y) at the point (a, b) in the direction of the unit vector i. Likewise, f y (a, b) is the slope of the function f(x, y) at the point (a, b) in the direction of the unit vector j. Alternatively, we can think of f x (a, b) as the slope of the curve at the intersection of the graph of z = f(x, y) with the plane y = b at the point x = a (see illustration below). Likewise, we can think of f y (a, b) as the slope of the curve at the intersection of the graph of z = f(x, y) with the plane x = a at the point y = b.

3 Example 2.. Find the slopes of the function f(x, y) = x 2 +y 2 at the point (2, 3) in the y and x directions. f x = 2x, so f x (2, 3) = 4, and f y = 2y, so f y (2, 3) = 6. Example 2.2. If f(x, y) = sin (x) x 2 + y find x We have and y x = cos (x)(x2 + y) 2x sin (x) and (x 2 + y) 2 y = sin (x) (x 2 + y) 2 3. Functions of more than two variables Partial derivatives can be define for functions of lots of variables in exactly the same way as we defined them for functions of two variables. Specifically, if y = f(x,..., x n ) is a function in n variables, we define the partial derivative f xi to be the slope in the direction of the x i axis. It is calculated in exactly the same way as f x - we simply consider all other variables as constants and differentiate with respect to the relevant one. Example 3.. Find a formula for f xi for arbitrary i given that f(x,...,x n ) = x + + x n. Notice that this equation is linear in all directions with a slope of, so for any i we have f xi =.

4 4. Higher Derivatives Just as with single variable, we can differentiate a function many times. However, with multivariable functions, there are lots of different variables we can differentiate with respect to, so we need to be careful to give suitable notation so we know when calculating higher derivatives which variable to differentiate with respect to and when. Result 4.. Suppose f is a function of 2 variables x and y. We define the second partial derivatives of f as follows: (i) The second partial derivative with respect to x: f xx = (f x ) x = x x = 2 f x x (ii) The mixed partial derivative with respect to y and then x : f xy = (f x ) y = x y = 2 f x y (iii) The mixed partial derivative with respect to x and then y : f yx = (f y ) x = y x = 2 f y x We define the second partial derivatives with respect to y in a similar way, and in general we can define the kth partial derivative of a function f(x,...,x n ) with respect to different variables in a similar way. Calculation is straight forward. Example 4.2. Calculate f xx, f yy, f xy and f yx for the following: (i) f(x, y) = x 2 y 2 + xy f xx = 2y 2, f yy = 2x 2, f xy = 4xy + and f yx = 4xy + (ii) f(x, y) = y x + y2 f xx = 2y x 3,f yy = 2, f xy = x 2 and f yx = x 2 Notice that in both these cases, we have f xy = f yx. Though this is not always true, it is true provided we add certain additional conditions. Specifically, we have the following result: Result 4.3. (Clairauts Theorem) If the functions f xy and f yx are continuous, then f yx = f xy. We finish with an example of how to determine partial derivatives when looking at graphs.

Example 4.4. You are stood at the point (, ) on top of a hill. What are the signs of the following partial derivatives? 5 3 2 0 3 2 x 0 2 3 y 2 3 f x - negative - when you walk in the x-direction, you will move down f y - negative - when you walk in the y-direction, you will move down f xx - negative - contours are getting closer together, so the hill is getting steeper, so the graph is concave down f yy - negative - contours are getting closer together, so the hill is getting steeper, so the graph is concave down f xy - positive - f xy is te rate of change of f x as y increase. If we fix and and move in the positive y-direction, then the contours in the x- direction are getting closer together. Notice also that the values of f x are positive, so the positive values of f x are getting larger, as y increases, so f xy is positive.