Lagrange multipliers October 2013
|
|
- Baldwin Simon Richardson
- 6 years ago
- Views:
Transcription
1 Lagrange multipliers October 2013
2
3 Example: Optimization with constraint. Example: Find the extreme values of f (x, y) = x + 2y on the ellipse 3x 2 + 4y 2 = 3. 3/ /2
4 Example: Optimization with constraint. Example: Find the extreme values of f (x, y) = x + 2y on the ellipse 3x 2 + 4y 2 = 3. 3/2 Maximum? 1 1 Minimum? 3/2
5 Example: Optimization with constraint. Example: Find the extreme values of f (x, y) = x + 2y on the ellipse 3x 2 + 4y 2 = 3. 3/2 Maximum? 1 1 Minimum? 3/2 Idea: f increases most rapidly when you move in the direction
6 Example: Optimization with constraint. Example: Find the extreme values of f (x, y) = x + 2y on the ellipse 3x 2 + 4y 2 = 3. 3/2 Maximum? 1 1 Minimum? 3/2 Idea: f increases most rapidly when you move in the direction 1, 2 which is f.
7 Example: Optimization with constraint. Example: Find the extreme values of f (x, y) = x + 2y on the ellipse 3x 2 + 4y 2 = 3. 3/2 Maximum? 1 1 Minimum? 3/2 Idea: f increases most rapidly when you move in the direction 1, 2 which is f. Constraint: 3x 2 + 4y 2 = 3, Objective function: f (x, y) = x + 2y
8 n f tangent line constraint f can be increased by moving right (clockwise) along curve
9 f n tangent line constraint f can be increased by moving left (counterclockwise) along curve
10 f n tangent line constraint f cannot be increased by moving in either direction along curve
11 constraint Idea: f has a critical point when f is parallel to the normal vector of the constraint.
12 Optimization with constraint: closer to rigorous. Same idea, closer to being rigorous: At a point on the constraint curve or surface, the possible directions of motion are given by the tangent plane. The rate of change of f in a given direction is the dot product of f with that tangent vector.
13 Optimization with constraint: closer to rigorous. Same idea, closer to being rigorous: At a point on the constraint curve or surface, the possible directions of motion are given by the tangent plane. The rate of change of f in a given direction is the dot product of f with that tangent vector. The direction in which f increases most rapidly is the projection of f onto the tangent plane.
14 Optimization with constraint: closer to rigorous. Same idea, closer to being rigorous: At a point on the constraint curve or surface, the possible directions of motion are given by the tangent plane. The rate of change of f in a given direction is the dot product of f with that tangent vector. The direction in which f increases most rapidly is the projection of f onto the tangent plane. The point is a critical point if and only if this projection is zero, i.e., f is orthogonal to the tangent plane.
15 Optimization with constraint: closer to rigorous. Same idea, closer to being rigorous: At a point on the constraint curve or surface, the possible directions of motion are given by the tangent plane. The rate of change of f in a given direction is the dot product of f with that tangent vector. The direction in which f increases most rapidly is the projection of f onto the tangent plane. The point is a critical point if and only if this projection is zero, i.e., f is orthogonal to the tangent plane. f is orthogonal to the tangent plane if and only if f is parallel to the tangent plane s normal vector. Recall that if the constraint is a level surface of a function g then the normal vector to the tangent plane is given by the gradient vector g.
16 Lagrange s Theorem. Theorem (Lagrange s Theorem) If P is a critical point of f (x, y) on the curve g(x, y) = k, and g P 0, then f P is parallel to g P : there is a scalar λ such that f P = λ g P.
17 Lagrange s Theorem. Theorem (Lagrange s Theorem) If P is a critical point of f (x, y) on the curve g(x, y) = k, and g P 0, then f P is parallel to g P : there is a scalar λ such that f P = λ g P. Theorem (Lagrange s Theorem, three variables) If P is a critical point of f (x, y, z) on the surface g(x, y, z) = k, and g P 0, then f P is parallel to g P : there is a scalar λ such that f P = λ g P.
18 Lagrange s Theorem. Theorem (Lagrange s Theorem) If P is a critical point of f (x, y) on the curve g(x, y) = k, and g P 0, then f P is parallel to g P : there is a scalar λ such that f P = λ g P. Theorem (Lagrange s Theorem, three variables) If P is a critical point of f (x, y, z) on the surface g(x, y, z) = k, and g P 0, then f P is parallel to g P : there is a scalar λ such that f P = λ g P. The scalar λ is called a Lagrange multiplier.
19 Lagrange s Theorem. Theorem (Lagrange s Theorem) If P is a critical point of f (x, y) on the curve g(x, y) = k, and g P 0, then f P is parallel to g P : there is a scalar λ such that f P = λ g P. Theorem (Lagrange s Theorem, three variables) If P is a critical point of f (x, y, z) on the surface g(x, y, z) = k, and g P 0, then f P is parallel to g P : there is a scalar λ such that f P = λ g P. The scalar λ is called a Lagrange multiplier. Usually looking for the point P or the value f (P), not λ.
20
21
22 Clicker Question: This plot shows the gradient vectors for a (hidden) function f (x, y) and a linear constraint. Which point is closest to the global min of f (x, y) on this constraint? A B C D receiver channel: 41 session ID: bsumath275
23 Example. f (x, y) = x + 2y g(x, y) = 3x 2 + 4y 2, g(x, y) = 3
24 Example. f (x, y) = x + 2y g(x, y) = 3x 2 + 4y 2, g(x, y) = 3 We have: f = 1, 2, g = 6x, 8y
25 Example. f (x, y) = x + 2y g(x, y) = 3x 2 + 4y 2, g(x, y) = 3 We have: f = 1, 2, g = 6x, 8y Solve 1 = 6xλ 2 = 8yλ 3x 2 + 4y 2 = 3
26 Example. f (x, y) = x + 2y g(x, y) = 3x 2 + 4y 2, g(x, y) = 3 We have: Solve f = 1, 2, g = 6x, 8y 1 = 6xλ 2 = 8yλ 3x 2 + 4y 2 = 3 = λ = 1 6x = 2 = 8y 6x = y = 3 2 x
27 Example. f (x, y) = x + 2y g(x, y) = 3x 2 + 4y 2, g(x, y) = 3 We have: Solve f = 1, 2, g = 6x, 8y 1 = 6xλ 2 = 8yλ 3x 2 + 4y 2 = 3 = λ = 1 6x = 2 = 8y 6x = y = 3 2 x = 3x 2 + 4( 3 2 x ) 2 = 3 = 12x 2 = 3
28 Example. f (x, y) = x + 2y g(x, y) = 3x 2 + 4y 2, g(x, y) = 3 We have: Solve f = 1, 2, g = 6x, 8y 1 = 6xλ 2 = 8yλ 3x 2 + 4y 2 = 3 = λ = 1 6x = 2 = 8y 6x = y = 3 2 x = 3x 2 + 4( 3 2 x ) 2 = 3 = 12x 2 = 3 = x = ± 1 2
29 Example. f (x, y) = x + 2y g(x, y) = 3x 2 + 4y 2, g(x, y) = 3 We have: Solve f = 1, 2, g = 6x, 8y 1 = 6xλ 2 = 8yλ 3x 2 + 4y 2 = 3 = λ = 1 6x = 2 = 8y 6x = y = 3 2 x = 3x 2 + 4( 3 2 x ) 2 = 3 = 12x 2 = 3 = x = ± 1 2 = y = ±3 4
30 Example, concluded. f = 1, 2, g = 6x, 8y Critical points: ( 1, ) ( 3 2 4, 1, ) f (1/2, 3/4) = 2, maximum f ( 1/2, 3/4) = 2, minimum 3/ /2
31 Why is Lagrange s Theorem true? Proof (two-variable case). Let r(t) be a parametrization of the curve g(x, y) = k, with r(0) = P and r (0) 0. (There exists such a parametrization, by the Implicit Function Theorem from real analysis.)
32 Why is Lagrange s Theorem true? Proof (two-variable case). Let r(t) be a parametrization of the curve g(x, y) = k, with r(0) = P and r (0) 0. (There exists such a parametrization, by the Implicit Function Theorem from real analysis.) By assumption t = 0 is a critical point of f (r(t)).
33 Why is Lagrange s Theorem true? Proof (two-variable case). Let r(t) be a parametrization of the curve g(x, y) = k, with r(0) = P and r (0) 0. (There exists such a parametrization, by the Implicit Function Theorem from real analysis.) By assumption t = 0 is a critical point of f (r(t)). Thus d dt f (r(t)) t=0 = 0.
34 Why is Lagrange s Theorem true? Proof (two-variable case). Let r(t) be a parametrization of the curve g(x, y) = k, with r(0) = P and r (0) 0. (There exists such a parametrization, by the Implicit Function Theorem from real analysis.) By assumption t = 0 is a critical point of f (r(t)). Thus d dt f (r(t)) t=0 = 0. The Chain Rule for Paths gives f P r (0) = 0.
35 Why is Lagrange s Theorem true? Proof (two-variable case). Let r(t) be a parametrization of the curve g(x, y) = k, with r(0) = P and r (0) 0. (There exists such a parametrization, by the Implicit Function Theorem from real analysis.) By assumption t = 0 is a critical point of f (r(t)). Thus d dt f (r(t)) t=0 = 0. The Chain Rule for Paths gives f P r (0) = 0. That is, f P is perpendicular to r (0), the tangent vector to the curve at P. That is, f P is parallel to the normal vector at P, which is g P.
36 Example. Example: Find the extreme values of f (x, y, z) = xy + z 2 on the unit sphere x 2 + y 2 + z 2 = 1.
37 Example. Example: Find the extreme values of f (x, y, z) = xy + z 2 on the unit sphere x 2 + y 2 + z 2 = 1. f = y, x, 2z g = 2x, 2y, 2z Solve for (x, y, z) in f = λ g:
38 Example. Example: Find the extreme values of f (x, y, z) = xy + z 2 on the unit sphere x 2 + y 2 + z 2 = 1. f = y, x, 2z g = 2x, 2y, 2z Solve for (x, y, z) in f = λ g: y = λ2x x = λ2y 2z = λ2z x 2 + y 2 + z 2 = 1
39 Example. Example: Find the extreme values of f (x, y, z) = xy + z 2 on the unit sphere x 2 + y 2 + z 2 = 1. f = y, x, 2z g = 2x, 2y, 2z Solve for (x, y, z) in f = λ g: y = λ2x x = λ2y 2z = λ2z x 2 + y 2 + z 2 = 1 = y = 4λ 2 y = y = 0 or 4λ 2 = 1
40 Example, continued. f (x, y, z) = xy + z 2 g(x, y, z) = x 2 + y 2 + z 2 = 1 f = y, x, 2z g = 2x, 2y, 2z y = λ2x, x = λ2y, 2z = λ2z, x 2 + y 2 + z 2 = 1
41 Example, continued. f (x, y, z) = xy + z 2 g(x, y, z) = x 2 + y 2 + z 2 = 1 f = y, x, 2z g = 2x, 2y, 2z y = λ2x, x = λ2y, 2z = λ2z, x 2 + y 2 + z 2 = 1 y = 0: = x = 0 = z = ±1
42 Example, continued. f (x, y, z) = xy + z 2 g(x, y, z) = x 2 + y 2 + z 2 = 1 f = y, x, 2z g = 2x, 2y, 2z y = λ2x, x = λ2y, 2z = λ2z, x 2 + y 2 + z 2 = 1 y = 0: = x = 0 = z = ±1 4λ 2 = 1: { = λ = ± 1 2 = z = 0 = x = ±y x 2 + y 2 = 1
43 Example, continued. f (x, y, z) = xy + z 2 g(x, y, z) = x 2 + y 2 + z 2 = 1 f = y, x, 2z g = 2x, 2y, 2z y = λ2x, x = λ2y, 2z = λ2z, x 2 + y 2 + z 2 = 1 y = 0: = x = 0 = z = ±1 4λ 2 = 1: { = λ = ± 1 2 = z = 0 = x = ±y x 2 + y 2 = 1 Six critical points: (0, 0, ±1), (±1/ 2, ±1/ 2, 0).
44 Example, concluded. f (0, 0, ±1/2) = 1 f (1/ 2, 1/ 2, 0) = f ( 1/ 2, 1/ 2, 0) = 1 2 f (1/ 2, 1/ 2, 0) = f ( 1/ 2, 1/ 2, 0) = 1 2
45 Example, concluded. f (0, 0, ±1/2) = 1 f (1/ 2, 1/ 2, 0) = f ( 1/ 2, 1/ 2, 0) = 1 2 f (1/ 2, 1/ 2, 0) = f ( 1/ 2, 1/ 2, 0) = 1 2 The maximum is 1, at (0, 0, ±1). The minimum is 1/2, at ±(1/ 2, 1/ 2, 0).
46 Multiple constraints. Example: Find the maximum of f (x, y, z) = x on the curve C given as the intersection of g(x, y, z) = 2x 2 + 7y 2 + 5z 2 = 3, h(x, y, z) = x + y + z = 0
47 Multiple constraints. Example: Find the maximum of f (x, y, z) = x on the curve C given as the intersection of g(x, y, z) = 2x 2 + 7y 2 + 5z 2 = 3, h(x, y, z) = x + y + z = 0 Answer: At critical point P, for some scalars λ, µ. f P = λ g P + µ h P
48 Multiple constraints. Example: Find the maximum of f (x, y, z) = x on the curve C given as the intersection of g(x, y, z) = 2x 2 + 7y 2 + 5z 2 = 3, h(x, y, z) = x + y + z = 0 Answer: At critical point P, for some scalars λ, µ. f P = λ g P + µ h P f = 1, 0, 0, g = 4x, 14y, 10z, h = 1, 1, 1
49 Multiple constraints, continued. f = 1, 0, 0, g = 4x, 14y, 10z, h = 1, 1, 1 Solve for (x, y, z) in f P = λ g P + µ h P : 1 = 4λx + µ 0 = 14λy + µ 0 = 10λz + µ 2x 2 + 7y 2 + 5z 2 = 3 x + y + z = 0
50 Multiple constraints, continued. f = 1, 0, 0, g = 4x, 14y, 10z, h = 1, 1, 1 Solve for (x, y, z) in f P = λ g P + µ h P : 1 = 4λx + µ 0 = 14λy + µ 0 = 10λz + µ 2x 2 + 7y 2 + 5z 2 = 3 x + y + z = 0 ( ) 6 5 = = two critical points: ±, ,
51 Multiple constraints, continued. f = 1, 0, 0, g = 4x, 14y, 10z, h = 1, 1, 1 Solve for (x, y, z) in f P = λ g P + µ h P : 1 = 4λx + µ 0 = 14λy + µ 0 = 10λz + µ 2x 2 + 7y 2 + 5z 2 = 3 x + y + z = 0 ( ) 6 5 = = two critical points: ±, , So, the maximum value of f (x, y, z) = x is. 59
52 III 3x + y/2 = 12 Clicker (a) I < II, Question: < III (b) III < II < I (c) II < III < I (d) This II < contour I < III plot of f (x, y) also shows the circle of radius 2 (e) centered III < I < II at (0, 0). If you are restricted to being on the circle, how many local maxes and mins does f (x, y) have? 154. This contour plot of f(x, y) also shows the circle of radius 2 centered at (0,0). If you are restricted to being on the circle, how many local maxes and mins does f(x, y) have? A. 1 max, 1 min B. 2 maxes, 2 mins C. 3 maxes, 3 mins D. None receiver channel: session ID: bsumath275
53 Clicker Question: Suppose (a, b, c) (0, 0, 0). How many local maxes and mins does the function f (x, y, z) = ax + by + cz have on the sphere x 2 + y 2 + z 2 = 1? A. 1 max, 1 min B. 2 maxes, 2 mins C. 3 maxes, 3 mins D. None E. Depends on a, b, c receiver channel: 41 session ID: bsumath275
Lagrange multipliers 14.8
Lagrange multipliers 14.8 14 October 2013 Example: Optimization with constraint. Example: Find the extreme values of f (x, y) = x + 2y on the ellipse 3x 2 + 4y 2 = 3. 3/2 Maximum? 1 1 Minimum? 3/2 Idea:
More informationConstrained Optimization and Lagrange Multipliers
Constrained Optimization and Lagrange Multipliers MATH 311, Calculus III J. Robert Buchanan Department of Mathematics Fall 2011 Constrained Optimization In the previous section we found the local or absolute
More informationwe wish to minimize this function; to make life easier, we may minimize
Optimization and Lagrange Multipliers We studied single variable optimization problems in Calculus 1; given a function f(x), we found the extremes of f relative to some constraint. Our ability to find
More informationMATH2111 Higher Several Variable Calculus Lagrange Multipliers
MATH2111 Higher Several Variable Calculus Lagrange Multipliers Dr. Jonathan Kress School of Mathematics and Statistics University of New South Wales Semester 1, 2016 [updated: February 29, 2016] JM Kress
More informationx 6 + λ 2 x 6 = for the curve y = 1 2 x3 gives f(1, 1 2 ) = λ actually has another solution besides λ = 1 2 = However, the equation λ
Math 0 Prelim I Solutions Spring 010 1. Let f(x, y) = x3 y for (x, y) (0, 0). x 6 + y (4 pts) (a) Show that the cubic curves y = x 3 are level curves of the function f. Solution. Substituting y = x 3 in
More information(c) 0 (d) (a) 27 (b) (e) x 2 3x2
1. Sarah the architect is designing a modern building. The base of the building is the region in the xy-plane bounded by x =, y =, and y = 3 x. The building itself has a height bounded between z = and
More informationREVIEW I MATH 254 Calculus IV. Exam I (Friday, April 29) will cover sections
REVIEW I MATH 254 Calculus IV Exam I (Friday, April 29 will cover sections 14.1-8. 1. Functions of multivariables The definition of multivariable functions is similar to that of functions of one variable.
More informationMAT203 OVERVIEW OF CONTENTS AND SAMPLE PROBLEMS
MAT203 OVERVIEW OF CONTENTS AND SAMPLE PROBLEMS MAT203 covers essentially the same material as MAT201, but is more in depth and theoretical. Exam problems are often more sophisticated in scope and difficulty
More information1. Suppose that the equation F (x, y, z) = 0 implicitly defines each of the three variables x, y, and z as functions of the other two:
Final Solutions. Suppose that the equation F (x, y, z) implicitly defines each of the three variables x, y, and z as functions of the other two: z f(x, y), y g(x, z), x h(y, z). If F is differentiable
More informationMath 113 Calculus III Final Exam Practice Problems Spring 2003
Math 113 Calculus III Final Exam Practice Problems Spring 23 1. Let g(x, y, z) = 2x 2 + y 2 + 4z 2. (a) Describe the shapes of the level surfaces of g. (b) In three different graphs, sketch the three cross
More informationMath 233. Lagrange Multipliers Basics
Math 233. Lagrange Multipliers Basics Optimization problems of the form to optimize a function f(x, y, z) over a constraint g(x, y, z) = k can often be conveniently solved using the method of Lagrange
More informationLagrange Multipliers. Lagrange Multipliers. Lagrange Multipliers. Lagrange Multipliers. Lagrange Multipliers. Lagrange Multipliers
In this section we present Lagrange s method for maximizing or minimizing a general function f(x, y, z) subject to a constraint (or side condition) of the form g(x, y, z) = k. Figure 1 shows this curve
More informationBounded, Closed, and Compact Sets
Bounded, Closed, and Compact Sets Definition Let D be a subset of R n. Then D is said to be bounded if there is a number M > 0 such that x < M for all x D. D is closed if it contains all the boundary points.
More informationMath 209 (Fall 2007) Calculus III. Solution #5. 1. Find the minimum and maximum values of the following functions f under the given constraints:
Math 9 (Fall 7) Calculus III Solution #5. Find the minimum and maximum values of the following functions f under the given constraints: (a) f(x, y) 4x + 6y, x + y ; (b) f(x, y) x y, x + y 6. Solution:
More informationMath 21a Homework 22 Solutions Spring, 2014
Math 1a Homework Solutions Spring, 014 1. Based on Stewart 11.8 #6 ) Consider the function fx, y) = e xy, and the constraint x 3 + y 3 = 16. a) Use Lagrange multipliers to find the coordinates x, y) of
More informationMath 233. Lagrange Multipliers Basics
Math 33. Lagrange Multipliers Basics Optimization problems of the form to optimize a function f(x, y, z) over a constraint g(x, y, z) = k can often be conveniently solved using the method of Lagrange multipliers:
More informationd f(g(t), h(t)) = x dt + f ( y dt = 0. Notice that we can rewrite the relationship on the left hand side of the equality using the dot product: ( f
Gradients and the Directional Derivative In 14.3, we discussed the partial derivatives f f and, which tell us the rate of change of the x y height of the surface defined by f in the x direction and the
More informationMath 241, Final Exam. 12/11/12.
Math, Final Exam. //. No notes, calculator, or text. There are points total. Partial credit may be given. ircle or otherwise clearly identify your final answer. Name:. (5 points): Equation of a line. Find
More informationMath 326 Assignment 3. Due Wednesday, October 17, 2012.
Math 36 Assignment 3. Due Wednesday, October 7, 0. Recall that if G(x, y, z) is a function with continuous partial derivatives, and if the partial derivatives of G are not all zero at some point (x 0,y
More informationTotal. Math 2130 Practice Final (Spring 2017) (1) (2) (3) (4) (5) (6) (7) (8)
Math 130 Practice Final (Spring 017) Before the exam: Do not write anything on this page. Do not open the exam. Turn off your cell phone. Make sure your books, notes, and electronics are not visible during
More informationGrad operator, triple and line integrals. Notice: this material must not be used as a substitute for attending the lectures
Grad operator, triple and line integrals Notice: this material must not be used as a substitute for attending the lectures 1 .1 The grad operator Let f(x 1, x,..., x n ) be a function of the n variables
More informationDirectional Derivatives. Directional Derivatives. Directional Derivatives. Directional Derivatives. Directional Derivatives. Directional Derivatives
Recall that if z = f(x, y), then the partial derivatives f x and f y are defined as and represent the rates of change of z in the x- and y-directions, that is, in the directions of the unit vectors i and
More informationSolution 2. ((3)(1) (2)(1), (4 3), (4)(2) (3)(3)) = (1, 1, 1) D u (f) = (6x + 2yz, 2y + 2xz, 2xy) (0,1,1) = = 4 14
Vector and Multivariable Calculus L Marizza A Bailey Practice Trimester Final Exam Name: Problem 1. To prepare for true/false and multiple choice: Compute the following (a) (4, 3) ( 3, 2) Solution 1. (4)(
More informationGradient and Directional Derivatives
Gradient and Directional Derivatives MATH 311, Calculus III J. Robert Buchanan Department of Mathematics Fall 2011 Background Given z = f (x, y) we understand that f : gives the rate of change of z in
More informationWinter 2012 Math 255 Section 006. Problem Set 7
Problem Set 7 1 a) Carry out the partials with respect to t and x, substitute and check b) Use separation of varibles, i.e. write as dx/x 2 = dt, integrate both sides and observe that the solution also
More information3.3 Optimizing Functions of Several Variables 3.4 Lagrange Multipliers
3.3 Optimizing Functions of Several Variables 3.4 Lagrange Multipliers Prof. Tesler Math 20C Fall 2018 Prof. Tesler 3.3 3.4 Optimization Math 20C / Fall 2018 1 / 56 Optimizing y = f (x) In Math 20A, we
More informationMath 213 Exam 2. Each question is followed by a space to write your answer. Please write your answer neatly in the space provided.
Math 213 Exam 2 Name: Section: Do not remove this answer page you will return the whole exam. You will be allowed two hours to complete this test. No books or notes may be used other than a onepage cheat
More informationf xx (x, y) = 6 + 6x f xy (x, y) = 0 f yy (x, y) = y In general, the quantity that we re interested in is
1. Let f(x, y) = 5 + 3x 2 + 3y 2 + 2y 3 + x 3. (a) Final all critical points of f. (b) Use the second derivatives test to classify the critical points you found in (a) as a local maximum, local minimum,
More informationUpdated: January 11, 2016 Calculus III Section Math 232. Calculus III. Brian Veitch Fall 2015 Northern Illinois University
Math 232 Calculus III Brian Veitch Fall 2015 Northern Illinois University 12.5 Equations of Lines and Planes Definition 1: Vector Equation of a Line L Let L be a line in three-dimensional space. P (x,
More informationMATH Lagrange multipliers in 3 variables Fall 2016
MATH 20550 Lagrange multipliers in 3 variables Fall 2016 1. The one constraint they The problem is to find the extrema of a function f(x, y, z) subject to the constraint g(x, y, z) = c. The book gives
More informationEquation of tangent plane: for implicitly defined surfaces section 12.9
Equation of tangent plane: for implicitly defined surfaces section 12.9 Some surfaces are defined implicitly, such as the sphere x 2 + y 2 + z 2 = 1. In general an implicitly defined surface has the equation
More informationWhat you will learn today
What you will learn today Tangent Planes and Linear Approximation and the Gradient Vector Vector Functions 1/21 Recall in one-variable calculus, as we zoom in toward a point on a curve, the graph becomes
More informationMATH. 2153, Spring 16, MWF 12:40 p.m. QUIZ 1 January 25, 2016 PRINT NAME A. Derdzinski Show all work. No calculators. The problem is worth 10 points.
MATH. 2153, Spring 16, MWF 12:40 p.m. QUIZ 1 January 25, 2016 PRINT NAME A. Derdzinski Show all work. No calculators. The problem is worth 10 points. 1. Evaluate the area A of the triangle with the vertices
More informationSolution of final examination
of final examination Math 20, pring 201 December 9, 201 Problem 1 Let v(t) (2t e t ) i j + π cos(πt) k be the velocity of a particle with initial position r(0) ( 1, 0, 2). Find the accelaration at the
More information14.5 Directional Derivatives and the Gradient Vector
14.5 Directional Derivatives and the Gradient Vector 1. Directional Derivatives. Recall z = f (x, y) and the partial derivatives f x and f y are defined as f (x 0 + h, y 0 ) f (x 0, y 0 ) f x (x 0, y 0
More informationDifferentiability and Tangent Planes October 2013
Differentiability and Tangent Planes 14.4 04 October 2013 Differentiability in one variable. Recall for a function of one variable, f is differentiable at a f (a + h) f (a) lim exists and = f (a) h 0 h
More informationPractice problems from old exams for math 233 William H. Meeks III December 21, 2009
Practice problems from old exams for math 233 William H. Meeks III December 21, 2009 Disclaimer: Your instructor covers far more materials that we can possibly fit into a four/five questions exams. These
More information14.6 Directional Derivatives and the Gradient Vector
14 Partial Derivatives 14.6 and the Gradient Vector Copyright Cengage Learning. All rights reserved. Copyright Cengage Learning. All rights reserved. and the Gradient Vector In this section we introduce
More informationHOMEWORK ASSIGNMENT #4, MATH 253
HOMEWORK ASSIGNMENT #4, MATH 253. Prove that the following differential equations are satisfied by the given functions: (a) 2 u 2 + 2 u y 2 + 2 u z 2 =0,whereu =(x2 + y 2 + z 2 ) /2. (b) x w + y w y +
More informationMAT175 Overview and Sample Problems
MAT175 Overview and Sample Problems The course begins with a quick review/overview of one-variable integration including the Fundamental Theorem of Calculus, u-substitutions, integration by parts, and
More information8(x 2) + 21(y 1) + 6(z 3) = 0 8x + 21y + 6z = 55.
MATH 24 -Review for Final Exam. Let f(x, y, z) x 2 yz + y 3 z x 2 + z, and a (2,, 3). Note: f (2xyz 2x, x 2 z + 3y 2 z, x 2 y + y 3 + ) f(a) (8, 2, 6) (a) Find all stationary points (if any) of f. et f.
More informationMath 21a Tangent Lines and Planes Fall, What do we know about the gradient f? Tangent Lines to Curves in the Plane.
Math 21a Tangent Lines and Planes Fall, 2016 What do we know about the gradient f? Tangent Lines to Curves in the Plane. 1. For each of the following curves, find the tangent line to the curve at the point
More informationMath 253, Section 102, Fall 2006 Practice Final Solutions
Math 253, Section 102, Fall 2006 Practice Final Solutions 1 2 1. Determine whether the two lines L 1 and L 2 described below intersect. If yes, find the point of intersection. If not, say whether they
More informationLagrangian Multipliers
Università Ca Foscari di Venezia - Dipartimento di Management - A.A.2017-2018 Mathematics Lagrangian Multipliers Luciano Battaia November 15, 2017 1 Two variables functions and constraints Consider a two
More information. Tutorial Class V 3-10/10/2012 First Order Partial Derivatives;...
Tutorial Class V 3-10/10/2012 1 First Order Partial Derivatives; Tutorial Class V 3-10/10/2012 1 First Order Partial Derivatives; 2 Application of Gradient; Tutorial Class V 3-10/10/2012 1 First Order
More informationMATH 200 (Fall 2016) Exam 1 Solutions (a) (10 points) Find an equation of the sphere with center ( 2, 1, 4).
MATH 00 (Fall 016) Exam 1 Solutions 1 1. (a) (10 points) Find an equation of the sphere with center (, 1, 4). (x ( )) + (y 1) + (z ( 4)) 3 (x + ) + (y 1) + (z + 4) 9 (b) (10 points) Find an equation of
More informationCalculus III. Math 233 Spring In-term exam April 11th. Suggested solutions
Calculus III Math Spring 7 In-term exam April th. Suggested solutions This exam contains sixteen problems numbered through 6. Problems 5 are multiple choice problems, which each count 5% of your total
More informationDirectional Derivatives and the Gradient Vector Part 2
Directional Derivatives and the Gradient Vector Part 2 Marius Ionescu October 26, 2012 Marius Ionescu () Directional Derivatives and the Gradient Vector Part October 2 26, 2012 1 / 12 Recall Fact Marius
More information6. Find the equation of the plane that passes through the point (-1,2,1) and contains the line x = y = z.
Week 1 Worksheet Sections from Thomas 13 th edition: 12.4, 12.5, 12.6, 13.1 1. A plane is a set of points that satisfies an equation of the form c 1 x + c 2 y + c 3 z = c 4. (a) Find any three distinct
More information1 Vector Functions and Space Curves
ontents 1 Vector Functions and pace urves 2 1.1 Limits, Derivatives, and Integrals of Vector Functions...................... 2 1.2 Arc Length and urvature..................................... 2 1.3 Motion
More informationThe Three Dimensional Coordinate System
The Three-Dimensional Coordinate System The Three Dimensional Coordinate System You can construct a three-dimensional coordinate system by passing a z-axis perpendicular to both the x- and y-axes at the
More informationLagrangian Multipliers
Università Ca Foscari di Venezia - Dipartimento di Economia - A.A.2016-2017 Mathematics (Curriculum Economics, Markets and Finance) Lagrangian Multipliers Luciano Battaia November 15, 2017 1 Two variables
More informationSection 4: Extreme Values & Lagrange Multipliers.
Section 4: Extreme Values & Lagrange Multipliers. Compiled by Chris Tisdell S1: Motivation S2: What are local maxima & minima? S3: What is a critical point? S4: Second derivative test S5: Maxima and Minima
More informationMath 206 First Midterm October 5, 2012
Math 206 First Midterm October 5, 2012 Name: EXAM SOLUTIONS Instructor: Section: 1. Do not open this exam until you are told to do so. 2. This exam has 8 pages including this cover AND IS DOUBLE SIDED.
More information13.7 LAGRANGE MULTIPLIER METHOD
13.7 Lagrange Multipliers Contemporary Calculus 1 13.7 LAGRANGE MULTIPLIER METHOD Suppose we go on a walk on a hillside, but we have to stay on a path. Where along this path are we at the highest elevation?
More informationName: Class: Date: 1. Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint.
. Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. f (x, y) = x y, x + y = 8. Set up the triple integral of an arbitrary continuous function
More informationWithout fully opening the exam, check that you have pages 1 through 11.
Name: Section: Recitation Instructor: INSTRUCTIONS Fill in your name, etc. on this first page. Without fully opening the exam, check that you have pages 1 through 11. Show all your work on the standard
More information1. Show that the rectangle of maximum area that has a given perimeter p is a square.
Constrained Optimization - Examples - 1 Unit #23 : Goals: Lagrange Multipliers To study constrained optimization; that is, the maximizing or minimizing of a function subject to a constraint (or side condition).
More informationMATH 19520/51 Class 10
MATH 19520/51 Class 10 Minh-Tam Trinh University of Chicago 2017-10-16 1 Method of Lagrange multipliers. 2 Examples of Lagrange multipliers. The Problem The ingredients: 1 A set of parameters, say x 1,...,
More informationChapter 3 Numerical Methods
Chapter 3 Numerical Methods Part 1 3.1 Linearization and Optimization of Functions of Vectors 1 Problem Notation 2 Outline 3.1.1 Linearization 3.1.2 Optimization of Objective Functions 3.1.3 Constrained
More information= w. w u. u ; u + w. x x. z z. y y. v + w. . Remark. The formula stated above is very important in the theory of. surface integral.
1 Chain rules 2 Directional derivative 3 Gradient Vector Field 4 Most Rapid Increase 5 Implicit Function Theorem, Implicit Differentiation 6 Lagrange Multiplier 7 Second Derivative Test Theorem Suppose
More informationCalculus III Meets the Final
Calculus III Meets the Final Peter A. Perry University of Kentucky December 7, 2018 Homework Review for Final Exam on Thursday, December 13, 6:00-8:00 PM Be sure you know which room to go to for the final!
More informationA small review, Second Midterm, Calculus 3, Prof. Montero 3450: , Fall 2008
A small review, Second Midterm, Calculus, Prof. Montero 45:-4, Fall 8 Maxima and minima Let us recall first, that for a function f(x, y), the gradient is the vector ( f)(x, y) = ( ) f f (x, y); (x, y).
More informationDirectional Derivatives and the Gradient Vector Part 2
Directional Derivatives and the Gradient Vector Part 2 Lecture 25 February 28, 2007 Recall Fact Recall Fact If f is a dierentiable function of x and y, then f has a directional derivative in the direction
More information13.1. Functions of Several Variables. Introduction to Functions of Several Variables. Functions of Several Variables. Objectives. Example 1 Solution
13 Functions of Several Variables 13.1 Introduction to Functions of Several Variables Copyright Cengage Learning. All rights reserved. Copyright Cengage Learning. All rights reserved. Objectives Understand
More informationTEST 3 REVIEW DAVID BEN MCREYNOLDS
TEST 3 REVIEW DAVID BEN MCREYNOLDS 1. Vectors 1.1. Form the vector starting at the point P and ending at the point Q: P = (0, 0, 0), Q = (1,, 3). P = (1, 5, 3), Q = (8, 18, 0). P = ( 3, 1, 1), Q = (, 4,
More informationOutcomes List for Math Multivariable Calculus (9 th edition of text) Spring
Outcomes List for Math 200-200935 Multivariable Calculus (9 th edition of text) Spring 2009-2010 The purpose of the Outcomes List is to give you a concrete summary of the material you should know, and
More informationMath (Spring 2009): Lecture 5 Planes. Parametric equations of curves and lines
Math 18.02 (Spring 2009): Lecture 5 Planes. Parametric equations of curves and lines February 12 Reading Material: From Simmons: 17.1 and 17.2. Last time: Square Systems. Word problem. How many solutions?
More informationThere are 10 problems, with a total of 150 points possible. (a) Find the tangent plane to the surface S at the point ( 2, 1, 2).
Instructions Answer each of the questions on your own paper, and be sure to show your work so that partial credit can be adequately assessed. Put your name on each page of your paper. You may use a scientific
More informationMATH 209, Lab 5. Richard M. Slevinsky
MATH 209, Lab 5 Richard M. Slevinsky Problems 1. Say the temperature T at any point (x, y, z) in space is given by T = 4 x y z 2. Find the hottest point on the sphere F = x 2 + y 2 + z 2 100 = 0; We equate
More informationOptimization problems with constraints - the method of Lagrange multipliers
Monday, October 12 was Thanksgiving Holiday Lecture 13 Optimization problems with constraints - the method of Lagrange multipliers (Relevant section from the textbook by Stewart: 14.8) In Lecture 11, we
More information21-256: Lagrange multipliers
21-256: Lagrange multipliers Clive Newstead, Thursday 12th June 2014 Lagrange multipliers give us a means of optimizing multivariate functions subject to a number of constraints on their variables. Problems
More informationLet and be a differentiable function. Let Then be the level surface given by
Module 12 : Total differential, Tangent planes and normals Lecture 35 : Tangent plane and normal [Section 35.1] > Objectives In this section you will learn the following : The notion tangent plane to a
More informationMATH 2400, Analytic Geometry and Calculus 3
MATH 2400, Analytic Geometry and Calculus 3 List of important Definitions and Theorems 1 Foundations Definition 1. By a function f one understands a mathematical object consisting of (i) a set X, called
More informationMultivariate Calculus Review Problems for Examination Two
Multivariate Calculus Review Problems for Examination Two Note: Exam Two is on Thursday, February 28, class time. The coverage is multivariate differential calculus and double integration: sections 13.3,
More information5 Day 5: Maxima and minima for n variables.
UNIVERSITAT POMPEU FABRA INTERNATIONAL BUSINESS ECONOMICS MATHEMATICS III. Pelegrí Viader. 2012-201 Updated May 14, 201 5 Day 5: Maxima and minima for n variables. The same kind of first-order and second-order
More informationPractice problems from old exams for math 233
Practice problems from old exams for math 233 William H. Meeks III October 26, 2012 Disclaimer: Your instructor covers far more materials that we can possibly fit into a four/five questions exams. These
More informationChapter 5 Partial Differentiation
Chapter 5 Partial Differentiation For functions of one variable, y = f (x), the rate of change of the dependent variable can dy be found unambiguously by differentiation: f x. In this chapter we explore
More informationPractice problems. 1. Given a = 3i 2j and b = 2i + j. Write c = i + j in terms of a and b.
Practice problems 1. Given a = 3i 2j and b = 2i + j. Write c = i + j in terms of a and b. 1, 1 = c 1 3, 2 + c 2 2, 1. Solve c 1, c 2. 2. Suppose a is a vector in the plane. If the component of the a in
More informationInverse and Implicit functions
CHAPTER 3 Inverse and Implicit functions. Inverse Functions and Coordinate Changes Let U R d be a domain. Theorem. (Inverse function theorem). If ϕ : U R d is differentiable at a and Dϕ a is invertible,
More informationMathematically, the path or the trajectory of a particle moving in space in described by a function of time.
Module 15 : Vector fields, Gradient, Divergence and Curl Lecture 45 : Curves in space [Section 45.1] Objectives In this section you will learn the following : Concept of curve in space. Parametrization
More informationVectors and the Geometry of Space
Vectors and the Geometry of Space In Figure 11.43, consider the line L through the point P(x 1, y 1, z 1 ) and parallel to the vector. The vector v is a direction vector for the line L, and a, b, and c
More informationBackground for Surface Integration
Background for urface Integration 1 urface Integrals We have seen in previous work how to define and compute line integrals in R 2. You should remember the basic surface integrals that we will need to
More informationMAC2313 Final A. a. The vector r u r v lies in the tangent plane of S at a given point. b. S f(x, y, z) ds = R f(r(u, v)) r u r v du dv.
MAC2313 Final A (5 pts) 1. Let f(x, y, z) be a function continuous in R 3 and let S be a surface parameterized by r(u, v) with the domain of the parameterization given by R; how many of the following are
More informationt dt ds Then, in the last class, we showed that F(s) = <2s/3, 1 2s/3, s/3> is arclength parametrization. Therefore,
13.4. Curvature Curvature Let F(t) be a vector values function. We say it is regular if F (t)=0 Let F(t) be a vector valued function which is arclength parametrized, which means F t 1 for all t. Then,
More informationImplicit Surfaces or Level Sets of Functions *
Implicit Surfaces or Level Sets of Functions * Surfaces in R 3 are either described as parametrized images F : D 2 R 3 or as implicit surfaces, i.e., as levels of functions f : R 3 R, as the set of points
More informationCSC418 / CSCD18 / CSC2504
2 2.1 Parametric There are multiple ways to represent curves in two dimensions: Explicit: y = f(x), given x, find y. The explicit form of a line is y = mx + b. representation what about vertical lines?
More informationLet s write this out as an explicit equation. Suppose that the point P 0 = (x 0, y 0, z 0 ), P = (x, y, z) and n = (A, B, C).
4. Planes and distances How do we represent a plane Π in R 3? In fact the best way to specify a plane is to give a normal vector n to the plane and a point P 0 on the plane. Then if we are given any point
More informationIn other words, we want to find the domain points that yield the maximum or minimum values (extrema) of the function.
1 The Lagrange multipliers is a mathematical method for performing constrained optimization of differentiable functions. Recall unconstrained optimization of differentiable functions, in which we want
More information. 1. Chain rules. Directional derivative. Gradient Vector Field. Most Rapid Increase. Implicit Function Theorem, Implicit Differentiation
1 Chain rules 2 Directional derivative 3 Gradient Vector Field 4 Most Rapid Increase 5 Implicit Function Theorem, Implicit Differentiation 6 Lagrange Multiplier 7 Second Derivative Test Theorem Suppose
More informationVolumes of Solids of Revolution Lecture #6 a
Volumes of Solids of Revolution Lecture #6 a Sphereoid Parabaloid Hyperboloid Whateveroid Volumes Calculating 3-D Space an Object Occupies Take a cross-sectional slice. Compute the area of the slice. Multiply
More informationQuiz 6 Practice Problems
Quiz 6 Practice Problems Practice problems are similar, both in difficulty and in scope, to the type of problems you will see on the quiz. Problems marked with a are for your entertainment and are not
More informationMath 241 Spring 2015 Final Exam Solutions
Math 4 Spring 5 Final Exam Solutions. Find the equation of the plane containing the line x y z+ and the point (,,). Write [ pts] your final answer in the form ax+by +cz d. Solution: A vector parallel to
More informationTrue/False. MATH 1C: SAMPLE EXAM 1 c Jeffrey A. Anderson ANSWER KEY
MATH 1C: SAMPLE EXAM 1 c Jeffrey A. Anderson ANSWER KEY True/False 10 points: points each) For the problems below, circle T if the answer is true and circle F is the answer is false. After you ve chosen
More informationKevin James. MTHSC 206 Section 15.6 Directional Derivatives and the Gra
MTHSC 206 Section 15.6 Directional Derivatives and the Gradient Vector Definition We define the directional derivative of the function f (x, y) at the point (x 0, y 0 ) in the direction of the unit vector
More informationDubna 2018: lines on cubic surfaces
Dubna 2018: lines on cubic surfaces Ivan Cheltsov 20th July 2018 Lecture 1: projective plane Complex plane Definition A line in C 2 is a subset that is given by ax + by + c = 0 for some complex numbers
More informationWorkbook. MAT 397: Calculus III
Workbook MAT 397: Calculus III Instructor: Caleb McWhorter Name: Summer 2017 Contents Preface..................................................... 2 1 Spatial Geometry & Vectors 3 1.1 Basic n Euclidean
More informationGeometric Queries for Ray Tracing
CSCI 420 Computer Graphics Lecture 16 Geometric Queries for Ray Tracing Ray-Surface Intersection Barycentric Coordinates [Angel Ch. 11] Jernej Barbic University of Southern California 1 Ray-Surface Intersections
More informationPlanes Intersecting Cones: Static Hypertext Version
Page 1 of 12 Planes Intersecting Cones: Static Hypertext Version On this page, we develop some of the details of the plane-slicing-cone picture discussed in the introduction. The relationship between the
More informationKey points. Assume (except for point 4) f : R n R is twice continuously differentiable. strictly above the graph of f except at x
Key points Assume (except for point 4) f : R n R is twice continuously differentiable 1 If Hf is neg def at x, then f attains a strict local max at x iff f(x) = 0 In (1), replace Hf(x) negative definite
More information