2.1 Partial Derivatives
|
|
- Rodger Tyler
- 6 years ago
- Views:
Transcription
1 .1 Partial Derivatives.1.1 Functions of several variables Up until now, we have only met functions of single variables. From now on we will meet functions such as z = f(x, y) and w = f(x, y, z), which are functions of two and three variables respectively. The domain of z = f(x, y) is the set of all points (x, y) at which f is defined and similarly w = f(x, y, z) is the set of all points (x, y, z) at which f is defined. For example, let f(x, y, z) = x + xy + y z. (1) We will find its value at the point (1,3,4). We get f(1, 3, 4) = 1 + (1)(3) = 11. () To find its domain, we notice that the first three terms are defined for all real numbers; however, the last term z is only defined is z. Hence the domain is everything on or above the z-axis. An easier function to understand is f(x, y) = x + y 4. (3) The argument of the square root must be positive, so the condition is x + y 4. This means that we must have x + y 4. (4) There, the domain of f is all points that reside on or outside the circle of radius..1. Graphs of functions of two variables A graph of a function of two variables is the graph of the function z = f(x, y). For example, let s sketch the graph of What is this shape? Well, we can rewrite this as z = 1 x y. (5) x + y + z = 1. (6) Written this way, it is clearly a hemisphere of radius 1, and is sketched in Figure 1. It is only a hemisphere because the square root imposes the restriction z since the argument of the square root must be positive. 1
2 y z x.5 1. Figure 1: z =.1..1 p 1 x y Level curves If there is a surface z = f (x, y), we define a level curve of height k to be the shape the surface makes on the plane z = k. If we project a series of such plots onto the xy-plane we get a contour plot. This is demonstrated in Figure. Here, the contour plot of z = x + y is shown for values z =, 1,, 3, 4. This is equivalent to plotting the curves x + y x + y x + y x + y x + y =, = 1, =, = 3, = 4. (7) The first of these is a point, and the rest are ellipses, which we can represent in the form x + y = k, which is equivalent to x y + = 1. k k (8) To draw the contour plot we simply draw the point and then the four ellipses with k = 1,, 3, 4.
3 y z k=, 1,, 3, 4 x Figure : A three dimensional surface and its level curves..1.3 Limits and continuity in functions of several variables In analogy to functions of a single variable, we define the limit of a function f (x, y, z) along a smooth curve C as (x, y, z) approaches (x, y, z ) (x(t ), y(t ), z(t )) to be lim f (x, y, z) = lim f (x(t), y(t), z(t)). t t (x,y,z) (x,y,z ) (9) A function f (x, y, z) is continuous at (x, y, z ) if f (x, y, z ) is defined and lim f (x, y, z) = f (x, y, z ). t t (1) To simplify the terminology, let s go to two variables now. If f (x, y) is continuous at every point in a region D, then it is continuous on D, and if it is continuous on the entire xy-plane, we say that f (x, y) is continuous everywhere. Similarly for three variables, but it s a bit harder to describe accurately. The properties of continuity are 1. If g(x) is continuous at x and h(y) is continuous at y, then f (x, y) = g(x)h(y) is continuous at (x, y ). 3
4 . If h(x, y) is continuous at (x, y ) and g(u) is continuous at u = h(x, y ), then f(x, y) = g(h(x, y)) is continuous at (x, y ), i.e composition of continuous function is continuous. 3. Sums, differences and products of continuous functions are continuous. 4. Quotients of two differentiable functions are continuous unless the denominator is zero..1.4 Partial derivatives If we have a function that depends on two or more variables, how do we treat derivatives? We might be interested in how the function changes with respect to only one of these variables. For example, we might be interested in how people s blood pressure depend on on age and on their career. If we just take a large sample of random people, it would be hard to see a pattern. But if we took the results of all accountants, we would get an idea how blood pressure varies with age; likewise if we took the results of all people aged forty, we would get an idea about how different careers affect blood pressure. The key thing here is that we had to fix one of the variables to see how the other changes. This is the idea behind partial derivatives. Take, for instance, z = f(x, y). Let us imagine that we can fix y at some value, say y = y. Then, the derivative at f(x, y ) in x is d dx f(x, y ). (11) In other words, treat y as a constant. Similarly, we could fix x and take a derivative in y. We define the partial derivatives as follows f(x + x, y) f(x, y) f x (x, y) = lim x x f y (x, y) = lim y f(x, y + y) f(x, y) y,. (1) The notation f x (x, y) means the partial derivative of z = f(x, y) with respect to x. Other notations are f x, z x. (13) 4
5 Often, we will want to find the partial derivative at a given point, say (x, y ). To do this, find the partial derivative and then substitute the values of the point (x, y ). This will be denoted f y, x=x,y=y f y (x,y ) Let us look at and example. Example: Let z = x sin y, and find z (π,π) and z (π,π). x, f y (x, y ). (14) y Solution: Similarly, z = x sin y x z x = π sin π =. (π,π) (15) z y = x cos y z x = π cos π = π. (π,π) (16) Higher order partial derivatives As with normal derivatives, we can of course have higher order derivatives, but now there can be mixed partials. We will use the following notations f xy (x, y) = f y x = f y x, (17) so in f xy (x, y) we differentiate in the variables from left to right: x then y. Similarly, we can have f xx (x, y) = f x, and so on. 5
6 Example: Find f xy (x, y) for f(x, y) = x (y y). Solution: f xy (x, y) = ( ( x (y y) )) y x = ( x(y y) ) y (18) = x(y 1) Slope You probably recall that the slope of a function is given by its derivative, slope = df. If a function has three variables, i.e three independent directions, dx it has three slopes. Therefore the function f(x, y, z) has slope df in the x- dx direction, slope df df in the y-direction, and slope in the z-direction. We will dy dz revisit this later when we discuss gradient One-dimensional wave equation If a string is oscillating in one dimension (up and down), the position of any point on the string depends on both a coordinate x and time t and can be described by a function u(x, t). Then, it can be shown that the wave equation is u t = u c x. (19) The constant c depends on the properties of the string. The wave equation also appears in Hooke s law and in a more general form in electromagnetic radiation Laplace s equation In three dimensions, Laplace s equation is f x + f y + f z =. () It appears in fluid dynamics and electrostatics for example. Example: Prove that φ = x 3 xy +xyz xz satisfies Laplace s equation. 6
7 Solution: φ x = 3x y + yz z φ x = 6x, φ = 4xy + xz y φ y = 4x, φ = xz xz z φ z = x. Hence f x + f y + f = 6x 4x x =, (1) z and so φ satisfies Laplace s equation Local Linear Approximation If a function f(x, y, z) is differentiable at a point, it can be approximated by a linear function. We consider the function at a point (x, y, z ), and consider shifting away to a nearby point (x = x + x, y = y + y, z = z + z), then we can approximate f(x, y, z) f(x, y, z )+f x (x, y, z ) x+f y (x, y, z ) y+f z (x, y, z ) z, () and since x = x x, y = y y and z = z z we define the local linear approximation to be L(x, y, z) = f(x, y, z )+f x (x, y, z )(x x )+f y (x, y, z )(y y )+f z (x, y, z )(z z ). (3) Example: Find the local linear approximation of f(x, y) = x α y β + yα x β (1, 1). at Solution: We need L(x, y) = f(1, 1) + f x (1, 1)(x 1) + f y (1, 1)(y 1), (4) and we have f(1, 1) = and then the derivatives are f x (x, y) = αx α 1 y β β yα x β+1 f x (1, 1) = α β, f y (x, y) = βx α y β 1 + α yα 1 x β f y (1, 1) = α + β, (5) 7
8 which gives us L(x, y) = + (α β)(x 1) + (α + β)(y 1). (6).1.5 The Chain Rule Remember that generally, a function f(x, y, z) depends on a parameter t via f(x(t), y(t), z(t)). Varying t will obviously change each of x, y and z. Recalling that the chain rule for a function v(u(t)) gives dv = dv du, we dt du dt define the chain rule for derivatives as df dt = f dx x dt + f dy y dt + f dz z dt. (7) We can take it a step further. If z = f(x, y) has variables that depend on two parameters u and v, i.e. x(u, v) and y(u, v), we have the chain rule for partial derivatives f u = f x x u + f y y u, f v = f x x v + f (8) y y v. Example: Use the chain rule to find z u and z v for z = x y ; x = u + v, y = u v. (9) Solution: z u = z x x u + z y y u = (xy)() + (x )(1) = x + 4xy = (u + v) + 4(u + v)(u v ), z v = z x x v + z y y v = (xy)(1) + (x )( v) = v(u + v) + (u + v)(u v ). (3) 8
9 .1.6 Directional derivatives and the gradient Directional derivatives If we consider a function at a given point f(x, y, z), there are obviously many different directions in which we could move away from the initial point. In general, any linear combination which is a unit vector (a + b + c = 1) u = ai + bj + ck, (31) if we fix the origin to be (x, y, z ). In terms of the arc length parameter s, then we express subsequent motion away from (x, y, z ) through the equations x = x + a s, y = y + b s, z = z + c s. (3) When we take s to, we recover the initial point. Then, differentiation with respect to s will give the slope in the direction of u when we set s =. In other words, we use s to test how a small change affects the function f at (x, y, z ). If we didn t set s = at the end, we would not find the derivative at (x, y, z ), but at a point an arc length s away in the relevant directions. As a result, we define the directional derivative of f in the direction of u to be D u f(x, y, z ) = d ds [f(x + a s, y + b s, z + c s)] s= = f x (x, y, z )a + f y (x, y, z )b + f z (x, y, z )c. (33) This can be regarded as the slope of the surface w = f(x, y, z) in the direction u The gradient Calculating directional derivative is made easier using the gradient. It is denoted by, which is called nabla, but generally read as del and is given by f(x, y, z) = f x (x, y, z)i + f y (x, y, z)j + f z (x, y, z)k. (34) Using this, we see that we can use it to express directional derivatives as D u f(x, y, z) = f(x, y, z) u. (35) 9
10 This is why it is called a gradient, because it can give the slope in any direction if the dot product with a unit vector is taken. Properties of the gradient are: 1. z = f(x, y, z) has its maximum slope in the direction of the gradient, and the maximum slope is f(x, y, z).. z = f(x, y, z) has its minimum slope in the direction opposite to that of the gradient, and the minimum slope is f(x, y, z). 3. If f = at a point, all directional derivatives are zero at that point. 4. Since level curves are curves of equal z = f(x, y), then the gradient is normal to the level curves. Therefore, on level curves, f T =. Example: Find the unit vector in the direction in which f(x, y) = 1 x y increases most quickly at P = (1, 1) and compute the rate of change in that direction. Solution: The direction in which f increases most is f (1,1) = f x i + f y j (1,1) = 4xi yj (1,1) = 4i j. (36) This has magnitude f = 4 + = 5, and so the unit vector is u = 5 i 1 5 j. (37) Finally, the rate of change is + f = Tangent planes and normal vectors We want to consider how to find the tangent plane to a surface. A tangent plane is intuitively the surface that contains all possible tangent lines of all 1
11 curves at a point P. At a point P = (x, y, z ), the surface F (x(t), y(t), z(t)) has value c = F (x, y, z ). We assume that the surface is continuous at P and that its partial derivatives are also continuous. Then, at the point P we have = F x (x, y, z )x (t ) + F y (x, y, z )y (t ) + F z (x, y, z )z (t ). (38) We now consider a curve C parameterised by r(t) = (x(t), y(t), z(t)), and we note that the tangent line to C runs parallel to r (t) = (x (t), y (t), z (t)). With this in mind we note that (38) may be rewritten as = (F x (x, y, z ), F y (x, y, z ), F z (x, y, z )) (x (t ), y (t ), z (t )), (39) which can be written as = F (x, y, z ) r (t ). (4) In other words, F (x, y, z ) is normal to the tangent line of the curve C at P, and indeed to the tangent line of any curve since C was arbitrary. We therefore define the tangent plane to be the plane with normal vector n = F (x, y, z ) = (F x (x, y, z ), F y (x, y, z ), F z (x, y, z )), (41) and the tangent plane is given by F x (x, y, z )(x x )+F y (x, y, z )(y y )+F z (x, y, z )(z z ) =, (4) since it is a plane that touches the surface F (x, y, z) the point (x, y, z ) in analogy to the tangent line. The normal line is the line that it parallel to the normal vector and has parametric form (r(t) = r + n t) x = x + F x (x, y, z )t, y = y + F y (x, y, z )t, z = z + F z (x, y, z )t. (43) A more useful form of (4) comes from considering z = f(x, y) at the point (x, y, f(x, y )) and gives the tangent plane as z = f(x, y ) + f x (x, y )(x x ) + f y (x, y )(y y ), (44) which is easier to understand as a generalisation of the tangent line. In this form, the normal vector is n = ( f x (x, y ), f y (x, y ), 1), (45) 11
12 x z y Figure 3: Tangent plane and normal line for z = (x + y ). since we would have F (x, y, z) = z f(x, y). The normal line may be written r(t) = r + t( f x (x, y )i f y (x, y )j + k). (46) These are the forms of the tangent plane and normal line that we will use for calculations. Notice that it is identical to the local linear approximation given in equation (3) for the surface z = f(x, y), which would read L(x, y) = f(x, y ) + f x (x, y )(x x ) + f y (x, y )(y y ). (47) This means that the graph of the local linear approximation z = L(x, y) is the tangent plane to the surface z = f(x, y) at the point (x, y ). Example: Find the tangent plane and normal line of the surface z = (x + y ) at the point (1, 1, ). Solution: First, we notice that indeed the point is (x, y, f(x, y )) since f(x, y ) =. The first step is to find the derivatives f f x =, (1,1) y =, (48) (1,1) 1
13 which we then use to find the equation for the tangent plane using equation (44) z = + ( )(x 1) + ( )(y 1) z = (1 x y). (49) The normal vector is given by equation (45) as n = (,, 1) and therefore the normal line is given from (46) as r = (1, 1, ) + t(,, 1) = (1 + t, 1 + t, t ). (5) Alternatively, we could have written F = z + x + y and the normal vector is given by equation (41) F = (F x (1, 1, ), F y (1, 1, ), F z (1, 1, )) = (,, 1). (51) The tangent plane using equation (4) is (x 1) + (y 1) + 1(z ( )) = x + y + z =. (5) We can rewrite this as z = (1 x y), (53) which is the same as before. Finally, the normal line from equation (46) is which is the same as before. x = 1 + t, y = 1 + t, z = + t, (54) Important note: When finding the equation of the normal line, some textbooks might give answers that have the opposite sign for t. However, this doesn t matter. A line parameterised by (1 + t, + t, t 1) is equivalent to a line parameterised by (1 t, t, 1 t) as it corresponds to a change of parameter t t. In fact we could change parameter by t a + bt for constants a and b and we would have the correct line, but it would be hard to recognise. In short, don t get too confused by the sign of t in the normal line if you see answers in textbooks, but for this course use equation (43) or (46). 13
14 .1.8 Minima and maxima of two functions Consider a function of two variables, f(x, y). Obviously, it varies in the two variables and just as for a function of a single variable we can define the concepts of minima and maxima. We state the definitions separately for clarity: f has a relative (or local) maximum at (x, y ) if f(x, y ) f(x, y) for all points that lie in some disk centered on (x, y ). f has a absolute (or global) maximum at (x, y ) if f(x, y ) f(x, y) for all point for which f is defined. f has a relative (or local) minimum at (x, y ) if f(x, y ) f(x, y) for all points that lie in some disk centered on (x, y ). f has a absolute (or global) minimum at (x, y ) if f(x, y ) f(x, y) for all point for which f is defined. Both minima and maxima are types of extrema, i.e. points for which the function takes an extreme value. Recall that a set is bounded if there is a box that can be drawn around the entire set of points. Also recall that a closed set contains its boundary, but an open set does not. Therefore, a disk including its boundary is closed and bounded, but an infinite line is clearly open since the endpoints are at infinity, and unbounded because no box can be bigger than infinite length. However, the interior of a disk, i.e. without the boundary is open but bounded. Extreme-Value Theorem If f(x, y) is continuous on a closed, bounded set, then it has an absolute maximum and an absolute minimum in that set Finding extrema The position of a stationary point is shown by the fact that the first derivatives vanish. In other words, there is a stationary point at (x, y ) if f x (x, y ) = f y (x, y ) =. In addition, a critical point is any point which is either a stationary point (i.e. all derivatives vanish) or where one or more of the derivatives doesn t exist. The Second Partial Derivative Test Let f(x, y) be a function with continuous second order partial derivatives in a disk centered around a critical 14
15 point (x, y ), and define D = f xx (x, y )f yy (x, y ) f xy (x, y ). (55) If D > and f xx (x, y ) > then f(x, y) has a relative minimum at (x, y ). If D > and f xx (x, y ) < then f(x, y) has a relative maximum at (x, y ). If D < then f(x, y) has a saddle point at (x, y ). If D = then no conclusion can be drawn. A saddle point is a stationary point that is not a relative or absolute extremum. An example is for f(x, y) = x y at the point (, ) and is shown in Figure 4. 5 z 5 x y Figure 4: Saddle point of f(x, y) = x y at the point (, ). Example: Find the critical points of f(x, y) = xy x 3 y and determine whether they are maxima, minima or saddle points. Solution: To find the critical points, we set f x (x, y) = and f y (x, y) =. This gives us y 3x =, x y =, (56) 15
16 and therefore from the second equation, we can rewrite the first equation as x 3x =, x(x 1/6) =, x = or x = 1/6. (57) The corresponding y values are then x = y = and x = 1/6 y = 1/1 and so the critical points are at (, ) and (1/6, 1/1). Next we find the second order partial derivatives and so f xx (x, y) = 6x, f yy (x, y) =, f xy (x, y) = 1, (58) D = ( 6x)( ) (1) = 1x 1. (59) At the point (, ), D = 1 < and therefore (, ) is a saddle point. At (1/6, 1/1), D = 1(1/6) 1 = 1 > and therefore it is either a minimum or maximum. We now check f xx (1/6, 1/1) = 6(1/6) = 1 < and therefore (1/6, 1/1) is a global maximum. 16
Math 148 Exam III Practice Problems
Math 48 Exam III Practice Problems This review should not be used as your sole source for preparation for the exam. You should also re-work all examples given in lecture, all homework problems, all lab
More informationFUNCTIONS OF SEVERAL VARIABLES AND PARTIAL DIFFERENTIATION
FUNCTIONS OF SEVERAL VARIABLES AND PARTIAL DIFFERENTIATION 1. Functions of Several Variables A function of two variables is a rule that assigns a real number f(x, y) to each ordered pair of real numbers
More informationMaxima and Minima. Terminology note: Do not confuse the maximum f(a, b) (a number) with the point (a, b) where the maximum occurs.
10-11-2010 HW: 14.7: 1,5,7,13,29,33,39,51,55 Maxima and Minima In this very important chapter, we describe how to use the tools of calculus to locate the maxima and minima of a function of two variables.
More informationSimilarly, the point marked in red below is a local minimum for the function, since there are no points nearby that are lower than it:
Extreme Values of Multivariate Functions Our next task is to develop a method for determining local extremes of multivariate functions, as well as absolute extremes of multivariate functions on closed
More informationDefinitions and claims functions of several variables
Definitions and claims functions of several variables In the Euclidian space I n of all real n-dimensional vectors x = (x 1, x,..., x n ) the following are defined: x + y = (x 1 + y 1, x + y,..., x n +
More information1. Vector Fields. f 1 (x, y, z)i + f 2 (x, y, z)j + f 3 (x, y, z)k.
HAPTER 14 Vector alculus 1. Vector Fields Definition. A vector field in the plane is a function F(x, y) from R into V, We write F(x, y) = hf 1 (x, y), f (x, y)i = f 1 (x, y)i + f (x, y)j. A vector field
More informationPartial Differentiation 1 Introduction
Partial Differentiation 1 Introduction In the first part of this course you have met the idea of a derivative. To recap what this means, recall that if you have a function, z say, then the slope of the
More informationThe Chain Rule, Higher Partial Derivatives & Opti- mization
The Chain Rule, Higher Partial Derivatives & Opti- Unit #21 : mization Goals: We will study the chain rule for functions of several variables. We will compute and study the meaning of higher partial derivatives.
More informationReview guide for midterm 2 in Math 233 March 30, 2009
Review guide for midterm 2 in Math 2 March, 29 Midterm 2 covers material that begins approximately with the definition of partial derivatives in Chapter 4. and ends approximately with methods for calculating
More informationCHAPTER 11 PARTIAL DERIVATIVES
CHAPTER 11 PARTIAL DERIVATIVES 1. FUNCTIONS OF SEVERAL VARIABLES A) Definition: A function of two variables is a rule that assigns to each ordered pair of real numbers (x,y) in a set D a unique real number
More informationMATH 8 FALL 2010 CLASS 27, 11/19/ Directional derivatives Recall that the definitions of partial derivatives of f(x, y) involved limits
MATH 8 FALL 2010 CLASS 27, 11/19/2010 1 Directional derivatives Recall that the definitions of partial derivatives of f(x, y) involved limits lim h 0 f(a + h, b) f(a, b), lim h f(a, b + h) f(a, b) In these
More informationTest Yourself. 11. The angle in degrees between u and w. 12. A vector parallel to v, but of length 2.
Test Yourself These are problems you might see in a vector calculus course. They are general questions and are meant for practice. The key follows, but only with the answers. an you fill in the blanks
More informationExam 2 Review Sheet. r(t) = x(t), y(t), z(t)
Exam 2 Review Sheet Joseph Breen Particle Motion Recall that a parametric curve given by: r(t) = x(t), y(t), z(t) can be interpreted as the position of a particle. Then the derivative represents the particle
More information11.7 Maximum and Minimum Values
Arkansas Tech University MATH 2934: Calculus III Dr. Marcel B Finan 11.7 Maximum and Minimum Values Just like functions of a single variable, functions of several variables can have local and global extrema,
More information[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
Midterm 2 review Math 265 Fall 2007 13.3. Arc Length and Curvature. Assume that the curve C is described by the vector-valued function r(r) = f(t), g(t), h(t), and that C is traversed exactly once as t
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 informationExam 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.
Exam 2 Summary Disclaimer: The exam 2 covers lectures 9-15, inclusive. This is mostly about limits, continuity and differentiation of functions of 2 and 3 variables, and some applications. The complete
More informationDifferentiable functions (Sec. 14.4)
Math 20C Multivariable Calculus Lecture 3 Differentiable functions (Sec. 4.4) Review: Partial derivatives. Slide Partial derivatives and continuity. Equation of the tangent plane. Differentiable functions.
More informationWESI 205 Workbook. 1 Review. 2 Graphing in 3D
1 Review 1. (a) Use a right triangle to compute the distance between (x 1, y 1 ) and (x 2, y 2 ) in R 2. (b) Use this formula to compute the equation of a circle centered at (a, b) with radius r. (c) Extend
More informationANSWER KEY. (a) For each of the following partials derivatives, use the contour plot to decide whether they are positive, negative, or zero.
Math 2130-101 Test #2 for Section 101 October 14 th, 2009 ANSWE KEY 1. (10 points) Compute the curvature of r(t) = (t + 2, 3t + 4, 5t + 6). r (t) = (1, 3, 5) r (t) = 1 2 + 3 2 + 5 2 = 35 T(t) = 1 r (t)
More informationName: ID: Section: Math 233 Exam 2. Page 1. This exam has 17 questions:
Page Name: ID: Section: This exam has 7 questions: 5 multiple choice questions worth 5 points each. 2 hand graded questions worth 25 points total. Important: No graphing calculators! Any non scientific
More informationPractice problems from old exams for math 233
Practice problems from old exams for math 233 William H. Meeks III January 14, 2010 Disclaimer: Your instructor covers far more materials that we can possibly fit into a four/five questions exams. These
More informationLecture 4 : Monday April 6th
Lecture 4 : Monday April 6th jacques@ucsd.edu Key concepts : Tangent hyperplane, Gradient, Directional derivative, Level curve Know how to find equation of tangent hyperplane, gradient, directional derivatives,
More information4 to find the dimensions of the rectangle that have the maximum area. 2y A =?? f(x, y) = (2x)(2y) = 4xy
Optimization Constrained optimization and Lagrange multipliers Constrained optimization is what it sounds like - the problem of finding a maximum or minimum value (optimization), subject to some other
More informationMATH 12 CLASS 9 NOTES, OCT Contents 1. Tangent planes 1 2. Definition of differentiability 3 3. Differentials 4
MATH 2 CLASS 9 NOTES, OCT 0 20 Contents. Tangent planes 2. Definition of differentiability 3 3. Differentials 4. Tangent planes Recall that the derivative of a single variable function can be interpreted
More informationSolutions to the problems from Written assignment 2 Math 222 Winter 2015
Solutions to the problems from Written assignment 2 Math 222 Winter 2015 1. Determine if the following limits exist, and if a limit exists, find its value. x2 y (a) The limit of f(x, y) = x 4 as (x, y)
More informationMath 233. Extrema of Functions of Two Variables Basics
Math 233. Extrema of Functions of Two Variables Basics Theorem (Extreme Value Theorem) Let f be a continuous function of two variables x and y defined on a closed bounded region R in the xy-plane. Then
More information14.4. Tangent Planes. Tangent Planes. Tangent Planes. Tangent Planes. Partial Derivatives. Tangent Planes and Linear Approximations
14 Partial Derivatives 14.4 and Linear Approximations Copyright Cengage Learning. All rights reserved. Copyright Cengage Learning. All rights reserved. Suppose a surface S has equation z = f(x, y), where
More informationMATH Review Exam II 03/06/11
MATH 21-259 Review Exam II 03/06/11 1. Find f(t) given that f (t) = sin t i + 3t 2 j and f(0) = i k. 2. Find lim t 0 3(t 2 1) i + cos t j + t t k. 3. Find the points on the curve r(t) at which r(t) and
More informationMath 5BI: Problem Set 1 Linearizing functions of several variables
Math 5BI: Problem Set Linearizing functions of several variables March 9, A. Dot and cross products There are two special operations for vectors in R that are extremely useful, the dot and cross products.
More informationMATH 234 THIRD SEMESTER CALCULUS
MATH 234 THIRD SEMESTER CALCULUS Fall 2009 1 2 Math 234 3rd Semester Calculus Lecture notes version 0.9(Fall 2009) This is a self contained set of lecture notes for Math 234. The notes were written by
More informationFunctions of several variables
Chapter 6 Functions of several variables 6.1 Limits and continuity Definition 6.1 (Euclidean distance). Given two points P (x 1, y 1 ) and Q(x, y ) on the plane, we define their distance by the formula
More informationVectorPlot[{y^2-2x*y,3x*y-6*x^2},{x,-5,5},{y,-5,5}]
hapter 16 16.1. 6. Notice that F(x, y) has length 1 and that it is perpendicular to the position vector (x, y) for all x and y (except at the origin). Think about drawing the vectors based on concentric
More informationMock final exam Math fall 2007
Mock final exam Math - fall 7 Fernando Guevara Vasquez December 5 7. Consider the curve r(t) = ti + tj + 5 t t k, t. (a) Show that the curve lies on a sphere centered at the origin. (b) Where does the
More informationChapter 16. Partial Derivatives
Chapter 16 Partial Derivatives The use of contour lines to help understand a function whose domain is part of the plane goes back to the year 1774. A group of surveyors had collected a large number of
More informationi + 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:
1 Directional Derivatives and Gradients Suppose we need to compute the rate of change of f(x, y) with respect to the distance from a point (a, b) in some direction. Let u = u 1 i + u 2 j be the unit vector
More informationSYDE 112, LECTURE 34 & 35: Optimization on Restricted Domains and Lagrange Multipliers
SYDE 112, LECTURE 34 & 35: Optimization on Restricted Domains and Lagrange Multipliers 1 Restricted Domains If we are asked to determine the maximal and minimal values of an arbitrary multivariable function
More informationLecture 19. Vector fields. Dan Nichols MATH 233, Spring 2018 University of Massachusetts. April 10, 2018.
Lecture 19 Vector fields Dan Nichols nichols@math.umass.edu MATH 233, Spring 218 University of Massachusetts April 1, 218 (2) Chapter 16 Chapter 12: Vectors and 3D geometry Chapter 13: Curves and vector
More informationMATH 105: Midterm #1 Practice Problems
Name: MATH 105: Midterm #1 Practice Problems 1. TRUE or FALSE, plus explanation. Give a full-word answer TRUE or FALSE. If the statement is true, explain why, using concepts and results from class to justify
More information14.2 Limits and Continuity
14 Partial Derivatives 14.2 Copyright Cengage Learning. All rights reserved. Copyright Cengage Learning. All rights reserved. Let s compare the behavior of the functions Tables 1 2 show values of f(x,
More informationMath 2411 Calc III Practice Exam 2
Math 2411 Calc III Practice Exam 2 This is a practice exam. The actual exam consists of questions of the type found in this practice exam, but will be shorter. If you have questions do not hesitate to
More informationSection 14.3 Partial Derivatives
Section 14.3 Partial Derivatives Ruipeng Shen March 20 1 Basic Conceptions If f(x, y) is a function of two variables x and y, suppose we let only x vary while keeping y fixed, say y = b, where b is a constant.
More informationIndependent of path Green s Theorem Surface Integrals. MATH203 Calculus. Dr. Bandar Al-Mohsin. School of Mathematics, KSU 20/4/14
School of Mathematics, KSU 20/4/14 Independent of path Theorem 1 If F (x, y) = M(x, y)i + N(x, y)j is continuous on an open connected region D, then the line integral F dr is independent of path if and
More informationLecture 15. Global extrema and Lagrange multipliers. Dan Nichols MATH 233, Spring 2018 University of Massachusetts
Lecture 15 Global extrema and Lagrange multipliers Dan Nichols nichols@math.umass.edu MATH 233, Spring 2018 University of Massachusetts March 22, 2018 (2) Global extrema of a multivariable function Definition
More informationUnit 7 Partial Derivatives and Optimization
Unit 7 Partial Derivatives and Optimization We have learned some important applications of the ordinary derivative in finding maxima and minima. We now move on to a topic called partial derivatives which
More informationReview Sheet for Math 230, Midterm exam 2. Fall 2006
Review Sheet for Math 230, Midterm exam 2. Fall 2006 October 31, 2006 The second midterm exam will take place: Monday, November 13, from 8:15 to 9:30 pm. It will cover chapter 15 and sections 16.1 16.4,
More informationFinal Exam Review Problems. P 1. Find the critical points of f(x, y) = x 2 y + 2y 2 8xy + 11 and classify them.
Final Exam Review Problems P 1. Find the critical points of f(x, y) = x 2 y + 2y 2 8xy + 11 and classify them. 1 P 2. Find the volume of the solid bounded by the cylinder x 2 + y 2 = 9 and the planes z
More informationES 111 Mathematical Methods in the Earth Sciences Lecture Outline 6 - Tues 17th Oct 2017 Functions of Several Variables and Partial Derivatives
ES 111 Mathematical Methods in the Earth Sciences Lecture Outline 6 - Tues 17th Oct 2017 Functions of Several Variables and Partial Derivatives So far we have dealt with functions of the form y = f(x),
More informationMath 32, October 22 & 27: Maxima & Minima
Math 32, October 22 & 27: Maxima & Minima Section 1: Critical Points Just as in the single variable case, for multivariate functions we are often interested in determining extreme values of the function.
More informationMATH 259 FINAL EXAM. Friday, May 8, Alexandra Oleksii Reshma Stephen William Klimova Mostovyi Ramadurai Russel Boney A C D G H B F E
MATH 259 FINAL EXAM 1 Friday, May 8, 2009. NAME: Alexandra Oleksii Reshma Stephen William Klimova Mostovyi Ramadurai Russel Boney A C D G H B F E Instructions: 1. Do not separate the pages of the exam.
More informationSection 3: Functions of several variables.
Section 3: Functions of several variables. Compiled by Chris Tisdell S1: Motivation S2: Function of two variables S3: Visualising and sketching S4: Limits and continuity S5: Partial differentiation S6:
More information33. Riemann Summation over Rectangular Regions
. iemann Summation over ectangular egions A rectangular region in the xy-plane can be defined using compound inequalities, where x and y are each bound by constants such that a x a and b y b. Let z = f(x,
More informationDiscussion 8 Solution Thursday, February 10th. Consider the function f(x, y) := y 2 x 2.
Discussion 8 Solution Thursday, February 10th. 1. Consider the function f(x, y) := y 2 x 2. (a) This function is a mapping from R n to R m. Determine the values of n and m. The value of n is 2 corresponding
More informationSection 15.3 Partial Derivatives
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
More informationCalculus II Fall 2014
Calculus II Fall 2014 Lecture 3 Partial Derivatives Eitan Angel University of Colorado Monday, December 1, 2014 E. Angel (CU) Calculus II 1 Dec 1 / 13 Introduction Much of the calculus of several variables
More informationB) 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
MTH 22 Exam Two - Review Problem Set Name Sketch the surface z = f(x,y). ) f(x, y) = - x2 ) 2) f(x, y) = 2 -x2 - y2 2) Find the indicated limit or state that it does not exist. 4x2 + 8xy + 4y2 ) lim (x,
More informationSOLUTIONS 2. PRACTICE EXAM 2. HOURLY. Problem 1) TF questions (20 points) Circle the correct letter. No justifications are needed.
SOLUIONS 2. PRACICE EXAM 2. HOURLY Math 21a, S03 Problem 1) questions (20 points) Circle the correct letter. No justifications are needed. A function f(x, y) on the plane for which the absolute minimum
More information11.2 LIMITS AND CONTINUITY
11. LIMITS AND CONTINUITY INTRODUCTION: Consider functions of one variable y = f(x). If you are told that f(x) is continuous at x = a, explain what the graph looks like near x = a. Formal definition of
More informationReview Problems. Calculus IIIA: page 1 of??
Review Problems The final is comprehensive exam (although the material from the last third of the course will be emphasized). You are encouraged to work carefully through this review package, and to revisit
More informationREVIEW SHEET FOR MIDTERM 2: ADVANCED
REVIEW SHEET FOR MIDTERM : ADVANCED MATH 195, SECTION 59 (VIPUL NAIK) To maximize efficiency, please bring a copy (print or readable electronic) of this review sheet to the review session. The document
More information18.3. Stationary Points. Introduction. Prerequisites. Learning Outcomes
Stationary Points 8.3 Introduction The calculation of the optimum value of a function of two variables is a common requirement in many areas of engineering, for example in thermodynamics. Unlike the case
More informationLECTURE 19 - LAGRANGE MULTIPLIERS
LECTURE 9 - LAGRANGE MULTIPLIERS CHRIS JOHNSON Abstract. In this lecture we ll describe a way of solving certain optimization problems subject to constraints. This method, known as Lagrange multipliers,
More information10.1 Curves defined by parametric equations
Outline Section 1: Parametric Equations and Polar Coordinates 1.1 Curves defined by parametric equations 1.2 Calculus with Parametric Curves 1.3 Polar Coordinates 1.4 Areas and Lengths in Polar Coordinates
More informationFunctions of more than one variable
Chapter 3 Functions of more than one variable 3.1 Functions of two variables and their graphs 3.1.1 Definition A function of two variables has two ingredients: a domain and a rule. The domain of the function
More informationExam 1 Study Guide. Math 223 Section 12 Fall Student s Name
Exam 1 Study Guide Math 223 Section 12 Fall 2015 Dr. Gilbert Student s Name The following problems are designed to help you study for the first in-class exam. Problems may or may not be an accurate indicator
More informationOn Surfaces of Revolution whose Mean Curvature is Constant
On Surfaces of Revolution whose Mean Curvature is Constant Ch. Delaunay May 4, 2002 When one seeks a surface of given area enclosing a maximal volume, one finds that the equation this surface must satisfy
More informationCalculus I Handout: Curves and Surfaces in R 3. 1 Curves in R Curves in R 2 1 of 21
1. Curves in R 2 1 of 21 Calculus I Handout: Curves and Surfaces in R 3 Up until now, everything we have worked with has been in two dimensions. But we can extend the concepts of calculus to three dimensions
More information14.6 Directional Derivatives
CHAPTER 14. PARTIAL DERIVATIVES 107 14.6 Directional Derivatives Comments. Recall that the partial derivatives can be interpreted as the derivatives along traces of f(x, y). We can reinterpret this in
More informationCalculus of Several Variables
Benjamin McKay Calculus of Several Variables Optimisation and Finance February 18, 2018 This work is licensed under a Creative Commons Attribution-ShareAlike 3.0 Unported License. Preface The course is
More informationThis exam contains 9 problems. CHECK THAT YOU HAVE A COMPLETE EXAM.
Math 126 Final Examination Winter 2012 Your Name Your Signature Student ID # Quiz Section Professor s Name TA s Name This exam contains 9 problems. CHECK THAT YOU HAVE A COMPLETE EXAM. This exam is closed
More informationPractice Problems: Calculus in Polar Coordinates
Practice Problems: Calculus in Polar Coordinates Answers. For these problems, I want to convert from polar form parametrized Cartesian form, then differentiate and take the ratio y over x to get the slope,
More informationExamples: Find the domain and range of the function f(x, y) = 1 x y 2.
Multivariate Functions In this chapter, we will return to scalar functions; thus the functions that we consider will output points in space as opposed to vectors. However, in contrast to the majority of
More informationMULTI-VARIABLE OPTIMIZATION NOTES. 1. Identifying Critical Points
MULTI-VARIABLE OPTIMIZATION NOTES HARRIS MATH CAMP 2018 1. Identifying Critical Points Definition. Let f : R 2! R. Then f has a local maximum at (x 0,y 0 ) if there exists some disc D around (x 0,y 0 )
More information2.1 BASIC CONCEPTS Basic Operations on Signals Time Shifting. Figure 2.2 Time shifting of a signal. Time Reversal.
1 2.1 BASIC CONCEPTS 2.1.1 Basic Operations on Signals Time Shifting. Figure 2.2 Time shifting of a signal. Time Reversal. 2 Time Scaling. Figure 2.4 Time scaling of a signal. 2.1.2 Classification of Signals
More informationUniversity of California, Berkeley Department of Mathematics 5 th November, 2012, 12:10-12:55 pm MATH 53 - Test #2
University of California, Berkeley epartment of Mathematics 5 th November, 212, 12:1-12:55 pm MATH 53 - Test #2 Last Name: First Name: Student Number: iscussion Section: Name of GSI: Record your answers
More informationCalculus IV Math 2443 Review for Exam 2 on Mon Oct 24, 2016 Exam 2 will cover This is only a sample. Try all the homework problems.
Calculus IV Math 443 eview for xam on Mon Oct 4, 6 xam will cover 5. 5.. This is only a sample. Try all the homework problems. () o not evaluated the integral. Write as iterated integrals: (x + y )dv,
More informationMATH Exam 2 Solutions November 16, 2015
MATH 1.54 Exam Solutions November 16, 15 1. Suppose f(x, y) is a differentiable function such that it and its derivatives take on the following values: (x, y) f(x, y) f x (x, y) f y (x, y) f xx (x, y)
More informationEXERCISES CHAPTER 11. z = f(x, y) = A x α 1. x y ; (3) z = x2 + 4x + 2y. Graph the domain of the function and isoquants for z = 1 and z = 2.
EXERCISES CHAPTER 11 1. (a) Given is a Cobb-Douglas function f : R 2 + R with z = f(x, y) = A x α 1 1 x α 2 2, where A = 1, α 1 = 1/2 and α 2 = 1/2. Graph isoquants for z = 1 and z = 2 and illustrate the
More informationCalculus 3 Exam 2 31 October 2017
Calculus 3 Exam 2 31 October 2017 Name: Instructions: Be sure to read each problem s directions. Write clearly during the exam and fully erase or mark out anything you do not want graded. You may use your
More informationEstimating Areas. is reminiscent of a Riemann Sum and, amazingly enough, will be called a Riemann Sum. Double Integrals
Estimating Areas Consider the challenge of estimating the volume of a solid {(x, y, z) 0 z f(x, y), (x, y) }, where is a region in the xy-plane. This may be thought of as the solid under the graph of z
More information47. Conservative Vector Fields
47. onservative Vector Fields Given a function z = φ(x, y), its gradient is φ = φ x, φ y. Thus, φ is a gradient (or conservative) vector field, and the function φ is called a potential function. Suppose
More informationLecture 19 - Partial Derivatives and Extrema of Functions of Two Variables
Lecture 19 - Partial Derivatives and Extrema of Functions of Two Variables 19.1 Partial Derivatives We wish to maximize functions of two variables. This will involve taking derivatives. Example: Consider
More information14.7 Maximum and Minimum Values
CHAPTER 14. PARTIAL DERIVATIVES 115 14.7 Maximum and Minimum Values Definition. Let f(x, y) be a function. f has a local max at (a, b) iff(a, b) (a, b). f(x, y) for all (x, y) near f has a local min at
More information7/26/2018 SECOND HOURLY PRACTICE I Maths 21a, O.Knill, Summer Name:
7/26/218 SECOND HOURLY PRACTICE I Maths 21a, O.Knill, Summer 218 Name: Start by printing your name in the above box. Try to answer each question on the same page as the question is asked. If needed, use
More informationLevel Curves, Partial Derivatives
Unit #18 : Level Curves, Partial Derivatives Goals: To learn how to use and interpret contour diagrams as a way of visualizing functions of two variables. To study linear functions of two variables. To
More informationMath for Economics 1 New York University FINAL EXAM, Fall 2013 VERSION A
Math for Economics 1 New York University FINAL EXAM, Fall 2013 VERSION A Name: ID: Circle your instructor and lecture below: Jankowski-001 Jankowski-006 Ramakrishnan-013 Read all of the following information
More information(d) If a particle moves at a constant speed, then its velocity and acceleration are perpendicular.
Math 142 -Review Problems II (Sec. 10.2-11.6) Work on concept check on pages 734 and 822. More review problems are on pages 734-735 and 823-825. 2nd In-Class Exam, Wednesday, April 20. 1. True - False
More informationMultivariate Calculus
Multivariate Calculus Partial Derivatives 1 Theory Recall the definition of the partial derivatives of a function of two variables, z = f(x, y): f x = lim f(x + x, y) f(x, y) x 0 x f y f(x, y + y) f(x,
More informationIndependence of Path and Conservative Vector Fields
Independence of Path and onservative Vector Fields MATH 311, alculus III J. Robert Buchanan Department of Mathematics Summer 2011 Goal We would like to know conditions on a vector field function F(x, y)
More informationMaxima and Minima. Chapter Local and Global extrema. 5.2 Continuous functions on closed and bounded sets Definition of global extrema
Chapter 5 Maxima and Minima In first semester calculus we learned how to find the maximal and minimal values of a function y = f(x) of one variable. The basic method is as follows: assuming the independent
More informationMath 2321 Review for Test 2 Fall 11
Math 2321 Review for Test 2 Fall 11 The test will cover chapter 15 and sections 16.1-16.5 of chapter 16. These review sheets consist of problems similar to ones that could appear on the test. Some problems
More information266&deployment= &UserPass=b3733cde68af274d036da170749a68f6
Sections 14.6 and 14.7 (1482266) Question 12345678910111213141516171819202122 Due: Thu Oct 21 2010 11:59 PM PDT 1. Question DetailsSCalcET6 14.6.012. [1289020] Find the directional derivative, D u f, of
More informationMathematics 205 HWK 19b Solutions Section 16.2 p750. (x 2 y) dy dx. 2x 2 3
Mathematics 5 HWK 9b Solutions Section 6. p75 Problem, 6., p75. Evaluate (x y) dy dx. Solution. (x y) dy dx x ( ) y dy dx [ x x dx ] [ ] y x dx Problem 9, 6., p75. For the region as shown, write f da as
More information11/1/2017 Second Hourly Practice 2 Math 21a, Fall Name:
11/1/217 Second Hourly Practice 2 Math 21a, Fall 217 Name: MWF 9 Jameel Al-Aidroos MWF 9 Dennis Tseng MWF 1 Yu-Wei Fan MWF 1 Koji Shimizu MWF 11 Oliver Knill MWF 11 Chenglong Yu MWF 12 Stepan Paul TTH
More information11/18/2008 SECOND HOURLY FIRST PRACTICE Math 21a, Fall Name:
11/18/28 SECOND HOURLY FIRST PRACTICE Math 21a, Fall 28 Name: MWF 9 Chung-Jun John Tsai MWF 1 Ivana Bozic MWF 1 Peter Garfield MWF 1 Oliver Knill MWF 11 Peter Garfield MWF 11 Stefan Hornet MWF 12 Aleksander
More informationMATH 20C: FUNDAMENTALS OF CALCULUS II FINAL EXAM
MATH 2C: FUNDAMENTALS OF CALCULUS II FINAL EXAM Name Please circle the answer to each of the following problems. You may use an approved calculator. Each multiple choice problem is worth 2 points.. Multiple
More informationMATH 255 Applied Honors Calculus III Winter Homework 1. Table 1: 11.1:8 t x y
MATH 255 Applied Honors Calculus III Winter 2 Homework Section., pg. 692: 8, 24, 43. Section.2, pg. 72:, 2 (no graph required), 32, 4. Section.3, pg. 73: 4, 2, 54, 8. Section.4, pg. 79: 6, 35, 46. Solutions.:
More informationMath 32A Discussion Session Week 9 Notes November 28 and 30, 2017
Math 3A Discussion Session Week 9 Notes November 8 an 30, 07 This week we ll explore some of the ieas from chapter 5, focusing mostly on the graient. We ll motivate this exploration with an example that
More informationDirectional Derivative, Gradient and Level Set
Directional Derivative, Gradient and Level Set Liming Pang 1 Directional Derivative Te partial derivatives of a multi-variable function f(x, y), f f and, tell us te rate of cange of te function along te
More information1. 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.
Math 234 WES WORKSHEET 9 Spring 2015 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. 2. Let f(x, y) = x 2 + y 2. Find all the points on the level
More information