Math 251 Final Review Pack

Of note changing the angle allows us to find half of

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Unformatted text preview: or a maximum For a minimum Where is either the local maximum or minimum value Definition (Global Extrema): f has an absolute or global extrema For a maximum For a minimum This is called the extreme maximum or minimum Theorem: If f has a local extreme at and the first order partial derivatives exists, then Definition (Critical/Stationary Points): The critical points are the points Or the derivatives do not exist Theorem: If is a local extreme point, then it is also a critical point 27 such that The Second Derivative Test Given that is a critical point of the function The second order partial derivatives of the function are continuous on a disk centered at Let We can arrive at some conclusions based on the value of D If , two cases arise When When , then , then If , then If is a local maximum is a local minimum is a saddle point , the test is inconclusive To find the local extremes and saddle points: 1) Find the critical points of the function 2) Use the second derivative to decide which of these points local extrema and/or saddle points Maximum and Minimum Values Recall in single variable calculus, for some closed interval and the function , f ,is continuous over this interval then f has a global maximum and a global minimum on Now for multivariable calculus, for some closed set (a set which contains its bounding points) , we arrive at the Extreme Value Theorem Theorem (Extreme Value Theorem): is closed and bound, and continuous then f attains an absolute (global) maximum and minimum ( respectively) at some points and in D is and To Find Absolute Maximum and Minimum 1) Find all the critical points (that are not on the boundary of D) and find the values for the function at these points 28 2) Find the extreme points on the boundary of D 3) The largest values found for 1) and 2) are the global maximum The smallest values found for 1) and 2) are the global minimum 29 14.8 Lagrange Multipliers Goal: Find the extreme points of a function subject to a constant. More generally: Goal: find the extreme values of subject to the constant So, how do we find them? - Suppose Take any curve has an extreme value at that is on the surface and passes through the point Evaluate along: - Since the function has an extreme value at then, Using the chain rule, we can evaluate at Using the gradient vector: This shows that these two vectors are orthogonal and that the gradient is also orthogonal to the tangent plane at the point to the surface Thus λ is defined as the lagrange multiplier Method of Legrange Multiplier (Find all the maximum and minimum points subject to the constraint ) Step 1: Find all such that Step 2: Evaluate all the points from the first step and the largest value yielded is the maximum and the smallest is the minimum Two constraints: Goal: Find the extreme values of subject to the constraints We recall that the gradient vector is orthogonal to the curve 30 and This means that the gradient vectors of the three curves must lie in the same plane Method: Step 1: Solve Step 2: Evaluate at each point 31 Chapter 15: Multiple Integrals 32 15.1 Double Integrals Over Rectangles Recall: Area and simple integrals: Now: Volumes and double integrals: Approximate the volume under the surface that lies above some subrectangle. Take the contributions of all the subrectangles: Definition (Double Integral): The double integral of f over is defin...
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This note was uploaded on 02/03/2014 for the course CMPT 150 taught by Professor Dr.anthonydixon during the Spring '08 term at Simon Fraser.

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