lecture11-09[1]

lecture11-09[1] - Math 18.02 (Spring 2009): Lecture 11...

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Unformatted text preview: Math 18.02 (Spring 2009): Lecture 11 Summary of Global max/min: Boundaries and infinity. Chain Rule. Differential. Vector Fields February 27 Last time: Second derivative test. Global max/min: Boundaries and infinity. Level curves and level surfaces. Today: Summary of Global max/min: Boundaries and infinity. Chain Rule. Differential. Vector Fields. Reading Material: From Simmos 19.4 and 19.6. 2 Summary of Global max/min: Boundaries and infinity Last time we considered the following exercise: Exercise 1. Consider the function f ( x, y ) = x 2 + y 2 defined on the domain W = { ( x, y ) / 4 x 2 + y 2 1 } . Compute global max and min of f on this domain. I would like to summarize the method we used to find the absolute (or global) max/min in this example by keeping the situation pretty general: Assume a function z = f ( x, y ) is given and with it a domain W with a boundary W which can be represented as a closed curve not intersecting itself. The absolute (or global) max/min are obtained in the following way: Find all critical points strictly inside W . Analyze them using the second derivative test and decide which ones are local max and which ones are local min. Assume that P 1 , ..., P n are local min and Q 1 , ..., Q m are local max. Now it is time to analyze what goes on on the bundary. Find parametric equations for it ~ r ( t ) = x ( t ) i + y ( t ) j, for t in [ a, b ] . Evaluate the function f ( x, y ) on the boundary by defining h ( t ) = f ( x ( t ) , y ( t )) ....
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lecture11-09[1] - Math 18.02 (Spring 2009): Lecture 11...

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