From Math 2220 Class 10 ample Higher Partials in Polar Cordinates - PowerPoint PPT Presentation

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- From Math 2220 Class 10 ample Higher Partials in Polar Cordinates Dr. Allen Back Gradient Estimate from Graph Method of Characteristics Sep. 19, 2014 Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Higher Partial Derivatives From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- Definition: A function f : U ⊂ R n → R is C 2 or twice ample Higher Partials continuously differentiable if f and all partial derivatives of first in Polar or second order exist and are continuous. Similarly f is C k for Cordinates Gradient other positive integers k . Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Higher Partial Derivatives From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample We earlier had a theorem that f a C 1 function implies f is Higher Partials in Polar Cordinates differentiable. Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Higher Partial Derivatives From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- Theorem: If f : U ⊂ R n → R is C 2 , then ample Higher Partials ∂ 2 f ∂ 2 f in Polar Cordinates = . ∂ x i ∂ x j ∂ x j ∂ x i Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Higher Partial Derivatives From Math 2220 Class 10 V1cc Higher Partial Derivatives The proof of this theorem is based on showing that the A Counterex- symmetric expression ample Higher Partials (∆ x , ∆ y ) → (0 , 0) [ f ( x 0 , y 0 ) − f ( x 0 + ∆ x , y 0 ) lim in Polar Cordinates Gradient − f ( x 0 , y 0 + ∆ y ) + f ( x 0 + ∆ x , y 0 + ∆ y )] / ∆ x ∆ y . Estimate from Graph Method of Characteristics exists and equals both f xy and f yx at ( x 0 , y 0 ) . Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Higher Partial Derivatives From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Higher Partial Derivatives From Math 2220 Class 10 The proof of this theorem is based on showing that the V1cc symmetric expression Higher Partial Derivatives (∆ x , ∆ y ) → (0 , 0) [ f ( x 0 , y 0 ) − f ( x 0 + ∆ x , y 0 ) lim A Counterex- ample Higher Partials − f ( x 0 , y 0 + ∆ y ) + f ( x 0 + ∆ x , y 0 + ∆ y )] / ∆ x ∆ y . in Polar Cordinates Gradient exists and equals both f xy and f yx at ( x 0 , y 0 ) . Estimate from Graph The basic tool is the mean value estimate Method of Characteristics Critical Points g ( x 0 + ∆ x ) − g ( x 0 ) = g ′ ( c )∆ x Why the First Derivative Test? for some c between x 0 and x 0 + ∆ x . Saddle Points Second Derivative Test

Higher Partial Derivatives From Math 2220 Class 10 The proof of this theorem is based on showing that the V1cc symmetric expression Higher Partial Derivatives (∆ x , ∆ y ) → (0 , 0) [ f ( x 0 , y 0 ) − f ( x 0 + ∆ x , y 0 ) lim A Counterex- ample − f ( x 0 , y 0 + ∆ y ) + f ( x 0 + ∆ x , y 0 + ∆ y )] / ∆ x ∆ y . Higher Partials in Polar Cordinates Gradient exists and equals both f xy and f yx at ( x 0 , y 0 ) . Estimate from Graph We would apply this to the function Method of Characteristics g ( x ) = f ( x , y 0 + ∆ y ) − f ( x , y 0 ) . Critical Points Why the First Derivative Test? since the numerator looks like g ( x 0 + ∆ x ) − g ( x 0 ) . Saddle Points Second Derivative Test

Higher Partial Derivatives From Math 2220 Class 10 V1cc Second partials show up in max-min as well as physical Higher Partial Derivatives modeling. For example the vibration of a string on a musical A Counterex- ample instrument is governed by the 1 dimensional wave equation Higher Partials in Polar ∂ 2 z ∂ 2 z ∂ x 2 = 1 Cordinates c 2 ∂ t 2 Gradient Estimate from Graph where z ( x , t ) is the displacement of the string (up or down) at Method of position x along the string and time t . The constant c is the Characteristics Critical Points speed of the vibration. Why the First Derivative Test? Saddle Points Second Derivative Test

Higher Partial Derivatives From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials Show that z ( x , t ) = A sin ( x − ct ) satisfies the wave equation in Polar Cordinates above. Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Higher Partial Derivatives From Math 2220 Class 10 V1cc Higher Partial Derivatives Or A sin ( x + ct ). A Counterex- ample Or indeed for any twice differentiable one variable function Higher Partials g ( u ), in Polar Cordinates z ( x , t ) = g ( x − ct ) Gradient Estimate from will also satisfy the wave equation. Graph Method of Why? Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Higher Partial Derivatives From Math 2220 Class 10 V1cc A typical application of mixed partials arises out of your homework problem 3.1 #22 where you show that the wave Higher Partial Derivatives equation A Counterex- ∂ 2 w ∂ x 2 = ∂ 2 w ample ∂ y 2 Higher Partials in Polar Cordinates becomes after a change of coordinates Gradient Estimate from Graph x = u + v , y + u − v Method of Characteristics ∂ 2 w Critical Points ∂ u ∂ v = 0 Why the First Derivative Test? Saddle Points Second Derivative Test

Higher Partial Derivatives From Math 2220 Class 10 V1cc A typical application of mixed partials arises out of your homework problem 3.1 #22 where you show that the wave Higher Partial Derivatives equation A Counterex- ∂ 2 w ∂ x 2 = ∂ 2 w ample ∂ y 2 Higher Partials in Polar Cordinates becomes after a change of coordinates Gradient Estimate from Graph x = u + v , y + u − v Method of Characteristics ∂ 2 w Critical Points ∂ u ∂ v = 0 Why the First Derivative Test? Saddle Points Second Derivative Test

Higher Partial Derivatives From Math 2220 Class 10 A typical application of mixed partials arises out of your V1cc homework problem 3.1 #22 where you show that the wave equation Higher Partial Derivatives ∂ 2 w ∂ x 2 = ∂ 2 w A Counterex- ∂ y 2 ample Higher Partials in Polar Cordinates ∂ 2 w Gradient x = u + v , y + u − v ∂ u ∂ v = 0 Estimate from Graph Method of implying the general solution Characteristics Critical Points w = f ( u ) + g ( v ) = f ( x − y ) + g ( x + y ) Why the First Derivative 2 2 Test? for any C 2 functions f and g . Saddle Points Second Derivative Test

A Counterexample From Math 2220 Class 10 V1cc Higher Partial Derivatives When f is not C 2 , f xy need not equal f yx even when both exist. A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

A Counterexample From Math 2220 Class 10 V1cc An example illustrating this is Higher Partial Derivatives � xy ( x 2 − y 2 ) if ( x , y ) � = (0 , 0) A Counterex- x 2 + y 2 f ( x , y ) = ample 0 if ( x , y ) = (0 , 0) Higher Partials in Polar Cordinates This function Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

A Counterexample From Math 2220 Class 10 An example illustrating this is V1cc � xy ( x 2 − y 2 ) Higher Partial if ( x , y ) � = (0 , 0) Derivatives x 2 + y 2 f ( x , y ) = 0 if ( x , y ) = (0 , 0) A Counterex- ample Higher Partials in Polar This function Cordinates is clearly C 2 for ( x , y ) � = (0 , 0) since it is a rational function Gradient Estimate from (quotient of polynomials) with nonzero denominator. Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

A Counterexample An example illustrating this is From Math 2220 Class 10 V1cc � xy ( x 2 − y 2 ) if ( x , y ) � = (0 , 0) f ( x , y ) = x 2 + y 2 Higher Partial 0 if ( x , y ) = (0 , 0) Derivatives A Counterex- ample This function Higher Partials in Polar satisfies f ( x , y ) = − f ( y , x ) implying Cordinates Gradient f xy = f yx Estimate from Graph Method of whenever x = y . (A good chain rule exercise . . . ) Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

A Counterexample From Math 2220 Class 10 An example illustrating this is V1cc � Higher Partial xy ( x 2 − y 2 ) if ( x , y ) � = (0 , 0) Derivatives f ( x , y ) = x 2 + y 2 0 if ( x , y ) = (0 , 0) A Counterex- ample Higher Partials in Polar This function Cordinates satisfies f x (0 , 0) = 0 and f y (0 , 0) = 0 since f vanishes along the Gradient Estimate from axes. Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test