Explanation Given: The following function is given: f ( x , y ) = ( x − y ) e x 2 − y 2 Let P = ( a , b ) be a critical point of f ( x , y ) . Let the second order derivatives f x x , f y y , f x y are continuous near P . Let D be the discriminant of the function f ( x , y ) and is defined as D ( x , y ) = f x x f y y − f x y 2 (a) If D > 0 , then f ( a , b ) is a local minimum or local maximum. (b) If D < 0 , then ( a , b ) is a saddle point. (c) If D = 0 , then the test is inconclusive. (d) If D > 0 and f x x ( a , b ) > 0 , then f ( a , b ) is a local minimum. (e) If D > 0 and f x x ( a , b ) < 0 , then f ( a , b ) is a local maximum. Calculation: Consider the following function: f ( x , y ) = ( x − y ) e x 2 − y 2 Differentiating, f x ( x , y ) = e x 2 − y 2 ( 1 + 2 x 2 − 2 x y ) f y ( x , y ) = e x 2 − y 2 ( − 1 + 2 y 2 − 2 x y ) For critical points, putting f x = 0 , f y = 0 e x 2 − y 2 ( 1 + 2 x 2 − 2 x y ) = 0 and e x 2 − y 2 ( − 1 + 2 y 2 − 2 x y ) = 0 ⇒ 1 + 2 x 2 − 2 x y = 0 , − 1 + 2 y 2 − 2 x y = 0 ( e x 2 − y 2 ≠ 0 ) 2 x

3rd Edition

ISBN: 9781464114885

Author: Jon Rogawski, Colin Adams

Publisher: W. H. Freeman

See similar textbooks

Question

Chapter 14.7, Problem 22E

To determine

The critical points of the function.

Whether they are local minimum, local maximum or saddle points using the Second Derivative Test.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 14.7, Problem 21E

Chapter 14.7, Problem 23E

Students have asked these similar questions

How to solve it explain it step by step please

2.3 Vibration amplitude of a machine plotted against time shown in Fig. T2.3. is described by sin(0.8+) In(t + 2) x(t) = t + 0.5 Here, t is in seconds and x is in millimeters 0.6 0.4 0.2 ° -0.2 mmm -0.4 0 10 20 30 40 50 60 70 Time [s] Fig. T2.3: Machine vibration amplitude variation against time Obtain the vibration velocity as a function of time, x(t), if it is measured by a vibration velocity sensor.

Example(1): (Adiabatic humidification and cooling of air). Air has to be humidified and cooled adiabatically in a honzontal spray chamber with recirculated water. The active part of the chamber is Im #2m #15 m long. Under the operating conditions, the coefficient of heat transfer is expected to be 1300 kcal/(hr)(m2)(°C). 200 m3/min of air at 60 °C and 1 atm pressure with a humidity of 0.018 kg water/kg dry air is to be blown through the spray chamber. Calculate the following (a) the temperature and hunudity of the exit air (b) make-up water to be supplied, windage and blow down are neglected (c) the expected gas-phase mass transfer coefficient, kya (d) the temperature and humidity of the exit air if an identical spray chamber is added in series with the existing one O

Chapter 14 Solutions

Calculus: Early Transcendentals

Ch. 14.1 - Prob. 1PQ Ch. 14.1 - Prob. 2PQ Ch. 14.1 - Prob. 3PQ Ch. 14.1 - Prob. 4PQ Ch. 14.1 - Prob. 5PQ Ch. 14.1 - Prob. 1E Ch. 14.1 - Prob. 2E Ch. 14.1 - Prob. 3E Ch. 14.1 - Prob. 4E Ch. 14.1 - Prob. 5E

Knowledge Booster

The critical points of the function. Whether they are local minimum , local maximum or saddle points using the Second Derivative Test.

The critical points of the function. Whether they are local minimum, local maximum or saddle points using the Second Derivative Test.

Want to see the full answer?

Chapter 14 Solutions

The critical points of the function.

Whether they are local minimum, local maximum or saddle points using the Second Derivative Test.