My Math Forum Classification of critical points

 Calculus Calculus Math Forum

 May 18th, 2018, 06:29 AM #1 Newbie   Joined: May 2018 From: London Posts: 1 Thanks: 0 Classification of critical points My question is to find critical points of: xy^2 - x^2 +3x—y^4+ze^(-z) and classify them. Please help, I have problems with finding the critical points, because the system of equation becomes big.
 May 18th, 2018, 07:23 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 Three equations is really that big a system! You have $\displaystyle f(x, y, z)= xy^2- x^2+ 3x- 4y^4+ ze^{-z}$. Setting the partial derivatives to 0: $\displaystyle f_x= y^2- 2x+ 3= 0$ $\displaystyle f_y= 2xy- 16y^3= 0$ $\displaystyle f_z= e^{-z}- ze^{-z}= 0$ The third equation involves only z: $\displaystyle e^{-z}(1- z)= 0$. Since $\displaystyle e^{-z}$ is never 0, z= 1. We now have $\displaystyle y^2- 2x+ 3= 0$ and $\displaystyle 2xy- 16y^3= y(2x- 16y^2)= 0$. From the second, either y= 0 or $\displaystyle 2x=16y^2$. If y= 0, the first equation becomes -2x+ 3= 0 so x= 3/2. x= 3/2, y= 0, z= 1 is one solution. If $\displaystyle 2x= 16y^2$ then $\displaystyle x= 8y^2$ and the first equation becomes $\displaystyle y^2- 2(8y^2)+ 3= -15y^2+ 3= 0$. $\displaystyle 15y^2= 3$ so $\displaystyle y^2= 1/5$and y is either $\displaystyle \frac{\sqrt{5}}{5}$ or $\displaystyle -\frac{\sqrt{5}}{5}$. In the either case $\displaystyle x= 8y^2= \frac{8}{5}$. So $\displaystyle x= \frac{8}{5}$, $\displaystyle y= \frac{\sqrt{5}}{5}$, $\displaystyle z= 1$ is another solution and [math]$\displaystyle x= \frac{8}{5}$, $\displaystyle y= -\frac{\sqrt{5}}{5}$, $\displaystyle z= 1$ is a third solution.

 Tags classification, critical, points

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post Dan3500 Calculus 1 August 17th, 2016 04:17 PM irvm Calculus 6 January 15th, 2016 05:54 AM mathkid Calculus 1 November 11th, 2012 06:34 PM Timk Calculus 3 November 29th, 2011 10:59 AM summerset353 Calculus 1 March 5th, 2010 01:50 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top