My Math Forum Critical Points, Partial Derivatives

 Calculus Calculus Math Forum

 May 5th, 2015, 03:59 PM #1 Senior Member     Joined: Dec 2014 From: The Asymptote Posts: 142 Thanks: 6 Math Focus: Certainty Critical Points, Partial Derivatives I'm partially deriving the equation $\displaystyle f(x,y)=x^3+x^2y+y^2-4y+2$ and want to find critical points. I'm at somewhat of a dilemma for my solutions of $x$ $\displaystyle f_{x}=0$, $\displaystyle x$ solutions are: $\displaystyle x_1=0$ $\displaystyle x_2=4$ $\displaystyle x_3=-1$ $\displaystyle f_{y}=0$ $\displaystyle , x$ solutions are: $\displaystyle x_1=4$ $\displaystyle x_2=-1$ I'm unsure as whether or not $\displaystyle x=0$ is ignored in this instance. When set in terms of $\displaystyle y$, $\displaystyle f_x=0$ when $\displaystyle x=0$ $\displaystyle f_y=2$ when $\displaystyle x=0$ My first order derivatives are correct, I have checked via second order derivatives and $\displaystyle f_{xy}=f_{yx}=2x$ Last edited by hyperbola; May 5th, 2015 at 04:06 PM.
May 5th, 2015, 08:23 PM   #2
Math Team

Joined: Jan 2015
From: Alabama

Posts: 3,264
Thanks: 902

Quote:
 Originally Posted by hyperbola I'm partially deriving the equation $\displaystyle f(x,y)=x^3+x^2y+y^2-4y+2$ and want to find critical points. I'm at somewhat of a dilemma for my solutions of $x$ $\displaystyle f_{x}=0$, $\displaystyle x$ solutions are: $\displaystyle x_1=0$ $\displaystyle x_2=4$ $\displaystyle x_3=-1$
$\displaystyle f_x= 3x^2+ 2xy$.
f_x(0, y)= 0
f_x(4, y)= 48+ 8y. Whether that is 0 or not depends on y.
f_x(-1, y)= 3- 2y. Whether that is 0 or not depends on y.

Quote:
 $\displaystyle f_{y}=0$ $\displaystyle , x$ solutions are: $\displaystyle x_1=4$ $\displaystyle x_2=-1$
$\displaystyle f_y= x^2- 2y+ 4$
$\displaystyle f_y(4, y)= 20- 2y$
whether that is 0 or not depends on y
$\displaystyle f_y(-1, y)= 5- 2y$
whether that is 0 or not depends on y.

Quote:
 I'm unsure as whether or not $\displaystyle x=0$ is ignored in this instance. When set in terms of $\displaystyle y$, $\displaystyle f_x=0$ when $\displaystyle x=0$ $\displaystyle f_y=2$ when $\displaystyle x=0$ My first order derivatives are correct, I have checked via second order derivatives and $\displaystyle f_{xy}=f_{yx}=2x$
Frankly, you don't seem to have any idea what you are doing! You are apparently finding values of x without giving any thought to y.

Yes, $\displaystyle f_x= 3x^2+ 2xy$ and $\displaystyle f_y= x^2- 2y+ 4$
set those equal to 0, $\displaystyle 3x^2+ 2xy= 0$ and $\displaystyle x^2- 2y+ 4= 0$ and find paired values of x and y that satisfy both equations.

From the second equation, $\displaystyle y= \frac{x^2+ 4}{2}$. Putting that into the first equation gives a single 3 degree equation for x. Then find the corresponding value of x.

 May 5th, 2015, 09:07 PM #3 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 522 Math Focus: Calculus/ODEs I get: $\displaystyle f_x(x,y)=x(3x+2y)\tag{1}$ $\displaystyle f_y(x,y)=x^2+2y-4\tag{2}$ Equating both to zero, from (2), we find: $\displaystyle 2y=4-x^2$ and then from (1), we obtain: $\displaystyle x\left(x^2-3x-4\right)=0$ Factor, and then you will have 3 values for $x$, and then use: $\displaystyle y=2-\frac{1}{2}x^2$ to get the corresponding $y$-values and you now have 3 critical points. Thanks from hyperbola
May 6th, 2015, 12:38 AM   #4
Senior Member

Joined: Dec 2014
From: The Asymptote

Posts: 142
Thanks: 6

Math Focus: Certainty
Quote:
 Originally Posted by Country Boy $\displaystyle f_x= 3x^2+ 2xy$. Frankly, you don't seem to have any idea what you are doing!
10/10 for positivity...

Thanks MarkFL, your replies are to the point with no extra rubbish on the side.

May 6th, 2015, 02:06 AM   #5
Senior Member

Joined: Dec 2014
From: The Asymptote

Posts: 142
Thanks: 6

Math Focus: Certainty
Quote:
 Originally Posted by Country Boy $\displaystyle f_x= 3x^2+ 2xy$. Putting that into the first equation gives a single 3 degree equation for x. Then find the corresponding value of x.
Naturally, I would've thought that both equations would give me an equal number of x values. Instead, I ended up with a 2nd degree (2 $x$ values) and a 3rd degree (3 $x$ values) polynomial. I've never before encountered this when solving simultaneous equations, hence the uncertainty.

Last edited by hyperbola; May 6th, 2015 at 02:08 AM.

 Tags critical, derivatives, partial, points

### critical points of a hyperbola

Click on a term to search for related topics.
 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post mathkid Calculus 1 November 11th, 2012 07:34 PM alexmath Calculus 8 April 26th, 2012 08:56 AM Timk Calculus 3 November 29th, 2011 11:59 AM summerset353 Calculus 1 March 5th, 2010 02:50 AM SSmokinCamaro Calculus 2 April 3rd, 2009 08:04 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top