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PostPosted: Thu, 10 May 2012 23:09:18 UTC 
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First of this is somewhat embarassing as I really shouldn't be strugling with this, however I'm having a hard time getting back into this after a break of about a year.

I'm trying to obtain the critical points of the function bellow and then determin their nature. I'm having no trouble with the differentiation however solving the first derivatives for x and y is driving me nuts. Once I have these values I know how to solve the rest of the problem.

A surface has the following equation:

z=x^3-3xy+8y^3+20

So:

dz/dx= 3x^2-3y = 0

dz/dy= -3x+24y^2 = 0

thus:

3x^2=3y so x^2=y

and

24y^2=3x so 8y^2=x

CP's:

x = (A)??? & (B)???
y = (A)??? & (B)???

Z= (A)x^3-3(A)x(A)y+8(A)y^3+20
&
Z= (B)x^3-3(B)x(B)y+8(B)y^3+20


Any help is greatly appreciated.

Craig


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PostPosted: Fri, 11 May 2012 00:20:18 UTC 
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Phoenix86 wrote:
First of this is somewhat embarassing as I really shouldn't be strugling with this, however I'm having a hard time getting back into this after a break of about a year.

I'm trying to obtain the critical points of the function bellow and then determin their nature. I'm having no trouble with the differentiation however solving the first derivatives for x and y is driving me nuts. Once I have these values I know how to solve the rest of the problem.

A surface has the following equation:

z=x^3-3xy+8y^3+20

So:

dz/dx= 3x^2-3y = 0

dz/dy= -3x+24y^2 = 0

thus:

3x^2=3y so x^2=y

and

24y^2=3x so 8y^2=x

CP's:

x = (A)??? & (B)???
y = (A)??? & (B)???

Z= (A)x^3-3(A)x(A)y+8(A)y^3+20
&
Z= (B)x^3-3(B)x(B)y+8(B)y^3+20


Any help is greatly appreciated.

Craig


You have a critical point when the tangent plane is "vertical", so first you find the equation for your tangent plane:

Write z=f(x,y) then the tangent plane is f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)+(z-f(x_0,y_0))=0

now $f_x(x_0,y_0)=-{\partial z\over\partial x}(x_0,y_0) and similarly, $f_y(x_0,y_0)=-{\partial z\over\partial y}(x_0,y_0), and the plane is vertical when both are equal to zero.

So all you need to do is decide when

x^2-y=0
8y^2-x=0

Using the second equation and squaring we get:

64y^4=x^2 and substituting into the first you get:

64y^4-y=0, so we get y=x=0 which works in the original set of equations, otherwise y\ne 0 so x\ne 0, this gives the other option as:

(4y)^3-1=0 so y={1\over 4} and x=\pm {1\over 2} from the first equation, and checking with the second equation, we see that only the positive possibility works.

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PostPosted: Fri, 11 May 2012 13:38:06 UTC 
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Shadow wrote:
You have a critical point when the tangent plane is "vertical".


Shouldn't that be "horizontal" instead?

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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PostPosted: Fri, 11 May 2012 13:44:34 UTC 
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outermeasure wrote:
Shadow wrote:
You have a critical point when the tangent plane is "vertical".


Shouldn't that be "horizontal" instead?


Probably, I had a mnemonic which is technically incorrect, but always lets me remember how it goes which involved a relation to vertical lines in the \mathbb{R}^2 case.

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PostPosted: Sat, 12 May 2012 22:37:02 UTC 
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Thanks for that, was starting to get kind of worried when I couldnt figure that out... seems kinda obvious now

Craig


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