Critical Point of Multivariable Function

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To locate the critical point of the function e^-(x^(2)+y^(2)+2x), the first step is to differentiate the function and set the derivatives fx and fy to zero. The derivatives are fx(x,y) = (-2x-2)e^(-x^(2)-y^(2)-2x) and fy = -2ye^(-x^(2)-y^(2)-2x). It's important not to take the logarithm of the equations since e^a is never zero; thus, the critical point occurs when -2x - 2 = 0. This leads to the conclusion that x = -1. The discussion emphasizes the correct approach to solving for critical points without misapplying logarithmic properties.
danny_manny
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Homework Statement



Locate the critical point of the function,

e^-(x^(2)+y^(2)+2x)

Homework Equations



none

The Attempt at a Solution


Ok first step differentiate the function and set it to zero for both fx and fy,

Fx(x,y) = (-2x-2)e^(-x^(2)-y^(2)-2x)
Fy = -2ye^(-x^(2)-y^(2)-2x)

Now I need to solve both equations simultaneously,

I get

ln(-2x-2)-x^(2)-y^(2)-2x = 0

and

ln(-2y)-x^(2)-y^(2)-2x = 0

and this is where I am stuck. :( any advice would be welcomed :D

thanks,
Dan
 
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You need to solve Fx(x,y)=0. You don't want to take the log of that. log(0) is undefined, it's NOT 0. The way to do this is to notice e^a is NEVER 0. So e^(-x^(2)-y^(2)-2x) is NEVER 0. The only way Fx(x,y) could be 0 is if (2x-2) is 0.
 
ah thank you.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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