How to find critical points for a differential equation?

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SUMMARY

The discussion focuses on finding critical points for the function f(x,y) = 3x^5y^2 - 30x^3y^2 + 60xy^2 + 150. Participants emphasize the importance of calculating the partial derivatives, specifically ∂f/∂x and ∂f/∂y, and setting them equal to zero to identify critical points. The correct partial derivatives are dx = 15x^4y^2 - 90x^2y^2 + 60y^2 and dy = 6yx^5 - 60x^3y + 10xy + 240. The discussion concludes that the critical points can be derived by solving these equations simultaneously, leading to four distinct (x,y) points along specific lines through the origin.

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  • Understanding of partial derivatives in multivariable calculus
  • Familiarity with algebraic factoring techniques
  • Knowledge of critical points in the context of differential equations
  • Ability to solve systems of equations
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  • Study the method for calculating partial derivatives in multivariable functions
  • Learn about the implications of critical points in optimization problems
  • Explore the concept of implicit differentiation in relation to finding critical points
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Students and educators in mathematics, particularly those studying calculus and differential equations, as well as anyone interested in understanding the behavior of multivariable functions and their critical points.

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Homework Statement



f(x,y)= 3x5y2 - 30x3y2 + 60xy2 +150

How do I find the critical points where this equation equals zero?

Homework Equations



None?

The Attempt at a Solution



I found the partial derivatives, but I'm stuck now and don't know where to go from here.

dx = 15x4y2 - 90x2y2 + 60y2
dy = 6x5 - 60x3y + 10xy + 240
 
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start by setting both equal to 0! then try quadratic factoring for the dx. the dy should be obvious.
 
also, you ask for critical points and where the equation equals zero...are you sure both occur simultaneously? i haven't checked but don't assume this to be the case
 
Okay so I did that. I got for dx, x = 3 +- sqrt(20)/2
The first part of dy should actually be 6yx^5.

After I have the dx value, I can sub them into the dy equation which gives approximately y = -0.1545 and y = -3.6099. Where do I go from here? There should be 4 separate (x,y) points.
 
Spell things out a little more carefully. What you quote a solution for I think is really saying

∂f/∂x = 0 at all points along (x/y)2 = those numbers, which is actually four lines through the origin I think.

Maybe there is a nice algebraic breakup of your ∂f/∂y but anyway what you are doing should lead you to the four points along the four lines.

Alternative approach is to find (dy/dx)f and its reciprocal and find the points where both are 0. Well same thing really, -(dy/dx)f is the ratio of your two derived polynomials.

Also I think the second of yours, what you call "dy" is mistaken and it works out fairly simply.

You might as well have taken the factor 3 out of your starting equation.
 
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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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