Implicit differentiation gives too many stationary points

In summary, the conversation discusses implicit differentiation and its application to finding stationary points of a function. The equation x2-5xy+3y2 = 7 is used as an example, and it is shown that the derivative dy/dx equals zero on the line y=(2/5)x, except at the origin. However, it is pointed out that this line does not intersect the hyperbola, meaning there are no values of x or y that satisfy both equations. This explains why there are no stationary points on the line, despite it appearing to have an infinite number of them. The conversation concludes with a helpful tip to confirm this by plugging in the line equation and solving for x, which gives imaginary solutions.
  • #1
nomadreid
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If, with y a function of x, I have the equation x2-5xy+3y2 = 7, then by implicit differentiation, I get that dy/dx = (2x-5y)/(5x-6y). This equals zero everywhere on the straight line y=(2/5)x except at the origin. This would seem to indicate stationary points everywhere on that line, which is hard to imagine, and anyway the graph of this equation seems to be a hyperbola, having no stationary points. Obviously I am missing something very basic here. Any help would be appreciated.
 
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  • #2
The line y = 2x/5 does not intersect your hyperbola, so there are no values of x or y that satisfy both the original equation and the stationary point equation. Hence, no stationary points. (Looking at the plot it might not be obvious that the curves don't intersect, so you can also plug y = 2x/5 back into your implicit equation and solve for x, which gives you imaginary solutions, confirming that the curves don't intersect).
 
  • #3
Ah. I should have spotted that. Thanks very much, Mute!
 

1. What is implicit differentiation and how does it relate to finding stationary points?

Implicit differentiation is a mathematical technique used to find the derivative of an equation that is not explicitly stated in terms of one variable. This is useful in finding stationary points, as they are points where the derivative is equal to zero.

2. Why does implicit differentiation sometimes give too many stationary points?

Implicit differentiation can give too many stationary points when the original equation has multiple solutions or when the derivative is not defined at certain points. This can lead to extraneous or incorrect solutions.

3. How can I determine which stationary points are valid when using implicit differentiation?

To determine which stationary points are valid, you can plug in the coordinates of each point into the original equation. If the equation is satisfied, then the point is a valid stationary point. If not, then it is an extraneous solution.

4. Are there any other methods for finding stationary points besides implicit differentiation?

Yes, there are other methods such as using the first or second derivative test, setting the derivative equal to zero and solving for the variable, or using other techniques such as optimization or graphing.

5. Can implicit differentiation be used for all types of equations?

No, implicit differentiation can only be used for equations that are not explicitly stated in terms of one variable. For example, it cannot be used for linear equations, as they are already in the form y=mx+b and do not require implicit differentiation.

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