Horizontal tangents via implicit differentiation

In summary, the points of horizontal tangent lines on the equation x2 + xy + y2 = 6 are (sqrt(2), -2sqrt(2)) and (-sqrt(2), 2sqrt(2)).
  • #1
atmega-ist
9
0

Homework Statement



Find the points (if any) of of horizontal tangent lines on :

x2 + xy + y2 = 6

Homework Equations



n/a

The Attempt at a Solution



So far I've concluded that I must find the points at which dy/dx = 0. I've solved for dy/dx and arrived at dy/dx = (-2x-y)/(x+2y)

I assume that I would just have to get a "0" in the numerator to satisfy the horizontal tangent but doing so gives me

-2x-y = 0 ==> y = -2x

This seems that there would be an infinite number of horizontal tangents (as long as the original denominator didn't equal "0") but the graph of the original equation, per Wolfram Alpha, seems to be an ellipse so I'm only looking for two solutions...

Have I missed a component of the concept or should I not be ending up with an ellipse?

Thank you
 
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  • #2
If you plug y = -2x into the original equation you don't get infinite points.
 
  • #3
I think I have it. In plugging in -2x for y in the original equation I get that x can be +/- sqrt(2) therefore y for x=sqrt(2) can be either -2sqrt(2) or sqrt(2) and y for x=-sqrt(2) can be either 2sqrt(2) or -sqrt(2).

Upon substitution of all possible pairs into the derivative, I've concluded that the only two points at which dy/dx=0 are: (sqrt(2), -2sqrt(2)) and (-sqrt(2), 2sqrt(2)).

Does this match what you have?

Thanks again.
 
  • #4
Yes, that is correct.
 

Related to Horizontal tangents via implicit differentiation

1. What is implicit differentiation?

Implicit differentiation is a method of finding the derivative of a function that is written in an implicit form, where the independent variable is not explicitly stated. This is commonly used to find the slope of a curve at a specific point.

2. How do you find horizontal tangents using implicit differentiation?

To find horizontal tangents using implicit differentiation, you first need to differentiate the function with respect to the independent variable. Then, you set the derivative equal to zero and solve for the independent variable. The resulting value(s) of the independent variable will correspond to the x-coordinate(s) of the point(s) where the slope of the curve is equal to zero, indicating a horizontal tangent.

3. Why is implicit differentiation used for finding horizontal tangents?

Implicit differentiation is used for finding horizontal tangents because it is a more general method that can be applied to functions that are not explicitly written in terms of the independent variable. This allows us to find horizontal tangents for a wider range of functions.

4. Can horizontal tangents be found using explicit differentiation?

No, horizontal tangents cannot be found using explicit differentiation. Explicit differentiation only works for functions that are explicitly written in terms of the independent variable, while implicit differentiation can be used for both explicit and implicit functions.

5. What is the significance of horizontal tangents in mathematics and science?

Horizontal tangents are important in mathematics and science because they represent points where the slope of a curve is equal to zero, indicating a change in direction. This can provide valuable information about the behavior of a function, such as the maximum or minimum points, or points of inflection.

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