Horizontal tangents via implicit differentiation


by atmega-ist
Tags: differentiation, horizontal, implicit, tangents
atmega-ist
atmega-ist is offline
#1
Oct11-10, 07:59 PM
P: 9
1. The problem statement, all variables and given/known data

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

x2 + xy + y2 = 6

2. Relevant equations

n/a

3. 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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Quinzio
Quinzio is offline
#2
Oct11-10, 08:07 PM
P: 558
If you plug y = -2x into the original equation you don't get infinite points.
atmega-ist
atmega-ist is offline
#3
Oct11-10, 08:35 PM
P: 9
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.

HallsofIvy
HallsofIvy is offline
#4
Oct12-10, 06:06 AM
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PF Gold
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Horizontal tangents via implicit differentiation


Yes, that is correct.


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