Horizontal tangents on the "Devil's Curve"

[Idealistic]

Find all horizontal tangents for:

(x[sup:377945o3]2[/sup:377945o3] + y[sup:377945o3]2[/sup:377945o3])[sup:377945o3]2[/sup:377945o3] = 2(x[sup:377945o3]2[/sup:377945o3] + y^2)

First I implicitly derived.

2(x[sup:377945o3]2[/sup:377945o3] + y^2)(2x + 2y) = 4x + 4y

First question. This curve is not a function so do I still have to put in my y primes? If I do, than is y a function of x? i.e, does y^2 become 2y, instead of 2yy'?

Next I set the equation equal to zero.

2(x[sup:377945o3]2[/sup:377945o3] + y[sup:377945o3]2[/sup:377945o3])(2x + 2y) - 4x - 4y = 0

4(x[sup:377945o3]2[/sup:377945o3] + y[sup:377945o3]2[/sup:377945o3])(x + y) - 4(x + y) = 0

4(x + y)(x[sup:377945o3]2[/sup:377945o3] + y[sup:377945o3]2[/sup:377945o3] - 1) = 0

So this equations equals zero when y = -x, or x^2 + y^2 = 1.

Second question. Do I just look at the graphs y = -x and x^2 + y^2 = 1 and see where they equal zero and those x, y coordinates are put into my original devil's equation? If so my answer are:

y = -x is zero at (0, 0) ** But there are infinite combinations of numbers on y = -x that would cause the derivative of my Devils Curve equations to equal zero. Essentially anywhere on the curve of y = -x.

x[sup:377945o3]2[/sup:377945o3] + y[sup:377945o3]2[/sup:377945o3] = 1 is zero at (1, 0) and (-1, 0) **Same application here as above, anywhere on the entrie circle would cause the derivative of the Devil's curve equation to equal zero.

Just a bit confused.

 

[Deleted member 4993]

Find all horizontal tangents for:

(x[sup:1rfp27fs]2[/sup:1rfp27fs] + y[sup:1rfp27fs]2[/sup:1rfp27fs])[sup:1rfp27fs]2[/sup:1rfp27fs] = 2(x[sup:1rfp27fs]2[/sup:1rfp27fs] + y^2) <<< Is this equation correct?

If it is then

x^2 + y^2 = 2

This is a circle [ with four removable discontinuities at (±sqrt(2) ,±sqrt(2)]- whose horizontal tangents are at (0,±sqrt(2))

First I implicitly derived.

2(x[sup:1rfp27fs]2[/sup:1rfp27fs] + y^2)(2x + 2yy') = 4x + 4yy'

4yy'(x^2+y^2) - 4yy' =4x - 4x(x^2+y^2)

y' = -x/y

You need to set y' = 0 --> giving you x = 0 and y = ±sqrt(2)

First question. This curve is not a function so do I still have to put in my y primes? If I do, than is y a function of x? i.e, does y^2 become 2y, instead of 2yy'?

Next I set the equation equal to zero.

Generally - Devil's curve is written as:


\(\displaystyle y^2(y^2-a^2) \, = \, x^2(x^2-b^2)\)

Which is very different from what you posted.