Consider the curve C given by $(x^2+y^2)^2=(x^2−y^2)$

Question

Consider the curve C given by $(x^2+y^2)^2=(x^2−y^2)$

1. Using implicit differentiation, find $dy/dx$ in terms of $x$ and $y$.
2. Find all points $(x,y)$ on $C$ such that the tangent line is horizontal. You may assume $(x,y)≠(0,0)$
3. What is the smallest value y can take on $C$? What is the largest value?

Hint: what should the slope of the tangent line be in each of those cases?

Answer 1

HINT

By implicit differentiation we have

$$(x^2+y^2)^2=(x^2−y^2)\implies 4x(x^2+y^2)dx+4y(x^2+y^2)dy=2xdx-2ydy$$

$$\implies \frac{dy}{dx}=\frac{x-2x(x^2+y^2)}{y+2y(x^2+y^2)}$$

Answer 2

Note: I write $x'$ and $y'$ instead of $dx$ and $dy$ to sort of imply that the derivative is taken with respect to some un-named variable.

From $(x^2+y^2)^2 =(x^2−y^2) $, since $(f^2)' = 2ff'$, the ID (implicit derivative) of the left side is

$\begin{array}\\ 2(x^2+y^2)(x^2+y^2)' &=2(x^2+y^2)(2xx'+2yy')\\ &=4(x^2+y^2)(xx'+yy')\\ &=4xx'(x^2+y^2)+4yy'(x^2+y^2)\\ \end{array} $

and the ID of the right side is $2xx'-2yy'$.

Equating these, dividing by 2, and then grouping the terms with $x'$ and $y'$,

$4xx'(x^2+y^2)+4yy'(x^2+y^2) =2xx'-2yy' $ or $2xx'(x^2+y^2)+2yy'(x^2+y^2) =xx'-yy' $ so that $xx'(2x^2+2y^2-1)+yy'(2x^2+2y^2+1) =0\\ $ or $yy'(2x^2+2y^2+1) =-xx'(2x^2+2y^2-1) $.

Therefore $\dfrac{dy}{dx} =\dfrac{y'}{x'} =\dfrac{-x(2x^2+2y^2-1)}{y(2x^2+2y^2+1)} $.

I'll let you handle the rest.