Points on $(x^2 + y^2)^2 = 2x^2 - 2y^2$ with slope of $1$
Let the curve in the plane defined by the equation: $(x^2 + y^2)^2 = 2x^2 - 2y^2$
How can i graph the curve in the plane and determine the points of the curve where $\frac{dy}{dx} = 1$.
My work:
First i found the roots of this equation with a change of variable $z = y^2$ and get:
and then i tried to graph the point $ x - y$ and $x + y $ but i stuck i can't graph this and find the point where the derivative is 1. Some help please.
5 Answers
$\begingroup$You've already got some good suggestions on the graphing part using polar coordinates (if you are familiar with that). For the other, implicitly differentiate the given relation to obtain ${dy\over dx}$:
\begin{align*} (x^2 + y^2)^2 &= 2x^2 - 2y^2\\ 2(x^2+y^2)(2x+2y{dy\over dx})&=4x-4y{dy\over dx}\\ 2x^3+2x^2y{dy\over dx}+2xy^2+2y^3{dy\over dx}&=2x-2y{dy\over dx}\\ (2x^2y+2y^3+2y){dy\over dx}&=2x-2x^3-2xy^2\\ {dy\over dx}&={2x-2x^3-2xy^2\over 2x^2y+2y^3+2y}\\ {dy\over dx}&={x(1-x^2-y^2)\over y(x^2+y^2+1)}\\ \end{align*}
Now solve ${dy\over dx}=1$ in terms of $x$ and $y$, keeping in mind you can use the original expression to simplify things along the way. Edit: This was much more involved than I anticipated; see below.
Here's the situation (note ${dy\over dx}$ is not defined at the origin):
So how to find these coordinates? As noted in another answer, the original relation can be expressed in polar coordinates as $r^2=2\cos(2t)$. Recall that the slope of the tangent line in polar coordinates is given by $$ {dy\over dx}={{dr\over dt}\sin t+r\cos t\over {dr\over dt}\cos t-r\sin t}, $$ so setting ${dy\over dx}=1$ and using $r^2=2\cos(2t)\implies {dr\over dt}={-2\sin(2t)\over r}$, we get \begin{align*} {dy\over dx}&=1\\ {dr\over dt}\sin t+r\cos t&={dr\over dt}\cos t-r\sin t\\ {-2\sin(2t)\over r}\sin t+r\cos t&={-2\sin(2t)\over r}\cos t-r\sin t\\ -2\sin(2t)\sin t+r^2\cos t&=-2\sin(2t)\cos t-r^2\sin t\\ r^2&={2\sin(2t)[\sin t-\cos t]\over \cos t+\sin t}\\ 2\cos(2t)&={2\sin(2t)[\sin t-\cos t]\over \cos t+\sin t}\\ \end{align*} Multiply on the right by ${\cos t-\sin t\over \cos t-\sin t}$, use $\cos^2t-\sin^2t=\cos(2t)$ and $\sin(2t)=2\sin t\cos t$ and rearrange to get \begin{align*} \sin(2t)&=-(\sin t+\cos t)^2\\ 2\sin t\cos t&=-1-2\sin t\cos t\\ 4\sin t\cos t&=-1\\ \sin(2t)&=-{1\over 2}\\ t&={7\pi\over 12},\ {11\pi\over 12}. \end{align*}
But $r^2=2\cos(2t)$ so, $t={11\pi/12}\implies r^2=2\cos(11\pi/6)=\sqrt{3}$. Thus, the polar coordinates of this point are $(3^{1/4},11\pi/12)$ which in rectangular coordinates is $$(3^{1/4}\cos(11\pi/12),3^{1/4}\sin(11\pi/12))\approx (-1.27,0.34)$$ as shown in the figure above.
(The fourth quadrant point of tangency can be found similarly.)
$\endgroup$ 0 $\begingroup$Maybe it would be better if you transform your equation to polar coordinates: Put $x = \rho \cos \varphi $ and $y = \rho \sin \varphi$, then
$$ (x^2+y^2)^2 = 2x^2 - 2y^2 $$
is
$$ \rho^2 = 2 \cos ( 2 \varphi) $$
$\endgroup$ 1 $\begingroup$I won't do the graph, just the algebra.
Implicit differentiation gives
$$2(x^2+y^2)(2x+2yy')=4x-4yy'$$
so setting $y'=1$ and simplifying leads to
$$(x^2+y^2)(x+y)=x-y\qquad(*)$$
If we now rewrite the equation for the curve as
$$(x^2+y^2)^2=2(x-y)(x+y)$$
we see that multiplying both sides of equation $(*)$ by $2(x+y)$ gives
$$2(x^2+y^2)(x+y)^2=2(x-y)(x+y)=(x^2+y^2)^2$$
Ignoring the point on the curve at $(0,0)$ (where the derivative is not defined), we can cancel a $(x^2+y^2)$ from the above, leaving $2(x+y)^2=x^2+y^2$, which simplifies to $x^2+y^2=-4xy$ (which will mean that one of $x$ and $y$ will have to be positive and the other one negative). Plugging this into the original equation for the curve gives $(-4xy)^2=2x^2-2y^2$, or
$$y^2={x^2\over1+8x^2}$$
Plugging this into the original equation for the curve gives
$$\left(x^2+{x^2\over1+8x^2}\right)^2=2x^2-2{x^2\over1+8x^2}$$
Simplifying this (and keeping in mind we're already ignoring $x=0$) produces
$$16x^4-24x^2-3=0$$
The real roots for this are
$$x=\pm\sqrt{3+2\sqrt3\over4}\approx\pm1.27$$
The corresponding values for $y$ (recalling that $-4xy$ must be positive) are
$$y=\mp\sqrt{-3+2\sqrt3\over4}\approx\mp.34$$
$\endgroup$ $\begingroup$I would try to solve this as an implicit derivative: $$\frac{d}{dx}(x^2 + y^2)^2 = \frac{d}{dx}(2x^2 - 2y^2)$$ $$2(x^2+y^2)(2x+2y*y') =4x-4y*y'$$ $$2 x^3+2x^2 y* y'+2 y^3 y'+2 x y^2 = 4x-4y*y'$$ $$2x^2y*y' + 2y^3y' + 4y*y' = 4x-2xy^2-2x^2$$ $$y'(2x^2y + 2y^3+4y) =4x-2xy^2-2x^2$$ $$y' = \frac{4x-2xy^2-2x^2}{2x^2y + 2y^3+4y} = 1$$ $$\implies 4x-2xy^2-2x^2 = 2x^2y + 2y^3+4y$$
$\endgroup$ 2 $\begingroup$For determining the graph is better plot it as Lemur says in polar coordinates and see what happen at different angles with $r$.
$\hskip2in$
For the points with derivative equal to one:
$$(x^2 + y^2)^2 = 2x^2 - 2y^2$$ $$x^4+y^4+2x^2y^2=2x^2-2y^2$$ $$4x^3+4y^3\frac{dy}{dx}+4xy^2+4x^2y\frac{dy}{dx}=4x-4y\frac{dy}{dx}$$ Setting $\frac{dy}{dx}=1$ $$4x^3+4y^3+4xy^2+4x^2y=4x-4y$$
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