Horizontal and Vertical Tangents!

**hossein** · July 10th 2007, 04:08 PM

http://img401.imageshack.us/img401/2243/havahx0.png

with the given equations, try to find out the coordinates of the horizontal and vertical tangents.
(I think have to get the derivative, and let the numerator equal to zero,, sub for y and then plog it back in the equation and solve for x.) im not sure though....

**topsquark** · July 10th 2007, 05:43 PM

Originally Posted by hossein

http://img401.imageshack.us/img401/2243/havahx0.png

with the given equations, try to find out the coordinates of the horizontal and vertical tangents.
(I think have to get the derivative, and let the numerator equal to zero,, sub for y and then plog it back in the equation and solve for x.) im not sure though....

First find the derivative:
$x^3 + y^3 - 3xy = 3$

By implicit differentiation:
$3x^2 + 3y^2 y^{\prime} - 3y - 3x y^{\prime} = 0$

$3y^2 y^{\prime} - 3x y^{\prime} = 3y - 3x^2$

$(3y^2 - 3x)y^{\prime} = 3y - 3x^2$

$y^{\prime} = \frac{3y - 3x^2}{3y^2 - 3x} = \frac{y - x^2}{y^2 - x}$

Now, where is $y^{\prime} = 0$ ? When
$y - x^2 = 0$

$y = x^2$

So put this into your original equation:
$x^3 + y^3 - 3xy = 3$

$x^3 + (x^2)^3 - 3x(x^2) = 3$

$x^6 - 2x^3 - 3 = 0$

$(x^3 - 3)(x^3 + 1) = 0$

So $x^3 = 3$ ==> $x = \sqrt[3]{3}$ ==> $y = (\sqrt[3]{3})^2 = \sqrt[3]{9}$
or $x^3 = -1$ ==> $x = -1$ ==> $y = (-1)^2 = 1$

So the points where $y^{\prime} = 0$ are $(-1, 1)$ and $(\sqrt[3]{3}, <br /> \sqrt[3]{9})$ . (Both of these are points in the original equation.)

We do the same process to find where the vertical tangents are. Where is $y^{\prime}$ undefined? Where
$y^2 - x = 0$

$y = \pm \sqrt{x}$

So put this into your original equation:
$x^3 + y^3 - 3xy = 3$

$x^3 + (\pm \sqrt{x})^3 - 3x(\pm \sqrt{x}) = 3$

$x^3 \pm \sqrt{x^3} \mp 3 \sqrt{x^3} - 3 = 0$

$x^3 \mp 2 \sqrt{x^3} - 3 = 0$

Let $z = \sqrt{x^3}$

Then
$z^2 \mp 2 z - 3 = 0$

For the "-" sign:
$z^2 - 2z - 3 = 0$

$(z - 3)(z + 1) = 0$

$z = 3$ ==> $x = \sqrt[3]{3^2} = \sqrt[3]{9}$ ==> $y = \pm \sqrt{\sqrt[3]{9}} = \pm \sqrt[6]{9}$
(I note that the "-" solution is not a point on the original equation.)
and
$z = -1$ ==> $x = \sqrt[3]{(-1)^2} = 1$ ==> $y = \pm \sqrt{1} = \pm 1$
(I note that the "+" solution is not a point on the original equation.)

So the "-" sign in the z quadratic equation gives points for the vertical asymptote at $(\sqrt[3]{9}, \sqrt[6]{9})$ and $(1, -1)$ .

For the "+" sign:
$z^2 + 2 z - 3 = 0$

$(z + 3)(z - 1) = 0$

$z = -3$ ==> $x = \sqrt[3]{-3}$ ==> $y = \pm \sqrt{\sqrt[3]{-3}}$
which isn't real.
and
$z = 1$ ==> $x = \sqrt[3]{1} = 1$ ==> $y = \pm \sqrt{1} = \pm 1$ .

So the "+" sign in the z quadratic equation gives points for the vertical asymptote at $(1, \pm 1)$ . I note that the "+" sign is not a solution of the original equation, so the vertical asymptote is at $(1, -1)$ .

Overall this gives us vertical asymptotes at $(\sqrt[3]{9}, \sqrt[6]{9})$ and $(1, -1)$ .

-Dan

**Soroban** · July 10th 2007, 05:45 PM

Hello, Hossein!

Your game plan is a good one . . . but the algebra is scary, isn't it?

Given: . $x^3 + y^3 - 3xy \:=\:3$ . [1]

Find the coordinates of the horizontal and vertical tangents.

Differentiate implicitly: . $3x^2 + 3y^2\frac{dy}{dx} - 3x\frac{dy}{dx} - 3y \;=\;0$
. . and we have: . $\frac{dy}{dx}\;=\;\frac{y-x^2}{y^2-x}$

We have a horizontal tangent when the numerator is 0.
. . That is: . $y-x^2\:=\:0\quad\Rightarrow\quad y \:=\:x^2$ . [2]

Substitute into [1]: . $x^3 + (x^2)^3 - 3x(x^2) \:=\:3\quad\Rightarrow\quad x^6 - 2x^3 - 3 \:=\:0$

. . Factor: . $(x^3 + 1)(x^3 - 3) \:=\:0$

. . and we have: . $\begin{array}{ccccc}x^3+1\:=\:0 & \Rightarrow & x^3\:=\:\text{-}1 & \Rightarrow & x \:=\:\text{-}1 \\<br /> x^3 - 3 \:=\:0 & \Rightarrow & x^3 \:=\:3 & \Rightarrow & x\:=\:\sqrt[3]{3} \end{array}$

. . Substitute into [2]: . $\begin{array}{ccc}x = -1 & \Rightarrow & y = 1 \\ x = \sqrt[3]{3} & \Rightarrow & y =\sqrt[3]{9} \end{array}$

Therefore, horizontal tangents at: . $(-1,\,1)$ and $(\sqrt[3]{3},\,\sqrt[3]{9})$

We have a vertical tangent when the denominator is zero.
. . That is: . $y^2 - x \:=\:0\quad\Rightarrow\quad x \:=\:y^2$ . [4]

Substitute into [1]: . $(y^2)^3 + y^3 - 3(y^2)y \:=\:3\quad\Rightarrow\quad y^6 - 2y^3 - 3 \:=\:0$

. . Factor: . $(y^3 +1)(y^3 - 3)\:=\:0$

. . and we have: . $\begin{array}{ccccc}y^3+1\:=\:0 & \Rightarrow & y^3 \:=\:\text{-}1 & \Rightarrow & y \:=\:\text{-}1 \\<br /> y^3-3\:=\:0 & \Rightarrow & y^3 \:=\:3 & \Rightarrow & y \:=\:\sqrt[3]{3} \end{array}$

. . Substitute into [4]: . $\begin{array}{ccc}y = -1 & \Rightarrow & x = 1 \\ y = \sqrt[3]{3} & \Rightarrow & x = \sqrt[3]{9} \end{array}$

Therefore, vertical tangents at: . $(1,\,-1)$ and $(\sqrt[3]{9},\,\sqrt[3]{3})$

Edit: Curses . . . too slow again!

**ThePerfectHacker** · July 10th 2007, 06:57 PM

We have,
$x^3+y^3 - 3xy = 3$

Take the derivative of both sides,

$3x^2 + 3y^2 \frac{dy}{dx} - 3y - 3x\frac{dy}{dx} = 0$

$\frac{dy}{dx} \left( y^2 - x \right) = y - 3x^2$

$\frac{dy}{dx} = \frac{y-x^2}{y^2-x}$

The horizontal tangents occur when $y-x^2 = 0 \implies y=x^2$ . Since the point also have to satisfy the equation of this curve (called "Folium of Descrates") it is where the parabola intersects the curve. See below.

Similarly the vertical tangents occur when $y^2 - x =0 \implies y=\pm \sqrt{x}$ which is where the tilted parabola intersects the Folium of Descrates.

**topsquark** · July 11th 2007, 05:15 AM

Originally Posted by ThePerfectHacker

We have,
$x^3+y^3 - 3xy = 3$

Take the derivative of both sides,

$3x^2 + 3y^2 \frac{dy}{dx} - 3y - 3x\frac{dy}{dx} = 0$

$\frac{dy}{dx} \left( y^2 - x \right) = y - 3x^2$

$\frac{dy}{dx} = \frac{y-x^2}{y^2-x}$

The horizontal tangents occur when $y-x^2 = 0 \implies y=x^2$ . Since the point also have to satisfy the equation of this curve (called "Folium of Descrates") it is where the parabola intersects the curve. See below.

Similarly the vertical tangents occur when $y^2 - x =0 \implies y=\pm \sqrt{x}$ which is where the tilted parabola intersects the Folium of Descrates.

How on Earth did you get the "graph" program to graph this thing?

-Dan

**ThePerfectHacker** · July 11th 2007, 06:35 AM

Originally Posted by topsquark

How on Earth did you get the "graph" program to graph this thing?

-Dan

Maybe you have an older version. There is a button called insert relation. Try that.

**topsquark** · July 11th 2007, 07:20 AM

Originally Posted by ThePerfectHacker

Maybe you have an older version. There is a button called insert relation. Try that.

Ooh! Thank you, I hadn't known about that one.

-Dan

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