1. ## implicit curve help

Hi,

This is driving me nuts... I've put in a good few hours and still stuck.. can't seem to find much on this online either.

OK, I'm trying to draw the curve of $\displaystyle x^3 + 2y^3 - 5xy = 0$

I used implicit differentiation to find

$\displaystyle \frac{dy}{dx} = \frac{5y - 3x^2}{6y^2-5x}$

I put this equal to zero to find critical points.. gives me
$\displaystyle y = \frac{3x^2}{5}$
I plug this back into the original equation to simplify to
$\displaystyle 27x^6 - 125x^3 = 0$ so I have $\displaystyle x =0, \frac {5}{3}$.

I need to find the turning points and to draw the curve.. Any help?

Thank you.

2. To find critical points, you need to use this formula

$\displaystyle d=f_{xx}*f{yy}-[f_{xy}]^2$

when

$\displaystyle d=0 \ \mbox{inconclusive}, \ d>0 \ \mbox{and} \ f_{xx}(x_0,y_0)>0 \ \mbox{rel. min}, \ \ d>0 \ \mbox{and} \ f_{xx}(x_0,y_0)< 0 \ \mbox{rel max}, \mbox{and} \ d<0 \ \mbox{saddle}$

3. Thanks for the reply (I've never seen that formula before...).. Looks like this is a tough one alright..

4. Originally Posted by minusb
Hi,

This is driving me nuts... I've put in a good few hours and still stuck.. can't seem to find much on this online either.

OK, I'm trying to draw the curve of $\displaystyle x^3 + 2y^3 - 5xy = 0$

I used implicit differentiation to find

$\displaystyle \frac{dy}{dx} = \frac{5y - 3x^2}{6y^2-5x}$

I put this equal to zero to find critical points.. gives me
$\displaystyle y = \frac{3x^2}{5}$
I plug this back into the original equation to simplify to
$\displaystyle 27x^6 - 125x^3 = 0$ so I have $\displaystyle x =0, \frac {5}{3}$.

I need to find the turning points and to draw the curve.. Any help?

Thank you.
So far so good you have found the x coordinates of the horizontal tangents.

Plugging them into $\displaystyle y = \frac{3x^2}{5}$ will give you the $(x,y)$ coordinates zero gives

$(0,0)$ and $\displaystyle \frac{5}{3} \implies \left(\frac{5}{3},\frac{5}{3} \right)$

Now if you set the denominator equal to zero you will get the coordinates of the vertical tangents.

$\displaystyle 6y^2-5x=0 \iff x=\frac{6y^2}{5}$

putting these back into the original gives

$\displaystyle y=0 \text{ or } y= \frac{5}{6}2^{\frac{2}{3}}$

Now putting these into $\displaystyle x=\frac{6y^2}{5}$

$\displaystyle (0,0) \text{ or } \left(\frac{5}{6}2^{\frac{2}{3}},\frac{5}{3}2^{\fr ac{1}{3}} \right)$

From here pick a few x values to so how the graph is behaving away from the loop.

for example $x=-2,-4$ and $x=3,5$ and solve for y to get some ordered pairs. Also once you have a few ordered pairs to can evaluate the derivative to get an idea of how fast the graph is changing at that point.

P.S the graph should look like this.

5. Thank you!