calculusderivativesimplicit-differentiation
With the equation $x^2+2xy-3y^2+16=0$, I need to find the coordinates of the points on the curve where $\frac{dy}{dx} = 0$. I think I have correctly used implicit differentiation to get $\frac{dy}{dx} = \frac{-x-y}{x-3y}$. And the multiplied by the denominator and subsituted zero to get, $0 = -x-y$, however, I'm not sure where to go from here.
If anyone could help that would be great. Thanks.
As OP already found, the expression for the slope of a tangent line to a point on the curve in question is
$$ y \ ' \ = \ \frac{y^2 \ - \ 3x^2}{3y^2 \ - \ 2xy} \ = \ \frac{y^2 \ - \ 3x^2}{y \ (3y \ - \ 2x)} \ \ . $$
Something useful to check for immediately is whether this rational function can be "indeterminate", that is, whether both numerator and denominator can be zero at the same time for a point on the curve. There are two cases to consider, due to the two factors in the denominator:
I -- $ \ y^2 \ = \ 3x^2 \ \ , \ \ y \ = \ 0 \ \ \Rightarrow \ \ x \ = \ 0 \ $ ; however, the point $ \ ( 0 , 0 ) \ $ does not lie on this curve
II -- $ \ y^2 \ = \ 3x^2 \ \ , \ \ 3y \ = \ 2x \ \ \Rightarrow \ \ \left( \frac{2}{3}x \right)^2 \ = \ 3 x^2 \ \ \Rightarrow \ \ x \ = \ 0 \ \ , \ \ y \ = \ 0 \ $ ; again, this is not a solution to the curve equation.
So the issue of "slope indeterminancy" does not occur for this curve, as it does for one of its "cousins", the folium of Descartes (discussed, for instance, here).
We can then locate the points at which horizontal tangents occur by inserting the relation $ \ y^2 \ = \ 3x^2 \ \Rightarrow \ y \ = \ \pm \ 3^{1/2}x \ $ into the equation for the curve (as mentioned by Graham Kemp), thus:
$$ x^3 \ + \ y^3 \ - \ xy^2 \ = \ 8 \ \ \Rightarrow \ \ x^3 \ + \ \left( \pm \ 3^{1/2}x \right)^3 \ - \ x \ \left( \pm \ 3^{1/2}x \right)^2 \ = \ 8 $$
$$ \Rightarrow \ \ \left( 1 \ \pm \ 3^{3/2} \ - \ 3 \right) \ x^3 \ = \ 8 $$
$$ \Rightarrow \ \ x \ = \ \left[ \frac{8}{-2 \ \pm \ 3 \sqrt{3}} \right]^{1/3} = \ \frac{2}{( -2 \ \pm \ 3 \sqrt{3} \ )^{1/3}} \ \approx \ -1.036 \ , \ 1.358 $$
$$ y \ = \ \pm \ \sqrt{3} \ \cdot \ x \ \rightarrow \ 1.794 \ , \ 2.352 \ \ . $$
[Upon insertion into the original curve equation, we see that the negative values for $ \ y \ $ are not solutions.] Thus, the horizontal tangent lines are found at
$$ \left( \frac{2}{( -2 \ - \ 3 \sqrt{3} \ )^{1/3}} \ , \ \frac{2 \ \sqrt{3}}{( -2 \ - \ 3 \sqrt{3} \ )^{1/3}} \right) \ \ , \ \ \left( \frac{2}{( -2 \ + \ 3 \sqrt{3} \ )^{1/3}} \ , \ \frac{2 \ \sqrt{3}}{( -2 \ + \ 3 \sqrt{3} \ )^{1/3}} \right) \ \ . $$
Vertical tangents occur at the points for which one of the factors in the denominator of the expression for $ \ y \ ' \ $ equals zero:
$$ y \ = \ 0 \ \ \Rightarrow \ \ x^3 \ + \ 0^3 \ - \ x \cdot 0^2 \ = \ 8 \ \ \Rightarrow \ \ x \ = \ 2 \ \ ; $$
$$ y \ = \ \frac{2}{3}x \ \ \Rightarrow \ \ x^3 \ + \ \left( \frac{2}{3}x \right)^3 \ - \ x \cdot \left( \frac{2}{3}x \right)^2 \ = \ 8 $$
$$ \Rightarrow \ \ \left( 1 \ + \ \frac{8}{27} \ - \ \frac{4}{9} \right) \ x^3 \ = \ \left( \frac{27 \ + \ 8 \ - 12}{27} \right) x^3 \ = \ 8 $$
$$ \Rightarrow \ \ x \ = \ \left( \frac{ 8 \ \cdot \ 27}{23} \right)^{1/3} \ = \ \frac{6}{23^{1/3}} \ \approx \ 2.110 \ \ \Rightarrow \ \ y \ = \ \frac{2}{3} \cdot \ \frac{6}{23^{1/3}} \ = \ \frac{4}{23^{1/3}} \ \approx \ 1.407 \ \ . $$
So the points where vertical tangents are found are
$$ ( \ 2, 0 \ ) \ \ \text{and} \ \ \left( \frac{6}{23^{1/3}} , \frac{4}{23^{1/3}} \right) \ \ . $$
A graph of the curve is presented below.
horizontal tangents are marked in green, vertical tangents, in red
An idea to avoid the cumbersome and annoying quotient rule: multiply by the common denominator
$$\frac{x^2}{x+y}=y^2+6\implies xy^2+6x+y^3+6y-x^2=0\implies$$
$$y^2+2xyy'+6+3y^2y'+6y'-2x=0\implies(2xy+3y^2+6)y'=2x-y^2-6\implies$$
$$y'=\frac{2x-y^2-6}{2xy+3y^2+6}$$
If nevertheless you want to use the quotient rule:
$$\frac{2x(x+y)-x^2}{(x+y)^2}-\frac{x^2}{(x+y)^2}y'=2yy'\implies$$
$$(2y(x+y)^2+x^2)y'=x^2+2xy\implies y'=\frac{x^2+2xy}{2y(x+y)^2+x^2}$$
Now, how come both expressions we got are equal?. Well, for one we can use, for example, that $\;\cfrac{x^2}{x+y} = y^2+6\;$ , so
$$\frac{x^2+2xy}{2y(x+y)^2+x^2}=\frac{2x-y^2-6}{2xy+3y^2+6}=\frac{2x-\frac{x^2}{x+y}}{2xy+2y^2+\frac{x^2}{x+y}}\iff$$
$$\frac{x^2+2xy}{2y(x+y)^2+x^2}=\frac{x^2+2xy}{2x^2y+4xy^2+2y^3+x^2}\iff$$
$$\frac{x^2+2xy}{2x^2y+4xy^2+2y^3+x^2}=\frac{x^2+2xy}{2x^2y+4xy^2+2y^3+x^2}\;\;\color{green}\checkmark$$
Best Answer
Thanks for all the help. Doing what you've said I have the coordinates as (2,-2) and (-2,2).