Probably missing something simple, but how do I find the gradient of: $3y^{2 }=2x^{3\ }+x^{2}$ at (0,0)?
I get derivative:
$6y\frac{dy}{dy} =6x^2 +2x$, and when I stick in (0,0) into this, it's undefined, but from the graph below, it looks like it should be a defined gradient. I'm confused! Where am I going wrong?
On desmos it plots as:
On
Note that $y=\pm\frac1{\sqrt3}\sqrt{2x^2+x^2}$. Then,
$$y’(x=0) =\lim_{x\to 0}\frac { 6x^2+2x}{6y} =\pm \lim_{x\to 0}\frac {\sqrt3( 3x^2+x)}{3\sqrt{2x^2+x^2}} =\pm \frac{\sqrt3}3$$
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Alternatively, near the origin $x=0$, the curve $3y^{2 }=2x^{3\ }+x^{2}$ simplifies to $3y^2=x^2$, or $y= \pm\frac1{\sqrt3}x$. Thus, the gradients are $\pm \frac1{\sqrt3}$.
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If we let $x = r \cos \theta$ and $y = r \sin \theta$ then substituting into $$ 3y^{2} = 2x^{3} + x^{2} $$ we get \begin{align*} 3r^{2} \sin^{2}\theta & = 2r^{3}\cos^{3}\theta + r^{2} \cos^{2}\theta \\ 3 \sin^{2}\theta & = \cos^{2}\theta (2r\cos\theta + 1) \\ r & = \frac{3 \tan^{2} - 1 }{2\cos\theta} \end{align*} so that \begin{align*} x & = \frac{3\tan^{2}\theta - 1}{2} \\ y & = \frac{3\tan^{2}\theta - 1}{2\cos\theta}\sin\theta \\ & = \frac{3\tan^{2}\theta - 1}{2}\tan{\theta} \\ & = x\tan\theta \end{align*} Now \begin{align*} \frac{dy}{dx} & = \frac{dy/d\theta}{dx/d\theta} \\ & = \frac{[x \tan\theta]^{\prime}}{x^{\prime}} \\ & = \frac{x^{\prime}\tan\theta + x\sec^{2}\theta}{x^{\prime}} \end{align*} So, at the point $(x,y) = (0,0)$ we can work out, for example, \begin{align*} x & = 0 \\ 3 \tan^{2}\theta - 1 & = 0 \\ \tan\theta & = \pm \frac{\sqrt{3}}{3} \end{align*} Finally, substituting $x = y = 0$ and $\tan\theta = \pm \frac{\sqrt{3}}{3}$ into $\frac{dy}{dx}$ we get $$ \frac{dy}{dx} = \frac{\pm\frac{\sqrt{3}}{3}x^{\prime} + 0}{x^{\prime}} = \pm \frac{\sqrt{3}}{3} $$ I'm pretty sure parameterisation and parametric differentiation were on the old spec so this should be within reach for an A Level student back then.
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This sort of self-intersecting "loop" curve is popular for implicit differentiation exercises, since the point of self-intersection will show an indeterminate slope for the tangent line to the curve there. In those exercises, a parameterization (different from the polar transformation presented by Shai) is generally supplied, so we might look at how to construct one.
Polynomial functions of a parameter $ \ t \ $ would be a convenient choice, since this would easily permit us to have values of the coordinate variable "run to infinity" as the parameter "goes to positive or negative infinity". We note that the equation of the curve, $ \ 3y^2 \ = \ 2x^3 + x^2 \ , \ $ indicates that the curve has symmetry about the $ \ x-$axis, with three intercepts given by $ \ x^2·(2x + 1) \ = \ 0 \ \Rightarrow \ x \ = \ -\frac12 \ , \ 0 \ $ (multiplicity 2) . Since $ \ 2x^3 + x^2 \ $ is constrained to non-negative values (the $ \ x-$axis acts like a "mirror" for the curve), this suggests that $ \ y \ $ may go to positive or negative infinity, while $ \ x \ $ only goes to positive infinity, making $ \ x \ = \ -\frac12 \ $ the minimum for that coordinate. (We can, naturally, also see this from a graph.)
So we could make the choice of an even function of $ \ t \ $ for the coordinate function $ \ x \ $ with a minimum value of $ \ -\frac12 \ \ , $ thus $ \ x(t) \ = \ t^2 - \frac12 \ \ . $ The three $ \ x-$intercepts suggest that we could choose a function with three symmetrically arranged zeroes, so that $ \ t \neq \ 0 \ $ for two of them; we look to make this choice so that we won't have $ \ t \ = \ 0 \ $ at the self-crossing point. A simple choice for the $ \ y-$coordinate then would be a cubic function with odd symmetry, $ \ y(t) \ = \ t^3 - t \ \ . $ This produces something like what we want:
We then need to make some adjustments to our coordinate functions. The $ \ x-$intercepts occur at $ \ t^2 - \frac12 \ = \ 0 \ \ , $ so $ \ t \ = \ 0 \ $ represents $ \ \left( -\frac12 \ , \ 0 \right)\ \ ; \ $ placing the self-intersection at the origin thus requires $ \ t \ = \ \pm \frac{1}{\sqrt2} \ \ . $ We can bring the $ \ y-$coordinate to zero by the choice
$$ \ t·( \ \alpha·t^2 \ - \ 1 \ ) \ \ = \ \ 0 \ \ \rightarrow \ \ \left(\pm \frac{1}{\sqrt2} \right) · \left( \ \alpha·\left[\pm \frac{1}{\sqrt2} \right]^2 \ - \ 1 \ \right) \ \ = \ \ 0 \ \ \Rightarrow \ \ \alpha \ \ = \ \ 2 \ \ . $$
Using the coordinate function $ \ y(t) \ = \ 2t^3 - t \ $ now gives us the curve as
It remains to "correct" the "vertical scale" of the loop. If we adopt $ \ y(t) \ = \ \beta·(2t^3 - t) \ \ , $ inserting our parametric functions into the original curve equation produces $$ 3y^2 \ \ = \ 2x^3 + x^2 \ \ \rightarrow \ \ 3·[ \ \beta·(2t^3 - t) \ ]^2 \ \ = \ \ 2·\left[ \ t^2 - \frac12 \ \right]^3 \ + \ \left[ \ t^2 - \frac12 \ \right]^2 $$ $$ \Rightarrow \ \ 12 ·\beta^2 ·t^6 \ - \ 12·\beta^2·t^4 \ + \ 3·\beta^2·t^2 \ \ = \ \ 2·t^6 \ - \ 2·t^4 \ + \ \frac12·t^2 \ \ \Rightarrow \ \ \beta^2 \ = \ \frac16 \ \ . $$
We at last have suitable parametric coordinate functions $ \ x(t) \ = \ t^2 - \frac12 \ \ , \ \ y(t) \ = \ \frac{1}{\sqrt6}·(2t^3 - t) \ \ , \ $ for which the graph appears below.
The first derivative at the origin takes on two values from the values of the parameter $ \ t \ $ there:
$$ \ \large{\frac{dy}{dx}|_{t \ = \ +\frac{1}{\sqrt2}} \ \ = \ \ \frac{dy/dt}{dx/dt}|_{t \ = \ +\frac{1}{\sqrt2}} \ \ = \ \ \frac{\frac{1}{\sqrt6}·(6t^2 - 1)}{2t}|_{t \ = \ +\frac{1}{\sqrt2}}} $$ $$ = \ \ \frac{\frac{1}{\sqrt6}·(6·\frac12 - 1)}{2·\frac{1}{\sqrt2}} \ \ = \ \ \frac{\frac{1}{\sqrt6}·2}{2·\frac{1}{\sqrt2}} \ \ = \ \ \sqrt{\frac26} \ \ = \ \ \frac{1}{\sqrt3} \ \ , $$ $$ \ \large{\frac{dy}{dx}|_{t \ = \ -\frac{1}{\sqrt2}} \ \ =} \ \ \normalsize{\frac{\frac{1}{\sqrt6}·2}{2·\left(-\frac{1}{\sqrt2}\right)} \ \ = \ \ -\frac{1}{\sqrt3} \ \ .} $$
The two different "slopes" at the origin can be thought of as representing the directions in which that point is traversed by the parameterized path at the differing values of $ \ t \ \ . $
Our expression for the first derivative also correctly describes the "end" of the loop at $ \ \left(-\frac12 \ , \ 0 \right) \ \ $ with
$$ \ \large{\frac{dy}{dx}|_{t \ = \ 0} \ \ =} \ \ \normalsize{\frac{\frac{1}{\sqrt6}·(6·0 - 1)}{2·0}} \ \ , $$
which, being undefined, indicates a "vertical tangent" there.
From $6yy'=6x^2+2x$ we get $$y'=\frac{x^2+\frac13x}{y}\Rightarrow y'^2=\frac{x^2(x+\frac13)^2}{\frac23x^3+\frac13x^2}= \frac{(x+\frac13)^2}{\frac23x+\frac13}.$$ Now plug in $x=0$ and solve for $y'(0)$.