Easiest way to show $(x^2+y^2-1)^3 = x^2y^3$ has a slope of 2 at (1,0)?

Question

What is the easiest way to show the function implicitly defined by the equation $(x^2+y^2-1)^3 = x^2y^3$ near $(1,0)$ has a slope of $2$ at this point?

Answer 1

First of all, let us take the cube root from both sides: \begin{align*} (x^{2} + y^{2} - 1)^{3} = x^{2}y^{3} & \Rightarrow x^{2} + y^{2} - 1 = x^{2/3}y\\\\ & \Rightarrow 2x + 2yy' = \frac{2y}{3x^{1/3}} + x^{2/3}y' \end{align*}

Since we are interested in the slope of the tangent line at the point $(1,0)$, one concludes that: \begin{align*} 2 + 0 = 0 + y' \Rightarrow y'(1) = 2 \end{align*}

just as desired.

Hopefully this helps!

Answer 2

If you have proven that the implicit function $y=f(x)$ exists and is differentiable at $(1,0)$, then the following solution has the benefit of (almost) always working, which is good for people (like me) who are not sufficiently imaginative to find the right trick for each situation. ;-)

$x=1+h$, we'll study the equation nearby $(h,y)=(0,0)$. We know that $y$ is equivalent to $y'h$ when $h \to 0$ if $y' \ne 0$, or $y=o(h)$ if $y'=0$. So $y=\mathcal{O}(h)$ when $h \to 0$.

$((1+h)^2+y^2-1)^3=(1+h)^2y^3$
$(1+2h+\mathcal{O}(h^2)+y^2-1)^3=(1+2h+\mathcal{O}(h^2))y^3$
$8h^3+\mathcal{O}(h^4)=y^3+\mathcal{O}(h^4)$
$y \sim 2h$ so $y'(1)=2$

Answer 3

In $\,\mathbb R^2$ your equation is equivalent to :

$x^2+y^2-1-\sqrt[3]{x^2}y=0\,.$

Hence , by using implicit differentiation , we get that :

$y’(x,y)=\dfrac{\frac2{3\sqrt[3]x}y-2x}{2y-\sqrt[3]{x^2}}\;.$

Consequently , $\;y’(1,0)=2\,.$