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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?

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    $\begingroup$ Try to apply implicit differentiation. $\endgroup$ Oct 25, 2022 at 16:20
  • $\begingroup$ Implicit differentiation yields $$y' =\frac{2x(x^2+y^2-1)^2-(2/3)xy^3}{x^2y^2-2y(x^2+y^2-1)^2}$$ which is undefined at $(1,0)$. $\endgroup$ Oct 25, 2022 at 16:26
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    $\begingroup$ Funny: the curve intersects the unit circle at four points; two of them ($(0, \pm 1)$) are honest cusps, but at $(\pm 1, 0)$ both sides go to zero at the same order and so miraculously it stays smooth. $\endgroup$
    – JBL
    Oct 25, 2022 at 17:07
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    $\begingroup$ @SelrachDunbar You should edit your question to include your efforts instead of posting a comment. $\endgroup$
    – jjagmath
    Oct 25, 2022 at 23:30
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    $\begingroup$ @ÁtilaCorreia my students will really appreciate seeing your solution to this made-up problem! I think it is neat that the graph of this equation is heart-shaped. Clearly, when you graph the curve (e.g. in Desmos) the tangent line has slope 2. Naturally I thought a direct attack with implicit differentiation would confirm this and so I teed this problem up as a fun in-class exercise. But alas, without your insight to take the cube root first, implicit differentiation fails to tell us what the slope at (1,0) is (as mentioned in my comment above). Thanks again! $\endgroup$ Oct 27, 2022 at 0:15

3 Answers 3

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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!

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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 $y'(1)=2$

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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 , $\;y’(1,0)=2\,.$

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