# Find slopes of tangent lines where $\frac{dy}{dx}$ has removable discontinuity.

A colleague of mine wrote a worksheet where she asked students to find the derivative $\frac{dy}{dx}$, where the points $(x,y)$ are defined by the "Heart Curve," $$(x^2+y^2-1)^3-x^2y^3=0.$$ She asks students about points where the derivative is well-defined, but I wondered about the points where the derivative isn't well-defined.

A plot of the graph shows sharp points at $(0,\pm1)$, and seems to be a smooth curve at the points $(\pm1,0)$, yet the derivative is undefined at points where $x=0$ or $y=0$. At points where $x=0$, there should be VERTICAL tangent lines (I have since ran the zoom in on a point test to find this), and at points where $y=0$ there should be one distinct tangent line. It seems slopes here should be $\pm2$. I have calculated $\frac{dy}{dx}$ as $$\frac{dy}{dx}=\frac{2xy^3-6x(x^2+y^2-1)^2}{6y(x^2+y^2-1)^2-3x^2y^2},$$ and I feel like there should be a way to remove singularities for $x=0$ or $y=0$ in the expression but I do not know exactly how.

I thought about maybe using partial derivatives and L'Hopital's Rule, but not sure. My goal is to discuss this with students in a single-variable calculus course, so I'd like a solution that doesn't use partial differentiation.

• Why do you think that the derivative is not define at $\left( { \pm 1,0} \right)$? :) I think it has no problems there! – Hosein Rahnama Oct 22 '15 at 14:07
• @H.R. you get $0/0$. – John Molokach Oct 22 '15 at 14:07
• Why it is not evident from the curve that the derivative is not defined at $\left( { \pm 1,0} \right)$? It is strange! On the contrary, you can expect from the curve that the derivative is not defined at $\left( {0, \pm 1} \right)$. – Hosein Rahnama Oct 22 '15 at 14:16

in proving that $(0,±1)$ has infinite slope,we have complicated differentiation but doable by single variable calculus.

taking cubic root for $(x^2+y^2−1)^3=x^2y^3$, we get $(x^2+y^2−1)=x^{2/3}y$. Rearrange to get

$$y^2+x^{2/3}y+(x^2-1)=0$$

using the quadratic formula for y, we get that

$$y = 1/2 \bigg(-x^{2/3}\pm\sqrt{x^{4/3}-4 x^2+4}\bigg)$$

which the positive square root represent the top part of the heart and the negative square root represent the bottom part of the heart.

now taking derivative on y, we get

$$y' = -\frac{2}{3 x^{1/3}} \pm \frac{1}{2} \cdot \frac{4 \frac{x^{1/3}}{3}-8 x}{(2 \sqrt{x^{4/3}-4 x^2+4)}}$$

finally taking the limit as $x\rightarrow 0^+$ , we have $y'\rightarrow \infty$;
as $x\rightarrow 0^-$ , we have $y'\rightarrow -\infty$

This part proves that $(0,±1)$ has infinite slope, thus not differentiable.

• You have to solve $$y^2-x^{2/3}y+(x^2-1)=0$$ for $y$. This implies some sign changes in your solution, but does not change it in an essential way. – Christian Blatter Oct 22 '15 at 18:23
• @ammon lam, you have also shown that the curve is differentiable at all points on the curve where $x\ne 0$. Perfect. – John Molokach Oct 22 '15 at 18:39
• @JohnMolokach: Is there any problem with my solution? :) I think it at least deserve a comment or maybe a vote up! :D – Hosein Rahnama Oct 23 '15 at 7:03

I think it's easier to move the second term to the right to get

$$(x^2+y^2-1)^3 = x^2y^3$$

Take the real cube root of both sides to obtain

$$x^2+y^2-1 = \sqrt{x^2}y$$

Then differentiate implicitly:

$$2x+2y\frac{dy}{dx} = \frac{2}{3\sqrt{x}}y + \sqrt{x^2}\frac{dy}{dx}$$ $$\left(2y - \sqrt{x^2}\right)\frac{dy}{dx} = \frac{2}{3\sqrt{x}}y - 2x$$ $$\frac{dy}{dx} = \frac{\frac{2}{3\sqrt{x}}y - 2x}{2y - \sqrt{x^2}}$$

and when evaluated at $y = 0$, we obtain

$$\frac{dy}{dx} = 2\sqrt{x}$$

which equals $\pm 2$ at $x = \pm 1$, respectively.

This solution works with implicit differentiation. The equation of the heart-curve is

$${({x^2} + {y^2} - 1)^3} - {x^2}{y^3} = 0\tag{1}$$

Using implicit differentiation, one can obtain the derivative to be

$${{dy} \over {dx}} = - {{6x{{\left( {{x^2} + {y^2} - 1} \right)}^2} - 2x{y^3}} \over {6y{{\left( {{x^2} + {y^2} - 1} \right)}^2} - 3\;{x^2}{y^2}}}\tag{2}$$

but from Eq.$(1)$ we can observe that

$${({x^2} + {y^2} - 1)^2} = {x^{{4 \over 3}}}{y^2}\tag{3}$$

Now combining $(2)$ and $(3)$ leads to

\eqalign{ & {{dy} \over {dx}} = - {{6x\left( {{x^{{4 \over 3}}}{y^2}} \right) - 2x{y^3}} \over {6y\left( {{x^{{4 \over 3}}}{y^2}} \right) - 3\;{x^2}{y^2}}} = - {{6{x^{{7 \over 3}}}{y^2} - 2x{y^3}} \over {6{x^{{4 \over 3}}}{y^3} - 3{x^2}{y^2}}} \cr & \,\,\,\,\,\,\,\,\, = - {{6{x^{{7 \over 3}}} - 2xy} \over {6{x^{{4 \over 3}}}y - 3{x^2}}} = - {{6{x^{{4 \over 3}}} - 2y} \over {6{x^{{1 \over 3}}}y - 3x}} \cr}\tag{4}

and in summary

$${{dy} \over {dx}} = - {{6x\root 3 \of x - 2y} \over {6\root 3 \of x y - 3x}}\tag{5}$$

Finally, we can use Eq.$(5)$ to conclude that the slope at $\left( { \pm 1,0} \right)$ is $\pm 2$ and also that the slope at $\left( {0, \pm 1} \right)$ goes to $\pm \infty$ depending on whether we are approaching the point from the left or right.

• Thanks for editing your answer to include all four points. I think your answer is a good one, and I don't mind the implicit differentiation (in fact that us what I used if your read the question), but I thought the prior answer was more clever and elementary in that one who hasn't learned implicit derivatives could still do it. – John Molokach Oct 23 '15 at 11:29
• @JohnMolokach: You are right! But the formula by implicit differentiation looks better and simpler! I think it's a neat way to go through these kind of problems since we cannot always solve for $y$! So this is a strategy which will work in more complicated cases! :) – Hosein Rahnama Oct 23 '15 at 12:16