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Using the implicit function $(x^2+y^2-1)^3=x^2y^3$ it can be shown that $y'=\frac{2xy^3-6x(x^2+y^2-1)^2}{6y(x^2+y^2-1)^2-3x^2y^2}$ but when I evaluate it for the point (1,0) I get $y'(1,0)=\frac{0}{0}$ even though the slope of the tangent line is 2 at that point.

Any ideas?

Garth

Garth
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2 Answers2

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Assuming you did all the algebra correctly, I haven't checked: You made an error in the algebra! Heh.

When you differentiated you got $$Q(x,y)y'+P(x,y)=0,\quad(*)$$where $P$ and $Q$ are two nasty polynomials. Then you converted that to $$y'=-\frac{P(x,y)}{Q(x,y)}.\quad(**)$$

But going from (*) to ($**$) is only valid if $Q(x,y)\ne0$. So you haven't actually shown that (**) holds, you've shown that $(**)$ holds at points where $Q\ne0$. Again assuming that you did everything else right, at $(1,0)$ equation (*) becomes $$0(2)=0,$$which is no problem since $0$ times $2$ does equal $0$. But $0(2)=0$ does not imply $2=0/0$.

David C. Ullrich
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  • Prof. Ullrich, welcome to MSE. Could you please take a look at my questions: http://math.stackexchange.com/questions/1263869/definition-functions-integrals-on-mathbb-rn-mathbb-r-mathbb-r and – Gary. Jun 22 '15 at 12:15
  • And http://math.stackexchange.com/questions/1306846/metrizability-models-of-non-standard-reals ? – Gary. Jun 22 '15 at 12:16
  • Don't think I can help much with those two questions, sorry. Looks like you're already getting reasonable feedback - given that the questions are a little vague (at least the first) you can't expect something definitive. Hmm. If $\mathbb{R}^{\mathbb N}$ means what I think it means it may or may not be helpful to point out that there's no problem defining something more or less like Lebesgue measure on $[0,1]^{\mathbb N}$. – David C. Ullrich Jun 22 '15 at 16:06
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Instead $(x^2+y^2-1)^3=x^2y^3$ we can take $x^2+y^2-1=x^{2/3}y$ near of $(1,0)$, hence \begin{align*} \left(y-\frac{1}{2}x^{2/3}\right)^2&=1-x^2+\frac{x^{4/3}}{4}\\ y&=-\left(1-x^2+\frac{x^{4/3}}{4}\right)^{1/2}+\frac{x^{2/3}}{2}\\ y'&=-\frac{1}{2}\left(1-x^2+\frac{x^{4/3}}{4}\right)^{-1/2}\left(-2x+\frac{x^{1/3}}{3}\right)+\frac{1}{3}x^{-1/3} \end{align*}

Then \begin{align*} y'(1)&=-\frac{1}{2}\left(1-1+\frac{1}{4}\right)^{-1/2}\left(-2+\frac{1}{3}\right)+\frac{1}{3}\\ &=-\frac{1}{2}(2)\left(-\frac{5}{3}\right)+\frac{1}{3}\\ &=2 \end{align*}

Ángel Mario Gallegos
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  • Thank you. I also came to this conclusion. It takes away the need to use the chain rule which forces the $y^2$ factor. Much appreciated. – Garth Jun 21 '15 at 20:44