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What's the general method to find the slope of a curve at the origin if the derivative at the origin becomes indeterminate. For Eg--

What is the slope of the curve $x^3 + y^3= 3axy$ at origin and how to find it because after following the process of implicit differentiation and plugging in $x=0$ and $y=0$ in the derivative we get $0/0$.

Actually this question has been asked by me before and a sort of satisfactory answer that I got was

" For small $x$ and $y$, the values of $x^3$ and $y^3$ will be much smaller than $3axy$, so the zeroes of the function will be approximately where the zeroes of $0=3axy$ are -- that is, near the origin the curve will look like the solutions to that, which is just the two coordinate axes. So the curve will cross itself at the origin, passing through the origin once horizontally and once vertically. (This is also why implicit differentiation can't work at the origin -- the solution set simply doesn't look like a straight line there under any magnification)."

Edit

If I approximate the function by saying that at (0,0) , the behavior is dominated be 3axy term as x^3 and y^3 are very small and then 3axy=0 and then tangents are x=0 and y=0 . Is doing so (saying x=0 and y=0) linear Approximation only. Because I am approximating the curve with a straight line at origin . But linear Approximation is 1st derivative (1st term of Taylor series). This cannot be right because Taylor series can't be formed where derivative doesn't exist*.

And if this is right then the function is approximately given by 3axy=0 at (0,0). But how does this give the tangent at (0,0).How shall I go about ?

Edit:

Is the answer give right because the solpe does exist.

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The curve of equation $x^3+y^3=3axy$ is a folium of Descartes, and has a double point at $(x,y)=(0,0)$. So at this point the curve does not have a well defined tangent and, as noted in the answer of Hagen, it has no derivative nor slope.

But, you can ask about the slope of the two tangents at this point, and you can find an answer using the parametric equation of the curve: $$ x=\frac{3at}{1+t^3} \qquad y=\frac{3at^2}{1+t^3} $$

Using this you can find: $\dot x=\frac{dx}{dt}$ and $\dot y=\frac{dy}{dt}$ and calculate the slope $$ \frac{dy}{dx}=\frac{\dot y}{\dot x} $$ or $$ \frac{dx}{dy}=\frac{\dot x}{\dot y} $$ for the two values of $t$ that corresponds to the two different passages through the origin.

Note that one of these values is $t=0$ but the other is for $t \to \infty$ so one of the two slopes have to be evaluated as a limit.

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  • $\begingroup$ Sorry for a late question . I was referring back to this question and had a doubt. 1st your method uses parametric method. Is there any other way out. I mean that I an trying that as x and y tend to 0 the values of the function is governed by 3axy because x^3 and y^3 get very same. So is I approximate the function near (0,0) I am essentially using " The Normal Linear Approximation". But is that valid because it is like forming a Taylor series ( 1st order series) where the derivative doesn't exist. Or is my argument flawed here $\endgroup$ – Shashaank Mar 11 '17 at 14:14
  • $\begingroup$ The key fact here is that $(0,0)$ is a double point and here there is not a derivative. Using the parametric equation we can ''separate'' the two ''arms'' of the function that pass through the point and treat each as a function that has a tangent. $\endgroup$ – Emilio Novati Mar 11 '17 at 15:30
  • $\begingroup$ So does that mean that my Approximation is wrong ? Because I always try to use only x , y even in line integrals , avoiding parametric form. Is this Approximation of function being governed by 3axy (which is dominant term) wrong. Though I have not used Taylor series (in any way) but the idea is that I have approximated the function at (0,0) as 3axy. Now what I don't know is that whether this is right or not. I know that we cannot form Taylor series where derivative doesn't exist but what I don't know is whether such an Approximation ( seeming like Taylor series) is right or not. $\endgroup$ – Shashaank Mar 11 '17 at 15:55
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It's as simple of that: No (simple) curve, no derivative, no slope.

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  • $\begingroup$ Ok but how should I find the tangeny line to the above curve at origin $\endgroup$ – Shashaank May 12 '16 at 7:36
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Use l'Hopital's rule.$ $ This rule, the proof of which is very complex, states that if you get 0/0 or infinity/infinity for some limit function $h(x)=f(x)/g(x)$, you can take the derivative of the top and bottom functions and recalculate. Thus, you would do $f'(x)/g'(x)$. If this result is 0/0 or infinity over infinity, try the next derative, and the next, and so forth. In your particular case, it can be confusing because the implicit function contains both y and x, so which do you take the derivative with respect to? The correct way is to take the derivative with respect to x because you are ultimately solving for dy/dx, even if you use implicit. When differentiating with resepect to x, remember to treat y as a constant. In this case, you should get 6x/constant after one application of l'Hops, with x being 0. Thus, the slope at x=0 is indeed 0. ps. the derivative is "technically" undefined even though it can be determined, so this is just how to get the slope.

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