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The curve $y=x^{1/3}$ is smooth everywhere even though $dy/dx$ does not exist at $x=0$. Why?

In general; Wherever $dy/dx$ does not exist on a curve, how can I show that it could still be smooth at those places?

Jostein Trondal
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2 Answers2

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The issue here is that if you have a continuously differentiable function $f(x,y)$, its level set $f(x,y)=c$ will be smooth curves whenever one of $\dfrac{\partial f}{\partial x}$ and $\dfrac{\partial f}{\partial y}$ is nonzero at each point of the curve.

Your curve is the graph $y=x^{1/3}$, and this can be rewritten as the $0$-level curve of $f(x,y)=y^3-x$ (this remark is equivalent to @Pratyush Sarkar's answer). Note that another example of your phenomenon is the curve $f(x,y)=x^2+y^2-1$. You might start with $y=\sqrt{1-x^2}$ and $dy/dx$ doesn't exist at $x=\pm 1$, and yet the circle is certainly a smooth curve; here we have $\dfrac{\partial f}{\partial x}(\pm1,0) \ne 0$, even though $\dfrac{\partial f}{\partial y}(\pm1,0) = 0$. Recall that the formula for implicit differentiation tells us that $$\text{when } \dfrac{\partial f}{\partial y} \ne 0, \text{ we have } \frac{dy}{dx} = -\frac{\partial f/\partial x}{\partial f/\partial y}\,.$$

Ted Shifrin
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  • According to my calculator, the dy/dx when x is equal to zero is 100. Is there any simple explanation for this? – recursive recursion Jan 11 '14 at 18:01
  • For what function? And what are you having your calculator do, precisely? – Ted Shifrin Jan 11 '14 at 18:31
  • For the function x ^ 1/3. I'm having it graph the derivative of this function. Where x = 0, y is equal to 100. – recursive recursion Jan 11 '14 at 18:45
  • What happens if you ask it to graph $\frac13 x^{-2/3}$? I can't tell if it's doing derivatives numerically or if it just has some numerical bugs in it. P.S. This shows the importance of thinking and learning to do calculations by hand, not just relying on calculators. – Ted Shifrin Jan 11 '14 at 19:02
  • All points are the exact same, except when x = 0. In the former function, it is 100, while in the latter, it is undefined. This is a TI-84 plus, and I don't know how it calculates derivatives. – recursive recursion Jan 11 '14 at 19:05
  • I do not use graphing calculators, so I am no expert here. But I'm guessing it does not compute the derivative formally, and the numerical errors are just overwhelming it when $x$ is near $0$. The evidence is that it knows what to do correctly if you feed in the derivative function yourself. Moral of the story: Dump the calculator. – Ted Shifrin Jan 11 '14 at 19:23
  • Except hyperbolas work perfectly well on this calculator, as does calculating the derivatives of these hyperbolas in question... – recursive recursion Jan 11 '14 at 19:25
  • What happens if you try the derivative of $\sqrt{1-x^2}$ at $x=1$? The circle and the hyperbola should work the same. But $\sqrt{x^2-1}$ at $x=1$ is ok with me. Does it even know how to do one-sided derivatives? – Ted Shifrin Jan 11 '14 at 19:41
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The curve $\gamma_1(x) = (x, x^{\frac{1}{3}})$ can be reparametrized as $\gamma_2(y) = (y^3, y)$. The second curve $\gamma_2$ is differentiable everywhere with ${\gamma_2}'(y) \neq (0, 0)$ for all $y$. So the curve is smooth.

Pratyush Sarkar
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