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Let $y=x^2/2$. Its parametric form is $r(t)=t\,\hat i+t^2/2\,\hat j$, and its evolute is

$$ c(t)=-t^3\,\hat i+\frac{3t^2+2}{2}\,\hat j.\tag{1} $$

Visually,

                                        

When I rewrite $(1)$ as a normal function, by letting $x=-t^3$, I get

$$ y=\frac{3x^{2/3}+2}{2}, $$

but the graph of this evolute is nothing like the one above. What am I doing wrong?

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1 Answer 1

up vote 2 down vote accepted

I think it would be clearer to write $|x|^{2/3}$. Other than that, your formulas are both right, but the plot is wrong – the $y$ coordinate of the evolute at $x=1$ should be $5/2$, not $3/2$, and you can see with the naked eye that the curve in your plot doesn't reflect the centres of curvature of the parabola; there are normals to the parabola that don't even cross that curve on the right side of the parabola.

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I put the absolute values and got what I was looking for. However, while solving this algebraically, I did not see the need for them. This is confusing to me. –  Josué Sep 26 '12 at 19:36
    
@Josué: Solving $x=-t^3$ for $t$ yields $t=(-x)^{1/3}$. Substituting that into $t^2$ yields $\left((-x)^{1/3}\right)^2$. This is well-defined because the third root of a negative number is well-defined, and using $(-x)^{1/3}=-x^{1/3}$ you can write it as $\left(-x^{1/3}\right)^2=\left(x^{1/3}\right)^2=\left(|x|^{1/3}\right)^2$. But if you now combine the exponents into $2/3$, you have to either use the version with the absolute value, or specify what you mean by taking a negative number to the power $2/3$. –  joriki Sep 26 '12 at 19:58

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