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Question: $$y=x^{\dfrac23}$$ a) Prove that Length of the arc can not be found using the formula $L=\int_{a}^b\sqrt{1+(\frac{dy}{dx})^2}dx$ (call this Formula $F1$) for $x=-1$ to $x=8$.

b) And hence find the Length of arc for the given interval.

My Attempt:

I thought it is fairly easy to prove part (a.) and hence use $L=\int_{a}^b\sqrt{1+(\frac{dx}{dy})^2}dy$ (call this Formula $F2$) to find the answer to (b.).

Hence, I proceeded by showing that $y'=\frac23x^\frac{-1}3$ and since $y'$ is not defined for $x=0$ thus, I concluded (in the back of my mind) that the integral $F1$ shouldn't be defined for the given bounds. But then when I calculated the integrals, both $F2$ (Twice from $y=0$ to $y=1$ and once from $y=1$ to $y=4$) and $F1$ (From $x=-1$ to $x=8$) were defined.

But $F1$ also gives the answer written in the book($=\frac{13\sqrt13+80\sqrt10-16}{27}$). What am I doing wrong? Or is it the question itself?

The same question as this is posted here but nothing had been said regarding this matter.

Edit: $F2$ is giving the correct result after changing the bounds and evaluating as shown in comments by @Shubham.

Samarth
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    $y=1$ to $4$ are evidently the wrong limits since $f([-1,8])=[0,4]$. You will still not get the correct answer, this is because both $[-1,0],[0,1]$ map to $[0,1]$ so you will have to find $2\int_{y=0}^1F_2+\int_{y=1}^4F_2$. – Shubham Johri Jan 09 '21 at 04:32
  • I am sorry. I'll edit that. But the problem still holds, unfortunately. Why am I getting the right answer by evaluating $F1$ for the given bounds? – Samarth Jan 09 '21 at 04:37
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    It looks to me that the question expected the answer that since the derivative does not exist at $0$ the integral cannot be found. That is not always the case, for example https://math.stackexchange.com/questions/61880/question-about-an-integrable-singularity – Shubham Johri Jan 09 '21 at 04:49
  • Due to symmetry, you can first find the arc length from $x=0$ to $x=1$ and then from $x=0$ to $x=8$. Add up the answers. In both cases it is a simple improper integral that turns out clean.The Origin is a cusp. You can have a cusp at the "end" of an arc, but not in the "middle". – imranfat Jan 09 '21 at 04:51
  • @imranfat Can you please elaborate on why the cusp can not occur in the middle? – Samarth Jan 09 '21 at 04:57
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    Because there are instances that the part of the function on "one" side of the cusp adds a "negative" amount to the arc length. That may not be the case in this particular example, but when the curve is both above and under the x-axis (with for example the cusp at the origin), then this unwanted "cancellation" can happen. – imranfat Jan 09 '21 at 05:13

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