Finding the arc length of $ x^2=(y-4)^3$ I nee to find the arc length of the following function, between P and Q:
$$x^2 = (y-4)^3, \quad P(1, 5), \;Q(8, 8)$$
I started and got $x = (y-4)^{3/2}$ but I'm not sure if I'm on the right path.
 A: It is a rational cubic that can be parametrised by
$$\begin{cases}x=t^3\\y=4+t^2\end{cases}$$
The points are obtained for $t=1$  and $t=2$ respectively, hence
$$\ell=\int_1^2\sqrt{9t^4+4t^2}\,\mathrm d\mkern1.5mu t=\int_1^2 2t\sqrt{\Bigl(\dfrac{3t}2\Bigr)^2+1}\,\mathrm d\mkern1.5mu t.$$
The integral can be calculated with the $\,\tan \theta=\dfrac{3t}2\,$ substitution.
A: I suggest you to write your equation in the parametric form 
\begin{equation}
X=\{t,4+t^{\frac{2}{3}} \}
\end{equation}
And then compute thar arc length as
\begin{equation}
l=\int^{a}_{b} \sqrt{x^{'}(t)+y^{'}(t)} dt
\end{equation}
where a and b are values of t such that $X(a)=P$ and  $X(b)=Q$. Thus $a=1$ and $b=8$. Then
\begin{equation}
l=\int^{1}_{8} \sqrt{x^{'}(t)+y^{'}(t)} dt= \frac{1}{3^3}(80 \sqrt{10}-13\sqrt{13})
\end{equation}
A: The original curve $x^2=(y-4)^3$, $x\in[1,8],\ y\in[5,8]$ 
can be expressed in parametric form
\begin{align}
x(t)&=t,\\
y(t)&=t^{2/3}+4.
\end{align}
Then the arclength 
\begin{align}
L&=\int_{1}^{8}\sqrt{x^\prime(t)^2+y^\prime(t)^2}dt\\
&=\int_{1}^{8}\sqrt{1+(2/3 t^{-1/3})^2}dt
\end{align}
which after substitution $t=u^3$ becomes
\begin{align}
L&=\int_1^2 \sqrt{9 u^2+4}\, u\, du\\
&=\frac{1}{27} \left. (9 u^2+4)^{3/2}\right|_{u=1}^2,
\end{align}
which gives a rather interesting answer
\begin{align}
L&=\frac{1}{27} (40\sqrt{40}-13\sqrt{13})
&\approx 7.6337.
\end{align}
The arclength can be also approximated between the two linear lengths:

