Find arc length of a curve $y^2 = x^3$ I need to find the arc length L of the curve $y^{2} = x^{3}$ from point (0,0) to point (4,8). How can I do this?
 A: Arc length is given by $$\int_a^b\sqrt{1+(y')^2}dx$$
We can graph $y^2 = x^3$ to see what we are working with:

Since we are interested in the length of the curve for $y \geq 0$ (between (0,0, and (4, 8)) we are interested only in the portion of the curve in the first quadrant, and so we can express $$y = \sqrt{x^3},\; (x\geq 0)$$
Compute $y'$:
$$y' = \frac 32 x^{1/2} = \frac 32 \sqrt x$$ and use the formula above to compute arc length, with bounds then, from $a = 0, b = 4$.
$$\int_0^4 \sqrt{1 + \left(\frac 32\sqrt x\right)^2}\,dx = \int_0^4 \sqrt{1 + \frac 94 x}\,dx$$
A: Hint: Parameterize the curve by $x=t^{2/3}$ and $y=t$ for $t\in[0,8]$.  Then apply the formula
$$L=\int_a^b\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt.$$
A: Hint: 

$$ s = \int_{a}^{b} \sqrt{1+y'^2} dx .$$

A: Set $y=f(x)=x^{3/2}$. Then, the length you need is the length of the graph of $f$ from $x=0$ to $x=4$. This is, by definition, given by
$$
\ell=\int_0^4\sqrt{1+(f'(x))^2}\,dx=\int_0^4\sqrt{1+\frac{9}{4}x}\,dx
$$  
