For a National Board Exam Review:
Find the length of an arc in the first quadrant of the semi-cubical parabola $y^2 = x^3$ from the point where $x=0$ to the point where $x = 5/9$
Answer is $19/27$.
Let's get the $dy/dx$ form of the function;
$$y^2 = x^3$$
$$\frac{dy}{dx} = \frac{3x^2}{y}$$
to get $ds$
$$ds = \sqrt{1 + \left(\frac{3x^2}{y}\right)^2}$$
Length is therefore
$$\int^{5/9}_0 \sqrt{1 + \left(\frac{3x^2}{y}\right)^2} dx = 0.598727$$
What am I doing wrong? Is it possible problem set is wrong?
Best Answer
You should express $y$ first. $y=x^{3/2}$. Then $$ y' = \frac32 x^{1/2} $$ and $$ \int_0^{5/9} \sqrt{1 + y'^2}\,dx = \int_0^{5/9} \sqrt{1 + \frac94 x}\,dx = \left.\frac{8}{27}\left(1 + \frac94x\right)^{\!3/2}\right|_{x=0}^{x=5/9} = \frac{19}{27} $$