Question

Find the length of the curve $4y^2=x^3$ between $x=0$ and $x=1$.

Answer 1

Given the curve $4y^2=x^3$
Differentiating with respect to $x$,
$8y\frac{dy}{dx}=3x^2$
$\Rightarrow \frac{dy}{dx}=\frac{3x^2}{8y}$
$\Rightarrow \sqrt{1+{\left(\frac{dy}{dx}\right)}^2}=\sqrt{1+\frac{9x^4}{64y^2}}$
$=\sqrt{1+\frac{9x^4}{16\times 4y^2}}$
$=\sqrt{1+\frac{9x^4}{16x^3}}=\sqrt{1+\frac{9x}{16}}$
The curve is symmetrical about x-axis.
Length of the curve $=2{\int }_0^1\sqrt{1+{\left(\frac{dy}{dx}\right)}^{2}dx}$
$=2{\int }_0^1\sqrt{1+\frac{9x}{16}}dx$
$=2\times {\left[\frac{{(1+\frac{9x}{16})}^{3/2}}{\frac{9}{16}\times \frac{3}{2}}\right]}_0^1=\frac{64}{27}{\left[{(1+\frac{9x}{16})}^{3/2}\right]}_0^1$
$=\frac{64}{27}[{(1+\frac{9}{16})}^{3/2}-1]=\frac{64}{27}[{\left(\frac{25}{16}\right)}^{3/2}-1]$
$=\frac{64}{27}[\frac{125}{64}-1]=\frac{64}{27}\left[\frac{125-64}{64}\right]$
$=\frac{64}{27}\left[\frac{61}{64}\right]=\frac{61}{27}$