exercise 2. find the length of the curve.\n1. $9x^{2}=4y^{3}$ from $(0,0)$ to $(2\\sqrt{3},3)$

Explanation:

Step1: Recall arc - length formula

The arc - length formula for a curve $y = f(x)$ from $x=a$ to $x = b$ is $L=\int_{a}^{b}\sqrt{1+(y')^{2}}dx$. First, we need to express $y$ as a function of $x$. Given $9x = 4y^{\frac{3}{2}}$, we can solve for $y$: $y=\left(\frac{9x}{4}\right)^{\frac{2}{3}}$. Then find the derivative $y'$. Differentiating $y=\left(\frac{9x}{4}\right)^{\frac{2}{3}}$ using the power rule $(u^{n})'=nu^{n - 1}u'$, where $u=\frac{9x}{4}$ and $n=\frac{2}{3}$. So $y'=\frac{2}{3}\left(\frac{9x}{4}\right)^{-\frac{1}{3}}\times\frac{9}{4}=\frac{3}{2}\left(\frac{4}{9x}\right)^{\frac{1}{3}}$.

Step2: Calculate $(y')^{2}$

$(y')^{2}=\frac{9}{4}\left(\frac{4}{9x}\right)^{\frac{2}{3}}$. Then $1+(y')^{2}=1 + \frac{9}{4}\left(\frac{4}{9x}\right)^{\frac{2}{3}}$. The limits of integration: when $x = 0$, the lower - limit; when $y=\sqrt{3}$, from $9x = 4y^{\frac{3}{2}}$, we substitute $y=\sqrt{3}$ and get $9x=4(\sqrt{3})^{\frac{3}{2}}$, $x=\frac{4\times3^{\frac{3}{4}}}{9}$. The arc - length formula becomes $L=\int_{0}^{\frac{4\times3^{\frac{3}{4}}}{9}}\sqrt{1+\frac{9}{4}\left(\frac{4}{9x}\right)^{\frac{2}{3}}}dx$. Another way: start from $9x = 4y^{\frac{3}{2}}$, we can express $x$ as a function of $y$: $x=\frac{4}{9}y^{\frac{3}{2}}$. Then $x'=\frac{4}{9}\times\frac{3}{2}y^{\frac{1}{2}}=\frac{2}{3}y^{\frac{1}{2}}$. The arc - length formula for $x = g(y)$ from $y = c$ to $y = d$ is $L=\int_{c}^{d}\sqrt{1+(x')^{2}}dy$. Here $c = 0$ and $d=\sqrt{3}$. $(x')^{2}=\frac{4}{9}y$. Then $1+(x')^{2}=1+\frac{4}{9}y$.

Step3: Evaluate the integral

$L=\int_{0}^{\sqrt{3}}\sqrt{1 + \frac{4}{9}y}dy$. Let $u=1+\frac{4}{9}y$, then $du=\frac{4}{9}dy$ and $y=\frac{9}{4}(u - 1)$. When $y = 0$, $u = 1$; when $y=\sqrt{3}$, $u=1+\frac{4}{9}\sqrt{3}$. The integral becomes $\frac{9}{4}\int_{1}^{1+\frac{4}{9}\sqrt{3}}u^{\frac{1}{2}}du$. Using the power - rule for integration $\int u^{n}du=\frac{u^{n + 1}}{n+1}+C(n\neq - 1)$, we have $\frac{9}{4}\times\frac{2}{3}\left[u^{\frac{3}{2}}\right]_{1}^{1+\frac{4}{9}\sqrt{3}}=\frac{3}{2}\left[\left(1+\frac{4}{9}\sqrt{3}\right)^{\frac{3}{2}}-1\right]$.

Answer:

$\frac{3}{2}\left[\left(1+\frac{4}{9}\sqrt{3}\right)^{\frac{3}{2}}-1\right]$