find the exact length of the curve.\n$x = \\frac{1}{3}\\sqrt{y}(y - 3),\\ 1 \\leq y \\leq 4$

Explanation:

Step1: First, rewrite the function

Rewrite $x=\frac{1}{3}\sqrt{y}(y - 3)=\frac{1}{3}(y^{\frac{3}{2}}-3y^{\frac{1}{2}})$.

Step2: Differentiate with respect to y

Using the power - rule $\frac{d}{dy}(y^n)=ny^{n - 1}$, we have $x^\prime=\frac{1}{3}(\frac{3}{2}y^{\frac{1}{2}}-\frac{3}{2}y^{-\frac{1}{2}})=\frac{1}{2}(y^{\frac{1}{2}}-y^{-\frac{1}{2}})$.

Step3: Calculate $1+(x^\prime)^2$

[ \begin{align*} 1+(x^\prime)^2&=1+\frac{1}{4}(y - 2 + y^{-1})\ &=\frac{1}{4}(4 + y-2 + y^{-1})\ &=\frac{1}{4}(y + 2 + y^{-1})\ &=\left(\frac{1}{2}(y^{\frac{1}{2}}+y^{-\frac{1}{2}})\right)^2 \end{align*} ]

Step4: Use the arc - length formula

The arc - length formula for a curve $x = x(y)$ from $y = a$ to $y = b$ is $L=\int_{a}^{b}\sqrt{1+(x^\prime)^2}dy$. Here, $a = 1$, $b = 4$, and $\sqrt{1+(x^\prime)^2}=\frac{1}{2}(y^{\frac{1}{2}}+y^{-\frac{1}{2}})$. [ \begin{align*} L&=\int_{1}^{4}\frac{1}{2}(y^{\frac{1}{2}}+y^{-\frac{1}{2}})dy\ &=\frac{1}{2}\left[\frac{2}{3}y^{\frac{3}{2}}+2y^{\frac{1}{2}}\right]_{1}^{4}\ &=\frac{1}{2}\left[\left(\frac{2}{3}(4)^{\frac{3}{2}}+2(4)^{\frac{1}{2}}\right)-\left(\frac{2}{3}(1)^{\frac{3}{2}}+2(1)^{\frac{1}{2}}\right)\right]\ &=\frac{1}{2}\left[\left(\frac{2}{3}\times8 + 4\right)-\left(\frac{2}{3}+2\right)\right]\ &=\frac{1}{2}\left[\left(\frac{16}{3}+4\right)-\left(\frac{2 + 6}{3}\right)\right]\ &=\frac{1}{2}\left[\frac{16 + 12}{3}-\frac{8}{3}\right]\ &=\frac{1}{2}\times\frac{16+12 - 8}{3}\ &=\frac{1}{2}\times\frac{20}{3}\ &=\frac{10}{3} \end{align*} ]

Answer:

$\frac{10}{3}$