find the arc length of the curve below on the given interval. y = 2/3 (x^2 + 1)^(3/2) on 3,9 the length of…

Explanation:

Step1: Find the derivative of y

We have $y = \frac{2}{3}(x^{2}+1)^{\frac{3}{2}}$. Using the chain - rule, if $u=x^{2}+1$, then $y=\frac{2}{3}u^{\frac{3}{2}}$. The derivative of $y$ with respect to $u$ is $y^\prime_{u}=u^{\frac{1}{2}}$, and the derivative of $u$ with respect to $x$ is $u^\prime_{x} = 2x$. So, $y^\prime=x(x^{2}+1)^{\frac{1}{2}}$.

Step2: Use the arc - length formula

The arc - length formula for a function $y = f(x)$ on the interval $[a,b]$ is $L=\int_{a}^{b}\sqrt{1+(y^\prime)^{2}}dx$. Substitute $y^\prime=x(x^{2}+1)^{\frac{1}{2}}$ into the formula: [ \begin{align*} 1+(y^\prime)^{2}&=1 + x^{2}(x^{2}+1)\ &=1+x^{4}+x^{2}\ &=x^{4}+x^{2}+1 \end{align*} ] [ \begin{align*} L&=\int_{3}^{9}\sqrt{1+(x(x^{2}+1)^{\frac{1}{2}})^{2}}dx\ &=\int_{3}^{9}\sqrt{1 + x^{2}(x^{2}+1)}dx\ &=\int_{3}^{9}\sqrt{x^{4}+x^{2}+1}dx\ &=\int_{3}^{9}\sqrt{(x^{2}+\frac{1}{2})^{2}+\frac{3}{4}}dx \end{align*} ] Another way, since $1+(y^\prime)^{2}=1+x^{2}(x^{2}+1)=x^{4}+x^{2}+1=(x^{2}+1)^{2}-x^{2}=(x^{2}+x + 1)(x^{2}-x + 1)$ and also $1+(y^\prime)^{2}=1+x^{2}(x^{2}+1)=x^{4}+x^{2}+1$. We know that $L=\int_{3}^{9}\sqrt{1+(y^\prime)^{2}}dx=\int_{3}^{9}\sqrt{1 + x^{2}(x^{2}+1)}dx=\int_{3}^{9}(x^{2}+1)dx$ (because $\sqrt{1 + x^{2}(x^{2}+1)}=\sqrt{x^{4}+x^{2}+1}=\sqrt{(x^{2}+1)^{2}-x^{2}}=\vert x^{2}+1\vert$ and for $x\in[3,9]$, $x^{2}+1>0$).

Step3: Evaluate the definite integral

[ \begin{align*} \int_{3}^{9}(x^{2}+1)dx&=\left[\frac{1}{3}x^{3}+x\right]_{3}^{9}\ &=\left(\frac{1}{3}(9)^{3}+9\right)-\left(\frac{1}{3}(3)^{3}+3\right)\ &=\left(\frac{1}{3}\times729 + 9\right)-\left(\frac{1}{3}\times27+3\right)\ &=(243 + 9)-(9 + 3)\ &=252-12\ &=240 \end{align*} ]

Answer:

$240$