Finding the arc length of $r(t)=ti+j+(\frac{1}{6}t^3+\frac{1}{2}t^{-1})$

Question

How would I find the arc length of the following curve from $t=0$ to $t=2$

$r(t)=ti+j+(\frac{1}{6}t^3+\frac{1}{2}t^{-1})$

I took the first derivative and got

$r(t)'=(1+\frac{1}{2}t^2-\frac{1}{2}\frac{1}{t^2})$

then I factored it out a $\frac{1}{2}$

and got $\frac{1}{2}(t^2-\frac{1}{t^2}+2)$ I then squared it and got $\frac{1}{4}(t^4-\frac{1}{t^4}+2)$

I then took the square root and integral but I find myself stuck. On solving the integral.

$\int_0^2\sqrt{\frac{1}{4}(t^4-\frac{1}{t^4}+2)}$

Answer 1

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\vec{\rm r}\pars{t} = t\,{\bf i} + {\bf j} + \pars{{1 \over 6}\,t^{3} + \half\,t^{-1}}{\bf k}}$

\begin{align} {\cal L}&\equiv\int_{t = 0}^{t = 2}\verts{\dd\vec{\rm r}\pars{t}} =\int_{0}^{2}\verts{\totald{\vec{\rm r}\pars{t}}{t}}\,\dd t =\int_{0}^{2} \root{1^{2} + 0^{2} + \bracks{{1 \over 6}\,3t^{2} + \half\pars{-t^{-2}}}^{2}}\,\dd t \\[3mm]&=\int_{0}^{2} \root{1 + {1 \over 4}\,t^{4} - \half + {1 \over 4t^{4}}}\,\dd t =\half\int_{0}^{2}{\root{t^{8} + 2t^{4} + 1} \over t^{2}}\,\dd t =\half\int_{0}^{2}{t^{4} + 1 \over t^{2}}\,\dd t \end{align} $\large\tt\mbox{The integral diverges !!!}$.

Answer 2

You need to find $\vert r'\left(t\right)\vert$.

Answer 3

HINT: Given a vector $r(t) = <f(t)$, $g(t)$, $h(t)>$. The arc length is $$L = \int_a^b\sqrt{(f'(t))^2 + (g'(t))^2 + (h'(t))^2}dt$$ The components have to be squared seperately.

Answer 4

Your mistake is $r(t)'=(1+\frac{1}{2}t^2-\frac{1}{2}\frac{1}{t^2})$ should be $r(t)'=(1,0,\frac{1}{2}t^2-\frac{1}{2}\frac{1}{t^2})$ and so $|r'(t)|=\sqrt{r'.r'}$