Notes: `x=sqrt(t) , y=3t-1 , 0

Thursday, 10 November 2016

$\displaystyle{x}=\sqrt{{{t}}},{y}={3}{t}-{1},{0}$

Arc length of a curve C described by the parametric equations x=f(t) and y=g(t), $\displaystyle{a}\le{t}\le{b}$ , where f'(t) and g'(t) are continuous on [a,b] and C is traversed exactly once as t increases from a to b , then the length of the curve C is given by:

$\displaystyle{L}={\int_{{a}}^{{b}}}\sqrt{{{{\left(\frac{{\left.{d}{x}}}{{\left.{d}{t}}}\right)}}^{{2}}+{{\left(\frac{{\left.{d}{y}}}{{\left.{d}{t}}}\right)}}^{{2}}}}{\left.{d}{t}\right.}$

Given parametric equations are : $\displaystyle{x}=\sqrt{{{t}}},{y}={\left({3}{t}-{1}\right)},{0}\le{t}\le{1}$

$\displaystyle{x}=\sqrt{{{t}}}$

$\displaystyle\Rightarrow\frac{{\left.{d}{x}}}{{\left.{d}{t}}}=\frac{{1}}{{2}}{{\left({t}\right)}}^{{\frac{{1}}{{2}}-{1}}}$

$\displaystyle\frac{{\left.{d}{x}}}{{\left.{d}{t}}}=\frac{{1}}{{2}}{{t}}^{{-\frac{{1}}{{2}}}}$

$\displaystyle\frac{{\left.{d}{x}}}{{\left.{d}{t}}}=\frac{{1}}{{{2}\sqrt{{{t}}}}}$

$\displaystyle{y}={3}{t}-{1}$

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{t}}}={3}$

Now let's evaluate the length of the arc by using the stated formula,

$\displaystyle{L}={\int_{{0}}^{{1}}}\sqrt{{{{\left(\frac{{1}}{{{2}\sqrt{{{t}}}}}\right)}}^{{2}}+{{3}}^{{2}}}}{\left.{d}{t}\right.}$

...

$\displaystyle{L}={\int_{{a}}^{{b}}}\sqrt{{{{\left(\frac{{\left.{d}{x}}}{{\left.{d}{t}}}\right)}}^{{2}}+{{\left(\frac{{\left.{d}{y}}}{{\left.{d}{t}}}\right)}}^{{2}}}}{\left.{d}{t}\right.}$

Given parametric equations are : $\displaystyle{x}=\sqrt{{{t}}},{y}={\left({3}{t}-{1}\right)},{0}\le{t}\le{1}$

$\displaystyle{x}=\sqrt{{{t}}}$

$\displaystyle\Rightarrow\frac{{\left.{d}{x}}}{{\left.{d}{t}}}=\frac{{1}}{{2}}{{\left({t}\right)}}^{{\frac{{1}}{{2}}-{1}}}$

$\displaystyle\frac{{\left.{d}{x}}}{{\left.{d}{t}}}=\frac{{1}}{{2}}{{t}}^{{-\frac{{1}}{{2}}}}$

$\displaystyle\frac{{\left.{d}{x}}}{{\left.{d}{t}}}=\frac{{1}}{{{2}\sqrt{{{t}}}}}$

$\displaystyle{y}={3}{t}-{1}$

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{t}}}={3}$

Now let's evaluate the length of the arc by using the stated formula,

$\displaystyle{L}={\int_{{0}}^{{1}}}\sqrt{{{{\left(\frac{{1}}{{{2}\sqrt{{{t}}}}}\right)}}^{{2}}+{{3}}^{{2}}}}{\left.{d}{t}\right.}$

$\displaystyle{L}={\int_{{0}}^{{1}}}\sqrt{{\frac{{1}}{{{4}{t}}}+{9}}}{\left.{d}{t}\right.}$

$\displaystyle{L}={\int_{{0}}^{{1}}}\sqrt{{\frac{{{1}+{36}{t}}}{{{4}{t}}}}}{\left.{d}{t}\right.}$

$\displaystyle{L}={\int_{{0}}^{{1}}}\frac{\sqrt{{{1}+{36}{t}}}}{\sqrt{{{4}{t}}}}{\left.{d}{t}\right.}$

$\displaystyle{L}={\int_{{0}}^{{1}}}\frac{{1}}{{2}}\frac{\sqrt{{{1}+{36}{t}}}}{\sqrt{{{t}}}}{\left.{d}{t}\right.}$

Take the constant out,

$\displaystyle{L}=\frac{{1}}{{2}}{\int_{{0}}^{{1}}}\frac{\sqrt{{{1}+{36}{t}}}}{\sqrt{{{t}}}}{\left.{d}{t}\right.}$

Let's first evaluate the indefinite integral: $\displaystyle\int\frac{\sqrt{{{1}+{36}{t}}}}{\sqrt{{{t}}}}{\left.{d}{t}\right.}$

Use integral substitution: $\displaystyle{u}={6}\sqrt{{{t}}}$

$\displaystyle{d}{u}={6}{\left(\frac{{1}}{{2}}\right)}{{t}}^{{\frac{{1}}{{2}}-{1}}}{\left.{d}{t}\right.}$

$\displaystyle{d}{u}=\frac{{3}}{\sqrt{{{t}}}}{\left.{d}{t}\right.}$

$\displaystyle\int\frac{\sqrt{{{1}+{36}{t}}}}{\sqrt{{{t}}}}{\left.{d}{t}\right.}=\int\frac{{1}}{{3}}\sqrt{{{1}+{{u}}^{{2}}}}{d}{u}$

$\displaystyle=\frac{{1}}{{3}}\int\sqrt{{{1}+{{u}}^{{2}}}}{d}{u}$

Use the following standard integral from the integration tables:

$\displaystyle\int\sqrt{{{{u}}^{{2}}+{{a}}^{{2}}}}{d}{u}=\frac{{1}}{{2}}{\left({u}\sqrt{{{{u}}^{{2}}+{{a}}^{{2}}}}+{{a}}^{{2}}{\ln}{\left|{u}+\sqrt{{{{a}}^{{2}}+{{u}}^{{2}}}}\right|}\right)}+{C}$

$\displaystyle=\frac{{1}}{{3}}{\left[\frac{{1}}{{2}}{\left({u}\sqrt{{{{u}}^{{2}}+{1}}}+{\ln}{\left|{u}+\sqrt{{{1}+{{u}}^{{2}}}}\right|}\right)}\right]}+{C}$

$\displaystyle=\frac{{1}}{{6}}{\left({u}\sqrt{{{1}+{{u}}^{{2}}}}+{\ln}{\left|{u}+\sqrt{{{1}+{{u}}^{{2}}}}\right|}\right)}+{C}$

Substitute back: $\displaystyle{u}={6}\sqrt{{{t}}}$

$\displaystyle=\frac{{1}}{{6}}{\left({6}\sqrt{{{t}}}\sqrt{{{1}+{36}{t}}}+{\ln}{\left|{6}\sqrt{{{t}}}+\sqrt{{{1}+{36}{t}}}\right|}\right)}+{C}$

$\displaystyle{L}=\frac{{1}}{{2}}{{\left[\frac{{1}}{{6}}{\left({6}\sqrt{{{t}}}\sqrt{{{1}+{36}{t}}}+{\ln}{\left|{6}\sqrt{{{t}}}+\sqrt{{{1}+{36}{t}}}\right|}\right)}\right]}_{{0}}^{{1}}}$

$\displaystyle{L}=\frac{{1}}{{12}}{{\left[{6}\sqrt{{{t}}}\sqrt{{{1}+{36}{t}}}+{\ln}{\left|{6}\sqrt{{{t}}}+\sqrt{{{1}+{36}{t}}}\right|}\right]}_{{0}}^{{1}}}$

$\displaystyle{L}=\frac{{1}}{{12}}{\left\lbrace{\left[{6}\sqrt{{{37}}}+{\ln}{\left|{6}+\sqrt{{{37}}}\right|}\right]}-{\left[{\ln}{\left|{1}\right|}\right]}\right\rbrace}$