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

Tuesday, 9 September 2014

$\displaystyle{x}=\frac{{1}}{{3}}\sqrt{{{y}}}{\left({y}-{3}\right)},{1}$

Arc length (L) of the function x=h(y) on the interval [c,d] is given by the formula,

$\displaystyle{L}={\int_{{c}}^{{d}}}\sqrt{{{1}+{{\left(\frac{{\left.{d}{x}}}{{\left.{d}{y}}}\right)}}^{{2}}}}{\left.{d}{y}\right.}$ , if x=h(y) and c $\displaystyle\le$ y $\displaystyle\le$ d,

Now let's differentiate the function with respect to y,

$\displaystyle{x}=\frac{{1}}{{3}}\sqrt{{{y}}}{\left({y}-{3}\right)}$

$\displaystyle\frac{{\left.{d}{x}}}{{\left.{d}{y}}}=\frac{{1}}{{3}}{\left\lbrace\sqrt{{{y}}}\frac{{d}}{{\left.{d}{y}}}{\left({y}-{3}\right)}+{\left({y}-{3}\right)}\frac{{d}}{{\left.{d}{y}}}\sqrt{{{y}}}\right\rbrace}$

$dx/dy=1/3{sqrt(y)(1)+(y-3)1/2(y)^(1/2-1)}$

$\displaystyle\frac{{\left.{d}{x}}}{{\left.{d}{y}}}=\frac{{1}}{{3}}{\left\lbrace\sqrt{{{y}}}+\frac{{{y}-{3}}}{{{2}\sqrt{{{y}}}}}\right\rbrace}$

$\displaystyle\frac{{\left.{d}{x}}}{{\left.{d}{y}}}=\frac{{1}}{{3}}{\left\lbrace\frac{{{2}{y}+{y}-{3}}}{{{2}\sqrt{{{y}}}}}\right\rbrace}$

$\displaystyle\frac{{\left.{d}{x}}}{{\left.{d}{y}}}=\frac{{1}}{{3}}{\left\lbrace\frac{{{3}{y}-{3}}}{{{2}\sqrt{{{y}}}}}\right\rbrace}$

$\displaystyle\frac{{\left.{d}{x}}}{{\left.{d}{y}}}=\frac{{1}}{{3}}{\left({3}\right)}\frac{{{y}-{1}}}{{{2}\sqrt{{{y}}}}}$

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

Plug in the above derivative in the arc length formula,

$\displaystyle{L}={\int_{{1}}^{{4}}}\sqrt{{{1}+{{\left(\frac{{{y}-{1}}}{{{2}\sqrt{{{y}}}}}\right)}}^{{2}}}}{\left.{d}{y}\right.}$

$\displaystyle{L}={\int_{{1}}^{{4}}}\sqrt{{{1}+\frac{{{{y}}^{{2}}-{2}{y}+{1}}}{{{4}{y}}}}}{\left.{d}{y}\right.}$

$\displaystyle{L}={\int_{{1}}^{{4}}}\sqrt{{\frac{{{4}{y}+{{y}}^{{2}}-{2}{y}+{1}}}{{{4}{y}}}}}{\left.{d}{y}\right.}$

$\displaystyle{L}={\int_{{1}}^{{4}}}\sqrt{{\frac{{{{y}}^{{2}}+{2}{y}+{1}}}{{{4}{y}}}}}{\left.{d}{y}\right.}$

$\displaystyle{L}={\int_{{1}}^{{4}}}{\left(\frac{{1}}{{2}}\right)}\sqrt{{\frac{{{\left({y}+{1}\right)}}^{{2}}}{{y}}}}{\left.{d}{y}\right.}$

$\displaystyle{L}=\frac{{1}}{{2}}{\int_{{1}}^{{4}}}\frac{{{y}+{1}}}{\sqrt{{{y}}}}{\left.{d}{y}\right.}$

Now let's compute first the indefinite integral by applying integral substitution,

Let $\displaystyle{u}=\sqrt{{{y}}}$

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

$\displaystyle\frac{{{d}{u}}}{{\left.{d}{y}}}=\frac{{1}}{{{2}\sqrt{{{y}}}}}$

$\displaystyle\int\frac{{{y}+{1}}}{\sqrt{{{y}}}}{\left.{d}{y}\right.}=\int{\left({{u}}^{{2}}+{1}\right)}{2}{d}{u}$

...

Arc length (L) of the function x=h(y) on the interval [c,d] is given by the formula,

$\displaystyle{L}={\int_{{c}}^{{d}}}\sqrt{{{1}+{{\left(\frac{{\left.{d}{x}}}{{\left.{d}{y}}}\right)}}^{{2}}}}{\left.{d}{y}\right.}$ , if x=h(y) and c $\displaystyle\le$ y $\displaystyle\le$ d,

Now let's differentiate the function with respect to y,

$\displaystyle{x}=\frac{{1}}{{3}}\sqrt{{{y}}}{\left({y}-{3}\right)}$

$dx/dy=1/3{sqrt(y)(1)+(y-3)1/2(y)^(1/2-1)}$

$\displaystyle\frac{{\left.{d}{x}}}{{\left.{d}{y}}}=\frac{{1}}{{3}}{\left\lbrace\sqrt{{{y}}}+\frac{{{y}-{3}}}{{{2}\sqrt{{{y}}}}}\right\rbrace}$

$\displaystyle\frac{{\left.{d}{x}}}{{\left.{d}{y}}}=\frac{{1}}{{3}}{\left\lbrace\frac{{{2}{y}+{y}-{3}}}{{{2}\sqrt{{{y}}}}}\right\rbrace}$

$\displaystyle\frac{{\left.{d}{x}}}{{\left.{d}{y}}}=\frac{{1}}{{3}}{\left\lbrace\frac{{{3}{y}-{3}}}{{{2}\sqrt{{{y}}}}}\right\rbrace}$

$\displaystyle\frac{{\left.{d}{x}}}{{\left.{d}{y}}}=\frac{{1}}{{3}}{\left({3}\right)}\frac{{{y}-{1}}}{{{2}\sqrt{{{y}}}}}$

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

Plug in the above derivative in the arc length formula,

$\displaystyle{L}={\int_{{1}}^{{4}}}\sqrt{{{1}+{{\left(\frac{{{y}-{1}}}{{{2}\sqrt{{{y}}}}}\right)}}^{{2}}}}{\left.{d}{y}\right.}$

$\displaystyle{L}={\int_{{1}}^{{4}}}\sqrt{{{1}+\frac{{{{y}}^{{2}}-{2}{y}+{1}}}{{{4}{y}}}}}{\left.{d}{y}\right.}$

$\displaystyle{L}={\int_{{1}}^{{4}}}\sqrt{{\frac{{{4}{y}+{{y}}^{{2}}-{2}{y}+{1}}}{{{4}{y}}}}}{\left.{d}{y}\right.}$

$\displaystyle{L}={\int_{{1}}^{{4}}}\sqrt{{\frac{{{{y}}^{{2}}+{2}{y}+{1}}}{{{4}{y}}}}}{\left.{d}{y}\right.}$

$\displaystyle{L}={\int_{{1}}^{{4}}}{\left(\frac{{1}}{{2}}\right)}\sqrt{{\frac{{{\left({y}+{1}\right)}}^{{2}}}{{y}}}}{\left.{d}{y}\right.}$

$\displaystyle{L}=\frac{{1}}{{2}}{\int_{{1}}^{{4}}}\frac{{{y}+{1}}}{\sqrt{{{y}}}}{\left.{d}{y}\right.}$

Now let's compute first the indefinite integral by applying integral substitution,

Let $\displaystyle{u}=\sqrt{{{y}}}$

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

$\displaystyle\frac{{{d}{u}}}{{\left.{d}{y}}}=\frac{{1}}{{{2}\sqrt{{{y}}}}}$

$\displaystyle\int\frac{{{y}+{1}}}{\sqrt{{{y}}}}{\left.{d}{y}\right.}=\int{\left({{u}}^{{2}}+{1}\right)}{2}{d}{u}$

$\displaystyle={2}\int{\left({{u}}^{{2}}+{1}\right)}{d}{u}$

$\displaystyle={2}{\left(\frac{{{u}}^{{3}}}{{3}}+{u}\right)}$

substitute back u= $\displaystyle\sqrt{{{y}}}$ and add a constant C to the solution,

$\displaystyle={2}{\left(\frac{{{y}}^{{\frac{{3}}{{2}}}}}{{3}}+\sqrt{{{y}}}\right)}+{C}$

$L=[1/2{2(y^(3/2)/3+sqrt(y)}]_1^4$

$\displaystyle{L}={{\left[\frac{{{y}}^{{\frac{{3}}{{2}}}}}{{3}}+\sqrt{{{y}}}\right]}_{{1}}^{{4}}}$

$\displaystyle{L}={\left[\frac{{{4}}^{{\frac{{3}}{{2}}}}}{{3}}+\sqrt{{{4}}}\right]}-{\left[\frac{{{1}}^{{\frac{{3}}{{2}}}}}{{3}}+\sqrt{{{1}}}\right]}$