Notes: Find the total length of the graph of the astroid `x^(2/3) + y^(2/3) = 4`

Tuesday, 7 June 2016

Find the total length of the graph of the astroid $\displaystyle{{x}}^{{\frac{{2}}{{3}}}}+{{y}}^{{\frac{{2}}{{3}}}}={4}$

We will use formula for arc length of curve $\displaystyle{y}={f{{\left({x}\right)}}}$ for $\displaystyle{a}\leq{x}\leq{b}.$

$\displaystyle{L}={\int_{{a}}^{{b}}}\sqrt{{{1}+{y}{'}^{{2}}}}{\left.{d}{x}\right.}$

Differentiating the given equation yields

$\displaystyle\frac{{2}}{{3}}{{x}}^{{-\frac{{1}}{{3}}}}+\frac{{2}}{{3}}{{y}}^{{-\frac{{1}}{{3}}}}{y}'={0}$

From this we get

$\displaystyle{y}'=-\frac{{{y}}^{{\frac{{1}}{{3}}}}}{{{x}}^{{\frac{{1}}{{3}}}}}$

Looking at the image below, we can see that the astroid is made up of four identical arches. Therefore, if we calculate length of one such arc we will know the total length of the graph.

Let us therefore, calculate the arc in the first quadrant. Lower bound...

We will use formula for arc length of curve $\displaystyle{y}={f{{\left({x}\right)}}}$ for $\displaystyle{a}\leq{x}\leq{b}.$

$\displaystyle{L}={\int_{{a}}^{{b}}}\sqrt{{{1}+{y}{'}^{{2}}}}{\left.{d}{x}\right.}$

Differentiating the given equation yields

$\displaystyle\frac{{2}}{{3}}{{x}}^{{-\frac{{1}}{{3}}}}+\frac{{2}}{{3}}{{y}}^{{-\frac{{1}}{{3}}}}{y}'={0}$

From this we get

$\displaystyle{y}'=-\frac{{{y}}^{{\frac{{1}}{{3}}}}}{{{x}}^{{\frac{{1}}{{3}}}}}$

Looking at the image below, we can see that the astroid is made up of four identical arches. Therefore, if we calculate length of one such arc we will know the total length of the graph.

Let us therefore, calculate the arc in the first quadrant. Lower bound of integration will obviously be 0, while the upper bound will be the point where the curve touches $\displaystyle{x}$ -axis (where $\displaystyle{y}={0}$ ).

$\displaystyle{{x}}^{{\frac{{2}}{{3}}}}={4}$

$\displaystyle{x}={{4}}^{{\frac{{3}}{{2}}}}$

$\displaystyle{x}={8}$

Hence, the quarter of the total length is

$\displaystyle\frac{{1}}{{4}}{L}={\int_{{0}}^{{8}}}\sqrt{{{1}+{{\left(-\frac{{{y}}^{{\frac{{1}}{{3}}}}}{{{x}}^{{\frac{{1}}{{3}}}}}\right)}}^{{2}}}}{\left.{d}{x}\right.}=$

$\displaystyle{\int_{{0}}^{{8}}}\sqrt{{{1}+\frac{{{y}}^{{\frac{{2}}{{3}}}}}{{{x}}^{{\frac{{2}}{{3}}}}}}}{\left.{d}{x}\right.}=$

$\displaystyle{\int_{{0}}^{{8}}}\sqrt{{\frac{{{{x}}^{{\frac{{2}}{{3}}}}+{{y}}^{{\frac{{2}}{{3}}}}}}{{{x}}^{{\frac{{2}}{{3}}}}}}}{\left.{d}{x}\right.}=$

Notice that the numerator is equal to the left side of the given equation which is equal to 4.

$\displaystyle{\int_{{0}}^{{8}}}\frac{{2}}{{{x}}^{{\frac{{1}}{{3}}}}}{\left.{d}{x}\right.}={\int_{{0}}^{{8}}}\sqrt{{\frac{{4}}{{{x}}^{{\frac{{2}}{{3}}}}}}}{\left.{d}{x}\right.}={2}{\int_{{0}}^{{8}}}{{x}}^{{-\frac{{1}}{{3}}}}{\left.{d}{x}\right.}={2}\cdot\frac{{3}}{{2}}{{x}}^{{\frac{{2}}{{3}}}}{{\mid}_{{0}}^{{8}}}=$