Question

Find the exact length of the polar curve.

\({\rm{r}} = {{\rm{\theta }}^{\rm{2}}}{\rm{,0}} \le {\rm{\theta }} \le {\rm{2\pi }}\)

Short answer

The exact length of the polar curve is \(\frac{{\rm{8}}}{{\rm{3}}} \cdot \left( {{{\left( {{{\rm{\pi }}^{\rm{2}}}{\rm{ + 1}}} \right)}^{{\rm{3/2}}}}{\rm{ - 1}}} \right)\).

Step-by-step solution

Differentiate the polar curve

Given curve is\({\rm{r}} = {{\rm{\theta }}^{\rm{2}}}\)

Differentiating based on power rule,

\(\frac{{{\rm{dr}}}}{{{\rm{d\theta }}}} = {\rm{2}} \cdot {{\rm{\theta }}^{{\rm{2 - 1}}}}\)

\(\frac{{{\rm{dr}}}}{{{\rm{d\theta }}}} = {\rm{2\theta }}\)

The formula for the length of the curve is given by

\(L = \int_{{{\rm{\theta }}_{\rm{1}}}}^{{{\rm{\theta }}_{\rm{2}}}} {\sqrt {{{\rm{r}}^{\rm{2}}}{\rm{ + }}{{\left( {\frac{{{\rm{dr}}}}{{{\rm{d\theta }}}}} \right)}^{\rm{2}}}} } {\rm{d\theta }}\)

\(\begin{aligned}{c}L = \int_{\rm{0}}^{{\rm{2\pi }}} {\sqrt {{{\left( {{{\rm{\theta }}^{\rm{2}}}} \right)}^{\rm{2}}}{\rm{ + }}{{\left( {{\rm{2\theta }}} \right)}^{\rm{2}}}} } {\rm{d\theta }}\\ = \int_{\rm{0}}^{{\rm{2\pi }}} {\sqrt {{{\rm{\theta }}^{\rm{4}}}{\rm{ + 4}}{{\rm{\theta }}^{\rm{2}}}} } {\rm{d\theta }}\end{aligned}\)

Further simplify as follows:

\(\begin{aligned}{c}\sqrt {{{\rm{\theta }}^{\rm{4}}}{\rm{ + 4}}{{\rm{\theta }}^{\rm{2}}}} = \sqrt {{{\rm{\theta }}^{\rm{2}}}\left( {{{\rm{\theta }}^{\rm{2}}}{\rm{ + 4}}} \right)} \\ = \sqrt {{{\rm{\theta }}^{\rm{2}}}} \sqrt {{{\rm{\theta }}^{\rm{2}}}{\rm{ + 4}}} \\ = {\rm{\theta }}\sqrt {{{\rm{\theta }}^{\rm{2}}}{\rm{ + 4}}} \end{aligned}\)

Therefore, integration over \(\left( {{\rm{0,2\pi }}} \right){\rm{,\theta }}\)is positive and \(\sqrt {{{\rm{\theta }}^{\rm{2}}}} = \left| {\rm{\theta }} \right| = {\rm{\theta }}\)

\(\begin{aligned}{c}L = \int_{\rm{0}}^{{\rm{2\pi }}} {\rm{\theta }} \sqrt {{{\rm{\theta }}^{\rm{2}}}{\rm{ + 4}}} {\rm{d\theta }}\\ = \int_{\rm{0}}^{{\rm{2\pi }}} {\sqrt {{{\rm{\theta }}^{\rm{2}}}{\rm{ + 4}}} } \left( {{\rm{\theta d\theta }}} \right)\end{aligned}\)

Integrating over the limit.

Substitute \({{\rm{\theta }}^{\rm{2}}}{\rm{ + 4}} = {\rm{u}}\)into the aboce equation and simplify.

\(\begin{aligned}{c}L = \int_{\rm{4}}^{{\rm{4}}{{\rm{\pi }}^{\rm{2}}}{\rm{ + 4}}} {\sqrt {\rm{u}} } \left( {\frac{{\rm{1}}}{{\rm{2}}}{\rm{du}}} \right)\\ = \frac{{\rm{1}}}{{\rm{2}}}\int_{\rm{4}}^{{\rm{4}}{{\rm{\pi }}^{\rm{2}}}{\rm{ + 4}}} {{{\rm{u}}^{{\rm{1/2}}}}} {\rm{du}}\\ = \frac{{\rm{1}}}{{\rm{2}}}\left( {\frac{{{{\rm{u}}^{\left( {{\rm{1/2}}} \right){\rm{ + 1}}}}}}{{\left( {{\rm{1/2}}} \right){\rm{ + 1}}}}} \right)_{\rm{4}}^{{\rm{4}}{{\rm{\pi }}^{\rm{2}}}{\rm{ + 4}}}\\ = \frac{{\rm{1}}}{{\rm{2}}}\left( {\frac{{{{\rm{u}}^{{\rm{3/2}}}}}}{{{\rm{3/2}}}}} \right)_{\rm{4}}^{{\rm{4}}{{\rm{\pi }}^{\rm{2}}}{\rm{ + 4}}}\end{aligned}\)

Further simplify the above equation gives,

\(\begin{aligned}{c}L = \frac{{\rm{1}}}{{\rm{2}}}\left( {\frac{{\rm{2}}}{{\rm{3}}}{{\rm{u}}^{{\rm{3/2}}}}} \right)_{\rm{4}}^{{\rm{4}}{{\rm{\pi }}^{\rm{2}}}{\rm{ + 4}}}\\ = \frac{{\rm{1}}}{{\rm{2}}}\left( {\frac{{\rm{2}}}{{\rm{3}}} \cdot {{\left( {{\rm{4}}{{\rm{\pi }}^{\rm{2}}}{\rm{ + 4}}} \right)}^{{\rm{3/2}}}} - \frac{{\rm{2}}}{{\rm{3}}}{\rm{ \times }}{{\rm{4}}^{{\rm{3/2}}}}} \right)\frac{{\rm{1}}}{{\rm{3}}} \cdot {\left( {{{\rm{2}}^{\rm{2}}}} \right)^{{\rm{3/2}}}} \cdot {\left( {{{\rm{\pi }}^{\rm{2}}}{\rm{ + 1}}} \right)^{{\rm{3/2}}}}{\rm{ - }}\frac{{\rm{1}}}{{\rm{3}}}{\rm{ \times }}{\left( {{{\rm{2}}^{\rm{2}}}} \right)^{{\rm{3/2}}}}\\ = \frac{{\rm{1}}}{{\rm{3}}} \cdot {{\rm{2}}^{\rm{3}}}{\left( {{{\rm{\pi }}^{\rm{2}}}{\rm{ + 1}}} \right)^{{\rm{3/2}}}} - \frac{{\rm{1}}}{{\rm{3}}} \times {{\rm{2}}^{\rm{3}}}\\ = \frac{{\rm{8}}}{{\rm{3}}}\left( {{{\left( {{{\rm{\pi }}^{\rm{2}}}{\rm{ + 1}}} \right)}^{{\rm{3/2}}}} - {\rm{1}}} \right)\end{aligned}\)

Therefore, the exact length of the polar curve is \(\frac{{\rm{8}}}{{\rm{3}}} \cdot \left( {{{\left( {{{\rm{\pi }}^{\rm{2}}}{\rm{ + 1}}} \right)}^{{\rm{3/2}}}}{\rm{ - 1}}} \right)\)