Chapter 5 - Applications Of The Definite Integral In Geometry, Science, And Engineering - 5.4 Length Of A Plane Curve - Exercises Set 5.4 - Page 376: 26

$$L = \frac{{80\sqrt {10} - 13\sqrt {13} }}{{27}}$$

$$\eqalign{ & x = {\left( {1 + t} \right)^2},\,\,\,\,\,y = {\left( {1 + t} \right)^3},\,\,\,\,\left( {0 \leqslant t \leqslant 1} \right) \cr & {\text{Calculate the derivatives }}\frac{{dx}}{{dt}}{\text{ and }}\frac{{dy}}{{dt}} \cr & \frac{{dx}}{{dt}} = \frac{d}{{dt}}\left[ {{{\left( {1 + t} \right)}^2}} \right] = 2\left( {1 + t} \right) \cr & \frac{{dy}}{{dt}} = \frac{d}{{dt}}\left[ {{{\left( {1 + t} \right)}^3}} \right] = 3{\left( {1 + t} \right)^2} \cr & {\text{Use the arc length formula }}L = \int_a^b {\sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2}} dt} \cr & L = \int_0^1 {\sqrt {{{\left[ {2\left( {1 + t} \right)} \right]}^2} + {{\left[ {3{{\left( {1 + t} \right)}^2}} \right]}^2}} dt} \cr & L = \int_0^1 {\sqrt {4{{\left( {1 + t} \right)}^2} + 9{{\left( {1 + t} \right)}^4}} dt} \cr & L = \int_0^1 {\sqrt {{{\left( {1 + t} \right)}^2}\left[ {4 + 9{{\left( {1 + t} \right)}^2}} \right]} dt} \cr & L = \int_0^1 {\left( {1 + t} \right)\sqrt {4 + 9{{\left( {1 + t} \right)}^2}} dt} \cr & L = \frac{1}{{18}}\int_0^1 {18\left( {1 + t} \right)\sqrt {4 + 9{{\left( {1 + t} \right)}^2}} dt} \cr & {\text{Integrating}} \cr & L = \frac{1}{{18}}\left[ {\frac{{2{{\left[ {4 + 9{{\left( {1 + t} \right)}^2}} \right]}^{3/2}}}}{3}} \right]_0^1 \cr & L = \frac{1}{{27}}\left[ {{{\left( {4 + 9{{\left( {1 + t} \right)}^2}} \right)}^{3/2}}} \right]_0^1 \cr & L = \frac{1}{{27}}\left[ {{{\left( {4 + 9{{\left( {1 + 1} \right)}^2}} \right)}^{3/2}} - {{\left( {4 + 9{{\left( {1 + 0} \right)}^2}} \right)}^{3/2}}} \right] \cr & L = \frac{1}{{27}}\left[ {{{\left( {40} \right)}^{3/2}} - {{\left( {13} \right)}^{3/2}}} \right] \cr & L = \frac{{80\sqrt {10} - 13\sqrt {13} }}{{27}} \cr} $$