Chapter 8 - Section 8.1 - Arc Length - 8.1 Exercises - Page 565: 1

$L = 4\sqrt 5 $

$$\eqalign{ & {\text{Let }}y = 3 - 2x,{\text{ }} - 1 \leqslant x \leqslant 3 \cr & {\text{Differentiating}} \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {3 - 2x} \right] \cr & \frac{{dy}}{{dx}} = - 2 \cr & {\text{Use the arc lenght formula }}\left( {{\text{see page 561}}} \right) \cr & L = \int_a^b {\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} dx} \cr & {\text{Substituting the limits and the derivative of the function}} \cr & L = \int_{ - 1}^3 {\sqrt {1 + {{\left( { - 2} \right)}^2}} dx} \cr & L = \int_{ - 1}^3 {\sqrt 5 dx} \cr & L = \sqrt 5 \int_{ - 1}^3 {dx} \cr & {\text{Integrating}} \cr & L = \sqrt 5 \left[ x \right]_{ - 1}^3 \cr & L = \sqrt 5 \left( {3 - \left( { - 1} \right)} \right) \cr & L = \sqrt 5 \left( {3 + 1} \right) \cr & L = 4\sqrt 5 \cr & \cr & {\text{Checking by the distance formula}} \cr & {\text{We have the points}} \cr & y\left( { - 1} \right) = 3 - 2\left( { - 1} \right) = 5{\text{ }} \to {\text{ }}\underbrace {\left( { - 1,5} \right)}_{\left( {{x_1},{y_1}} \right)} \cr & y\left( 3 \right) = 3 - 2\left( 3 \right) = - 3{\text{ }} \to {\text{ }}\underbrace {\left( {3, - 3} \right)}_{\left( {{x_2},{y_2}} \right)} \cr & {\text{The distance formula between two points }}\left( {{x_1},{y_1}} \right){\text{ and }}\left( {{x_2},{y_2}} \right) \cr & {\text{is: }}d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \cr & d = \sqrt {{{\left( {3 - \left( { - 1} \right)} \right)}^2} + {{\left( { - 3 - 5} \right)}^2}} \cr & d = \sqrt {{{\left( 4 \right)}^2} + {{\left( { - 8} \right)}^2}} \cr & d = \sqrt {80} \cr & d = \sqrt {\left( {{4^2}} \right)\left( 5 \right)} \cr & d = 4\sqrt 5 \cr} $$