Chapter 11 - Vectors and the Geometry of Space - 11.1 Exercises - Page 756: 67

$$\eqalign{ & \left( {\text{a}} \right) \pm \frac{{\left\langle {1,6} \right\rangle }}{{\sqrt {37} }} \cr & \left( {\text{b}} \right) \pm \frac{{\left\langle { - 6,1} \right\rangle }}{{\sqrt {37} }} \cr} $$

$$\eqalign{ & f\left( x \right) = {x^2},{\text{ point }}\left( {3,9} \right) \cr & *{\text{First calculate the slope }}m{\text{ at the point }}\left( {3,9} \right) \cr & {\text{Differentiating}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {{x^2}} \right] \cr & f'\left( x \right) = 2x \cr & {\text{Evaluate the derivative at the given point }}\left( {3,9} \right) \cr & f'\left( 3 \right) = 2\left( 3 \right) \cr & f'\left( 3 \right) = 6 \cr & {\text{The slope at the point }}\left( {3,9} \right)\,{\text{is }}m = 6 \cr & \cr & \left( a \right){\text{The vector is parallel to the graph of }}f{\text{ then the slope }} \cr & {\text{is }}{m_2} = m \cr & {\text{Write the slope as a fraction}}{\text{, recall that }}m = \frac{{\Delta y}}{{\Delta x}},{\text{ so}} \cr & \frac{{\Delta y}}{{\Delta x}} = 6 \to \frac{{\Delta y}}{{\Delta x}} = \frac{6}{1} \cr & {\text{Let consider the vector }}{\bf{v}} = \left\langle {\Delta x,\Delta y} \right\rangle \cr & {\bf{v}} = \left\langle {1,6} \right\rangle \cr & {\text{Let the unit vector }}{{\bf{u}}_1} = \pm \frac{{\bf{v}}}{{\left\| {\bf{v}} \right\|}} \cr & {{\bf{u}}_1} = \pm \frac{{\left\langle {1,6} \right\rangle }}{{\left\| {\left\langle {1,6} \right\rangle } \right\|}} = \pm \frac{{\left\langle {1,6} \right\rangle }}{{\sqrt {1 + 36} }} = \pm \frac{{\left\langle {1,6} \right\rangle }}{{\sqrt {37} }} \cr & \cr & \left( b \right){\text{The vector is perpndicular to the graph of }}f{\text{ then }} \cr & {\text{the slope is the same}}{\text{. }}{m_2} = - \frac{1}{m} \cr & {\text{The slope of the vector is }}m = - \frac{1}{6} \cr & {\text{Write the slope as a fraction}}{\text{, recall that }}m = \frac{{\Delta y}}{{\Delta x}},{\text{ so}} \cr & \frac{{\Delta y}}{{\Delta x}} = - \frac{1}{6} \to \frac{{\Delta y}}{{\Delta x}} = \frac{1}{{ - 6}} \cr & {\text{Let consider the vector }}{\bf{v}} = \left\langle {\Delta x,\Delta y} \right\rangle \cr & {\bf{v}} = \left\langle { - 6,1} \right\rangle \cr & {\text{Let the unit vector }}{{\bf{u}}_2} = \pm \frac{{\bf{v}}}{{\left\| {\bf{v}} \right\|}} \cr & {{\bf{u}}_2} = \pm \frac{{\left\langle { - 6,1} \right\rangle }}{{\left\| {\left\langle { - 6,1} \right\rangle } \right\|}} = \pm \frac{{\left\langle { - 6,1} \right\rangle }}{{\sqrt {36 + 1} }} = \pm \frac{{\left\langle { - 6,1} \right\rangle }}{{\sqrt {37} }} \cr & \cr & {\text{Graph}} \cr} $$