Answer
(a) the slope of $f$ at this position is ${D_{\bf{u}}}f\left( {1,2} \right) = 2$
The corresponding angle of inclination is $\psi \simeq 63.43^\circ $
(b) the slope of $f$ at this position is ${D_{\bf{u}}}f\left( {1,2} \right) = 3$
the corresponding angle of inclination is $\psi \simeq 71.57^\circ $
(c) the slope of $f$ at this position is ${D_{\bf{u}}}f\left( {1,2} \right) = \frac{5}{2}\sqrt 2 $
The corresponding angle of inclination is $\psi \simeq 74.21^\circ $
(d) the steepest slope is ${D_{\bf{u}}}f\left( {1,2} \right) \simeq 3.6$
Work Step by Step
(a) We are given $z = f\left( {x,y} \right) = {x^2} + {y^2} - y$. So, the gradient is
$\nabla f = \left( {{f_x},{f_y}} \right) = \left( {2x,2y - 1} \right)$
At the point $\left( {1,2,3} \right)$, the gradient is $\nabla {f_{\left( {1,2} \right)}} = \left( {2,3} \right)$.
If we head due East, the unit vector in this direction is ${\bf{u}} = \left( {1,0} \right)$.
Using Eq. (3), the slope of $f$ at this position is given by
${D_{\bf{u}}}f\left( {1,2} \right) = \nabla {f_{\left( {1,2} \right)}}\cdot{\bf{u}} = \left( {2,3} \right)\cdot\left( {1,0} \right) = 2$
By Eq. (6), we have
$\tan \psi = {D_{\bf{u}}}f\left( {1,2} \right) = 2$
So, the corresponding angle of inclination is
$\psi = {\tan ^{ - 1}}2 \simeq 63.43^\circ $
(b) If we head due North, the unit vector in this direction is ${\bf{u}} = \left( {0,1} \right)$.
Using Eq. (3), the slope of $f$ at this position is given by
${D_{\bf{u}}}f\left( {1,2} \right) = \nabla {f_{\left( {1,2} \right)}}\cdot{\bf{u}} = \left( {2,3} \right)\cdot\left( {0,1} \right) = 3$
By Eq. (6), we have
$\tan \psi = {D_{\bf{u}}}f\left( {1,2} \right) = 3$
So, the corresponding angle of inclination is
$\psi = {\tan ^{ - 1}}3 \simeq 71.57^\circ $
(c) If we head due North-East, the unit vector in this direction is
${\bf{u}} = \left( {\cos 45^\circ ,\sin 45^\circ } \right) = \left( {\frac{1}{2}\sqrt 2 ,\frac{1}{2}\sqrt 2 } \right)$
Using Eq. (3), the slope of $f$ at this position is given by
${D_{\bf{u}}}f\left( {1,2} \right) = \nabla {f_{\left( {1,2} \right)}}\cdot{\bf{u}} = \left( {2,3} \right)\cdot\left( {\frac{1}{2}\sqrt 2 ,\frac{1}{2}\sqrt 2 } \right) = \frac{5}{2}\sqrt 2 $
By Eq. (6), we have
$\tan \psi = {D_{\bf{u}}}f\left( {1,2} \right) = \frac{5}{2}\sqrt 2 $
So, the corresponding angle of inclination is
$\psi = {\tan ^{ - 1}}3 \simeq 74.21^\circ $
(d) At the point $\left( {1,2,3} \right)$, $\nabla {f_{\left( {1,2} \right)}} = \left( {2,3} \right)$ points in the direction of maximum rate of increase. In such case the unit vector is
${\bf{u}} = \frac{{\nabla {f_{\left( {1,2} \right)}}}}{{||\nabla {f_{\left( {1,2} \right)}}||}} = \frac{{\left( {2,3} \right)}}{{||\left( {2,3} \right)||}} = \left( {\frac{2}{{\sqrt {13} }},\frac{3}{{\sqrt {13} }}} \right)$
Therefore, the steepest slope is
${D_{\bf{u}}}f\left( {1,2} \right) = \nabla {f_{\left( {1,2} \right)}}\cdot{\bf{u}}$
${D_{\bf{u}}}f\left( {1,2} \right) = \left( {2,3} \right)\cdot\left( {\frac{2}{{\sqrt {13} }},\frac{3}{{\sqrt {13} }}} \right) = \sqrt {13} \simeq 3.6$
