Answer
(a) the slope of $f$ at this position is ${D_{\bf{u}}}f\left( {2,1} \right) = 3$
The corresponding angle of inclination is $\psi \simeq 71.57^\circ $
(b) the slope of $f$ at this position is ${D_{\bf{u}}}f\left( {2,1} \right) = 4\sqrt 2 $
The corresponding angle of inclination is $\psi \simeq 79.98^\circ $
(c) the slope of $f$ at this position is ${D_{\bf{u}}}f\left( {2,1} \right) = - 4\sqrt 2 $
The corresponding angle of inclination is $\psi \simeq - 79.98^\circ $
(d) the steepest slope is ${D_u}f\left( {2,1} \right) \simeq 5.83$
The compass direction measured in degrees from East that we would head is $120.96^\circ $.
Work Step by Step
(a) We are given $z = f\left( {x,y} \right) = xy + {y^3} - {x^2}$. So, the gradient is
$\nabla f = \left( {{f_x},{f_y}} \right) = \left( {y - 2x,x + 3{y^2}} \right)$
At the point $\left( {2,1, - 1} \right)$, the gradient is $\nabla {f_{\left( {2,1} \right)}} = \left( { - 3,5} \right)$.
If we head due West, 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
${D_{\bf{u}}}f\left( {2,1} \right) = \nabla {f_{\left( {2,1} \right)}}\cdot{\bf{u}} = \left( { - 3,5} \right)\cdot\left( { - 1,0} \right) = 3$
By Eq. (6), we have
$\tan \psi = {D_{\bf{u}}}f\left( {2,1} \right) = 3$
So, the corresponding angle of inclination is
$\psi = {\tan ^{ - 1}}3 \simeq 71.57^\circ $
(b) If we head due North-West, the unit vector in this direction is
${\bf{u}} = \left( {\cos 135^\circ ,\sin 135^\circ } \right) = \left( { - \frac{1}{2}\sqrt 2 ,\frac{1}{2}\sqrt 2 } \right)$
Using Eq. (3), the slope of $f$ at this position is given by
${D_{\bf{u}}}f\left( {2,1} \right) = \nabla {f_{\left( {2,1} \right)}}\cdot{\bf{u}} = \left( { - 3,5} \right)\cdot\left( { - \frac{1}{2}\sqrt 2 ,\frac{1}{2}\sqrt 2 } \right) = 4\sqrt 2 $
By Eq. (6), we have
$\tan \psi = {D_{\bf{u}}}f\left( {2,1} \right) = 4\sqrt 2 $
So, the corresponding angle of inclination is
$\psi = {\tan ^{ - 1}}4\sqrt 2 \simeq 79.98^\circ $
(c) If we head due South-East, the unit vector in this direction is
${\bf{u}} = \left( {\cos \left( { - 45^\circ } \right),\sin \left( { - 45^\circ } \right)} \right) = \left( {\frac{1}{2}\sqrt 2 , - \frac{1}{2}\sqrt 2 } \right)$
Using Eq. (3), the slope of $f$ at this position is given by
${D_{\bf{u}}}f\left( {2,1} \right) = \nabla {f_{\left( {2,1} \right)}}\cdot{\bf{u}} = \left( { - 3,5} \right)\cdot\left( {\frac{1}{2}\sqrt 2 , - \frac{1}{2}\sqrt 2 } \right) = - 4\sqrt 2 $
By Eq. (6), we have
$\tan \psi = {D_{\bf{u}}}f\left( {2,1} \right) = - 4\sqrt 2 $
So, the corresponding angle of inclination is
$\psi = {\tan ^{ - 1}}\left( { - 4\sqrt 2 } \right) \simeq - 79.98^\circ $
(d) At the point $\left( {2,1, - 1} \right)$, $\nabla {f_{\left( {2,1} \right)}} = \left( { - 3,5} \right)$ points in the direction of maximum rate of increase. For maximum rate of increase, the unit vector is
${\bf{u}} = \frac{{\nabla {f_{\left( {2,1} \right)}}}}{{||\nabla {f_{\left( {2,1} \right)}}||}} = \frac{{\left( { - 3,5} \right)}}{{||\left( { - 3,5} \right)||}} = \left( {\frac{{ - 3}}{{\sqrt {34} }},\frac{5}{{\sqrt {34} }}} \right)$
Therefore, the steepest slope is
${D_u}f\left( {2,1} \right) = \nabla {f_{\left( {1,2} \right)}}\cdot{\bf{u}}$
${D_u}f\left( {2,1} \right) = \left( { - 3,5} \right)\cdot\left( {\frac{{ - 3}}{{\sqrt {34} }},\frac{5}{{\sqrt {34} }}} \right) = \sqrt {34} \simeq 5.83$
To realize this steepest slope, we need to head in the direction of ${\bf{u}} = \left( {\frac{{ - 3}}{{\sqrt {34} }},\frac{5}{{\sqrt {34} }}} \right)$.
Now, we find the angle of ${\bf{u}}$ measured from $x$-axis:
$\tan \theta = \frac{5}{{\sqrt {34} }}/\frac{{ - 3}}{{\sqrt {34} }} = - \frac{5}{3}$
Notice that the direction of ${\bf{u}}$ is in the second quadrant. So,
$\theta = \pi + {\tan ^{ - 1}}\left( { - \frac{5}{3}} \right) \simeq 2.11 \simeq 120.96^\circ $
Thus, the compass direction measured in degrees from East that we would head is $120.96^\circ $.
