Answer
(a) $\nabla f = \left( {\frac{y}{{{x^2} + {y^2}}}, - \frac{x}{{{x^2} + {y^2}}}} \right)$
(b) ${D_{\bf{u}}}f\left( {1,1} \right) = 0$, ${\ \ \ }$ ${D_{\bf{u}}}f\left( {\sqrt 3 ,1} \right) = \frac{{\sqrt 2 }}{8} - \frac{{\sqrt 6 }}{8}$
(c) Setting ${\tan ^{ - 1}}\frac{x}{y} = c$, where $c$ are constants, we show that the lines $y = mx$ for $m \ne 0$ are level curves for $f$.
(d) We verify that $\nabla {f_P}$ is orthogonal to the level curve through $P$ for $P = \left( {x,y} \right) \ne \left( {0,0} \right)$.
Work Step by Step
(a) We have $f\left( {x,y} \right) = {\tan ^{ - 1}}\frac{x}{y}$ and ${\bf{u}} = \left( {\frac{{\sqrt 2 }}{2},\frac{{\sqrt 2 }}{2}} \right)$.
Using Theorem 2 of Section 7.8:
${f_x} = \frac{1}{{{{\left( {x/y} \right)}^2} + 1}}\cdot\left( {\frac{1}{y}} \right) = \frac{y}{{{x^2} + {y^2}}}$
${f_y} = \frac{1}{{{{\left( {x/y} \right)}^2} + 1}}\cdot\left( { - x{y^{ - 2}}} \right) = - \frac{x}{{{x^2} + {y^2}}}$
So, the gradient of $f$:
$\nabla f = \left( {{f_x},{f_y}} \right) = \left( {\frac{y}{{{x^2} + {y^2}}}, - \frac{x}{{{x^2} + {y^2}}}} \right)$
(b) Using Eq. (3) of Theorem 3, we calculate:
1. ${D_{\bf{u}}}f\left( {1,1} \right)$
${D_{\bf{u}}}f\left( {1,1} \right) = \nabla {f_{\left( {1,1} \right)}}\cdot\left( {\frac{{\sqrt 2 }}{2},\frac{{\sqrt 2 }}{2}} \right)$
${D_{\bf{u}}}f\left( {1,1} \right) = \left( {\frac{1}{2}, - \frac{1}{2}} \right)\cdot\left( {\frac{{\sqrt 2 }}{2},\frac{{\sqrt 2 }}{2}} \right) = 0$
2. ${D_{\bf{u}}}f\left( {\sqrt 3 ,1} \right)$
${D_{\bf{u}}}f\left( {\sqrt 3 ,1} \right) = \nabla {f_{\left( {\sqrt 3 ,1} \right)}}\cdot\left( {\frac{{\sqrt 2 }}{2},\frac{{\sqrt 2 }}{2}} \right)$
${D_{\bf{u}}}f\left( {\sqrt 3 ,1} \right) = \left( {\frac{1}{4}, - \frac{{\sqrt 3 }}{4}} \right)\cdot\left( {\frac{{\sqrt 2 }}{2},\frac{{\sqrt 2 }}{2}} \right) = \frac{{\sqrt 2 }}{8} - \frac{{\sqrt 6 }}{8}$
(c) The level curves for $f$ are obtained by setting ${\tan ^{ - 1}}\frac{x}{y} = c$, where $c$ are constants. So,
$\frac{x}{y} = \tan c$, ${\ \ \ }$ $y = \frac{x}{{\tan c}}$
Let $m = \frac{1}{{\tan c}} \ne 0$.
Hence, the lines $y = mx$ for $m \ne 0$ are level curves for $f$.
(d) From part (a), we obtain the gradient of $f$:
$\nabla f = \left( {{f_x},{f_y}} \right) = \left( {\frac{y}{{{x^2} + {y^2}}}, - \frac{x}{{{x^2} + {y^2}}}} \right)$
From part (b), we obtain the level curves for $f$ are lines $y = mx$ for $m \ne 0$. Setting $x \equiv t$, these lines can be parametrized by
${\bf{r}}\left( t \right) = t\left( {1,m} \right)$, ${\ \ \ }$ for $t>0$ and $m \ne 0$
Notice that
1. the level curves do not pass through the origin, that is, the points on the line are $P = \left( {x,y} \right) \ne \left( {0,0} \right)$.
2. the direction vectors of the lines are ${\bf{n}} = \left( {1,m} \right)$.
Now, we compute $\nabla f\cdot{\bf{n}}$
$\nabla f\cdot{\bf{n}} = \left( {\frac{y}{{{x^2} + {y^2}}}, - \frac{x}{{{x^2} + {y^2}}}} \right)\cdot\left( {1,m} \right)$
$ = \frac{y}{{{x^2} + {y^2}}} - \frac{{mx}}{{{x^2} + {y^2}}}$
Since $y = mx$, therefore $\nabla f\cdot{\bf{n}} = 0$. Hence, $\nabla {f_P}$ is orthogonal to the level curve through $P$ for $P = \left( {x,y} \right) \ne \left( {0,0} \right)$.
