Answer
(a) $||\nabla {f_P}|| = \sqrt {10} $
(b) the rate of change of $f$ in the direction $\nabla {f_P}$ is $||\nabla {f_P}|| = \sqrt {10} $.
(c) the rate of change of $f$ in the direction of a vector making an angle of $45^\circ $ with $\nabla {f_P}$ is $\sqrt 5 $.
Work Step by Step
(a) We have $f\left( {x,y} \right) = x{{\rm{e}}^{{x^2} - y}}$ and $P = \left( {1,1} \right)$.
The gradient of $f$ is $\nabla f = \left( {{{\rm{e}}^{{x^2} - y}} + 2{x^2}{{\rm{e}}^{{x^2} - y}}, - x{{\rm{e}}^{{x^2} - y}}} \right)$.
The gradient of $f$ at $P$ is $\nabla {f_P} = \left( {3, - 1} \right)$. So,
$||\nabla {f_P}|| = \sqrt {{3^2} + {{\left( { - 1} \right)}^2}} = \sqrt {10} $
(b) Since $\nabla {f_P}$ points in the direction of maximum rate of increase. The rate of change of $f$ in the direction $\nabla {f_P}$ is $||\nabla {f_P}|| = \sqrt {10} $.
(c) Using Theorem 4, the rate of change of $f$ in the direction of a unit vector making an angle $\theta $ with $\nabla {f_P}$ is
${D_{\bf{u}}}f\left( P \right) = ||\nabla {f_P}||\cos \theta $
${D_{\bf{u}}}f\left( P \right) = \sqrt {10} \cos 45^\circ = \sqrt 5 $
So, the rate of change of $f$ in the direction of a vector making an angle of $45^\circ $ with $\nabla {f_P}$ is $\sqrt 5 $.
