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\centerline
{
\bf Solutions to the ``QUIZ'' for Lecture 8
}
\bigskip

{\bf 1.} Find the directional derivative of the function
$f(x,y,z)=xy^2z^3$ at the point $(2,1,1)$ in the direction
$\langle 2,-1,-1 \rangle$.

{\bf Solution:} We first find the {\bf gradient} $\g f$.
$$
\g f = \langle \, f_x \, , \, f_y \, , \, f_z \, \rangle \quad.
$$
Since $f_x=y^2z^3$, $f_y=2xyz^3$, $f_z=3xy^2z^2$, we have
$$
\g f = \langle \, y^2z^3 \, , \, 2xyz^3 \, , \, 3xy^2z^2 \, \rangle \quad.
$$

{\bf Next} we plug-in the point $(2,1,1)$, i.e. plug-in $x=2,y=1,z=1$.
$$
\g f = \langle \, 1^2 \cdot 1^3 \, , \, 2 \cdot 2 \cdot 1 \cdot 1^3 \, , \, 3 \cdot 2 \cdot 1^2 \cdot 1^2 \, \rangle=
\langle \, 1 \, , \, 4 \, , \, 6 \, \rangle \quad .
$$

{\bf Next} we have to find the {\bf unit-vector} in the given direction  $\langle 2,-1,-1 \rangle$.
Take the magnitude:
$$
||\langle 2,-1,-1 \rangle||=\sqrt{2^2+(-1)^2+(-1)^2}=\sqrt{6} \quad .
$$
To get the unit vector ${\bf u}$  you divide by the magnitude:
$$
{{\langle 2,-1,-1 \rangle} \over {\sqrt{6}}}
$$
Finally, the {\bf directional derivative} is $\g f \cdot u$, giving
$$
\langle \, 1 \, , \, 4 \, , \, 6 \, \rangle \cdot {{\langle 2,-1,-1 \rangle} \over {\sqrt{6}}}=
{{1 \cdot 2 + 4 \cdot (-1) + 6 \cdot (-1)} \over {\sqrt{6}}}=
{{-8} \over {\sqrt{6}}}=-{{8 \sqrt{6}} \over {6}}=-{{4 \sqrt{6}} \over {3}} \quad .
$$

{\bf Ans.}: $-{{4 \sqrt{6}} \over {3}}$.



{\bf 2.} Find the maximum rate of change of $f(x,y)=x^2+y^3$ at
the point $(2,1)$ and the direction in which it occurs.

{\bf Sol.} We first find the gradient $\g f=\langle \, f_x \, , f_y \, \rangle $. Since
$f_x=2x$, $f_y=3y^2$, we have
$$
\g f= \langle \, 2x \, , \, 3y^2 \, \rangle \quad.
$$
Now plug-in $x=2, y=1$ getting:
$$
\g f= \langle \, 2 \cdot 2 \, , \, 3 \cdot 1 ^2 \, \rangle =\langle \, 4 \, , \, 3 \, \rangle  \quad .
$$
{\bf The maximum rate of change} is the {\bf magnitude}
$$
|| \g f ||= ||\langle \, 4 \, , \, 3 \, \rangle||=\sqrt{4^2+3^2}=\sqrt{25}=5 \quad .
$$
The direction in which it occurs is the {\bf unit vector} along the direction of the gradient:
$$
{{\g f} \over {||\g f||}}={{\langle \, 4 \, , \, 3 \, \rangle} \over {5}}= \langle \, {{4} \over {5}} \, , \, {{3} \over {5}} \rangle .
$$

{\bf Ans.}: The maximum rate of change is $5$ and it occurs in the direction $\langle \, {{4} \over {5}} \, , \, {{3} \over {5}} \, \rangle $.

{\bf Note}: If you wrote that it occurs in the direction $\langle \, 4 \, , \, 3 \, \rangle$ that's correct too,
but it is nice to give the {\bf unit direction}.

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