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\centerline
{
\bf Dr. Z's Math251 Handout \#14.5  (2nd ed.)
[The Gradient and Directional Derivatives]
}

By Doron Zeilberger

{\bf Problem Type 14.5a}: Find the directional derivative
of the function $f(x,y,z)$ at the
point $(x_0,y_0,z_0)$ in the direction $\langle v_1,v_2,v_3 \rangle$.


%%General Problem


{\bf Example Problem 14.5a}: Find the directional derivative
of the function $f(x,y,z)=\ln(x^2+y^2+z^2)$ at the
point $(2,1,3)$ in the direction $\langle 1,2,2 \rangle$.

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{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} Find the {\bf gradient} $\Nabla f=\langle f_x,f_y,f_z \rangle$
by taking all the first partial derivatives.
Also find the unit vector in the direction of 
$\langle v_1,v_2,v_3 \rangle$ by dividing by its length.
&
{\bf 1.} 
$$
f_x={{2x} \over {x^2+y^2+z^2}} \,\, , \,\,
f_y={{2y} \over {x^2+y^2+z^2}} \,\, , \,\,
f_z={{2z} \over {x^2+y^2+z^2}} \quad .
$$
So
$$
\Nabla f = 
\langle {{2x} \over {x^2+y^2+z^2}},
{{2y} \over {x^2+y^2+z^2}},
{{2z} \over {x^2+y^2+z^2}} \rangle
$$
$\vert \langle 1,2,2 \rangle \vert=\sqrt{1^2+2^2+2^2}=3$,
so
$$
{\bf u}={{1} \over {3}} \langle 1,2,2 \rangle=
\langle {{1} \over {3}},{{2} \over {3}},{{2} \over {3}} \rangle
\quad .
$$
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{\bf 2.} Plug-in $x=x_0,y=y_0,z=z_0$ into
$\Nabla f \,\,$.
&
{\bf 2.} 
$$
\Nabla f (2,1,3)=
\langle {{2 \cdot 2} \over {2^2+1^2+3^2}},
{{2 \cdot 1} \over {2^2+1^2+3^2}},
{{2 \cdot 3} \over {2^2+1^2+3^2}}\rangle
$$
$$
=\langle {{2} \over {7}},
{{1} \over {7}},
{{3} \over {7}} \rangle \quad .
$$
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{\bf 3.} Take the dot product $\Nabla f . {\bf u} \,\,$ .
&
{\bf 3.} 
$$
\langle {{2} \over {7}},{{1} \over {7}},{{3} \over {7}} \rangle 
.
\langle {{1} \over {3}},{{2} \over {3}},{{2} \over {3}} \rangle
$$
$$
={{2} \over {7}}\cdot {{1} \over {3}}+
{{1} \over {7}}\cdot {{2} \over {3}}+
{{3} \over {7}}\cdot {{2} \over {3}}=
 {{10} \over {21}}
$$
{\bf Ans.:} The requested directional derivative  is
${{10} \over {21}}$.
\cr}

{\bf Problem Type 14.5b}: 
Find the maximum rate
of change of $f$ at the given point and the direction
in which it occurs.
$$
f(x,y)=Expression(x,y) \quad,\quad (x_0,y_0) \quad .
$$

{\bf Example Problem 14.5b}: Find the maximum rate
of change of $f$ at the given point and the direction
in which it occurs.
$$
f(x,y)= \sin(xy) \quad , \quad (1,0).
$$
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{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} Find the gradient 
$$
{\Nabla f}= \langle f_x,f_y \rangle
$$

&
{\bf 1.} 
$f_x=y\cos(xy),f_y=x\cos(xy)$.
So
$$
{\Nabla f}= \langle f_x,f_y \rangle=
\langle y\cos(xy),x\cos(xy) \rangle \quad .
$$
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{\bf 2.} Plug-in $x=x_0,y=y_0$ into
${\Nabla f}$.

&
{\bf 2.} 
$$
{\Nabla f} (1,0)= 
\langle 0 \cdot \cos(0), 1 \cdot \cos(0) \rangle =
\langle 0, 1 \rangle \quad .
$$
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{\bf 3.} 
The maximum rate of change of $f$ is simply the length
of ${\Nabla f}$ at the designated point.
The direction in which is occurs is that direction.
So find the unit vector in that direction.
&
{\bf 3.} 
$$
\vert \langle 0, 1 \rangle \vert = \sqrt{0^2+1^2}=1 \quad.
$$
$\langle 0, 1 \rangle$ is already a unit vector, so
the direction is $\langle 0, 1 \rangle$.

{\bf Ans.}: The maximum rate of change is $1$ in the direction
$\langle 0, 1 \rangle$ (or ${\bf j}$ ).


\cr}




  
{\bf A Problem from a Previous Final}

Let
$$
f(x,y,z)= -x^2+y^2+z^2-1 \quad.
$$

{\bf (a)} (2 points) Compute $\g f$.
\bigskip
{\bf (b)} (5 points) Find a normal to the level surface $f(x,y,z)=0$ at the
point $(1,1,1)$, and give an equation for the tangent plane
to that surface at that point.
\bigskip
{\bf (c)} (6 points) Compute the directional derivative of $f(x,y,z)$
at the point $(1,1,1)$ in the direction 
$\langle \, 1 \, , \, 2\, , \, 2 \, \rangle$ .


{\bf Ans.}: a) $\langle -2x,2y,2z\rangle$ ; b) $z=x-y+1$; c) $2$.

\end


directinal
