%ho145.tex: (2nd ed.) %%a Plain TeX file by Doron Zeilberger for Math 251(1 page) %\bigskip & \bigskip \cr %& %\cr %begin macros \def\g{\bigtriangledown} \def\L{{\cal L}} \baselineskip=14pt \parskip=10pt \def\Tilde{\char126\relax} \def\Nabla{\bigtriangledown} \font\eightrm=cmr8 \font\sixrm=cmr6 \font\eighttt=cmtt8 \magnification=\magstephalf \def\P{{\cal P}} \def\Q{{\cal Q}} \parindent=0pt \overfullrule=0in \def\frac#1#2{{#1 \over #2}} %end macros \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 directinal 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 directinal derivative of the function $f(x,y,z)=\ln(x^2+y^2+z^2)$ at the point $(2,1,3)$ in the direction $\langle 1,2,2 \rangle$. \bigskip \hrule \bigskip %\halign{#\hfill & \quad #\hfill \cr \halign{ \vtop{\hsize=15pc\noindent #\strut} \qquad & \vtop{\hsize=15pc\noindent #\strut} \cr {\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 $$ {\bf u}={{1} \over {3}} \langle 1,2,2 \rangle= \langle {{1} \over {3}},{{2} \over {3}},{{2} \over {3}} \rangle \quad . $$ \cr \bigskip & \bigskip \cr {\bf 2.} Plug-in $x=x_0,y=y_0,z=z_0$ into $\Nabla f \,\,$. & {\bf 2.} $$ \Nabla f (2,1,3)= \langle {{2 \cdot 2} \over {2^2+1^2+3^2}}, {{2 \cdot 1} \over {2^2+1^2+3^2}}, {{2 \cdot 3} \over {2^2+1^2+3^2}}\rangle $$ $$ =\langle {{2} \over {7}}, {{1} \over {7}}, {{3} \over {7}} \rangle \quad . $$ \cr \bigskip & \bigskip \cr {\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. $$ f(x,y)= \sin(xy) \quad , \quad (1,0). $$ \bigskip \hrule \bigskip %\halign{#\hfill & \quad #\hfill \cr \halign{ \vtop{\hsize=15pc\noindent #\strut} \qquad & \vtop{\hsize=15pc\noindent #\strut} \cr {\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 $$ {\Nabla f}= \langle f_x,f_y \rangle= \langle y\cos(xy),x\cos(xy) \rangle \quad . $$ \cr \bigskip & \bigskip \cr {\bf 2.} Plug-in $x=x_0,y=y_0$ into ${\Nabla f}$. & {\bf 2.} $$ {\Nabla f} (1,0)= \langle 0 \cdot \cos(0), 1 \cdot \cos(0) \rangle = \langle 0, 1 \rangle \quad . $$ \cr \bigskip & \bigskip \cr {\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