Solutions

1.
The directional derivative is given by Duf$\left(\vphantom{
x,y,z)}\right.$x, y, z)$\left.\vphantom{
x,y,z)}\right)$ = ∇f$\left(\vphantom{ x,y,z}\right.$x, y, z$\left.\vphantom{ x,y,z}\right)$u, where u is a unit vector in the direction of the outward normal to the sphere x2 + y2 + z2 = 9. The vector

$\displaystyle \left(\vphantom{ x^{2}+y^{2}+z^{2}}\right.$x2 + y2 + z2$\displaystyle \left.\vphantom{ x^{2}+y^{2}+z^{2}}\right)$ =  $\displaystyle \left[\vphantom{
\begin{array}{c}
2x \\
2y \\
2z
\end{array}
}\right.$$\displaystyle \begin{array}{c}
2x \\
2y \\
2z
\end{array}$$\displaystyle \left.\vphantom{
\begin{array}{c}
2x \\
2y \\
2z
\end{array}
}\right]$

is normal to the sphere x2 + y2 + z2 = 9, and at $\left(\vphantom{ 2,2,1}\right.$2, 2, 1$\left.\vphantom{ 2,2,1}\right)$ this normal is $\left(\vphantom{ 4,4,2}\right.$4, 4, 2$\left.\vphantom{ 4,4,2}\right)$. A unit vector in the same direction is given by

u = $\displaystyle \left[\vphantom{
\begin{array}{c}
4 \\
4 \\
2
\end{array}
}\right.$$\displaystyle \begin{array}{c}
4 \\
4 \\
2
\end{array}$$\displaystyle \left.\vphantom{
\begin{array}{c}
4 \\
4 \\
2
\end{array}
}\right]$÷$\displaystyle \left\Vert\vphantom{ \left[
\begin{array}{c}
4 \\
4 \\
2
\end{array}
\right] }\right.$$\displaystyle \left[\vphantom{
\begin{array}{c}
4 \\
4 \\
2
\end{array}
}\right.$$\displaystyle \begin{array}{c}
4 \\
4 \\
2
\end{array}$$\displaystyle \left.\vphantom{
\begin{array}{c}
4 \\
4 \\
2
\end{array}
}\right]$$\displaystyle \left.\vphantom{ \left[
\begin{array}{c}
4 \\
4 \\
2
\end{array}
\right] }\right\Vert$ =  $\displaystyle {\frac{{1}}{{6}}}$$\displaystyle \left[\vphantom{
\begin{array}{c}
4 \\
4 \\
2
\end{array}
}\right.$$\displaystyle \begin{array}{c}
4 \\
4 \\
2
\end{array}$$\displaystyle \left.\vphantom{
\begin{array}{c}
4 \\
4 \\
2
\end{array}
}\right]$ =  $\displaystyle \left[\vphantom{
\begin{array}{c}
2/3 \\
2/3 \\
1/3
\end{array}
}\right.$$\displaystyle \begin{array}{c}
2/3 \\
2/3 \\
1/3
\end{array}$$\displaystyle \left.\vphantom{
\begin{array}{c}
2/3 \\
2/3 \\
1/3
\end{array}
}\right]$

Hence, f$\left(\vphantom{ x,y,z}\right.$x, y, z$\left.\vphantom{ x,y,z}\right)$u evaluated at (2, 2, 1) is   - ${\frac{{2}}{{3}}}$.BITMAPSETProbSolvHint0.2006in0.243in0inq1

2.
A normal vector is given by

$\displaystyle \left(\vphantom{ \sqrt{x^{2}+y^{2}}+\left( x^{2}+y^{2}\right) ^{3/2}-z}\right.$$\displaystyle \sqrt{{x^{2}+y^{2}}}$ + $\displaystyle \left(\vphantom{ x^{2}+y^{2}}\right.$x2 + y2$\displaystyle \left.\vphantom{ x^{2}+y^{2}}\right)^{{3/2}}_{}$ - z$\displaystyle \left.\vphantom{ \sqrt{x^{2}+y^{2}}+\left( x^{2}+y^{2}\right) ^{3/2}-z}\right)$
                 = $\displaystyle \left(\vphantom{ \dfrac{1}{\sqrt{\left( x^{2}+y^{2}\right) }}x+3\...
...rt{\left( x^{2}+y^{2}\right) }}y+3\sqrt{\left( x^{2}+y^{2}\right) }y,-1}\right.$$\displaystyle {\dfrac{{1}}{{\sqrt{\left( x^{2}+y^{2}\right) }}}}$x + 3$\displaystyle \sqrt{{%
\left( x^{2}+y^{2}\right) }}$x,$\displaystyle {\dfrac{{1}}{{\sqrt{\left( x^{2}+y^{2}\right) }}}}$y + 3$\displaystyle \sqrt{{\left( x^{2}+y^{2}\right) }}$y, - 1$\displaystyle \left.\vphantom{ \dfrac{1}{\sqrt{\left( x^{2}+y^{2}\right) }}x+3\...
...rt{\left( x^{2}+y^{2}\right) }}y+3\sqrt{\left( x^{2}+y^{2}\right) }y,-1}\right)$
                 = $\displaystyle \left(\vphantom{ x\dfrac{1+3x^{2}+3y^{2}}{\sqrt{\left(
x^{2}+y^{...
...ght) }},y\dfrac{1+3x^{2}+3y^{2}}{\sqrt{\left(
x^{2}+y^{2}\right) }},-1}\right.$x$\displaystyle {\dfrac{{1+3x^{2}+3y^{2}}}{{\sqrt{\left(
x^{2}+y^{2}\right) }}}}$, y$\displaystyle {\dfrac{{1+3x^{2}+3y^{2}}}{{\sqrt{\left(
x^{2}+y^{2}\right) }}}}$, - 1$\displaystyle \left.\vphantom{ x\dfrac{1+3x^{2}+3y^{2}}{\sqrt{\left(
x^{2}+y^{...
...ght) }},y\dfrac{1+3x^{2}+3y^{2}}{\sqrt{\left(
x^{2}+y^{2}\right) }},-1}\right)$
                 = $\displaystyle {\dfrac{{1}}{{\sqrt{\left( x^{2}+y^{2}\right) }}}}$$\displaystyle \left(\vphantom{ x\left(
1+3x^{2}+3y^{2}\right) ,y\left( 1+3x^{2}+3y^{2}\right) ,-\sqrt{\left(
x^{2}+y^{2}\right) }}\right.$x$\displaystyle \left(\vphantom{
1+3x^{2}+3y^{2}}\right.$1 + 3x2 +3y2$\displaystyle \left.\vphantom{
1+3x^{2}+3y^{2}}\right)$, y$\displaystyle \left(\vphantom{
1+3x^{2}+3y^{2}}\right.$1 + 3x2 +3y2$\displaystyle \left.\vphantom{
1+3x^{2}+3y^{2}}\right)$, - $\displaystyle \sqrt{{\left( x^{2}+y^{2}\right) }}$$\displaystyle \left.\vphantom{ x\left(
1+3x^{2}+3y^{2}\right) ,y\left( 1+3x^{2}+3y^{2}\right) ,-\sqrt{\left(
x^{2}+y^{2}\right) }}\right)$

1.
Evaluate the expression $\left(\vphantom{ \dfrac{mM}{\sqrt{%
x^{2}+y^{2}+z^{2}}}}\right.$${\dfrac{{mM}}{{\sqrt{%
x^{2}+y^{2}+z^{2}}}}}$$\left.\vphantom{ \dfrac{mM}{\sqrt{%
x^{2}+y^{2}+z^{2}}}}\right)$. Then delete the first two rows of the vector, because m and M are constant parameters. BITMAPSETProbSolvHint0.2006in0.243in0inq3

$\displaystyle \left(\vphantom{ \dfrac{mM}{\sqrt{x^{2}+y^{2}+z^{2}}}}\right.$$\displaystyle {\dfrac{{mM}}{{\sqrt{x^{2}+y^{2}+z^{2}}}}}$$\displaystyle \left.\vphantom{ \dfrac{mM}{\sqrt{x^{2}+y^{2}+z^{2}}}}\right)$ 8pt
         = $\displaystyle \left(\vphantom{ -m\dfrac{M}{\left( \sqrt{\left( x^{2}+y^{2}+z^{2...
...frac{M}{\left( \sqrt{\left( x^{2}+y^{2}+z^{2}\right) }%
\right) ^{3}}z}\right.$ - m$\displaystyle {\dfrac{{M}}{{\left( \sqrt{\left( x^{2}+y^{2}+z^{2}\right) }%
\right) ^{3}}}}$x, - m$\displaystyle {\dfrac{{M}}{{\left( \sqrt{\left( x^{2}+y^{2}+z^{2}\right) }%
\right) ^{3}}}}$y, - m$\displaystyle {\dfrac{{M}}{{\left( \sqrt{\left( x^{2}+y^{2}+z^{2}\right) }%
\right) ^{3}}}}$z$\displaystyle \left.\vphantom{ -m\dfrac{M}{\left( \sqrt{\left( x^{2}+y^{2}+z^{2...
...frac{M}{\left( \sqrt{\left( x^{2}+y^{2}+z^{2}\right) }%
\right) ^{3}}z}\right)$ 8pt
         = -  $\displaystyle {\dfrac{{mM}}{{\left( x^{2}+y^{2}+z^{2}\right) ^{3/2}}}}$$\displaystyle \left(\vphantom{ x,y,z}\right.$x, y, z$\displaystyle \left.\vphantom{ x,y,z}\right)$

This gives the Newtonian gravitational force between two objects of masses m and M, with one object at the origin and the other at the point $\left(\vphantom{ x,y,z}\right.$x, y, z$\left.\vphantom{ x,y,z}\right)$.




Subsections