# ˆ ˆ ˆ ˆ ˆ ˆ ˆ 1 0 2 0 4 1 0 3 . 2 0 1 x y z r F x y z y = × = = +

```Lecture 12 – Appendix B: Some sample problems from Boas
Here are some solutions to the sample problems assigned for Chapter 6, sections 3 to
7.
&sect;6.3: 7

Solution: We want to consider the force F  2 xˆ  3 yˆ  zˆ acting at the point (1,5,2)
generating a torque in various ways. We first find the total torque and then take it’s
components along various directions.
a) The torque about the origin is given by
xˆ
  
  r F  1
yˆ
5
zˆ
2  xˆ  5  6   yˆ  4  1  zˆ  3  10   11xˆ  3 yˆ  13 zˆ.
2 3 1
b) The torque about the y-axis is given by using just the components in the x-z plane
as defined by
xˆ yˆ zˆ



 y axis  ry  axis  Fy axis  1 0 2  xˆ  0   yˆ  4  1  zˆ  0   3 yˆ.
2 0 1
This result is found most easily by taking the y component of the previous result,

 y  yˆ   3.
c) The torque about the line x 2  y 1  z  2  means we must take the component
of the torque about the unit vector parallel to this line, nˆ   2 xˆ  yˆ  2 zˆ 
we have

 2 xˆ  yˆ  2 zˆ  22  3  26
 n    nˆ  11xˆ  3 yˆ  13 zˆ   
 17.

3
3


Physics 227 Lecture 12 Appendix B
1
Autumn 2008
9 . Thus
&sect;6.3: 17
Solution: Consider the vector triple product (relevant to uniform motion in a circle)
   
a      r  . Using Eq. 8.4 we have
 
  
  
  
 

a       r      r        r     r     2 r .

  
Thus if the radius is perpendicular is the angular velocity, r   , r    0 , we have
   

a      r    2 r

a  ar rˆ, ar   2 r.
Thus the acceleration is toward the origin and, since the magnitude of the velocity is

v  v   r , we have ar  v 2 r
&sect;6.4: 5
Solution: We want to consider derivatives of a location vector defined by

ˆ . Thus the velocity and acceleration are given by
r  t   xˆ cos t  yˆ sin t  zt


v  t   r   xˆ sin t  yˆ cos t  zˆ,

v  t   sin 2 t  cos 2 t  1  2  constant,


a  t   
r   xˆ cos t  yˆ sin t ,

a  t   sin 2 t  cos 2 t  1  constant.
The motion is along a helix with constant angular velocity (in the x-y plane) and
constant longitudinal velocity (in the z direction).
Physics 227 Lecture 12 Appendix B
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Autumn 2008
&sect;6.4: 8
Solution: We want to consider general motion in spherical coordinates. We just take
derivatives and use Eq. 4.13 in Boas. We find

r  t   r  t  rˆ  t 

ˆ
 ˆ  rrˆ  rr
 ˆ  r
 r  rr

 ˆ  r
 ˆ  r
 ˆ
ˆ  2rr
 ˆ  rrˆ  
 
r  rr
rrˆ  2r
 ˆ  r
 ˆ  r 2 rˆ
 
rrˆ  2r

 

 
r  r 2 rˆ  2r  r ˆ.
&sect;6.6: 6
Solution: Here we practice some old ideas about normal vectors and tangent planes
with our new vector calculus technology. The normal to surface is given by the
gradient operating on the function (field) defining the surface and we find
 2
  x  y 2  z   2 xxˆ  2 yyˆ  zˆ


 N  3, 4, 25     x 2  y 2  z 
 3,4,25
 6 xˆ  8 yˆ  zˆ.
We can use this normal vector and the point of interest to find the (familiar) equations
of the tangent plane (the flat surface perpendicular to the normal vector at the point)
and the normal line (the line parallel to the normal vector through the point). We find
 

 r  r0   N  6  x  3  8  y  4    z  25   0
 6 x  8 y  z  25  tangent plane 

x  3 y  7 z  25


 normal line.
6
8
1
Physics 227 Lecture 12 Appendix B
3
Autumn 2008
&sect;6.6: 9
Solution: This exercise gives us more practice with the gradient operator.
a) We find the gradient to be


1,1,1
 2
   x  y2 z 
1,1,1
  2 xxˆ  2 yzyˆ  y 2 zˆ 
1,1,1
 2 xˆ  2 yˆ  zˆ.
b) The directional derivative is just the appropriate component of the gradient (at the
point using the appropriate unit vector) and we find


1,1,1

xˆ  2 yˆ  zˆ
xˆ  2 yˆ  zˆ 2  4  1 5
  2 xˆ  2 yˆ  zˆ  


.
1 4 1
6
6
6
c) Finally we find the normal line to be (recall the gradient gives the normal vector)

 
r  t   r0  Nt  1  2t  xˆ  1  2t  yˆ  1  t  zˆ
or
x 1 y 1 z 1


.
2
2
1
&sect;6.7: 4
Solution: Now we practice taking a divergence and a curl. We find
  
y z x
  V     yxˆ  zyˆ  xzˆ      0,
x y z
xˆ
  

  V     yxˆ  zyˆ  xzˆ  
x
y
yˆ
zˆ

y
z

z
x
 xˆ  0  1  yˆ  0  1  zˆ  0  1   xˆ  yˆ  zˆ.
Physics 227 Lecture 12 Appendix B
4
Autumn 2008
&sect;6.7: 14
Solution: Now we get to try evaluating the Laplacian to find

1
 
 x2  y2  z 2

2

  2
2
2
 2  2  2
  x
y
z

3 x2  y2  z 2 

x2  y2  z 2

 
5
111
x2  y2  z2

1
  2
  x  y2  z2

3




 0  x2  y 2  z 2  0 .


In this last expression the first terms comes from both derivatives operating on the
square root, while the second comes from one derivative on the square root and one
on the resulting factor in the numerator. This result is the very important result that,
away from the origin,
1
 2    0.
r
&sect;6.7: 19
Solution: Finally we consider the divergence of the radial unit vector and obtain the
useful result

 
ˆ
 r    



r

r
  

   xxˆ  yyˆ  zzˆ
ˆ
ˆ
ˆ
   x  y  z    2
y
z   x  y 2  z 2
  x

 
x
 
x  x 2  y 2  z 2

3


2
2
2
x y z

Physics 227 Lecture 12 Appendix B




  
 
y
z
 
 
 y  x 2  y 2  z 2  z  x2  y 2  z 2




x2  y2  z 2
2
2

 .
3
x2  y 2  z 2 r
x2  y2  z 2

5
Autumn 2008




```