MAT 342E – Differential Geometry
HOMEWORK IV
Assigned Date: 21.03.2007
Due Date: 04.04.2007
QUESTION 1
Student Number: 040020337
e-mail: beycan@mail.com
Name: Beycan
Show that the curvature along the binormal indicatrix is  b2 
 2  2
2
SOLUTION




b  b

b '  b   .n








b ' '   .n   .n  ( .t   .b )  .t .n  2 .b

3
b '  
3

t
 
 b ' b ' '  0

n


b


0   3t   2b
     2


 b ' b ' '   6   2 4

 

b
 b2 

 b ' b ' '

b '
3
 2  2
2
 6   2 4
 6   2 4


3
6

Surname: Kahraman
Grade:
MAT 342E – Differential Geometry
HOMEWORK IV
Assigned Date: 21.03.2007
Due Date: 04.04.2007
QUESTION 2
Student Number: 040020337
e-mail: beycan@mail.com
Name: Beycan
Surname: Kahraman

 


Show that the unit binormal of the involute  *    (c  s)t of    (s) is


*
 .b   .t
b 
(c  s)  *
SOLUTION
 *  * ' * ' '


b   *  * so let’s find a way to calculate  * ' and  * ' ' .
 ' ' '





 
 *    (c  s)t
 * '    (c  s)t  t


 * '  (c  s) .n








 * ' '     (c  s) n   2 (c  s)t   (c  s)b
 * ' '   .n  (c  s ).n  (c  s) .n


t
0

n
(c  s)

b
0
  2 (c  s )
   (c  s)
 (c  s)

 * ' * ' ' 




 * ' * ' ' 
 (c  s)    (c  s)  
2
2 2
2
3 2


  (c  s) 2  2 .t  (c  s) 2  3 .b
  2 (c  s)4  4  (c  s)4  6
 * ' * ' '  (c  s) 4  4 ( 2   2 )  (c  s) 2  2  2   2


 *  (c  s) 2  2 .t  (c  s) 2  3 .b  .t   .b
b 

(c  s ) 2  2  2   2
 2  2
But we have already find that,
2 2
* 

 2   2  (c  s)  *
(c  s)
Therefore, inserting it in the equation we get:



 *  .t   .b
 .t   .b
.
b  2

*
   2 (c  s) 
Grade:
MAT 342E – Differential Geometry
HOMEWORK IV
Assigned Date: 21.03.2007
Due Date: 04.04.2007
QUESTION 3
Student Number: 040020337
e-mail: beycan@mail.com
Name: Beycan
Surname: Kahraman




Find the equation of the involute of the helix  (t )  a cos t.e1  a sin t.e2  bt.e3 , a  0 , b any constant.
SOLUTION

First of all, we have to check if  (t ) is in its
arc-length parametrized form:




 (t )  a cos t.e1  a sin t.e2  bt.e3




 ' (t )  a cos t.e1  a sin t.e2  bt.e3
t
t

s    ' (  ) d   a 2  b2 d  a 2  b2 t
0
t
0
s
a 2  b2





s
s
.e1  a sin  2
.e  b 2
.e3
2  2
2
a b
 a b 
 a b 

Now let’s calculate t ,





a
s
a
s
.e 
t ( s)   ( s)   2
sin  2
cos 2
2
2  1
2
2
2
a b
a b
 a b 
 a b

s
 ( s)  a cos
2
2


b
.e2 
.e3
2
2
a b


The involute of the  (s) is



where c is a constant, then:
 *    (c  s)t



s
.e1  a sin  2
2
 a b 
 a b

s
 *  a cos
2
2


a
s
(c  s )  
sin  2
2
2
2
a b
 a b



s
.e2  b 2
.e3 
2
a b



a
s
.e1 
cos 2
2
2
2
a b

 a b


b
.e2 
.e3 
2
2
a b



The involute of the  (s) is




 
.e1 

 

s
bc
 .e 
.e3
2
2  2
2
2
a  b 
a b
 a (c  s )

s
 
sin  2
2
2
2
a b
 a b 
 a b
 *  a cos
s
2
2


s
a sin  2
2
 a b

 a (c  s )

 
cos
2
2
a b


where a  0 , b, c are constants.
Grade:
MAT 342E – Differential Geometry
HOMEWORK IV
Assigned Date: 21.03.2007
Due Date: 04.04.2007
QUESTION 4
Student Number: 040020337
e-mail: beycan@mail.com
Name: Beycan
Surname: Kahraman
a) Show that the distance between corresponding points on two Bertrand curves is constant.
b) Show that the angle between corresponding tangent lines on two Bertrand curves is constant.
c) Prove that the product of the torsions of two Bertrand curves is constant.
SOLUTIONS





a)  &  * are Bertrand curves (mates), if  *    .n where  is constant.





 * (s * )   (s)  .n



 * (s * )   (s)  .n


 * ( s * )   ( s)  
 * (s0* )   (s0 )   constant.


 
b) t * and t are both orthagonal to n  n * .
 


d (t * .t ) d (t * )  d (t )  *  *  ds*   *

t 
t  t .t
 t .t
ds
ds
ds
ds
 
  ds* *  
 
d (t *.t ) ds* * * 

 n .t  n.t * 
 n.t  n*.t *  0
ds
ds
ds
 
d (t *.t )
0
ds
 
t *.t  c
where c is constant.
   
t *.t  t * . t .cos  c 
cos  c

 is constant.
Grade:



c)  * '  (1   )t   .b





 * ' '  .t  (1   )t  .b   .b





 * ' '  .t  (1   ) .n  .b   2 .n




 * ' '  .t  (   2   2 )n  .b







 * ' ' '  .t  .t  (  2  2)n  (   2   2 )n  .b  .b








 * ' ' '  .t   .n  (  2   )n  (   2   2 )(  .t   .b )  .b  .n






 * ' ' '  .t  (    2 )n  (   2   2 )(  .t   .b )  .b




 * ' ' '  ( 2   3   2  )t  (    2 )n  (   2   3  )b







t

b

n
0
 * ' * ' '  1  

    2   2 


 

 * ' * ' '  (  2 2  2 3 ).t  (    2).n  (   2   2 )(1   )b



 * ' * ' '  (  2 2  2 3 ).t  (2    2).n  (  2 2   2  2 3  2 2 )b
 * ' * ' '  (  2 2  2 3 ) 2  (2    2) 2  (  2 2   2  2 3  2 2 ) 2
2


    

  2   3   2  
    2
   2   3  
  
( * ' , * ' ' , * ' ' ' )  (1   )(   2   2 )(   2   3  )  (    2 ) 
 [ (    2 )  (   2   2 )(  2   3   2  )]
  
( * ' , * ' ' , * ' ' ' ) 
1  
  
( * ' , * ' ' , * ' ' ' )
 .   .  *  * 2  ???
 ' ' '
*
0
2
2