Solutions To Differential Geometry Problems

 1) (a) Sketch the catenary: y = 3 cosh (x/2 -  ¼)

(b) If u²  is the parallel  u² =0.5  and h' =   c / Ö(u ²) ²  -  c² 

Compute the mean curvature H if:

H = -½ (h' / [ u ²(1 + h' ²)^1/2 ] + [h' / [ u ²(1 + h' ²)^3/2 ])

(c) If  K  = h = +1   for a positively curved minimal surface, find; k _u

Solution:

(a) 

(b) The form is:  f(x) = a cosh (bx - c)

Then from the graph:  a= 3, b =1/2,   c = ¼

h' =   c / Ö(u ²) ²  -  c²   =   0.25 /Ö(0.5) ²  -  (0.25)²

h' = 0.25 /(0.5)  -  (0.0625)=   0.571

  H = -½ (h' / [ u ²(1 + h' ²)^1/2 ] + [h' / [ u ²(1 + h' ²)^3/2 ])

= -½ (0.571 / [ 0.5 (1 + (0.571) ²)^1/2 ] + [0.571/ [ 0.5 (1 + (0.571) ²)^3/2 ])  =   -0.87  

c) K  + h / 2H  =  k _u

But: K  = h = +1   So:   k _u  =  2/ 2(-0.87)  =  -1.15

(2) Using a quantitative approach find a representation of the right helicoids which are isometric to a catenoid.

Hint: You may use:  (u¹) a cosh (x₃ /a ) as the representation of the meridian of the catenoid.

Solution:

If (u¹) a cosh (x₃ /a  ) represent the meridian of the catenoid then we can write;

x ( u¹ ,  u ² ) =  (u ² cos u² , u ² sin u² , a cosh^-1 (u²/ a)

ds²   =   (u ¹ )^3/2  / (u ¹) ²  -  a²  (du¹) ²/ a)  +  (u¹)²(du²)³

Further:   1 + h' ²    =  (u ¹)^3/2  / (u ¹) ²  -  a²

^{If  we use the appropriate substitution the left- hand side of the preceding eqn. becomes:}

^h4 (u ¹)² / ^h2 [(u ¹) ²  +  c ² ] - a² 

Finally integrate, setting  h = 1 and  a²  -   c ² =  k²        :

k ò  Ö{((u ¹) ²  +  c ²)/(u ¹) ² (u ¹) ² -  k² } du¹. 

Which yields, provided c = a:

x ( u^1* ,  u ^2* ) =  (u ^2* cos u^2* , u ^2* sin u^2* , cu^2* )

Where each asterisked quantity (*) denotes an arc length, e.g. on S* which is the same as that for the inverse image on the surface S. Hence, enabling a length -preserving or isometric mapping.   In this case, as the pitch increases (for c= a)  we obtain the right helicoid.  This shows that a catenoid, cut along a meridian, can be deformed into a right helicoid. See below: