## Thursday, November 3, 2022

### 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(22  -  c2

Compute the mean curvature H if:

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

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

Solution:

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

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

h' =   c / Ö(22  -  c2   =   0.25 /Ö(0.52  -  (0.25)2

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

H = -½ (h' / [ 2(1 + h' 2)1/2 ] + [h' / [ 2(1 + h2)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:  (u1a cosh (x3 /a ) as the representation of the meridian of the catenoid.

Solution:

If (u1a cosh (x3 /a  ) represent the meridian of the catenoid then we can write;

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

ds2   =   (u )3/2  / (12  -  a (du1) 2/ a)  +  (u1)2(du2)3

Further:   1 + h' 2    =  (u 1)3/2  (12  -  a2

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

h4 (1)2 h2 [(12  +  c 2 ] - a2

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

ò  Ö{((12  +  c 2)/(12 (12 -  k2 } du1

Which yields, provided c = a:

x ( u1* ,  2* ) =  (2* cos u2* , 2* sin u2* , cu2)

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: