of a given Surface on another given Surface. 1 1 L 



The differential equation, 



m = (K 2 + 1) dt 2 + %}t*du* = 0, gives the two integrals, 



log t ± i*J R77 • u = const. 

 We have, accordingly, the solution 



X + *Y -f(logt + i s /~ r uy 



X-iY=f(\ogt + iJ 1 ^ l .u); 



that is to say,/ denoting an arbitrary function, X is to be the 



real part of/ (log t + i *J R tn, and Y the imaginary 



part, leaving out the factor u 



Let an exponential quantity be taken for/ or let/w = he v 

 where ^ is constant and e the base of the hyperbolical loga- 

 rithms, and the most simple representation will be 



X = ht. cos s/-KT+i- u - Y = ht.sm*J g~| .a. 

 The application of the formulae of article 7, gives in this case 



rc = (K 2 +l)* 2 N=l 

 and $ being = $ ! v = h /, 

 * (logZ + zV^ . «) . $' (log t-ij^.u) = tf « 



consequently m = ■ - , and therefore constant. If now, 



besides, h is made = \/(K 2 + 1), the representation becomes a 

 perfect development. 



10. Let it next be required; to represent in a plane the 

 surface of a sphere whose radius = a. We put here 



x = a . cos t . sin #, ^ = « . sin £ . sin u 9 

 Z = a . cos w, by which we obtain 

 co = dtsmu? dt 2 + a 2 du% The differential equation 

 co = gives consequently 



d 2 :p t . -^- = 0, and its integration 



t ± i log . cotang ,\u =; const. 

 If we denote therefore again by f an arbitrary function, 

 X is to be put equal to the real, and i Y to the imaginary part 

 of f{t + i log cotang \ u). We shall adduce some parti- 

 cular cases of this general solution. If we choose for / a 

 linear function by putting fv ;=( kv, we shall have X = kt 

 Y = k log cotang \ iu 



This 



