The Dissolution of an Isotropic Solid. 157 



In the figure the angle at O is a right angle, so that in this 

 case = 0. From the last equation we may obtain an 

 expression for the undissolved area, for we shall have 



ONL - Oi\I K - Sr + ^Y?" -</.") (11) 



when for 0, must be written, if as in the figure the angle 

 between the extreme normals be a right angle. As a 

 particular example of the foregoing investigation, suppose 

 that we have a cylinder of which a section normal to its 

 length is the parabola 



Y- = 4aX 

 From the two following equations 



along with the equation to the parabola, we obtain 

 (.r-X)-^(X + «) = «^2 



X( 2^- + X - xY ='ay- (12) 



From these equations we may deduce 



X- 2a /y^ + , 



{x - 2a)- 



3 y^ 3 9 



substituting this value of X in (12), we obtain the following 

 as the equation to the curve cutting the normals to a 

 parabola at a constant distance. 



{~V- 



^j-c^ (x - 2ay- } 

 3 9 I 



w^^^ 



(x - lay x — 2a\ '-' , ^ 

 9 3 J 



This equation when expanded so as to get rid of radical 

 forms, is of the sixth degree ; it will also include an external 

 branch, cutting the normals produced externally at a distance 

 c. If in (13), be written for c, it will be found to include 

 the parabola itself; for we then may write the equation 

 in the form 



(/ - ^ax)\f- + (x - of) = o. 



