480 
ME. A. COHEN ON THE DIEEEEENTIAL COEFFICIENTS 
resultant couple is the complete sum of the axes of two component couples. Therefore, 
substituting for those axes the equivalent determinants, we see that the determinant of 
(P) + (P') to Q is the complete sum of the determinant of P to Q, and of the deter- 
minant of P' to Q. Thus we have 
det (P, Q)+det (F, Q)=det {(P)+(F), Q} (I.) 
And it may similarly be shown that 
det (P, Q)-det(F Q)=det {(P)-(F), Q} (II.) 
The same proposition follows also easily from a consideration of the linear expressions 
(II.) in section 24 for the projections of a determinant, and is in fact equivalent to a 
fundamental theorem concerning algebraic determinants, which theorem may be found 
in Salmon’s ‘ Lessons on Higher Algebra,’ section 19, page 9. 
27. The proposition proved in the last section will be found to be of constant use in 
explaining and shortening analytical processes in mechanics. One useful application can 
be made of it in proving “ the parallelogram of angular velocities.” For taking Q to he 
the radius vector of a particle, and P and P' to represent two axes and angular velocities 
of rotation then the formula (I.) of the last section translates itself at once by means of 
section 25 into the following proposition : — “ The linear velocity of a particle due to a 
rotation whose axis and angular velocity are represented by the line P, compounded 
■with the linear velocity due to a rotation similarly represented by the line P', is equiva- 
lent to the linear velocity due to a rotation represented by the complete sum of P and 
P'.” And this is evidently the same as the proposition called “ the parallelogram of 
angular velocities of rotation.” 
28. Let us next investigate the complete differential coefficient of det (P, Q). 
We will first premise that, if m be any numerical quantity, it follows evidently from 
the definition of a determinant, that 
det (mP, Q)=m det(P, Q)=det (P, mQ) (I.) 
Suppose now that P and Q after an interval of time At become respectively (P)-[-(AP), 
(Q)d-(AQ), the sign + here denoting the complete sum. Then the complete increment 
of det (P, Q) is 
det(P-f-AP, Q+AQ)— det (P, Q)= 
det (P+ AP, Q+ AQ)-det (P+AP, Q)-f det (P-f- AP, Q)-det (P, Q). 
But it follows from formula (II.) of section 26, that 
det (P+AP, Q+AQ)- det (P+AP, Q)= det (P+AP, AQ); 
and similarly, 
det (P+AP, Q)— det(P, Q)=det (AP, Q). 
Substitute, then, these values in the above equation (II.), and divide both sides by A^ by 
means of the above formula (I.), and finally let At diminish without limit. We thus 
obtain for the complete differential coefficient of det (P, Q) 
det (P, D,(Q))+det(D,(P), Q). 
