274 PROFESSOR TAIT ON THE 



It is easy to see, from the forms of F(z/), and of its first two differential co- 

 efficients, that the equation 



h 2 smy)=K^{yJ^) 



can hold for, at most, one finite positive value of y. 



56. As a particular, and very instructive case, let us suppose 



the case of oxygen and hydrogen. 



(a) First, assume the diameters to be equal. Then the integral of (10), 

 with limits as in (11), taken on the supposition that the flow is constant, is 



A f, J8 ,^ 1 / T'/ /W-W(1) | FWW-1^(|)/(|) | 16F(|) 



As remarked above, the definite integral is essentially negative. For so is 

 every expression of the form 



a — b , Aa — Bb, B 



J I OCT 



A-B^(A-B)2 b A 

 provided A, B, a, and b be all positive. When A and B are equal its value is 



I have made a rough attempt at evaluation of the integral, partly by calcu- 

 lation, partly by a graphic method. My result is, at best, an approximation, 

 for the various instalments of the quadrature appear as the relatively small 

 differences of two considerable quantities. Thus the three decimal places, to 

 which, from want of leisure, I was obliged to confine myself, are not sufficient 

 to give a very exact value. The graphical representations of my numbers were, 

 however, so fairly smooth that there seems to be little risk of large error. 

 The full curve in the sketch below shows (on a ten-fold scale) the values of 

 the integrand (with their signs changed), as ordinates, to the values of y as 

 abscissa. The area is about — 2'165. Hence we have 



l ^r = — ?V 3 ' 463 ' 



at irs z Jh t 



(b) Suppose next that the diameter of a Pj is three times that of a P 2 , but 

 s 

 form 



the semi-sum of the diameters is s as before. The definite integral takes the 



