Remarks on Several Subjects. 137 
° jl ae ue . . 
u asin hliog ?, sin. y, whose differential, put 
=o gives, 2 y=tang. y. 
Problem 11. Ina Catenarian curve, in which a is the 
parameter, « the absciss, y the ordinate, and z the arch; 
a"; y=a. hyp. log. === 
included by the vee and the eae line joining its ex- 
tremities u=2(a+<c) y—2az. If we suppose this to be 
a maximum when 2 is given or v=2, we shall have, by put- 
hale 5 (a+2x) aa uz". 
we have d= , and the area 
ting «=z; a=z. 
1—t ike 
— hyp. log. 
Bh sy oa ee 
i , , and v=z, hence m=2, 
22 
aes dea ables ee 
eslig wk (= mie ~. hyp. log. i= ce hess 
differential put =o & reduced gives hyp. log.! Hi 21 -+4,) 
ee aes 
whence we may find ¢ and then a, a, z. ‘This may be re- 
duced to the same form as Professor Fisher's solution, by 
+t _+—, which gives i= = log. Gi) "= 
2t i! 
2 log. , and the cone expression may be put 
2s? a 
putting s= 
ie as in his solu- 
under the form 
tion. 
We shall add the following problem, not embraced in 
Professor Fisher’s rules. 
Problem. Suppose u=x?+ azy and v=2x'y, a being a’ 
given numerical coefficient, and let it be required to find 
the value of ua maximum, when v is a constant quantity. 
_ The value of u is not homogeneus i in x, y, but it may be 
made so by substituting Y? for y and then proceeding as 
above, or we may at once put y=", ee which means we 
get u=a {Cl at) J whence m=3, n=6, and T= 
a a Tat 2 whose differential put =o gives 
v 
oy whence y=2?t= > or cal ay: 
a 
Vou. VIII.—No. 1. 18 
