FLUXIONS. 
y—brfzrx — xx, and the curve Is an ellipfis whofe femi- 
axis is AB. 
Ex. 15. To know whether a given curve line be con¬ 
cave or convex towards its axis, in any given point thereof. 
138. Let AB be the axis, BC any ordinate, take DB= 
BF, and draw the ordinates Dr, Fo, infinitely near BC ; 
draw en and Cg parallel to AB, 
and produce eC to k. Now by 
the nature of curvature, if the 
curve be concave towards the axis, 
in the point C, and the ordinates in- 
creafe, then the point 0 of the curve 
will fall between k and «■, and there¬ 
fore the increment go is lefs than 
«C, whence ok or go — gk, that is 
A D J5 f 
the radius of curvature is therefore -cc—, and 
x y x 
fuming the given quart- y ^ 
. aa an i.h \ 
tity —, 
2 
aax .. , 
or -— —yyz, and the 
fluent aaxzzy 2 z; but in 
A, x — z and y—b; 
therefore the fluent cor¬ 
rected is aax—aaz = 
_____ 
yy — bb X z, or aax — 
aa+yy—bb X V^ 2 -fy 2 > 
af- 
go — n C, will be negative ; but ok is the fecond moment 
of the ordinate, and that is as the fecond fluxion when 
F^ and De approach to and coincide with BC ; therefore 
if the curve be concave towards the axis, the fecond 
fluxion of the ordinate will be negative. And the con¬ 
trary will happen if the curve is convex towards the axis. 
Wherefore 
Let the axis AB=jr, ordinate B C—y, 
then compute the value of y by the 
nature of the curve, and fubftitute 
numbers for all the quantities if 
there be occafion, then if its value 
comes out negative it is concave in 
that point, if affirmative it is convex 
towards the axis. 
A D B F 
Ex. 16. A line being inflated into the form of a curve, 
and kept in that form by a force or preffiure afting perpen¬ 
dicularly upon every point of the curve ; to find the pro¬ 
portion between the tenfion of the curve, and the force 
adting perpendicularly upon it. 
1 39. Let AB, BC, be two equal particles of the curve, 
and let the fo.rce afting on the particle B reduce it to 
the pofition ABC; complete the pa- 
fff' ralleiogram ABCE, and draw AD, 
B /7 \ CD, perpendicular to AB, CB ; and 
C /x \ \ then 'h e f°nr points, A, B, C, D, 
will lie in a circle. Let AB or BC 
/ 
—z, AD or CD—r: now by mecha- 
nics the point B is added upon by 
three forces BA, BC, BE; BA or 
BC is the tenfion of the curve, and 
BE is the force adding perpendicu¬ 
larly again!! the curve, therefore thefe are to one ano¬ 
ther as BA to BE, or (by fimilar triangles) as ABD to z. 
Let now the points A, C, approach to B and coincide 
with it, and |BD becomes %r, the radius of curvature ; 
and z is the particle of the curve, whereon the prefTure 
adds : and therefore the force aiding perpendicularly on 
any particle of the curve is to the tenfion of the curve, 
as that particle of the curve is to the radius of curva¬ 
ture in that point. 
Cor. Therefore if the particle of the curve and its 
tenfion be given, the'force adting again!! that particle is 
reciprocally as the radius of curvature; that is, directly 
as the curvature. 
Ex. 17. EAF is a curve line fupporting a fluid ; to 
find the nature of the curve. 
14c. Let the axis of the figure CA —b, CT)=zx, 
DB=:jr, ABxzz, let z be given : by reafon of the 
fluidity of the water, the tenfion of the curve is 
equal in all points, and therefore by the foregoing 
example, the prefTure at B is reciprocally as the ra¬ 
dius of curvature in B. But by the laws of hydro- 
liatics the prefl'ure at B is as the height DB or y. and 
Vom VII. No. 445.. 
i-i 1 , • . a <* — bb 4 - yy X JV - , 
which reduced gives *= ■ ■ ■ y f or the 
\/2aabb + 2bb 2 4 
—d > 4 —2 aa^ ? 
equation of the curve fought. 
Cor. Draw EK perpendicular, and IK parallel to 
ED, and alfo the tangent El ; than if EK —p, El —q, 
sz=i area EAC, which is as its weight, then (by mecha¬ 
nics) p : q:: s : —— tenfion of the curve, whence (by 
P 
the foregoing example) fince yz= prefTure at B, it will 
be y. 
qs zy , , asx .. aa .. 
— : z : -4, and thence -— =rvz= — x, 
p x p 2 
and 
therefore —-—aa. 
P 
Prop. LXVI.— The nature of the refePling curve AMs, 
and the luminous point L, being given ; to find the focus F, or 
the concourfe of the neareft refcEled rays MF, «F. 
141. Take the particle of the curve M n infinitely 
fmall, and let C be the center and CM the radius of cur¬ 
vature of the arch Mu; and on ML, nl,, MF, wF, let 
fall the perpendiculars CE, Ce, CG, Cg ; alfo outlie 
centers L, F, deferibe the 
fmall arches Mr, no; then 
the little triangles M on, 
Mnr, are equal and fimilar, 
and Mo—nr. By the nature 
of reflection the angle LMC 
= CMF, and I.«C — C»F, 
whence CE=CG, and Cezzz 
Cg. Now if CE— Ce (that 
is, if L falls in E or e) then 
will CG—G^; that is, the 
point F will fall in G or ^ 
. when M and n coincide : but 
if Ce be lefs than CE, that 
is, if L falls below E, then will Cg be lefs than CG, 
and tlie interfedlion F will then fall above G towards 
M ; and the contrary. The triangles LEQ, LMr, and 
For, F'SG, are fimilar, and EQ=rCE— Ce— CG—Cg— SG ; 
and M r=no, and FG=MG-MF— ME—MF ; there¬ 
fore LM : LE :: (Mr : EQ :: no : SG :: FM : FG ::) 
LMxME 
FM : ME—MF. Whence MF: 
2LM-ME 
Ot’nerwife thus.—Draw the tangent MP, and the per¬ 
pendicular LP, and let LMtzry, LP —u, ME—<v, and 
yy y u y 
MC— -7, whence by fimilar triangles a>y — '—.— , or vn— 
u u 
uy, therefore MF=— ' 1 ’ —— 
2yu—uy 
1. Wherefore if CM be the radius of curvature. CE 
perpendicular to LM, and LP perpendicular to the tan¬ 
gent PM, and we make the diftance of the radiating 
point LM—y, ME=t', LP=?;; then compute the value 
6 P of 
