<48v>

☞ Note yt if there happen to bee in any equation either a \fraction or/ surde quantity or a Mechanichall one, (i:e: wch cannot bee Geometrically computed, but is expressed by ye {illeg} \area/ or length or gravity or content of some curve line or sollid, &c) To find in what proportion they \unknowne quantitys/ increase or decrease doe thus. |1| Take two letters ye one (as ξ) to signify yt quantity, ye other (as χ π) its motion of increase or decrease: And making an equation betwixt yt|ye| letter (ξ) & ye quantity signifyed by it, find thereby (by prop 7 if ye Equation \quantity/ bee Geometricall, or by some other meanes if it bee mechanicall) ye valor of ye other letter (π). |2| Then substituting ye letter (ξ) signifying yt quantity, into its place in ye maine Equation esteeme yt letter (ξ) as an unknowne quantity & performe ye worke of seaventh proposition; & into ye resulting Equation instead of those letters ξ & π substitute theire valors. And soe you have ye Equation required.

Example 1. To find p & q ye motions of x & y whose relation is, $yy=x\sqrt{aa-xx}$. first suppose $\xi =\sqrt{aa-xx}$, Or $\xi \xi +xx-aa=0$. & thereby find π ye motion of ξ, viz: (by prop 7) $2\pi \xi +2px=0$. Or $\frac{-px}{\xi }=\pi =\frac{-px}{\sqrt{aa-xx}}$. Secondly in ye Equation $yy=x\sqrt{aa-xx}$, writing ξ in stead of $\sqrt{aa-xx}$, the result is $yy=x\xi$, whereby find ye relation of ye motions p, q, & π: viz (by prop 7) $2qy=p\xi +x\pi$ . In wch Equation instead of ξ & π writing theire valors, ye result is, $2qy=p\sqrt{aa-xx}\frac{-pxx}{\sqrt{aa-xx}}$. wch was required \to/.

[wch equation multiplyed by $\sqrt{aa-xx}$, is $2qy\sqrt{aa-xx}=paa-2pxx$. & in stead of $\sqrt{aa-xx}$, writing its valor $\frac{yy}{x}$, it is $\frac{2{qy}^{3}}{x}=paa-2pxx$. Or $2{qy}^{3}=paax-2pxxx$. Which {sic} Which conclusion will also bee found by taking ye surde quantity out ye given Equation for both parts being squared it is ${y}^{4}=aaxx-{x}^{4}$. & therefore (by prop 7) $4{py}^{3}=2qaax-4{x}^{3}$, as before.]

☞ Note also yt it may bee more convenient $$setting all ye termes on one side of ye Equation)/ to put {illeg} every fractionall, irrationall & mechanicall terme, as also ye summe of ye rationall termes, equall severally to some letter: & then to find ye motions corresponding to each letter of those letters ye sume of wch motions is ye Equation required. Example ye 2d. If ${x}^{3}-ayy+\frac{{by}^{3}}{a+y}-xx\sqrt{ay+xx}=0$ is ye relation twixt x & y, whose motions p & q are required. I make ${x}^{3}-ayy=\tau$; $\frac{{by}^{3}}{a+y}=\phi$; & $-xx\sqrt{ay+xx}=\xi$. & ye motions of τ, φ, & ξ being called β, γ, & δ; ye first Equation {illeg} ${x}^{3}-ayy=\tau$, gives (by prop 7) $3pxx-2qay=\beta$. ye second ${by}^{3}=a\phi +y\phi$, gives $3qbyy=a\gamma +y\gamma +q\gamma$; Or $\frac{3qbyy-q\phi }{a+y}=\gamma =\frac{3qabyy+2qb{y}^{3}}{aa+2ay+yy}$. & ye Third $ay{x}^{4}+{x}^{6}=\xi \xi$, gives, ${qax}^{4}+4{payx}^{3}+6{px}^{5}=2\delta \xi$; Or $\frac{-qaxx-4payx-6{px}^{3}}{2\sqrt{ay+xx}}=\delta$. Lastly $\beta +\gamma +\delta =3pxx-2qay\frac{+3qabyy+2{qby}^{3}}{aa+2ay+yy}\phantom{\rule{10px}{0ex}}\frac{-qaxx-4payx-6{px}^{3}}{2\sqrt{ay+xx}}=0$, is ye Equation sought. Example 3d. If $x=ab\perp bc=\sqrt{ax-xx}$. be=y. & ye superficies abc=z suppose yt ax+xz−y3 $zz+axz-{y}^{4}=0$, is ye relation twixt x, y & z, whose motions are p, q, & r: & yt p & q are desired. The Equation $zz+axz-{y}^{4}=0$ gives (by prop 7), $2rz+rax+paz-4{qy}^{3}=0$. Now drawing dh∥ab⊥ad=1−bh. I consider ye superficies abhd=ab×bh=x×1=x, & abd=z doe increase in ye proportion of bh to bc: yt is, 1∶$\sqrt{ax-xx}$ ∷p∶r. Or $r=p\sqrt{ax-xx}$. Which valor of r being substituted into ye Equation $2rz+rax+paz-4{qy}^{3}=0$, gives $\overline{2pz+pax}×\sqrt{ax-xx}+paz-4{qy}^{3}=0$. wch was required. How to proceede in other cases (as when there are cube rootes, surde denominators, rootes within rootes (as $\sqrt{ax+\sqrt{aa-xx}}$ &c: in the equation) may bee easily bee deduced from what {ha}th bee{n} already said. <49r> October 1666. To resolve Problems by Motion these following Propositions are sufficient 1 If the body a in the Perimeter of ye cirkle or sphære adc moveth towards its center b, its velocity to each point (b, c, d /d, c, e,$$ of yt circumference is as ye chords (ad, ac, ae) drawne from that body to those points are.

2 If ye △s adc, aec, are alike \viz ad = ec &c/ (though in divers plaines) & 3 bodys move from the point a uniformely & in equall times ye first to d, the 2d to e, ye 3d to c; Then is the thirds motion compounded of ye \motion of the/ firsts & second.s motion.

3. All the points of a Body keeping Parallel to it selfe are in equall motion \velocity/.

4. If a body move onely circularly | angularly about some axis, ye motion \velocity/ of its points are as their distance from that axis.

5. The motions of all bodys are either parallel or angular, \Call these two simple motions, parallel & angular/ or mixed of ym both, after ye same manner yt the motion towards c (Prop 2) is compounded of that|os|e towards d & e. And in mixed motion any line may bee taken for ye axis (or if a line or superficies move in plano, any point in yt plane may bee taken for the center) of ye angular motion

6 If ye lines ae, ah being moved doe continually intersect; I describe ye trapezium abcd, & its diagonall ac: & say yt, ye proportion & position of these five lines ab, ad, ac, cb, cd, being determined by requisite data; shall designe ye proportion & position of these five motions; viz: of ye point a fixed in ye line ae & moveing towards b, of ye point a fixed in ye line ah & moveing towards d; of ye point a intersection point a moveing in ye plaine abcd towards c, (for those five lines are ever in ye same plaine, though ae & ah may chanch onely to touch that plaine in their intersection point a); of ye intersection point a moveing in ye line ae parallely to cb & according to ye order of ye letters c, b; & of ye intersection point a move{ing} in ye line ah parallely to cd & according to ye order of the {lett}ers c, d.

Note yt one of ye lines as ah (fig 3d & 4th) resting, ye points d & a are coincident, & ye point c shall bee in ye line ah if it bee streight (fig 3), otherwise in its tangent (fig 4th)

7. Haveing an equation expressing ye relation twixt two or more lines x, y, z &c: described in ye same time by two or more moveing bodys A, B, C, &c. the relation of their velocitys may bee thus found p, q, r, &c may bee thus found, viz: Set all ye termes on one side of ye Equation that they become equall to nothing. And first multiply each terme by so many times $\frac{p}{x}$ as x hath dimensions in yt terme. Secondly multiply each terme by so many times $\frac{q}{y}$ as y hath dimensions in it. Thirdly (if there be 3 unknowne quantitys) multiply each terme by so many times $\frac{r}{z}$ as z hath dimensions in yt terme. (& if there bee still more unknowne quantitys doe like to every unknowne quantity). The summe of all these products shall bee equall to nothing. wch Equation gives ye relation of ye velocitys p, q, r, &c Or thus. Translate all ye termes to one side of ye equation, & multiply them being ordered according to x by this progression, {illeg} $\frac{3p}{x}.\frac{2p}{x}.\frac{p}{x}.0.\frac{-p}{x}.\frac{-2p}{x}.\frac{-3p}{x}.\frac{-4p}{x}.$ &c. or being ordered by ye dimensions of y multiply them by this,: $\frac{3q}{y}.\frac{2q}{y}.\frac{q}{y}.0.\frac{-q}{y}.\frac{-2q}{y}.$ &c. The sume of these products shall bee equall to nothing, which equation gives ye relation of their velocitys p, q, &c.

<49v>

Or more Gera Generally ye Equation may bee multiplyed by ye terme of these progressions $\frac{ap+4bp}{x}.\frac{ap+3bp}{x}.\frac{ap+2bp}{x}.\frac{ap+bp}{x}.\frac{ap}{x}.\frac{ap-bp}{x}.\frac{ap-2bp}{x}$ &c. And $\frac{aq+2bq}{y}.\frac{aq+bq}{y}.\frac{aq}{y}.\frac{aq-bq}{y}$ &c. (a & b signifying any two numbers whither rationall or irration\all/

8. If two Bodys A & B, by their velocitys p & q describe ye lines x & y. & an Equation bee given expressing ye relation twixt one of ye lines x, & y ratio $\frac{q}{p}$ of their motions q & p; To find ye other line y.

Could this bee ever done all problems whatever might bee resolved. But by ye following rules it may bee very often done. |(Note yt ±m & ±n are \logarithmes or/ numbers signifying ye dimensions of x.)|

ffirst get yt valor of $\frac{q}{p}$. Which if it bee rationall & its Denominator consist of but one terme: Multiply yt valor by x & divide each terme of it by ye logarithme of x in yt terme \ye quote shall bee ye valor of y/. As if ${ax}^{\frac{m}{n}}=\frac{q}{r}$ . Then is $\frac{na}{m+n}{x}^{\frac{m+n}{n}}=y$. Or if ${ax}^{\frac{m}{n}}=\frac{q}{p}$. Then is $\frac{na}{n+m}{x}^{\frac{n+m}{n}}=y$. (Soe if $\frac{a}{x}={ax}^{\frac{\overline{1}}{1}}=\frac{q}{p}$. Then is $\frac{a}{0}{x}^{0}=y$. soe yt y is infinite. But note yt in this case x & y increase in ye same proportio\n/ yt {illeg} numb{illeg}|er|s & thi|thei|r logarithmes doe, y being like a logarithme added to an infinite number $\frac{a}{0}$. But if x bee diminished \by c/, as if $\frac{a}{c+x}$=$\frac{q}{p}$, y is also diminished by ye infinite number $\frac{a}{0}{c}^{0}$ & becomes finite like a logarithme of ye number x. & so x being given, y may bee mechanichally found by a Table of logarithmes, as shall bee hereafter showne.)

Secondly. But if ye denominator of ye valor of $\frac{q}{p}$ consist of more termes yn one, it may bee reduced to such a forme yt ye denominator of each ꝑte \of it/ shall have but one terme, unlesse yt ꝑte bee $\frac{a}{c+x}$: Soe yt y may bee yn found by ye precedent rule. Which reduction is thus performed, viz: 1st, If neither ye numerrator nor \neither ye numerator nor/ denominator bee \the denominator bee not a+bx, nor all the \its/ termes of/ \its termes not/ multiplyed in all their termes by x \or xx, or x3, &c/; Increase or diminish x untill ye last terme of ye Denominator vanish. 2dly, And when all ye termes in ye Denominator are multiplyed by x, xx, or x3 &c: Divide ye numerator by ye Denominr (as in Decimall numbers) untill ye Quotient consist of such ꝑts none of wch whose Denominators are so multiplyed by x, x2 & begin ye Division in those termes in wch x is of its fewest dimensions \unlesse ye Denominator be a+bx /. If yn ye termes in ye Denominator \valor/ of $\frac{q}{p}$ bee such as was before required ye valor of y may bee found by ye first ꝑte of this Prop: onely it must bee so much diminished or increased as it was before increased or diminished by increasing or diminishing x. But if t{illeg}y \the/ denominator \of any terme/ consist of two \more/ termes yn one, in some of wch x is of more yn {illeg} \one/ dimension \unlesse yt terme bee $\frac{a}{c+x}$ ./ First find those ꝑts of y's valor wch correspond to ye other ꝑts of $\frac{q}{p}$ its valor. & yn seeke yt of y's valor belon {illeg} by ye preceding reductions \&c/: seeke ye ꝑte of y's valor answering to this ꝑte of $\frac{q}{p}$ its valor.

Example 1st. If $\frac{xx}{ax+b}=\frac{q}{p}$. Then by Division tis $\frac{x}{a}-\frac{b}{aa}\frac{+bb}{{a}^{3}x+aab}=\frac{xx}{ax+b}=\frac{q}{p}$. (as may appeare by multiplication.) Therefore (by 1st ꝑte of this Prop:) tis $\frac{xx}{2a}\frac{-bx}{aa}+⃞\frac{bb}{{a}^{3}x+aab}=y$. ($⃞\frac{bb}{{a}^{3}x+aab}$ signifys yt ꝑte of ye valor of y wch is correspondent to ye terme $\frac{bb}{{a}^{3}x+aab}$ of ye valor of $\frac{q}{p}$, wch will may bee found by a Table of logarithmes as may hereafter appeare.)

Example 2d. If $\frac{{b}^{3}}{-aa+xx}=\frac{q}{p}$. I suppose x=a+c, & consequently $\frac{{b}^{3}}{2az+zz}=\frac{q}{p}$. <50r> =$\frac{q}{p}$ & by Division, $\frac{{b}^{3}}{2az}\frac{-{b}^{3}}{4aa+2az}=\frac{q}{p}$.

Example 2d. If $\frac{{x}^{3}}{aa-xx}=\frac{q}{p}$. I suppose x=z−a. Or $\frac{{z}^{3}-3azz+3aaz-{a}^{3}}{2az-zz}=\frac{q}{p}$. And by Division $\frac{-aa}{2z}-z+a\frac{+aa}{4a-2z}=\frac{q}{p}$, (as may appeare by multiplication.) {It} consequently by ye 1st ꝑte of ye Prop.) $az-\frac{zz}{2}-\square \frac{aa}{2z}+\square \frac{aa}{4a-2z}-aa+\frac{aa}{2}+\square \frac{aa}{2a}-\square \frac{aa}{4a-2a}$ $=y=az-\frac{aa}{2}-\frac{zz}{2}-\square \frac{aa}{2x+2a}+\square \frac{aa}{2a-2x}=\frac{-xx}{2}+\square \frac{+aa}{2a-2x}+\square \frac{-aa}{2a+2x}=y$. And substituteing x+a into ye place of z, tis $x\frac{-aa}{2x+2a}\frac{+aa}{2a-2x}=\frac{q}{p}=\frac{{x}^{3}}{aa-xx}$ . And Therefore (by ꝑte 1st of Prop 8) $\frac{xx}{2}+\square \frac{-aa}{2x+2a}+\square \frac{aa}{2a-2x}=y$.

But sometimes The last terme of ye Denominator cannot bee taken away, (as if ye Denominr bee aa+xx. or a4+x4 or a4+bbxx+x4 &c) And then it will bee necessary to have in readinesse some examples wth such Denominators to wch all other cases of like denomination may bee by Division reduced. As if $\frac{cx}{a+bxx}=\frac{q}{p}$ . Make bxx=z, Then is $\square \frac{c}{2ab+2bz}=y$.

$\frac{cxx}{a+b{x}^{3}}=\frac{q}{p}$. Make bx3=z, Then is $\square \frac{c}{3ba+3bz}=y$.

$\frac{c{x}^{3}}{a+b{x}^{4}}=\frac{q}{p}$. Make bx4=z, Then is $\square \frac{c}{4ba+4bz}=y$. &c. In Generall if $\frac{c{x}^{n-1}}{a+b{x}^{n}}=\frac{q}{p}$. Make bxn=z, & yn is $\square \frac{c}{nba+nbz}=y$. Also if $\frac{c}{a+bxx}=\frac{q}{p}$. Make $\sqrt{\frac{c}{a}}-\sqrt{\frac{c}{a+bxx}}=z$ & yn is $\frac{cx}{a+bxx}+\square 2\sqrt{\frac{2z\sqrt{ac}}{b}-\frac{azz}{b}}=y$. That is, if $\sqrt{\frac{c}{bx}-\frac{a}{b}}=\frac{q}{p}$; I make x=zz, & $\square 2\sqrt{\frac{c}{b}-\frac{a}{b}zz}=y$. Or if $\frac{c}{a+bxx}=\frac{q}{p}$. Make $\sqrt{\frac{c}{a+bxx}}=z=CB$, $2\sqrt{\frac{c}{b}-\frac{a}{b}zz}=y=BD$ & □=CDV=y Thirdly If ye valor of $\frac{q}{p}$ is irrationall being a square{illeg} roote, The simplest cases may bee reduced to these following examples.

1. If $\frac{c{x}^{n}}{x}\sqrt{a+b{x}^{n}}=\frac{q}{p}$. Then $\frac{2ac+2bc{x}^{n}}{3nb}\sqrt{a+b{x}^{n}}=y$.

2. If $\frac{c{x}^{2n}}{x}\sqrt{a+b{x}^{n}}=\frac{q}{p}$. Then $\frac{6bbc{x}^{2n}+2abc{x}^{n}-4aac}{15nbb}\sqrt{a+b{x}^{n}}=y$

3. If $\frac{c{x}^{3n}}{x}\sqrt{a+b{x}^{n}}=\frac{q}{p}$. Then $\frac{30{b}^{3}c{x}^{3n}+6abbc{x}^{2n}-8aabc{x}^{n}+16{a}^{3}c}{105n{b}^{3}}\sqrt{a+b{x}^{n}}=y$.

4. If $\frac{c{x}^{4n}}{x}\sqrt{a+b{x}^{n}}=\frac{q}{p}$. Then $\frac{210{b}^{4}c{x}^{4n}+30a{b}^{3}c{x}^{3n}-36aabbc{x}^{2n}+48{a}^{3}bc{x}^{n}-96{a}^{4}c}{945n{b}^{4}}\sqrt{a+b{x}^{n}}=y$.

5. If $\frac{c{x}^{5n}}{x}\sqrt{a+b{x}^{n}}=\frac{q}{p}$. Then $\frac{1890{b}^{5}c{x}^{5n}+210a{b}^{4}c{x}^{4n}-240aa{b}^{3}c{x}^{3n}+288{a}^{3}bbc{x}^{2n}-384{a}^{4}bc{x}^{n}+768{a}^{5}c}{10395n{b}^{5}}\sqrt{a+b{x}^{n}}=y$.

1. If $\frac{c{x}^{n}}{x\sqrt{a+b{x}^{n}}}=\frac{q}{p}$. Then $\frac{2c}{nb}\sqrt{a+b{x}^{n}}=y$.

2. If $\frac{c{x}^{2n}}{x\sqrt{a+b{x}^{n}}}=\frac{q}{p}$. Then $\frac{2bc{x}^{n}-4ac}{3nbb}\sqrt{a+b{x}^{n}}=y$.

3. If $\frac{c{x}^{3n}}{x\sqrt{a+b{x}^{n}}}=\frac{q}{p}$. Then $\frac{6bbc{x}^{2n}-8abc{x}^{n}+16aac}{15n{b}^{3}}\sqrt{a+b{x}^{n}}=y$.

4. If $\frac{c{x}^{4n}}{x\sqrt{a+b{x}^{n}}}=\frac{q}{p}$. Then is $\frac{30{b}^{3}c{x}^{3n}-36abbc{x}^{2n}+48aabc{x}^{n}-96{a}^{3}c}{105n{b}^{4}}\sqrt{a+b{x}^{n}}=y$.

5. If $\frac{c{x}^{5n}}{x\sqrt{a+b{x}^{n}}}=\frac{q}{p}$. Then $\frac{210{b}^{4}c{x}^{4n}-240a{b}^{3}c{x}^{3n}+288aabbc{x}^{2n}-384{a}^{3}bc{x}^{n}+768{a}^{4}c}{945n{b}^{5}}\sqrt{a+b{x}^{n}}=y$.

<50v>

1. If $\overline{\frac{b}{x}-\frac{a}{{2x}^{n+1}}}\sqrt{a{x}^{n}+b{x}^{2n}}=\frac{q}{p}$. Then $\frac{a+b{x}^{n}}{n{x}^{n}}\sqrt{a{x}^{n}+b{x}^{2n}}=y$.

1. If $\frac{3a{x}^{n}+6b{x}^{2n}}{x}\sqrt{a{x}^{n}+b{x}^{2n}}=\frac{q}{p}$. Then is $\frac{2a{x}^{n}+2b{x}^{2n}}{n}\sqrt{a{x}^{n}+b{x}^{2n}}=y$.

2. If $\frac{-15aa{x}^{n}+48bb{x}^{3n}}{x}\sqrt{a{x}^{n}+b{x}^{2n}}=\frac{q}{p}$. Then is $\frac{-10a+12b{x}^{n}}{n}×\overline{a{x}^{n}+b{x}^{2n}}×\sqrt{a{x}^{n}+b{x}^{2n}}=y$.

3. If $\frac{105{a}^{3}{x}^{n}+480{b}^{3}{x}^{4n}}{x}\sqrt{a{x}^{n}+b{x}^{2n}}=\frac{q}{p}$. Then $\frac{70aa-84ab{x}^{n}+96bb{x}^{2n}}{n}×\overline{a{x}^{n}+b{x}^{2n}}×\sqrt{a{x}^{n}+b{x}^{2n}}=y$.

4 $\frac{-945{a}^{4}{x}^{n}+5760{b}^{4}{x}^{5n}}{x}\sqrt{a{x}^{n}+b{x}^{2n}}=\frac{q}{p}$. −630a3+

1. If $\frac{a{x}^{n}+2b{x}^{2n}}{x\sqrt{a{x}^{n}+b{x}^{2n}}}=\frac{q}{p}$. Then is $\frac{2}{n}\sqrt{a{x}^{n}+b{x}^{2n}}=y$.

2. $\frac{3a{x}^{2n}+4b{x}^{3n}}{x\sqrt{a{x}^{n}+b{x}^{2n}}}=\frac{q}{p}$. Then is $\frac{2{x}^{n}}{n}\sqrt{a{x}^{n}+b{x}^{2n}}=y$.

3. If $\frac{15aa{x}^{2n}-24bb{x}^{4n}}{x\sqrt{a{x}^{n}+b{x}^{2n}}}=\frac{q}{p}$. Then is $\frac{10a{x}^{n}-8b{x}^{2n}}{n}\sqrt{a{x}^{n}+b{x}^{2n}}=y$.

4. If $\frac{105{a}^{3}{x}^{2n}-192{b}^{3}{x}^{5n}}{x\sqrt{a{x}^{n}+b{x}^{2n}}}=\frac{q}{p}$. Then $\frac{70aa{x}^{n}-56ab{x}^{2n}+48bb{x}^{3n}}{n}\sqrt{a{x}^{n}+b{x}^{2n}}=y$

1|2| If, $\frac{c}{x}\sqrt{a{x}^{n}+b{x}^{2n}}=\frac{q}{p}$. Make xn=zz. And yn is $\square \frac{2c}{n}\sqrt{a+bzz}=y$.

2|3|. If, $\frac{c{x}^{n}}{x}\sqrt{a{x}^{n}+b{x}^{2n}}=\frac{q}{p}$. Make xn=z. Then is $\square \frac{c}{n}\sqrt{az+bzz}=y$.

3|4|. If, $\frac{c{x}^{2n}}{x}\sqrt{a{x}^{n}+b{x}^{2n}}=\frac{q}{p}$. Make xn=z. Then, $\frac{2ac{x}^{n}+2bc{x}^{2n}}{6nb}\sqrt{a{x}^{n}+b{x}^{2n}}-\square \frac{ac}{2nb}\sqrt{az+bzz}=y$.

4|5|. If $\frac{c{x}^{3n}}{x}\sqrt{a{x}^{n}+b{x}^{2n}}=\frac{q}{p}$. Make xn=z. yn is $\frac{12bc{x}^{n}-10ac}{48nbb}×\overline{a{x}^{n}+b{x}^{2n}}×\sqrt{a{x}^{n}+b{x}^{2n}}+\square \frac{5aac}{16nbb}\sqrt{az+bzz}=y$.

1 If $\frac{c}{x{x}^{n}}\sqrt{a{x}^{n}+b{x}^{2n}}=\frac{q}{p}$. Make xn=zz. yn is $\frac{-2ac-2bc{x}^{n}}{na{x}^{n}}\sqrt{a{x}^{n}+b{x}^{2n}}+\square \frac{4bc}{na}\sqrt{a+bzz}=y$.

1 If $\frac{c{x}^{n}}{x\sqrt{a{x}^{n}±b{x}^{2n}}}=\frac{q}{p}$. Make $\sqrt{a±b{x}^{n}}=z$. yn is $\frac{2c}{na}\sqrt{a{x}^{n}±b{x}^{2n}}\mp \square \frac{4c}{na}\sqrt{\frac{zz-a}{±b}}=y$

2 If $\frac{c{x}^{2n}}{x\sqrt{a{x}^{n}+b{x}^{2n}}}=\frac{q}{p}$. Make $\sqrt{a+b{x}^{n}}=z$. yn is $\square \frac{2c}{nb}\sqrt{\frac{zz-a}{b}}=y$.

3 If $\frac{c{x}^{3n}}{x\sqrt{a{x}^{n}+b{x}^{2n}}}=\frac{q}{p}$. Make $\sqrt{a+b{x}^{n}}=z$. yn is $\frac{c{x}^{n}}{2nb}\sqrt{a{x}^{n}+b{x}^{2n}}-\square \frac{3ac}{2nbb}\sqrt{\frac{zz-a}{b}}=y$.

If $\frac{c{x}^{n}}{x}\sqrt{a+b{x}^{n}+d{x}^{2n}}=\frac{q}{p}$. Make xn=z. Then is $\square \frac{c}{n}\sqrt{a+bz+d{z}^{2}}=y$.

If $\frac{c{x}^{n}\sqrt{d+e{x}^{n}}}{x\sqrt{a+b{x}^{n}}}=\frac{q}{p}$. Make xn=z $\sqrt{a+b{x}^{n}}=z$. Then is $\square \frac{2c\sqrt{db-ae+ezz}}{nb\sqrt{b}}=y$

If $\frac{-c{x}^{2n}\sqrt{a+b{x}^{n}}}{x\sqrt{a-3b{x}^{n}}}=\frac{q}{p}$. Then is $\frac{ac+bc{x}^{n}}{6nbb}\sqrt{aa-2ab{x}^{n}-3bb{x}^{2n}}=y$.

If $\frac{c{x}^{3n}\sqrt{3a+b{x}^{n}}}{x\sqrt{-a+b{x}^{n}}}=\frac{q}{p}$. Make $\overline{\frac{2a}{b}+{x}^{n}}\sqrt{a+b{x}^{n}}=z$. Then is $\square \frac{2c\sqrt{bbzz-4{a}^{3}}}{3nbb}=y$.

<52r>

1. $\frac{a}{b+cz:}$$\frac{a}{b+cz}$::Note 2. yt$\sqrt{az+bzz:}$is to$\sqrt{az+bzz:}$::3.$\sqrt{a+bzz:}$$\sqrt{a+bzz}$:: as ye ordinately applyed line bc in some of ye Conick sections: is to its corresponding superficies abc, ye axis ab being in like manner related to z. But all those areas (& consequently □$\frac{a}{b+cz}$, □$\sqrt{az+bzz}$, □$\sqrt{a+bzz}$) may bee Mechanichally found either by a Table of logarithmes or signes & Tangents. And I have beene therefore hitherto content to suppose ym knowne, as ye basis of most of ye precedent propositions.

Note also yt if ye val|C| Valor of $\frac{q}{p}$ consists of severall ꝑts each ꝑt must bee considered severally, as if: $\frac{a{x}^{3}-bbxx}{{c}^{4}}=\frac{q}{p}$ . Then is □$\frac{a{x}^{3}}{{c}^{4}}=\frac{a{x}^{4}}{4{c}^{4}}$ . & □$\frac{bbxx}{{c}^{4}}=\frac{bb{x}^{3}}{3{c}^{4}}$ . Therefore □$\frac{a{x}^{3}-bbxx}{{c}^{4}}=\frac{a{x}^{4}}{4{c}^{4}}-\frac{bb{x}^{3}}{3{c}^{4}}=y$.

Note also yt of ye valor denominator of ye valor of $\frac{q}{p}$ consist of both rationall & surde quantitys or of two {illeg} or more surde quantitys First take those surde quantitys out of ye denominator, & yn seeke (y) by ye pe|r|ecedent {illeg} theoremes

But this eighth Proposition may bee ever thus resolved mechanichally. viz: Seeke ye Valor of $\frac{q}{p}$ as if you were resolving ye equation in Decimall numbers either by Division or Extraction of rootes or Vieta's Analyticall resolution of powers; This operation may bee continued at pleasure, ye farther ther better. & from each terme ariseing from this operation may bee deduced a ꝑte of ye valor of y, (by ꝑte ye 1st of this prop).

Example 1. If $\frac{a}{b+cx}=\frac{q}{p}$ . Then by division is $\frac{q}{p}=\frac{a}{b}-\frac{acx}{bb}\frac{+accxx}{{b}^{3}}-\frac{a{c}^{3}{x}^{3}}{{b}^{4}}+\frac{a{c}^{4}{x}^{4}}{{b}^{5}}-\frac{a{c}^{5}{x}^{5}}{{b}^{6}}+\frac{a{c}^{6}{x}^{6}}{{b}^{7}}$ &c. And consequently $y=\frac{ax}{b}-\frac{acxx}{2bb}+\frac{acc{x}^{3}}{3{b}^{3}}-\frac{a{x}^{3}{x}^{4}}{4{b}^{4}}+\frac{a{c}^{4}{x}^{5}}{5{b}^{5}}$ &c.

Example 2. If $\sqrt{aa-xx}=\frac{q}{p}$ . Extract ye roote & 'tis $\frac{q}{p}=a-\frac{xx}{2a}-\frac{{x}^{4}}{8{a}^{3}}-\frac{{x}^{6}}{16{a}^{5}}-\frac{5{x}^{8}}{128{a}^{7}}-\frac{7{x}^{10}}{256{a}^{9}}-\frac{21{x}^{12}}{1024{a}^{11}}$ &c (as may appeare by squareing both ꝑts). Therefore (by 1st ꝑte of Prop 8) $y=ax-\frac{{x}^{3}}{6a}-\frac{{x}^{5}}{40{a}^{3}}-\frac{{x}^{7}}{112{a}^{5}}-\frac{5{x}^{9}}{1152{a}^{7}}-\frac{7{x}^{11}}{2816{a}^{9}}$ &c.

Example 3. If $\frac{{q}^{3}}{{p}^{3}}\ast -ax\frac{q}{p}-{x}^{3}=0$

<52v>

But ye Demonstracons of hath beene said must not bee wholly omitted.

Prop 7 Demonstrated. Lemma. If two bodys A, B, move uniformely ye oneother from ab to c, d, e, f,g, h, k, l, &c: in ye same time. Then are ye lines ac,bg, & cd,gh, & de,hk, & ef,kl, &c: as their velocitys p.q. And though they move not uniformely yet{illeg} are ye infinitely little lines wch each moment they describe, as their velocitys wch they have while they describe ym. As if ye body A wth ye velocity p describe ye infinitely little line \(cd=)/p×o in one moment, in yt moment ye body B wth ye velocity q, will describe ye line \(gh=)/q×o. For p:q::po:qo. Soe yt if ye described lines bee (ac=)x, & (bg=)y, in one moment, they will bee (ad=)x+po, & (bh=)y+qo in ye next.

Demonstr: Now if ye equation expressing ye relation twixt ye lines x & y bee x3−abx+a3−dyy=0. I may substitu{illeg}|t|e x+po & y+qo into ye place of x & y; because (by ye lemma) they as well as x & y, doe signify ye lines described by ye bodys A & B. By doeing so there results x3+3poxx+3ppoox+p3o3−dyy−2dqoy−dqqoo=0−abx−abpo+a3. But x3−abx+a3−dyy=0 (by supp). Therefore there remaines onely 3p{illeg}|o|xx+3ppoox+p3o3−abpo−2dqoy−dqqoo=0. Or dividing it by o tis 3px2+3ppox+p3oo−abp−2dqy−dqqo=0. Also those termes are infinitely little in wch o is yn those in wch tis not. Therefore b|o|mitting them there rests 3pxx−abp−2dqy=0. The like may bee done in all other equations.

Hence I observe. First yt those termes ever vanish in wch o is not \are not multiplyed by o/, they being ye propounded equation. Secondly those termes also vanish in wch o is of more yn one dimension, because they are infinitely lesse yn those in wch o is but of one dimension. Thirdly ye rem still remaining termes, being divided by o will have yt forme wch, by ye 1st rule in Prop 7th, they should have (as may ꝑtly appeare by Mr Oughtreds ye second termes of Mr Oughtreds latter Analiticall table).

After ye sàme manner may this \7th/ Prop: bee demonstrn: there being 3 or more unknowne quantitys (x, \y, z, &c/

Prop 8th is ye Converse of this 7th Prop. & may bee therefore Ab Analytically demonstrar|t|ed by it.

Prop 1st Demonstrated. If some body A move in ye \right/ line gafc from g towards c. From any point d draw df⊥ac. & call, df=a. fg=x, dg=y. Then is aa+xx−yy=0. Now by Prop 7th, may ye proportion of (p) ye velocity of yt body towards f; to (q) e|i|ts velocity towards d bee found viz 2xp−2yq=0. Or x∶y∷q∶p. That is gf∶gd∷ its velocity to d: its velocity towards f or c. & {illeg} when ye body A is at a, yt is when ye points g & a are coincident then is ac∶ad∷ad∶af∷ velocity to c∶ velocity to d.

Prop 2d, Demonstrated.     From ye points d & e draw df⊥ac⊥ge. And let ye firsts bodys velocity (it moveing to d) bee called ad, ye seconds to e bee ae, & ye 3ds toward c bee ac. Then shall ye firsts velocity towards c bee af (by Prop 1): |&| The seconds towards c is ag, (prop 1). but af af=gc ({illeg}|for| △adcsimilis/=\△aec, & △adf {illeg}=△gec. by sup). Therefore ac=ag+gc =ag+af. That is ye velocity of ye third body towards c is compo equall to ye \summ of the/ velocitys of ye first & second body towards c.

<53v>

## The former Theorems Applyed to Resolving of Problems.

Prob. 1. To draw Tangents to crooked lines.

Seeke (by prop 7th; or 3d, 4th & 2d, &c) ye motions of those streight lines to wch ye crooked line is cheifely referred, & wth what velocity they increase or decrease: & they shall give (by prop 6t, or 1st or 2d) ye motion of ye point describing ye crooked line; wch motion is in its tangent.

|Tangents to Geometricall lines.|

Example 1. If ye crooked line fac is described by ye intersection of two lines cb & dc ye one \ye one/ moveing parallely, ye one viz: cb∥ad, & insisting upon ab, ye other \&/ dc∥ab; & insisting soe yt if ab=x, & bc=y=ad, Their relation is x4−3y+10ax3+ayxx−2y3x+a4−y4=0. To draw ye tangent r|h|cr; I call p ye velocity of cb yt is ye velocity of ye increase of ab & q ye velocity of dc, yt is of ye increasing of {illeg} ad or cb. And so I have (by Prop 7) this Equation 4px3−9p{illeg}y{illeg}+3opa+{illeg}xx+2payx−2py3−3qx3+qaxx−6qyyx−4qy3=0. Now because ye crooked line fac\{illeg}/ is described by ye intersection point of ye lines dc & cb, I consider yt ye point c fixed in ye line cb moves towards e parallely to ab (for so doth ye line cb (by supp:) & consequently all its points): also ye point c fixed in ye line dc moves towards g parallely to ad (by sup): therefore I draw ce∥ab & cg∥ad, makeing {yn} |&| in such proportions as ye motions they designe \& so draw er∥cb, & gr∥dc, & ye diagonall cr,/ (by Prop 6), yt is, if ye velocity of ye line cb, (yt is ye celerity of ye increasing of ab, or dc; or ye velocity of ye point c from d) bee called p, & ye velocity of ye line cd bee called q; I make ce∶gc∷p∶q(∷ce∶er∷hb∶cb.) \& ye point c shall move in the diagonall line cr (by prop 6) wch is therefore the required tangent/ Now ye relation of p & q may bee found by ye foregoing Equation (p signifying ye increase of x, & q of y) to bee 4px3−9pyxx+30paxx+2payx−2py3−3qx3+qaxx−6qyyx−4qy3=0. |by Prop 7| And therefore $hb=\frac{py}{q}=\frac{3y{x}^{3}-ayxx+6{y}^{3}x+4{y}^{4}}{4{x}^{3}-9yxx+30axx+2ayx-2{y}^{3}}$. wch determines ye tangent hc

☞ Hence may bee observed this Generall Theorem for Drawing Tangnts to crooked lines thus referred to streight ones; yt is, to such lines in wch y=bc is ordinately{sic} applyed to x=ab at any given angle abc. viz: Multiply the termes of ye Equation ordered according to ye dimensions of y, by any Arimeticall {sic} progressiō wch product shall bee ye Numerator: Againe change ye signes of ye Equation & ordering it according to x, multiply ye termes by any Arithmeticall progression & ye product divided by x shall bee ye Denominator of ye valor of (hb, yt is, of) x produced from y to ye tangent hc.

As if $rx-\frac{rxx}{q}=yy$ . 2.1.0.Then firstyy$+\frac{r}{q}xx$=0−rx0.-1.-2., produceth 2yy, \or, 2rx−2$\frac{r}{q}$xx./ Secondly 2.1.0.Secondly$\frac{-r}{q}xx$+rx−yy produceth $rx-\frac{2r}{q}xx$ . Therefore $\frac{2yy}{r-\frac{2r}{q}x}=bh$ . Or else $\frac{2rx-\frac{2r}{q}xx}{r-\frac{2r}{q}x}=bh=\frac{2qrx-2rxx}{qr-2rx}$ .

Example 2. If ye crooked line chm bee describe by ye intersection of two lines ac, bc circulating about their centers a & b, soe yt if ac=x, & bc=y; their relation is x3−abx+cyy=0. To draw ye tangent ec I consider yt ye point c moves to fixed in ye line bc moves towards f in ye line cf⊥bc (for ye tangnt to a circle is perpendicular to its radius). also ye point c fixed in ye line {illeg}|a|c moves towards {illeg}|d| in ye line cd⊥ac & from those lines c{illeg}|d| & c{illeg}|f| &|I| draw two others de∥cg & e{illeg}|f|∥bc wch must bee in such proportion one to another as ye motions represented by ym (prop 6), yt is (prop 6) as ye <54r> as ye {sic} motions{sic} of ye intersection point c moveing in ye lines ca & cb to or from ye centers a & b; yt is, as ye velocity of ye increase or decrease of ca=cx & cb={yn} (ye celerity of ye increase or decrease of x being called p, & of y being q), de∶ef∷p∶q. Then shall ye diagonall ce bee ye required tangent. Or wch is ye same, (for △ecn|g|=△ecd, & △ecf=△ecg\n/,) I produce ac & bc to g & n, so yt cg∶cn∷p∶q. & yn draw n{e} ne⊥bn, & ge⊥n|a|g; & ye tane|g|ent diagonall \ce/ to their intersection point e. Now ye relation of p & q may bee found by ye given Equation to bee, 3pxx−pab+2qcy=0 (by prop 7) Or 2cy∶ab−3xx∷p∶q∷cg∶cn, wchdetermins ye tangent ce.

But note yt if p, or q be negative cg or cn must bee drawn from c towards a or b, but from a or b if affirmative.

Hence tis easy to pronounce a s|T|heorems for Tangents in such like cases & ye like may bee done in all other cases however Geometricall lines bee referred to streight ones.

Tangents to Mechanichall lines

Example ye 3d. If ye Quadratrix kbf is described by ye intersection \(b)/ of ye two lines {illeg}|hb| & ap, ye one hp∥ma moving uniformly from k to a, whilest ye other ap circulates from k to m about ye center a. I d|D|raw ye circle gbl wth ye Rad. ab; & make ab⊥bd=bf bf= bl=bd⊥ab∥de; & to ye intersection point e of ye lines am & ed draw yt tan eb e{illeg} wch shall touch ye Quadratrix in b. For suppose ye motion of ye point p fixed in ye line ap, {illeg} \towards m/ to bee pm, yn ye motion of ye point b fixed in ye line ab, \towards d/ is bl\=bd,/ (prop 4), & ye motion of ye line bh towards ca, & therefore of ye point b fixed in it towards c (prop 3) is ha=bc (by supp): Also ce∥bh & ed∥ap (sup). Therefore (by Prop 6) is ye diagonall {illeg} eb ye motion of ye intersection point b of those two lines ap & hb, moves in ye diagonall eb{illeg}, & consequently eb toucheth ye Quadratrix in b.

Example 4th

<54v>

Prob 2d. To find ye quantity of crookednesse of lines.

Lemma. The crookednesse of \equall parts of/ circles are as their diameters reciprocally. For the crookednesse of a whole circle (acdea, bfghmb) amounts to 4 right angles. Therefore there is not more crookednesse in one whole circle acdea yn in another bfghmb. Suppose ye perimeter acde=bfgh. Then tis, \ar∶br∷/ bfgh=acde∶bfghmb∷ crookednesse of bfgh∶crookednesse of bfghmb=crookednesse of acdea.

Lemma 2. That point of a Crooked Line's Perpendicular wch is in least motion is ye center required of ye circle of equall /whose\ crookednesse wth ye given is required.

Resolution. This may bee done by drawing perpendiculars to 2 curved lines described by ye motion of any two points fixed in ye Tangent, wch two perpendiculars shall intersect in ye center of a Circle whose crookednesse was{illeg} required. One of those curve lines & it perpendiculars may bee ye given crooked line & ye perpendicular to it at its given point.

But if y bee ordinately applyed to x at right angles ye best way is

Resolution. \Find/ That point of \fixed in/ ye crooked line's perpendicular wch is |yn| in least motion, \for it/ is ye center of a circle wch passing through{illeg} ye given point is of equall crookednesse wth ye line at yt given point. To find wch point I observe yt Now, since ye crooked line's tangent & perpendicular &c: (at yt moment) circulate about yt center; I observe, 1st yt every point fixed in ye Tangnt or Perpendicular, or whose position to ym is {illeg} determined, doth describe a curve line to wch ye right line drawne from yt center is perpendicular, & is also ye radius of a circle of equall crookedness wth it: 2dly yt ye motion of every such point is as its distance from yt center; & so are ye motions of ye intersection point, in wch any radius drawn from yt center intersects two parallel |lines.|

Example 1. If cb=y is ordinately applyed to ab=x at a right angle abc, nc being tangent & mc perpendicular to ye curve line ac: I draw seeke ye motion of two points c & d fixed in ye perpendicular cd; or (wch is better & to ye same purpose) I draw cg∥ab, & seeke ye motions of ye two \intersection/ points c & d in wch ye perpendicular cd intersects those \fixed/ lines cfg, & abdk: & yn draw cg & dk in such proportion as those motions are, & ye line{sic} gkm drawn by their ends shall intersect ye perpendicular cd in ye required center m: mc being ye radius of a circle of equall crookednesse wth ye curve line ac at ye point c. Now, Making △cegfe= & like △ncdbc, suppose ye motion of ye line cb & consequently of ye points c & b fixed in it & moveing towards f f & k, to bee p=nb=cf: Then is ce=cn ye motion of ye point c in wch ye lines ne & bc intersect, yt is \bc intersects ye tangent ne/ (by Prop 6), yt is of ye point c in|fix|e{illeg}|d| in ye perpendicular cd, & moveing in ye tangent ne: & therefore cg=nd={sic}p+bd(=p+v if bd=v), is ye motion of ye intersection point c towards g in wch point ye perp: cd intersects cg \(by Prop 6)/. If also ye motion of ye intersection point d from ye point b {illeg} (yt is ye velocity of ye increase of v) bee called r, yn is dk=nb+r=p+r soe yt, cg−dk∶cd∷cg∶cm; yt is, $v-r:cd=\sqrt{yy+vv}\colon\colon p+v:cm=\frac{\overline{p+v}×\sqrt{yy+vv}}{v-r}$. Also $v-:y\colon\colon p+v:ck=\frac{py+vy}{v-r}$. Lastly {illeg} ye motion of ye point c from b, (yt is ye velocity wth wch y=cb increaseth) will bee q=cb=y.

As if ye nature of ye crooked line bee {illeg} x3−axy+ayy=0. Then is $nb=\frac{axy-2ayy}{3xx-ay}=p$ (by examp: 1. Prob: 1.), & $\frac{3xxy-ayy}{ax-2ay}=v=bd$ (for nb∶bc∷bc∶bd) Soe that 3xxy−ayy+\/axv+2b|a|yv=0. & therefore (by Prop 7) 6pxy−pavqay\2/qay+3qxx+2qb|a|v−rax+\2/rb|a|y/=0\ <55r> & substituting ye $\frac{axy-2ayy}{3xx-ay}$, & $\frac{3xxy-ayy}{ax-2ay}$, & y, into ye places of p, v, & q in ye this equation The product will bee $\frac{6axxyy-12ax{y}^{3}}{3xx-ay}-3ayy+3yxx\frac{+6xxyy-2a{y}^{3}}{x-2y}-rax+2ray=0$. Or 6aax3yy $\frac{9y{x}^{5}-6ayy{x}^{3}-12a{y}^{3}xx+24a{y}^{4}x+3aa{y}^{3}x-4{a}^{2}{y}^{4}}{3a{x}^{4}-12ay{x}^{3}-{a}^{2}yxx+12ayyxx+4{a}^{2}yyx-4{a}^{2}{y}^{3}}=r$. And therefore $\frac{-18yy{x}^{4}+24a{y}^{3}xx-24a{y}^{4}x-2aa{y}^{3}x+2aa{y}^{4}}{3a{x}^{4}-12ay{x}^{3}-aayxx+12ayyxx+4aayyx-4aa{y}^{3}}=v-r=\frac{2aa{y}^{4}-2aa{y}^{3}x-24a{y}^{4}x+24a{y}^{3}xx-18yy{x}^{4}}{3{x}^{3}-6yxx-ayx+2ayyinax-2ay}$. Also $p+v=\frac{axy-2ayy}{3xx-ay}+\frac{3xxy-ayy}{ax-2ay}=\frac{9{x}^{4}y-6axxyy+5aa{y}^{3}-4aaxyy+aaxxy}{3xx-ay×ax-2ay}$. Soe yt $ck=\frac{py+vy}{v-r}=\frac{9{x}^{4}y-6axxyy+5aa{y}^{3}-4aaxyy+aaxxyin3{x}^{3}-6xxy-axy+2ayy}{2aa{y}^{3}-2aaxyy-24ax{y}^{3}+24axxyy-18{x}^{4}yin3xx-ay}$. That is $ck=\frac{9{x}^{5}-6a{x}^{3}y+aa{x}^{3}-6aaxxy+13aaxyy+12axxyy-10aa{y}^{3}-18{x}^{4}y}{2aayy-2aaxy-24axyy+24axxy-18{x}^{4}}$. |Which Equation gives ye point k & consequently ye point m. for km∥abk.|

☞ But in such cases where y is ordinately applyed to x at right angles, From ye consideration of ye Equation $\frac{py+vy}{v-r}=ck$ ; Or rather $\frac{py+vy}{r-v}=ck$ : may ye following Theoreme bee pronoundced. To wch purpose let X signify {illeg}|t|he given Equation, yt is, all ye algebraicall termes (expressing ye nature of ye given line) considered as equall to nothing & not some of ym to others. Let X· signify those termes multiplyed by ordered according to ye dimensions of x & yn multiplyed by any arithmeticall progressiō Let X· signify ye {illeg}|tho|se termes ordered according to ye dimensions of y & yn multiplyed by any Arithmeticall progression. Let :X signify those termes ordered by x & |yn| multiplyed by any two arithmetia|c|all progressions one of ym being greater yn ye other by a terme. Let X: signify those termes ordered by y & yn multiplied by any two Arith: Progr: one of ym being greater yn ye other \differing/ by a terme. Let ·X· signify those termes ordered according to x, & yn multiplyed by ye greater of ye progressions wch multiplyed :X; & yn ordered by y & multiplyed by ye greater progression wch multiplyed X:. Then \(observing yt all these progressions have the same difference & proceede ye same way in respect of ye dimension or {sic} x & y)/ will ye 3 Theorems bee

1st. $\frac{·X·XX·yy+X·X·X·xx}{-·X·XX:y+2·XX··X·y-X·X·:Xy}=ck=\frac{·X·Xyy+X·X·xx}{-\frac{·X·XX:y}{X·}+2·X·X·y-X·:Xy}$ .

2d. $\frac{·X·X·Xyy+·XX·X·xx}{-·X·XX:x+2·XX··X·x-X·X·:Xx}=km=bl=\frac{·X·Xyy+X·X·xx}{-·XX:x+2X··X·x-\frac{X·X·:Xx}{·X}}$.

3d. $\frac{·X·Xyy+X·X·xxin\sqrt{·X·Xyy+X·X·xx}}{-·X·XX:yx+2·XX··X·yx-X·X·:Xyx}=cm=$ {illeg}radio circuli æqualis curvitatis cum curva ac in puncto c.

As if ye line bee x3−axy+ayy=0. Then is ·X=3x3−axy. X·=−axy+2ayy. :X=6x3=3×2.1×0.0 x−1. :X=6x3=x3axy+ayy. 0x−1.1×0.2×1.X:=2ayy=x3axy+ayy & 3×0.1×1.0×2.·X·=−axy=x3axy+ayy. Which valors of ·X, X· &c being substitu{illeg}|te|d into their places in ye first rule, ye result is; $\frac{9{x}^{6}yy-6a{x}^{4}{y}^{3}+5aaxx{y}^{4}+aa{x}^{4}yy-4aa{x}^{3}{y}^{3}}{\frac{-18a{x}^{6}{y}^{3}+12aa{x}^{4}{y}^{4}-2{a}^{3}xx{y}^{5}}{2ayy-axy}+2aaxx{y}^{3}-12a{x}^{3}{y}^{3}}=ck$ . Which being conveniently reduced is
$\frac{9{x}^{5}-6a{x}^{3}y+aa{x}^{3}-6aaxxy+13aaxyy+12axxyy-10aa{y}^{3}-18{x}^{4}y}{18{x}^{4}-24axxy+24axyy+2aaxy-aayy}=ck$ . As was found before. <55v> $ck=\frac{1}{2}a+3y-\frac{3ay}{2x}\frac{+ay-3xx}{2a-6x}$ .

Or suppose ye line is a Conick section whose nature $rx+\frac{rxx}{q}=yy$ . Then is ·X=rx+$\frac{2r}{q}{x}^{2}$ 210 =rx$\frac{r}{q}$ $\frac{r}{q}$ xx+rxyy . 0312 X·=−2yy=$\frac{r}{q}$ xx+rxyyy . 2×1.1×0.0x−1. :X=$\frac{2r}{q}$xx=$\frac{r}{q}$ xx+rxyy . 0x−1.0x−1.1×0.2×1. X:=−2yy=$\frac{r}{q}$ xx+rxyy . 2×01×00×2 ·X·=0=$\frac{r}{q}$ xx+rxyy . Which valors of ·X, ·X, :X, &c being substituted into their places in ye first Theoreme, give $\frac{rrxxyy+4\frac{rr}{q}{x}^{3}yy+\frac{4rr}{qq}{x}^{4}yy+4xx{y}^{4}}{-rrxxy-\frac{4rr}{q}{x}^{3}y-\frac{4rr}{qq}{x}^{4}y+\frac{4r}{q}xx{y}^{3}}=ck=\frac{qqrry+4qrrxy+4rrxxy+4qq{y}^{3}}{4qryy-qqrr-4qrrx-4rrxx}$ . Or (since 4qryy−4qrrx−4rrxx=0) tis, $-ck=y+\frac{4xy}{q}+\frac{4xxy}{qq}+\frac{4{y}^{3}}{rr}=y+\frac{4{y}^{3}}{qr}+\frac{4{y}^{3}}{rr}=y+\frac{4r{y}^{3}+4q{y}^{3}}{qrr}$ . So by ye second Theoreme tis $\frac{rrxxyy+\frac{4r}{q}xx{y}^{4}+4xx{y}^{4}}{2r{x}^{3}yy+\frac{4r}{q}{x}^{3}yy\frac{-\frac{8r}{q}{y}^{4}{x}^{3}}{rx+\frac{2r}{q}xx}}=km=bl=\frac{r}{2}+\frac{2yy}{q}+\frac{2yy}{r}+\frac{rx}{q}+\frac{4xyy}{qr}+\frac{4xyy}{qq}$ . Or $bl=\frac{r}{2}$ +$\frac{3rx}{q}$ +2x+$\frac{3rx}{q}+\frac{6xx}{q}+\frac{6rxx}{qq}+\frac{4{x}^{3}}{qq}+\frac{4r{x}^{3}}{{q}^{3}}$ . And so by ye 3d Theoreme, tis $cm=\frac{rrxxyy+\frac{4rr}{q}{x}^{3}yy+\frac{4rr}{qq}{x}^{4}yy+4xx{y}^{4}}{2rr{x}^{3}{y}^{3}+8\frac{rr}{q}{x}^{4}{y}^{3}+\frac{8rr}{qq}{x}^{5}{y}^{3}-8\frac{r}{q}{x}^{3}{y}^{5}}×\sqrt{rrxxyy+\frac{4rr}{q}{x}^{3}yy+\frac{4rr}{qq}{x}^{4}yy+4xx{y}^{4}}$ . Or $cm=\frac{qqrr+4qrrx+4rrxx+4qqyyin\sqrt{qqrr+4qrrx+rrxx+4qqyy}}{2{q}^{3}rr}$ . Or $cm=\frac{qrr+4qyy+4ryy}{2qqrr}\sqrt{qqrr+4qqyy+4qryy}$ .

This Theoreme may bee thus Demonstrated

Note yt ye curvity of any curve whose ordinates are applyed at \inclined from right to/ oblique angles is to ye curvity of the same curve whose ordinates are at right angles, as the curvity of an Ellipsis whose ordinates are in the sa a circle whose ordinates are in like manner inclined so as to make it becom an Ellipsis.

Prob: To find ye points of curves where they have a given degree of curvity.

<56r>

Prob 3d. To find ye points distinguishing twixt ye concave & convex
portions of crooked lines.

Resolution. The lines are not crooked at those points: & therefore ye radius cm determining ye crookednesse at yt point must bee infinitely greate. To wch purpose I put ye denominator of its valor (in rule ye 3d) to bee equall to nothing, & {sic} so have this Theoreme ·X·XX:−2·XX··X·+X·X·:X=0. Or better ꝑhaps $\frac{·XX:}{X·}-2·X·+\frac{X·:X}{·X}=0$ .

Example, was this point to \be/ found in ye Concha whose nature is +bb x4+2bx3+ccxx−2bccx−bbcc=0. +yy . Then is ·X=2x4+2bx3+2bccx+2bbcc. X·=2xxyy=X: :X=2x4−4bccx−bbbcc. ·X·=0. Which valors subrogated into ye Theoreme, they produce 2x4+2bx3+2bccx+2bbcc+$\frac{\overline{2{x}^{4}-4bccx-6bbcc}×2xxyy}{2{x}^{4}+2b{x}^{3}+2bccx+2bbcc}=0$ . And subrogating −bb −x4−2bx3−ccxx +2bccx+bbcc ye valor of xxyy into its stead & \twice/ reducing ye Equation by ye divisor x+b. Tis x3+bcc+$\frac{\overline{{x}^{4}-2bccx-3bbcc}×\overline{cc-xx}}{{x}^{3}+bcc}=0$ . Or x4+4bx3+3bbxx−2bccx−2bbcc=0. Which being againe reduced by x+b=0. Tis x3+3bxx−2bcc=0. See Geometr: Chart: pag: 259.

<56v>

Probl: 4. To \find/ ye points at wch lines are most or least crooked.

Resol: At those points ye afforesaid radius cm neither increaseth nor decreaseth. So yt ye center m in yt moment doth absolutely rest, & therefore neither ye line bk nor al doth increase or diminish, yt is, ck & bl doe soe much increase or diminish as x|y| & x (lb & ab) doe diminish or increase. Or in a word the point m resteth. Find therefore the motion of al or cm or lm & suppose it t nothing.

Thus in the Concha to find the point of least crookednes beyond ye point of reflection, having substituted ye valors of ·X, X· &c (exprest in the precedent problem) into this The computation is too tedious

Thus to find the point of least crookednesse in ye curve x3=\cc/yy By the rule in prob 2 I make ·X=3x3. X·=−ccy. :X=6x3. X:=0 & ·X·equals;0 & thence obteine $ck=\frac{3}{2}y+\frac{cc}{6x}$, or $=\frac{3{x}^{3}}{2cc}+\frac{cc}{6x}$ wch is least when $\frac{9{x}^{3}}{2cc}-\frac{cc}{6x}=0$ or $x=\sqrt{4}:\frac{c4}{27}$. And then therefore happens the greatest crookednesse

In like mane|n|er if ye curve be xxy=a3 The rule gives ·X=2a3 X·=a3 {=} :X={illeg}\−6a3/ X:;=0=·X· And thence $\frac{2y}{3}+\frac{xx}{6y}=ck$ which is least when $y=a\sqrt{\frac{1}{2}}$ .

Soe if ye curve bee x3=byy. Th|e|i|n| is ·X=3x3. X·=−\2/byy. :X=6x3 X:=−2yyc. ·X·=0. And therefore ck=3y+$\frac{4xx}{3y}$ , which {illeg} \hath no/ least when x=$\frac{4b}{27}$ nor the curve any least crookednesse.

<57r>

Prob 5t. To find ye nature of ye crooked line whose area is expressed by any given equation.

That is, ye nature of ye area being given to find ye nature of ye crooked line whose area it is.

Resol. If ye relation of ab=x, & abc=y bee given & ye relation of ab=x, & bc=q bee required (bc being ordinately applyed at right angles to ab). Make de∥ab⊥ad∥be=1. & yn is □abed=x. Now supposing ye line cbe by parallel motion from ad to describe ye two superficies ae=x, & abc=y; The velocity wth wch they increase will bee, as be to bc: yt is, ye motion by wch x increaseth being be=p=1, ye motion by wch y increaseth will bee bc=q. which therefore may bee found by prop: 7th. viz: $\frac{-·Xy}{X·x}=q=bc$ .

Example 1. If 4rx3=9yy /$\frac{2x}{3}\sqrt{rx}=y$ \. Or −4rx3+9yy=0. Then is $\frac{12rxx}{18y}=q=\sqrt{rx}$ . Or rx=qq & therefore abc is ye Parabola whose area \abc/ is $\frac{2x}{3}\sqrt{rx}=\frac{2qx}{3}$ .

Example 2d. If x3−ay+xy=0. Then is $\frac{3xx+y}{-x+a}=q$ . Or $\frac{3axx-2{x}^{3}}{aa-2ax+xx}=q=bc$

Example 3d. If $\frac{na}{n+m}{x}^{\frac{n+m}{n}}-y=0$ . Then is $a{x}^{\frac{m}{n}}=q$ . Or if axm=bxn; yn is $\frac{ma{x}^{m-1}}{nb{x}^{n-1}}=q$ .

Note yt by this probleme may bee gathered a Catalogue of all those lines wch can bee squared. And therefore it will not bee necessary to shew how this Probleme may bee resolved in other cases in wch q is not ordinately applye{d} to x at right angles.

Prob 6. The nature of any Crooked line being given, to find other lines whose areas may bee compared to ye area of yt given line{illeg}.

Resol: Suppose ye given line to be ac & its area abc=s, ye sought line df & its area def=t; & yt bc=z is ordinately applyed to ab=x, & ef=v to de=y, soe yt ∠abc=∠def; & yt ye velocitys of y wth wch ab & de increase (yt is, ye velocity of ye points b & e, ofr of ye lines bc & ef moving from a & d) bee called p & q. Then may ye ordinately applyed lines bc & ef multiplyed by their velocitys p & q, (yt is pz & qv) signify ye velocitys wth wch ye areas abc\=s/ & def\=t/ increase. Now ye relation of ye areas s & t (taken at pleasure) gives ye relation of ye motions pz & qv describing those areas, by Proposition ye 7th.; Also ye relation of ye lines x & y ab=x & de=y (taken at pleasure) gives ye relation of p & q, by Prop 7th: wch two equations, together with ye given {illeg} Equation expressing ye nature of ye line ac, give ye relation of de=y & ef=v yt is ye \desired/ nature of ye line df.

Example 1. As if ax+bxx={illeg}zz is ye nature of ye line ac: & at pleasure I assume s=t {&}|t|o be ye relation of {illeg} ye areas abc & def; & {illeg}x=yy to bee ye relation of ye lines ab & de. Then is pz=qv (prop 7), & p=2qy (by prop 7). Therefore 2yz=v (by ye 2 last Equations), & $\frac{vv}{4yy}$ =zz=ax+bxx=ayy+by4. or vv=4ay4+4by6. & $v=2yy\sqrt{a+byy}$ : Wch is ye nature of ye line def whose area def is equall to ye area abc, supposing $\sqrt{ab}=de=\sqrt{x}=y$ .

<57v>

Example 2. If ax+bxx=zz as before; & I assume, as+bx=t, & x=yy. Then is apz+bp=qv (by prop 7), & p=2qy. 2azy+2by=v=2by+2ay$\sqrt{ax+bxx}$ . Or v=2by+2ayy$\sqrt{a+byy}$ ; The required nature of ye line def.

Example 3d. If ax+bxx=xx as {if} $\frac{4c}{a}\sqrt{xx-a}=z$ : & at pleasure I assume $\frac{2c}{a}\sqrt{ay+yy}-s=t$ , & {illeg} xx−a=y. Then is 4ccay+ccyy=aass+2aast+aatt, And (by prop 7) 4ccaq+8ccyq=2aaspz+2aatpz+2aasqv+2aatqv=$\overline{2aapz+2aaqv}×\frac{2c}{a}\sqrt{ay+yy}$ . & (by prop 7) 2px=q. Therefore $8cax+16cyx=\overline{4az+8avx}×\sqrt{ay+yy}$ . But $\frac{4c}{a}\sqrt{xx-a}=z=\frac{4c}{a}\sqrt{y}$ . Therefore & $x=\sqrt{a+y}$ , Therefore $\overline{8ca+16cy}\sqrt{a+y}=\overline{16c\sqrt{y}+8av\sqrt{a+y}}×\sqrt{ay+yy}$ . That is 8 ca+2cy=2cy+$av\sqrt{ay+yy}$ . Or $v=\frac{c}{\sqrt{ay+yy}}$ .

Example 4th. If ax+bxx=zz as before & I assume ss=t, & x=yy. Then is 2spz=qv, & p=2qy (by prop 7) Therefore $4syz=v=4sy\sqrt{ax+bxx}=4syy\sqrt{a+byy}$ . Where note yt in this case ye area line $v=4syy\sqrt{a+byy}$ is a Mechanicall one because s ye area of ye line ax+bxx=zz canott bee Geometrically found. The like is to bee observed in other such like cases.

Probl: 7. The Nature of any Crooked line being given to find its area when it may bee. Or more generally, two crooked lines being given to fin{g}|d| the relation of their areas, when it may bee.

Resol: Reassuming ye figure of the 5t probleme in wch ye given line is ac, & supposing its area y & {x}|th|e other area x to bee described \by ye line ce moving from a/ as was there taught: {illeg} Suppose the motion describing x bee p=be=1, Then ye motion describing y is cb=q=$\frac{q}{1}=\frac{q}{p}$ . Now having ye relation twixt ab{illeg}=abed=x, & bc=q=$\frac{q}{1}=\frac{q}{p}$ . (By supp:); I may (some times viz when it may bee) find

Resolution. In ye figure of ye fift probleme Let abc=y represent ye area of ye given line acf; cb=q ye motion describing yt area; abed=x another area wch is equall to ye basis ab=x of yt given line acf, (viz: supposing ab∥de⊥be∥ad=1); & be=p=1 ye motion describing yt other area. Now haveing (by supp) ye relation twixt ab=x=abed, & bc=q=$\frac{q}{1}=\frac{q}{p}$ given, I seeke ye area abc=y by ye Eight propo/sition\

Example 1. If ye nature of ye line bee, $\frac{ax}{\sqrt{aa-xx}}=bc=\frac{q}{p}$ . I looke in ye tables of ye Eight proposition for ye Equation corresponding to this Equati{ō} wch I find to bee $\frac{c{x}^{n}}{x\sqrt{a+b{x}^{n}}}=\frac{q}{p}$ , (For if instead of c, a, b, n I write a, aa, −1, 2, it will be $\frac{ax}{\sqrt{aa-xx}}=\frac{q}{p}$ .) And against it is ye equation $\frac{2c}{nb}\sqrt{a+b{x}^{n}}=y$. And substituting a, aa, −1, 2 into ye places of c, a, b, n it will bee, $-a\sqrt{aa-{x}^{n}}=y=abc$ ye required area.

Example 2. If $\sqrt{\frac{{x}^{3}}{a}}-\frac{eeb}{x\sqrt{ax-xx}}=bc=\frac{q}{p}$ . T|B|ecause there are two termes in ye valor of bc I consider them severally & first I find ye area correspondent to ye terme $\sqrt{\frac{{x}^{3}}{a}}$ , or $\frac{1}{{a}^{\frac{1}{2}}}{x}^{\frac{3}{2}}$ ; To bee $\frac{2}{5{a}^{\frac{1}{2}}}{x}^{\frac{5}{2}}$ , or $\frac{2}{5}\sqrt{\frac{{x}^{5}}{a}}$ , \by prop: 8. part 1./ Secondly to find ye area corresponding to ye other terme $\frac{eeb}{x\sqrt{ax-xx}}$ I looke ye Equation (in prop 8. part 3) corresponding to it wch is $\frac{c{x}^{n}}{x\sqrt{a+c{x}^{n}}}=\frac{q}{p}$, (for if instead of c, a, b, n, I write eeb, −1, a, −1, it will bee $\frac{eeb{x}^{\overline{1}}}{x\sqrt{-1+a{x}^{\overline{1}}}}$ , Or $\frac{eeb}{x\sqrt{-xx+ax}}$ ): Against which is ye Equation $\frac{2c}{nb}\sqrt{a+b{x}^{n}}=y$ . In {sic} <58r> In which writing eeb, −1, a, −1, instead of c, a, b, n, the result will bee $\frac{2eeb}{-a}\sqrt{-1+a{x}^{\overline{1}}}$ ; Or, $\frac{-2eeb}{ax}\sqrt{-xx+ax}=y$ wch is the the {sic} area corresponding to yt {illeg}|ot|her terme. Lastly \Now/ to see how these areas stand related one to another I draw ye annexed scheme, in which is ab=x, bE=q=$\frac{q}{p}$, adg ye fkeg the given curved line, & adg ye curve line described by one of its parts $\sqrt{\frac{{x}^{3}}{a}}=bd$ ab=x. $bd=\sqrt{\frac{{x}^{3}}{a}}$ . $de=\frac{eeb}{x\sqrt{ax-xx}}$ . & $be=\sqrt{\frac{{x}^{3}}{a}}-\frac{eeb}{x\sqrt{ax-xx}}=\frac{q}{p}=\frac{q}{p}=q$ . Soe that $abd=\frac{2}{5}\sqrt{\frac{{x}^{5}}{a}}$ is ye superficies corresponding to ye first terme b{illeg}|d|, wch because it is affirmative must bee extended (or lye) from ye side ye from ye line bd towards a. Also ye other superficies correspondent to ye 2d terme de, being negative must lye on ye other side bd from a, wch is therefore gde $gde=\frac{-2eeb}{ax}\sqrt{ax-xx}$ . Lastly if x=ab=r Then is $abd=\frac{2}{5}\sqrt{\frac{{r}^{5}}{a}}$ . & $gde=\frac{-2bee}{ar}\sqrt{ar-rr}$ . And if aB=x=s, yn is $\frac{2}{5}\sqrt{\frac{{s}^{5}}{a}}=aBD$ , And $gDE=\frac{-2bee}{as}\sqrt{as-ss}$ . Soe yt, $\frac{2}{5}\sqrt{\frac{{s}^{5}}{a}}-\frac{2}{5}\sqrt{\frac{{r}^{5}}{a}}=bBDd$ . & $\frac{-2bee}{as}\sqrt{as-ss}+\frac{2bee}{ar}\sqrt{ar-rr}=DdeE$ & substracting DdeE from bBDd there remaines bBEe{illeg} $bBEe=\frac{2\sqrt{{s}^{5}}-2\sqrt{{r}^{5}}}{5\sqrt{a}}+\frac{2bee}{as}\sqrt{as-ss}-\frac{2bee}{ar}\sqrt{ar-rr}=bBEe$ . wch is ye required Area of ye given line &cfkeEg. Where note yt takging \for/ ye quantitys r=ab & s=aB taking any numbers you may thereby finde ye area bBdD correspond to their difference bB.

Note yt sometimes one parte of ye Area may bee Affirmative & ye other negative. as if a2B=r, & ab=s. Then is b2B2Ee=kbe−k2B2B|E|={sic}$\frac{2\sqrt{{s}^{5}}-2\sqrt{{r}^{5}}}{5\sqrt{a}}+\frac{2bee}{as}\sqrt{as-ss}-\frac{2bee}{ar}\sqrt{ar-rr}$ =b2B2Ee=kbe−k2B2E.

<59r>

Prob 8|9|. To find such crooked lines whose lengths may bee found &also to find theire lengths.

Lemma 1. If to any crooked immovable line acg the streight line dcmσ moves to & fro perpendicularly every point of ye said line cdmσ (as γ, θ, σ, &c) shall describe a curve line (as βγ, δθ, λσ, &c) all which will bee perpendicular to ye said line cdmσ, & also parallel one to another & to ye line acg.

2. If acg bee not a circle, there may bee drawne some curve line βδmλ, wch ye moving line cdmδ will always touch in some point or other (as at {illeg} m) & to wch therefore all ye curve lines βγ, δθ, λσ, &c: are perpendiculars.

3. Soe that every point (γ, θ, m, σ, &c:) of ye line cdmσ, when it begineth or ceaseth to touch ye curve line βδmλ, doth yn move perpendicularly to or from it: & therefore ye line γdmσ doth not at all slide upon ye curve line βδ{illeg}|m|λ, but exactly measure it by applying it selfe to it point by point: & therefore ye correspondent parts of ye said lines are equall (viz: βm=γm. δm=θm. δλ=θσ. &c).

Resol. Take any Equation for ye nature of ye crooked line acg, & by ye 2d probleme find ye center m of its crookedness at c. That point m is a point of ye required curve line βδ{illeg}mλ. For yt point m whereat cdmσ ye perpendicular to acg doth touch ye curve{illeg} line βδmλ is lesse moved yn any other point of ye said perpendicular, bee|i|ng as it were ye hinge & center upon | about wch ye ye perpendicular turneth | moveth at that moment.

Example 1. If acg is a Parabola whose nature is {xx} −rr=yy (supposing ab=x⊥{illeg}bc=y) is al rx=yy. By Theoreme ye 1st of Problem 2d I find cb+lm=y+$\frac{4{y}^{3}}{rr}$ . By ye 2d Theoreme, bl=2x+$\frac{1}{2}$r. And by ye 3d Theoreme $\frac{\overline{r+4x}\sqrt{r+4x}}{2r}=cm$. Soe yt supposing ab=$\frac{1}{2}$r. bl=3x=z. $lm=\frac{4{y}^{3}}{rr}=v$. The relation twixt v & z will bee 27rvv=16z3, (for r4vv=16y6=16y3x3=$\frac{16{z}^{3}{r}^{3}}{27}$) wch is ye nature of ye required line βδmλ. And since cγ=aβ=$\frac{1}{2}$r. Therefore $\gamma m=\frac{r+4x}{2r}\sqrt{rr+4xr}-\frac{1}{2}r=\frac{3r+4z}{18r}\sqrt{9rr+12rz}-\frac{r}{2}$. is ye length of its ꝑte βδm.

Example 2. Soe if aa=xy, is ye nature of ye line acg. By ye afforesaid Theoreme I find, cb+lm=$\frac{xx+yy}{2y}$; &, $bl=\frac{xx+yy}{2x}$; {&} whereby ye nature of ye curve line βδmλ is determined. And lastly I find $cm=\frac{xx+yy}{2aa}\sqrt{xx+yy}$ which determine its length.

<59v>

Prob: 10. Any curve line being given to find other lines whose lengths may be compared \to its length or/ to its area, & to compare ym.

Resolution. Take any Equation for ye relation twixt ad & ye perpendicular {illeg} cd=y (whither yt relation bee expressed by an Geometricall Equation or whither it bee ye same wch some streight line beares to a curved one or to its superficies &c). Then (by prop 7) find ye relation twixt ye increa{sic}e or decrease (p & q) of ye lines ad=x, & dc=y. & say (by prop 6) yt {illeg} q∶p∷dc∶dn=$\frac{py}{q}$ . And soe is ye triangle dnc (rectanguled at c) given & consequently ye Nature of ye curve line acg to wch dc is perpendicular, & cn a tangent.

Now ye nature of ye center (m) of ye perpendiculars motion (wch gives ye nature & length of ye required curve line βm) may be found as in ye 2d or 9th Probleme, But more conveniently thus. Draw any fixed line he∥ad⊥ah=a=ef⊥ad. Also call fd=v. & ye increase \or decrease/ of (fd) call r. And ye increase or decrease of ye motions p & q, call γ β & γ Now considering yt ye motion (p+r) of ye intersection point e in ye line he is to ye motion (p) of ye intersection point (d) in ye line (ad), as (em) is to (dm) (see prob 2); That is yt ye difference (r) of those motions is to ye motion (p) of ye point (b) soe is (ed) to (dm): First I find ye valor of v. viz $-\sqrt{pp-qq}$ ∶q∷cn∶cd∷∷ {sic}ef=a∶fd=v=$\frac{-aq}{\sqrt{pp-qq}}$ . Or aaqq+vvqq−vvpp=0. Secondly by this Equation I find ye valor of r, viz: (by Prob 7), 2aaqγ+2vvqγ+2rvqq−2rvpp−2vvpβ=0. Thirdly $r=\frac{aaq\gamma +vvq\gamma -vvp\beta }{ppv-qqv}:p\colon\colon ed=\frac{pv}{q}:dm=\frac{{p}^{4}zz-ppqqzz}{aaqq\gamma +vvqq\gamma -vvpq\beta }$ . Lastly supposing the motion p to bee uniforme yt its increase or decrease β may vanish, & also substituting ppzz ye valor of aaqq+vvqq in its stead in ye Denominator of dm, ye result will bee $\frac{pp-qq}{\gamma }=dm=\frac{1-qq}{\gamma }$ , if p=1. And to find ye lines dl & ml, say −p∶q∷dm∶dl=$\frac{-q+{q}^{3}}{\gamma }$ . & $p:\sqrt{pp-qq}\colon\colon dm:ml=\frac{1-qq\sqrt{1-qq}}{\gamma }$

Soe yt by the equation expressing ye relation twixt x & y first I se{illeg}|find|e q & yn γ. wch two give me $\frac{1-qq}{\gamma }=dm$ &c.

Example 1. If ye relation twixt x & y bee supposed to bee yy−ax=0. Then (by Prop 7) I find, first 2qy−ap=0, Or 2qy−a=0. & secondly 2γy+2qq=0. And substituting these valors of $q=\frac{a}{2y}$ & γ=$\frac{-qq}{y}=\frac{-aa}{4{y}^{3}}$ in their stead in ye Equatio $\frac{1-qq}{\gamma }=dm$ , &c: The results will bee $\frac{aay-4{y}^{3}}{aa}=dm$ . $\frac{-aa+4yy}{2a}=dl$ . & $\frac{aa-4yy}{2aa}\sqrt{4yy-aa}=lm$ . (And adding cd=−y to dm, & ad=x to dl ={illeg}$\frac{yy}{a}$ to dl, ye result is ad{illeg} cm=$\frac{-4{y}^{3}}{aa}$ . & al=$\frac{3yy}{a}-\frac{a}{2}$ .) Wch determine ye nature & crookednesse of ye line βm. (For if aβ={illeg}$\frac{a}{4}$ & βl =z=al−ab=al−a aβ=$\frac{a}{4}$ . βl=z. lm=v. The relation twixt z & v will bee 16z3=27rvv. The length of βm being $m\gamma =\frac{4{y}^{3}}{aa}-\frac{a}{2}=\frac{3a+4z}{18a}\sqrt{9aa+12az}-\frac{1}{2}a$ . {I} as before was found).

Example ye 2d. If x=ad⊥dk=$\frac{aa}{x}$ , is ye nature of ye crooked line {illeg} (ye Hyperbola) gkw: And I would find other crooked lines (βm) whose lengths may bee compared wth ye area sdkg (calling as=b⊥gs) of that crooked line gkw. I call <60r> I call {sic} yt area sdkg=ξ, & its motion θ. Now since 1∶dk∷{illeg}p∶θ, (see prob 5. & ye Note on prop 7 in Example 3); Therefore θ{illeg} dk×p=dk=θ=$\frac{aa}{x}$ . This being knowne I take at pleasure any Equation, in wch ξ is, for ye valor of cd=y.

As 1st suppose ay=ξ, That (by prop 7) gives aq=θ=$\frac{aa}{x}$ , Or qx=a{illeg}; wch also (by prop 7) gives γx+qp=0. Which valors of q=$\frac{a}{x}$ , & γ=$\frac{-q}{x}=\frac{-a}{xx}$ ; by helpe of ye Theorems $\frac{1-qq}{\gamma }=dm$ &c: doe give $\frac{xx-aa}{-a}=dm$ . $\frac{xx-aa}{x}=dl$ . & $\frac{xx-aa}{-ax}\sqrt{xx-aa}=lm$ =lm {sic}. wch determines ye nature of ye required curve line βm. The length of yt portion of it wch is intercepted twixt ye point m & the curve line acg being −cm=cd+dm=$\frac{\xi +xx-aa}{-a}$ , Or mc=$\frac{\xi +xx-aa}{a}$ .

☞ [Note yt in this case although ye area sdkg=ξ cannot bee Geometrically found & therefore ye line acg is a Mechanicall one yet ye desired line βm is a Geometricall one. And ye like will happen in all other such like cases, when |in| ye Equation taken at pleasure to expresse ye relation twixt x, y, & ξ; neither x, y, nor ξ doe multiply \or divide/ one another, nor it selfe, nor is in any denominator or roote, except x wch may multiply it selfe & bee in \sometimes/ denominators & rootes, when y or ξ are not in those one|fra|ctions or rootes{sic} \& herein onely doth this excell the precedent 9th probleme/. Such is this Equation aξ−aby+ax3+$\frac{axx}{ax-xx}-5xx\sqrt{ab-xx}=0$ . &c: But not this ξξ=a3y. nor ξ=xy. &c]

Secondly suppose ξ=xy. yt (by prop 7) give θ={q}y+xq=$aax$ , Or xy+xxq=aa, & yt (by prop 7) gives y+qx+2qx+γxx=0. Which two valors of $\frac{aa\prime xy}{xx}=q$ , And $\frac{y+3qx}{\prime xx}=\gamma$ ; by meanes of ye Equatio Theoremes $\frac{1-qq}{\gamma }=dm$ , {illeg} $\frac{{q}^{3}-q}{\gamma }=dl$ , & $\frac{1-qq}{\gamma }\sqrt{1-qq}=lm$ ; doe determine ye nature \& length/ of ye desired curve line βm.

Example ye 3d. In like manner to find curve lines whose lengths may bee compared to ye length \gk/ of ye said curve (Hyperbola) gkw. Call gk=ξ, & its motion θ. Now, drawing kh ye tangent to gkw at k, I condsider that ad=x & gk=ξ doe increase in ye proportion of dh to kh; yt is, dh∶kh∷p∶θ. Now finding (by prob 1) yt dh=−x, & $kh=\frac{-a}{x}\sqrt{aa+xx}$ ; therefore is $\frac{kh×p}{dh}=\frac{kh}{dh}=\frac{a\sqrt{aa+xx}}{xx}=\theta$ . Which being found, I take any equation, in wch ξ is, for the valor of cd=y. & |yn| worke as in ye precedent Example.

Note yt by this &|or| ye Ninth Probleme may bee gathered a Catalogue of whatever lines, whose lengths can bee Geometrically found.

<60v>

Prob {illeg} 12. To find ye Length of any given crooked line when{illeg} it may /bee.\

Resolution. The length of any streight line to wch ye curve line is cheifly related being called (x), ye length of ye curve line (y), & theire motions (p & q) first (by prob 1) get ye an equation expressing ye relation twixt x & $\frac{q}{p}$, & yn seeke ye valor of (y) by ye Eight proposition. [Or find a curve line whose area is equall to ye length of ye given line, by Prob 11. And then find that area by Prob 7.]

<61r>

Prob 11. To find curve lines whose Areas shall bee Equall (or have any other given relation) to ye length of any given Curve line drawn into a given |right line|

Resolution. The length of any streight line, to wch ye given curve line is cheifely referred, being called x, ye length of ye curve line y, |&| their motions of increase p & q. Get The valor of $\frac{q}{p}$ (te|fo|und by ye first probleme) being ordinately applyed at right angles to x, gives ye nature of a curve line whose area is equall to (y) ye length of ye curve line.

And this th|L|ine thus found gives (by prob 6) other lines. whose areas have any other given relation to ye length (y) of ye given curve line

<61v>

Prob 13. To find ye nature of a Crooked line whose length is expressed
by any given Equation, (when it may bee done).

Resolution. Suppose ab=x, bc=y, ac=z. & their motions p, q, r. And let ye relation twixt x & z bee supposed given. Then (by prop 8) finding the relation twixt p & r make $\overline{rrr-pp}=q$ . (For drawing cd tangent to ac at c & de⊥cb⊥ab: ye lines de, ec, dc, shall bee as p, q, r. but $\sqrt{dc×dc-de×de}=ec$ , & therefore $\sqrt{rr-pp}=q$ ). Lastly, ye ratio twixt x & $\frac{q}{p}$ being thus knowne, seeke y (by prop 8). Which relation twixt ab=x & bc=y determines ye nature of ye crooked line ac=z.

<62r>

## Of Gravity.

Definitn. 1. I call yt point ye center of Motion in any Body, wch always rests when or howsoever yt Body circulates wthout progressive motion. It would a{illeg}|l|ways bee ye same wth ye center of Gravity were ye Rays of Gravity parallel & not converging towards ye center of ye Earth.

Lemma 1 \Def: 2 And ye right lines passing through yt point I call ye axes of Motion or Gravity./

Lemma 1 The place & distance of Bodys is determinded by their center of Gravity. Which is ye middle point of a right line circle &|o|r Parallelogram:

Lemma 2 Those weights doe equiponderate whose quantitys are reciprocally proportionall to their distances from the common axis of Gravity, supposing their centers of Gravity were to bee in ye same plaine wth yt common axis of Gravity.

Prob To find ye center of Gravity in rectilinear plaine figures 1. In ye Triangle acd make ab=bc, & cd=fd. & draw db, & af, their intersection point (e) is its center of Gravity.

In ye Trapezium abdc, draw ad & cb. Ioyne ye centers of Gravity e & h, f & g of ye opposite triangles acb & dcb, bad & {illeg} adc wth ye lines eh, fg. Their intersection point n is ye center of Gravity in ye Trapezium. (And so of Pentagons, hexagones &c)

Prop|b|: To find such plaine figures wch are equiponderate to any given plaine figure in respect of an axis of Gravity in any given position.

Resol That ye natures & positions of ye given curvilinear plaine (gbc,) & sought plaine (bde) bee such yt they may equiponderate in respect of ye axis (ak;) I suppose x=ab⊥bc=z, & y=ad⊥de=v to bee either perpendicular or parallel or coincident to ye said axis ak: And ye motions whereby x & y doe increase or decrease (i:e: ye motions of bc, & de to or from ye point a) I call p & q. Now ye ordinatly applyed lines bc=z, & de=v, multiplyed into their motion p & q (yt is, pz, & qv) may signify ye infinitly little parts of those areas (acb, & lde) wch each moment they describe; wch infinitly little parts doe equiponderate (by Lemma 1 & 2), if they multiplied by {illeg} their distances from ye axis ak doe make equall products. (yt is; pxz=qyv, in fig 1: pxz=$\frac{1}{2}$qvv in fig 2: $\frac{1}{2}$pzz=qv×fm, in fig 3; supposing dm=me. &) And if all ye respective infinitly little parts doe /equiponderate ye superficies must do so too.\

Now therefore, (ye relation of x & y|z| being given by ye nature of ye curve line cg,) I take at pleasure any Equation for ye relation twixt x & y, & thereby (by prop 7) find p & q, & so by ye ꝑcedent Theorem find ye relation twixt y & v, for ye nature of ye sought are plaine lde.

Exam: 1. If cg (fig 2) is an Hyperbola, soe yt aa=xz. & I suppose 2x={illeg}y. yn is 2p={illeg}q (prop 7). & {illeg} paa=pxz=$\frac{1}{2}$qvv=$\frac{1}{2}$pvv. Or aa=vv. or a=v=de. Soe yt le is a streight line, & lde a parallelogram, wch equiponderates wth ye Hyperbola cgkabc (infinitly extended towards gk) if 2ab=ad. al×al=ab×bc.

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Example 2. If cg (fig 3) is a circle whose nature is, $\sqrt{aa-xx}=z$ {illeg}. |&| I suppose at pleasure 3aax−x3=\6/aay. Then (by prop 7) I find 3aap−3xxp=6aaq. And therefore $\frac{1}{2}$paa−$\frac{1}{2}$pxx=$\frac{pzz}{2}$ ={illeg}qv×fm=(if fd=$\frac{1}{2}$a,) {illeg}$\frac{qvv+qav}{2}=\frac{vv+av}{2}×q=\frac{vv+av}{2}×\frac{aap-xxp}{2aa}$ . {illeg} Or $aa×\overline{paa-pxx}=\frac{vv+av}{2}×aap-xxp$ . Or 2aa=vv+av. Or $v=-\frac{1}{2}a♉\sqrt{\frac{9aa}{4}}=a$ ; {illeg}. Soe that le is a streight line & alde a parallelogram

Example 3. If abcg is a parallelogram (fig 4) whose nature is, a=y a=z. & I suppose at pleasure x=yy−b. Then (by prop 7) tis p=2qy. Therefore $\frac{1}{2}$aap=aaqy=$\frac{1}{2}$pzz=qvy. Or aa=v. Soe yt aed is a parallelogram.

Or if I suppose at pleasure, x=y3−b. Then is (prop 7) p=3qyy. & therefore $\frac{1}{2}$aap=$\frac{3}{2}$ aaqyy=$\frac{1}{2}$pzz=qvy. Or 3aay=2v. ||

Or if I suppose x=y4−b. yn is p=4qy3 & 2aayy=v. so that aed is a Parabola. [Soe if xx=y5. yn is 2px=5qy4. &, $\frac{5{a}^{2}qy\sqrt{y}}{4}=\frac{5aaq{y}^{4}}{4x}$ =$\frac{1}{2}$ap=$\frac{1}{2}$pzz=qvy. Or 5aaqy$\sqrt{y}$ =4qvy. & 25a4y=16vv. soe yt aed is a p|P|arabola.]

Example {3}|4| If gbc (fig 1) is an Hyperbola whose nature is xx−aa={qy} zz. & I suppose x=y+{illeg}|b|. Then (by prop 7) is p=q. Therefore $px\sqrt{xx-aa}$ =pxz=qyv=pyv. Or $\overline{y+b}×\sqrt{yy+2by+bb-aa}=x\sqrt{xx-aa}=yv$ . Or $\overline{yy+2by+bb}$ in $\overline{yy+2by+bb-aa}=yyvv$ . &c.

Or if I suppose xx=2y. Then is 2px=2q. Therefore $q\sqrt{xx-aa}$ =pxz=qyv. Or (xx−aa=)2y−aa=yyvv.

Or if I suppose xx=yy+aa. Then (prop 7) is 2px+\=/2qy{illeg}. Therefore qyz=pxz=qyv. Or $x=\sqrt{aa}$ $\sqrt{xx-aa}=$ $y=\sqrt{xx-aa}=z=v$ . & y=v; so yt aed is a triangle

Note yt This Probleme may bee resolved a{illeg}|lt|hough the lines have a x, z, y, v, & ak \have/ any other given inclination one to another, but the ꝑcedent cases may suffice

Note also yt if I take a Parallelogram for ye knowne superficies (as in ye 3d Example I may thereby gather a Catalogue of all such curvilinear superficies whose weight in respect of ye axis, may bee knowne.

Note also I might have shewn how to find lines whose wh weights in respect of any axis are not onely equall but have also any other given proportion one to another. And yn have made two Problems instead of this, as I did in Probl: 5 & 6: 9 & 10.

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Prob 15. To find ye Gravity of any given plaine in respect of any given axis, given in position
when it may bee done.

Resol: Suppose ek to bee ye Axis of Gravity, acb the given plaine, cb=y, & db=z to bee ordinatly applyed at any angles to ab=x. Bisect cb at m & draw mn⊥ek. Now, since [cb×mn] is ye gravity of ye line [cb], (by lem 1 & 2); if I make cb×mn=db=z, every line db shall designe ye Gravity of its correspondent line cb, yt is, ye superficies adb shall designe ye Gravity of ye superficies acb. Soe yt finding ye quantity of yt superficies adb (by prob 7) I find ye gravity of ye sup{illeg}e|er|ficies acb.

Example 1 If ac is a Parabola; soe yt, ab{illeg} bc={4}, & z=d{a}∥k parallelTo;ak=axis rx=yy, & ye axis ak is ∥ dcb. &, {illeg} nb⊥ak, & yt, ab∶nb∷d∶e. Then is bc×nb=y×$\frac{e×ab}{d}=\frac{ex}{d}\sqrt{rx}=z$ . Or eerx3=ddzz, is ye nature of ye curve line ad. whose area (were{illeg} abd a right angle would be $\frac{2e}{5d}{x}^{\frac{5}{2}}{y}^{\frac{1}{2}}=\frac{2e}{5d}\sqrt{r{x}^{5}}$ but now it) is $\frac{2ee}{5dd}\sqrt{r{x}^{5}}$ , (by prob 7) wch is ye weight of ye area acb in respect of ye axis ak.

Examp: 2 If ac is a Circle

Prob 16. To find ye Axes of Gravity i|o|f any Plaines

Resol. Find ye quantity of ye Plaine (by Prob 7) \wch call A/ & ye quantity of its gravity in respect of any axis (by prob 15) wch call B{illeg}. & parallell to yt axis draw a line whose distance from it shall bee {illeg} $\frac{A}{B}$ . That line shall bee an Axis of Gravity of ye given plain

Or If you y cannot find ye quantity of the plane: Then find its gravitys in respect of two divers axes (AB & AC) wch gravitys call B{illeg} & C & D. & through |(A)| ye intersection of those axes draw a line AD wth this condition yt ye distances (DB & DC) of any one of its points (D) from the said axes (AB & AC), bee in such proportion as \to/ the gravitys of the plane. That line (AD) shall bee an axis of gravity of ye said plane EF.

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Prob 17 To find ye Center of Gravity of any Plaine, when it may bee

Resol Find two axes of Gravity &|b|y the precedent Prop, & their common intersection is ye Center of Gravity desired. If ye figure have any knowne Diameter that {illeg} \{illeg}/ may bee taken for one of its axes of Gravity.