(x + d)2 +d2: - 2d2. d. d d2, or x2 - 2dx d/2, or x= dNo2 Hence (x-d)2 = 2d2, or x Whence this Construction.-Draw two lines AB and AD at right angles. Bisect the angle BAD by the line AC, on which lay AEd, the given difference. Draw EF perpendicular to AC, meeting AB in F; then EF= AE =d, and AF=√AE + EF2√2dd2. Now, make FB FEd; then AB = d2+d=x. Draw BC perpendicular to AB; then, since the angle BAC = half a right angle, BC AB. Complete the square ABCD, and it will be the one required. 2 Demonstration by geometry.—Join CF; then the right-angled triangles CEF and CBF are equal in all their parts, and we have CE CB. Hence AC-CB AC CE AE the given difference. Q. E. D. Calculation.-AF=√2 AE2 = AE√√2; AB=AF÷FB=AF +AE AE/2+AE= the side of the square. Also AC-AE + CEAE + AB=AE√2+2AE= AE (√2 + 2). PROBLEM X.—In a right-angled triangle are given one leg, and the difference of the segments made by a perpendicular let fall from the right angle on the hypothenuse, to determine the triangle. Analysis by algebra -Let BAC be the right-angled triangle, and AF perpendicular to BC. Take FD FC; then AB, and BD= BF — CF, the difference of the segments, are given. Now, put AB a, BD=b, and BF=x; then CF-x—b, and BC=BF÷ CF = x + x -b: 2x b. By IV. 23,* BF:BA::BA:BC; that is, xa::a: 2x — b. Hence 2x2 - bx a2, or x2 X Whence this b 2 11 a2 Construction.--On the given leg AB describe a semicircle, which bisect in H, and join AH and HB; then AH2 HB2 = AB2: a2. (See Problem VIII., "Algebraic Equations.") Lay HM b 2 the coefficient of x; on it describe a circle of which the centre is * VI. 8, cor. For (IV. 30*) BE (BI L, and draw BELI; then BI= I L C F D M E B. 2 apply BF=BI; draw BFC to meet the perpendicular AC, and join AF; then will BAC be the required triangle, having AF perpendicular to BC. Demonstration by geometry.—We have to prove that BD = 2 HM 2 IE. We have AB22 BH' (IV. 30+) 2 BIX BH2=(IV. 30†) × 2 BI2 BI × 2 IE - 2 BF2 BF X 2 IE. Also, AB2 (IV. 231) BFX BC= BFX (BF+FC) + BFX (BF FD) = BF x (BF BF BF (2 BF × BE=2BI(BI — IE) BD) ÷ 2 BF2 — BF × BD. Hence 2 BF2 BD) = BF × BD=2 BF2. BF x 2 IE. Wherefore BD2 IE 2 HM — the given difference. Calculation.—We have BL=√ BH2 + HL2 Also, BF AB2 BD2 + 2 16 BI=BL+LH; then BF: BA::BA: BC, and AC ✔BC AB2. (See next problem.) Limit. The given difference must be less than AB, and consequently HM must be less than the radius AO. PROBLEM XII.-Given, the hypothenuse of a right-angled triangle, and the line bisecting one of the acute angles and terminating in the opposite leg, to determine the triangle. Analysis by algebra.-Let ABC represent the required triangle, in which the hypothenuse AB, and the line BD bisecting the angle ABC, are given. On AB describe the semicircle ACB; produce BD to meet it in F; and join AF; then (III. 18, cor. 18) angle FAC = FBC=FBA. Hence arcs AF and FC are equal, and the triangles AFD, AFB, and BCD are similar. * III. 36. † III. 36. † VI. 8, cor. ? III. 21. Now, put AB=a, BD=b, BF=x, and AF=y. Then, in similar triangles AFD and AFB, we have BF (x): AF(y) :: AF (y): FD (x —b); hence y2x2-bx. Also, a2= x2 + y2 = x2 + x2 equation (see Problem VIII., "Algebraic Equations") the following Construction.--Take H, the middle point of the semicircle AHB, and join AH and HB; then AH2 or HB2 — 1⁄2 AB2 b lay HP 2 the given bisecting line BD. On HP describe a circle PEHI, of which the centre is L. Draw BELI, and apply BFBI. Make the arc FC = FA, and join ADC and BC, and ABC will be the required triangle. For (IV. 30*) BI (x):BH:: b BH:BE (BI — HP); that is, BI × BE — BH2, or x × (x Demonstration geometrically.--Since the arcs AF and FC are equal by construction, the angles FBA and FBC are equal, and BD bisects the angle ABC. It remains to prove only that BD=2 PH 2 IE. In the similar triangles FAD and FBA, we have BF: AF:: AF: FD. Hence AF2 BF × FD BF × (BF — BD) BF2 BD × BF; then AB2 BF2 + AF2 —— BF12 + BF2 — BD × BF Also, AB22 BH' = (IV. 30†) 2 BI × × = 2 BF2 — BD x BF. = Wherefore we have 2 BF2 · BD × BF= 2 BF2 — BF × 2 IE. Whence BD2 IE 2 PH= the given bisecting line. Q. E. D. Calculation.--In triangle BHL, we have BL=√BH2 + HL2 AB2 BD2 2 + 16 Then BF=BI=BL+LH. Also, AF AB2 BF2; and FD = BF BD. Then, by similar triangles BAF, FAD, and DBC, we have BF: BA::AF: AD, and AD:AF::DB:BC, and AD: DF::DB:DC; then we have AC AD + DC. NOTE. This problem and the preceding one depend upon the same principles, and result in like equations, as may be seen. ANALYSIS BY THE DIFFERENTIAL CALCULUS. PROBLEMS ANALYZED BY THE DIFFERENTIAL CALCULUS AND CONSTRUCTED AND DEMONSTRATED BY PLANE GEOMETRY. PRELIMINARY REMARKS.-The differential of x is written d.x, or simply dx; of y, dy; of z, dz; the letter d being merely a symbol for differential. To find the differential of any power of an unknown quantity, multiply by the index denoting the power, diminish this index by unity, and multiply by the differential of the root, and the product of these three quantities will be the differential required. A constant or known quantity has no differential; and when a quantity is a maximum or a minimum, its differential is 0. Examples.—1. xo being equal to 1, a constant quantity, it has no differential; that is, d.xo = 0. 2. The differential of x1, or d.x1=1 × x1 1 × dxdx.* 1 × dx 3. d.x2 = 2 × x1× dx=2xdx; and d.xy=xdy + ydx. 4. d.x3-3 x x x dx = 3x dx; and d.xy2x2 × d.y2 + y2 × d.x2 = 2x2ydy + 2y2xdx. 5. d.ax1 =ax d.x1- 6. d.3ax5=15ax1dx. 7. d. (ax2 + 4x3 + 5x1) - 2axdx + 12x2dx + 20x3dx 12x2 + 20x3) dx. (2αx + 8. d. (by2+3y+a-4)= 2bydy + 3dy = (2by + 3) dy. The constants, 3 and 4, have no differential. * The division of powers of the same quantity or root, as is shown in algebra, is effected by subtracting the index of the divisor from the index of the dividend to NOTE 1.--The term employed for this process is to differentiate, differentiating, or differentiation, according to the connection in which the term is used. NOTE 2.—In problems involving the calculus, the letter d is not usually employed by mathematicians to denote a quantity. NOTE 3.-The quantity which do is multiplied by is called the differential co-efficient, as 2by + 3 in Ex. 8. PROBLEM I.-In a plane triangle are given the base and the vertical angle, to construct it such that the sum of the other two sides shall be a maximum; that is, the greatest amount possible. Analysis by the differential calculus.*—Let ABC represent the required triangle. On AC, produced if necessary, let fall the perpendicular BD. Put the given angle ACB=0, the given side AB = a, the side AC x, and BC y. Then CD: y cos 0. By IV. 12,† AC2+ BC-2AC x DC-AB2; that is, x2 + y2 — 2xy cos 0 — a2. By the problem, x+y=max. By differentiating, AC2 A student who has not yet studied the calculus may omit the analysis by that method, and just employ the results for the construction, and observe how beautifully the geometrical demonstration verifies the result obtained by the calculus, and how entirely the different processes of mathematical investigation harmonize with each other. Then when he studies the calculus he will feel increased confidence in the results of its determinations. † II. 13. |