squaring these quantities and adding, we find N2 (1+a2 + b2) (1 + a2 + b2 )2 + (x'—a)2 + (y'—ß)2 + z2 — 2N.2′+ (x'—¿) + (y'—ß) 1+ a2 + b2 N2 - 1+ a2 + b2 • If we let fall from the origin two straight lines perpendicular on these planes, the angle contained by the straight lines will be the same as the angle contained by the planes, let the equation to these straight lines be 1+ aa' + bb' cos. = √(1+a2 + b2) (1 + a2 + b2) But, in order that the straight lines may be perpendicular to the given planes, we must have A + a = 0, B + b = 0, A' + a′ = 0, B′ + b = 0 Substituting therefore, the values of a, b, d, b', derived from these equations, we find that the expression of the cosine of the angle between the two planes, 1+AA' + BB' cos. @= √(1 + A2 + B2) (1 + A22 + B/2) In order to find the angle which any plane makes with the co-ordinate planes, we have only to suppose that one of the above planes assumes in succession the position of the different co-ordinate planes, thus let us suppose, that (2) is the plane of xz, then its equation becomes y= 0, so that, A' = 0, C' = 0 and therefore, if we denate by the symbols (xz), (yz), (xy), the angles which the given plane makes with the planes xz, yz, xy, we have cos. (xx) = cos. (yz) = cos. (xy) = and have and cos. = cos. (xz) cos. (x'z') + cos. (xy) cos. (x'y') + cos. (yz) cos. (y'z') To find the angle (4) contained by a plane and straight line in space. The angle sought is that which the straight line makes with its projection on the plane. If from any point in the given straight line we let fall a perpendi cular upon the plane, the angle contained between these two straight lines will be the complement of the required angle. The equations of the line let fall perpendicular on the plane will be of the But in order that this may be perpendicular to the given plane, we must have A+ α = 0 B+6=0 Now, the cosine of the angle contained by the two straight lines, is It appears 1 + da + b'b √(1 + a2 + b2) (1 + a'2 +b2) from what has been said above, that, in the present case = 90o e, and.. cos. = sin. 6. Substituting therefore for a', U, these values in terms of A and B, we find DIFFERENTIAL CALCULUS. INTRODUCTION We have already seen that any binomial of the form (a + b)" may be expanded in a series of the form whatever may be the value of n; but although this theorem is very extended in its application, algebraists have invented another method for the developement of quantities more simple in its first principles and more general in its application. This is called the method of indeterminate co-efficients: and we now proceed to explain its nature. In order to convey an idea of this method, let it be required to develope the expression a+ba in a series ascending by powers of x. It is manifest that such a developement is possible, for a a a+b'x may be put under the form a (a' + b'x)−1, and by applying the binomial theorem to this expression, we shall obtain a series ascending regularly by powers of x. Let us then assume a = A + Bx + Cx2 + Dx3 + Ex2 + Fx5 +.......................................... (1) a+b'x ... where A, B, C, D, . . . . are quantities involving a, a', b', but independent of r, coefficients whose value we are required to determine and which for that reason are called indeterminate co-efficients, although in strict propriety of language they ought rather to be styled co-efficients to be determined. In order to ascertain the value of these co-efficients, let us multiply the two members of the equation (1) by a' + b'x, arranging the result according to powers of x and transposing a, we find 0= Aa―a+(Ba'+Ab')x+(Ca'+Bb')x2+(Da'+Cb')x3+(Ea'+Db')x*+...(2) We may here remark, that, if we suppose the values of A, B, C, D, . properly determined, the equation (1) must hold good whatever may be the value of x, and in like manner equation (2) also. Let us suppose then x = 0, this last equation becomes when x 0, must preserve the same value, whatever may be the value assigned to x, for, by hypothesis, A is independent of x, therefore whatever may be the value of x, equation (2) reduces itself to 0 = (Ba' + Ab') x + (Ca + Bb') x2 + (Da' + Cb') x3 + (Ea' + Db') x2+ . . or dividing both sides by x 0 = Bď + Ab+ (Ca' + Bb') x + (Da' + Cb') x2+(Ea' + Db') x3...... (3) Since this equation must hold good whatever may be the value assigned to x, let us suppose x = 0, hence Since B must preserve its value whatever may be the value of x, we may suppress in (3) the first term Ba' + Ab', which disappears by the value of B, and dividing both members of the equation by x it becomes 0 = Ca + Bb' + (Da' + Cb') x + (Ea' + Db') x2 + Let us again make x = 0, we have We at once perceive that each co-efficient is formed by multiplying the one If we reflect upon the reasoning employed in the process we have just executed, we shall at once perceive that the fundamental principle of the method of indeterminate co-efficients consists in this, that, If an equation of the form 0 = A + Bx + Сx2 + Dx3 + (A, B, C, D, ... being co-efficients independent of x) hold good whatever be the value of x, then each of the co-efficients must separately be equal to 0. In fact, since the co-efficients are independent of x, if we can determine their value by making particular suppositions with regard to the value of x, these values must still be the same whatever value we may afterwards assign to 2. But by making x = 0, we find A = 0 and both members of the equation being divided by x, it becomes making x = 0 in this new equation, we find B=0 and the original equation is reduced, after dividing both members by x to A = 0, B = 0, C = 0, D = 0, separately, and in this manner, we obtain as many equations as there are co-efficients A, B, C, D, to be determined. ... This principle may be enunciated under a different form. hold good whatever be the value of x, the co-efficients of the terms affected by the same power of x, in the two members of the equation, are respectively equal to each other. For if we transpose all the terms in the second member of the equation, the equation will be of the same form as that given above, from whence we may conclude that and a - a' — 0, b — b'′ = 0, c — c' = 0, d — d' = 0... a = a', b = b, c = c', d=d... we shall have several examples of the application of this principle in what follows. To expand a3. a may be put under the form 1 + (a − 1) .. a2 = {1 + (a — 1)}* Expanding by the binomial theorem x − 1) x = 1 + x (a− 1) + * (x — 1) (a−1)2 + 7 Arranging according to powers of ø where =1+{(a−1)—{(a—1)2+}(a—1)3—{(a—1)*+...}x+Px2+Qx3+. =1+ px + qx2 + rx3 + p = (a1) § (a − 1)2 + † (a — 1)3 — † (a − 1)2 + . . . . = (1 + px + qx2 + ra3 + . . .) × (1 + px + qx2 + ræ3 + ...) Performing the multiplication = 1 + 2p. x + (p2 + 2g) x2 + 2 (pq + 1') x3 + ... (2) Equations (1) and (2) are identical, therefore comparing the co-efficients of homologous terms |