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V d? V (5) Transform = 0 into a function of p and d x2 dy e, having given x = r cos 0, y = q sin 0. dv dV sin cos O dV + do V d’ V cosO d’V cosA d V sino e dy* dr p2 dᎾ 2 sin 0 cos e d'V dV + po? dr do dy dr n + + r dr T 202 dy The expression for đV may be deduced from that of dy dx by putting O for 0. We then get 2 da V d? V sin d V sine Ꮎ dᏙ = cos 0 + + dx d go's d Ꮎ 7 dr 2 sin 0 cos dᏤ dV dr de de Adding these together, đV đV dy 1 d V 1 dv dx? dyi dr2 goz do r dr dv d V dy (6) Transform + + d x2 dy do2 into a function of r, 0, and , having given X = r cos , y = r sin o sin , % = r sin 0 cos 0. A slight artifice will enable us to do this with considerable facility. Assume p=r sin e, so that y = p sin , x = p cos , ps = p sin 0, Q = p cos . + + + = 0. =0 + Taking first the two variables y and %, we find as in the preceding example d’V đV V 1 đV dy dz dpi 'pi do p2 dop de In exactly the same way, the equations of condition being similar, we find + + Also, as in the first part of the last example, + 1 d V 1 DV cot Ꮎ dᏙ + p de po dr go2 dᎾ " Adding these three expressions, d'V div d’V dy do d x2 đ? V 1 TV 1 d°Ꮩ 2 dV cot Ꮎ d A V + + + + 0. dr2 po de p dg po dr 22 d Ꮎ By substituting for p its value, and making some obvious reductions, this becomes (sino O d.cos e = 0. sin o do This important equation is the basis of the Mathematical Theories of Attraction and Electricity. The artifice here used is given by Mr A. Smith in the Cambridge Mathematical Journal, Vol. 1. p. 122. (7) Transform the double integral Sfany"-1 dy dx into one where u and v are the independent variables, x, y, u, v being connected by the equations W + y = U, Y = UV. Therefore dy dx = ududv, and Slam-lyu-1 dy dx = Sfum +--) (1 – v)m-\yn-Idu dv. This transformation is given by Jacobi in Crelle's Journal, Vol. XI. p. 307: it is of great use in the investigation of the values of definite integrals. (8) Transform the double integral Sfex*+y* dx dy into one where r and are the independent variables, having given X = r cos 0, y = r sin 0, Sfero + yo dx dy = - Sférer dr de. (9) Having given 20 = r cos 0, y = r sin o sin , * = r sin 0 cos 0, transform the triple integral SSS V dæ dy dx into a function of r, e, and p. Using the same artifice as in Ex. 6, we find SSSV då dy dz = SSS V pol dr sin 0 de do. This is a very important transformation, being that from rectangular to polar co-ordinates in space. V = 1, SS du dy dz is the expression for the volume of any solid referred to rectangular co-ordinates: and it becomes SSS=2 dr sin 0 do do when referred to polar co-ordinates. If we suppose + (10) Having given z a function of x and y determined by the equation y? 1, a? 62 it is required to transform dx Sidx dy {1 + dx into a function of O and when x = a sin 8 cos , y = b sino sin , and consequently x = c cos 0. In this case = Hence dx dy dx dy = ab sin 0 cos 0, de' do do de dx dy dx dy - bc (sin 6)? cos , de' do do de dx dx da dx = a c (sin ) sin . de do do de Substituting these values in the general expressions for dr dz and dx dy, we find dx' dy' dz Ssdæ dy {1+ = Ssdo do sin 0 {ab? (cos()? +(c sin )' (a’ sin’p + bocosop)}4. Ivory, Phil. Trans. 1809. CHAPTER IV. ELIMINATION OF CONSTANTS AND FUNCTIONS BY MEANS OF DIFFERENTIATION. Ex. (1) y' = ax + b.........(1). To eliminate b, differentiate, when we have dy 2y = a.........(2). dx To eliminate a, substitute its value given by (2) in (1); dy then y* = 2xy + 6. d x To eliminate both a and b, differentiate (2) again ; then dy dy y dx + = 0. dir? (4) Eliminate a and b from the equation y - a mo? – bæ = 0; dạy 2 dy 2y the result is + = 0. dixo « PreviousContinue »