ELEMENTARY PROBLEMS TREATED NON-ELEMENTARY
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1 ELEMENTARY PROBLEMS TREATED NON-ELEMENTARY DANIEL SITARU Abstract. Witout entering into details regarding Fermat s teorem for functions aving multiple variables, te teorem of bilinear forms and Fubini s formula for calculus of double and triple integrals we can consider known teir results and we can apply tem to te class or to te circle of pupils to te following types of problems: Problem 1: Find te minimum of te expression: E(x, y) = x + y x 4y + 1; x, y R { E x = x E y = y 4 ; { E x = E y = { x = 1 y = min E(x, y) = E(1, ) = = 5 E x = ; E y = ; E xy = ( ) H E (1, ) = 1 = > ; = 4 > (1, ) minimum point Problem Find te minimum of te expression: E(x, y, z) = x + y + z x 4y + 6z + E x = x E y = y 4 E z = z + 6 E x E x = ; E y = E z = = ; E xy = ; E x = 1 ; y = z = xz = E yx = ; E y = ; E yz = E zx = ; E zy = ; E z = H E (1,, ) = 1 = > ; = 4 > ; = 8 > (1,, ) minimum point min E(x, y, z) = E(1,, ) = 16 1
2 DANIEL SITARU Problem For wic value m >, te area of te set: { A = (x, y) m x m; y x + 6 } x is minimum? Find te minimum of te area. Let be f : (, ) R; f(m) = area(a) = dxdy = m m ( x+ 6 x dy ) dx = m + m f (m) = m = 1 min(a) = f(1) = 9 Problem 4 Prove tat te volume of te cube aving te side a is V = a. Let be C(x, y, z) = {(x, y, z) x a, y a, z a}. a a ) ) a a ) a V = dxdydz = dz dy dx = ady dx = = a a dx = a Problem 5 Prove tat te volume of te cuboid aving te sides a, b, c is V = abc. Let be P D(x, y, z) = {(x, y, z) x a, y b, z c} a b ) ) c a ) b V = dxdydz = dz dy dx = cdy dx = = a bcdx = abc Problem 6 Prove tat te volume of regular rectangular prism aving te base side a and te eigt is V = a. Let be P (x, y, z) = {(x, y, z) x a, y a, z } a a ) ) a ) a V = dxdydz = dz dy dx = dy dx = = a adx = a
3 ELEMENTARY PROBLEMS TREATED NON-ELEMENTARY Problem 7 Prove tat te lateral area and te volume of te rigt circular cylinder aving te radius r and eigt are given by te formulas: A l = πrg; V = πr A f = π V f = π f : [, ] R; f(x) = r; f (x) = f (x)dx = π f(x) 1 + f (x)dx = π r dx = πr x = πr r 1 + dx = πrx = πr = πrg Problem 8 Prove tat te lateral area and te volume of te rigt circular cone aving te radius r,, generator g are given by te formulas: A l = πrg; V = πr
4 4 DANIEL SITARU x y 1 OA : 1 r 1 = ; OA : ya xr = ; OA : y = r x; f : [, ] R; f(x) = r x V f = π A f = π f (x) = r ; 1 + f (x) = 1 + r = + r f r x πr (x)dx = π dx = x f(x) 1 + f rx g xdx = π = πr = g πr dx = g = πr x = πrg Problem 9 Prove tat te lateral area and te volume of te circular truncated cone aving te radii R, r, eigt, generator g are given by te formulas: A l = πg(r + r); V = π (R + r + Rr) x y 1 AB : r 1 R 1 = AB : xr + y r Rx = ; AB : y = x(r r) + r; AB : y = R r x + r f : [, ] R; f(x) = R r x + r; f (x) = R r ; 1 + f (R r) (x) = 1 + = g [ (R r ) x V f = π f (x)dx = π (R r)r + x + r ]= (R r) = π x π(r r)r + x +πr x = π(r r) +π(r)r+πr = π (R Rr + r + Rr r + r ) = π (R + Rr + r ) A f = π f(x) 1 + f (x)dx = π ( R r x + r ) g dx =
5 = πg ELEMENTARY PROBLEMS TREATED NON-ELEMENTARY 5 ( R r x +rx ) = πg ( R r ) ( R r ) +r = π +r = πg(r+r) Problem 1 Prove tat te area and te volume of te spere aving te radius R are given by te formulas: A = 4πR ; V = 4πR f : [, r] R; f(x) = r x ; f (x) = V f = π f (x)dx = π x r x ; 1 + f (x) = r r x (r x )dx = πr x r π x = = πr (r + r) π (r () ) = πr π r = 4πr r A f = π f(x) 1 + f (x)dx = π r x r r x dx = = πrx r ( ) = πr r () = 4πr Problem 11 Prove tat te lengt of te circle aving te radius R is given by te formula: L = πr
6 6 DANIEL SITARU f : [, r] R; f(x) = r x ; 1 + f (x) = 1 + x r x = r r x r r L f = 1 + f (x)dx = r x dx = r arcsin x r ( ) = r arcsin 1 arcsin( 1) = r ( π = r + π ) = πr; L semicircle = πr L circle = πr Problem 1 Prove tat te aria of te disc aving te radius R is given by te formula: A semidisc = f(x)dx = A = πr r x dx = πr ; A disc = πr References [1] C. Năstăsescu, C. Niţă, G. Rizescu, Algebra. Manual for te 9t grade. Didactical and Pedagocial Publising House, Bucarest, 1997 [] Daniel Sitaru, Claudia Nănuţi, Matematics for Olympiads. Ecko - Print Publising House, Drobeta Turnu - Severin, 14. Matematics Department, Teodor Costescu National Economic College, Drobeta Turnu - Severin, MEHEDINTI. address: dansitaru6@yaoo.com
IOAN ŞERDEAN, DANIEL SITARU
Romanian Mathematical Magazine Web: http://www.ssmrmh.ro The Author: This article is published with open access. TRIGONOMETRIC SUBSTITUTIONS IN PROBLEM SOLVING PART IOAN ŞERDEAN, DANIEL SITARU Abstract.
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