Math 120 Answers for Homework 13


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1 Mth 12 Answers for Homework In this problem we will use the fct tht if m f(x M on n intervl [, b] (nd if f is integrble on [, b] then (* m(b f dx M(b. ( The function f(x = 1 + x 3 is n incresing function on [, 2], in fct it is incresing on ( 1,. This follows either from (i the fct tht x 3 (nd so lso 1 + x 3 is n incresing function nd tht the squre root function is incresing, nd so the composite is incresing; or (ii f (x = 3x3 2 for ll x > 1. 1+x 3 This mens tht the minimum vlue of f(x on [, 2] occurs when x =, nd the mximum vlue of f(x on [, 2] occurs when x = 2. So, on the intervl [, 2] we hve 1 = f( f(x f(2 = 3. Therefore by (* 2 = 1 (2 2 f(x dx 3 (2 = 6. (b If f on n intervl [, b] then (using m = in (*, = (b f(x dx. (c If f g on [, b] then f g on [, b]. By prt (b this mens tht ( f(x g(x dx. Since f(x g(x dx = ( f(x dx ( f(x dx g(x dx, or g(x dx, this mens tht f(x dx g(x dx. (d For ny number A, A A A. This follows from the definition of bsolute vlue. If A then A = A, nd so A A = A nd A = A A. If A then A = A (which is positive, nd A = ( A = A A, nd A A = A. I.e., in either cse A A A. Since this is true for numbers, it is lso true for functions. I.e., for ny x, using A = f(x in the inequlity bove, then f(x f(x f(x. 1
2 (e Combining prts (c nd (d, we get or f(x dx f(x dx f(x dx f(x dx f(x dx f(x dx. If we let C = f(x dx, nd D = f(x dx, then wht we ve shown bove is tht C D nd D C (which is the sme s D C. By Homework 2, 4(, this mens tht C D, or substituting the definitions of C nd D, tht f(x dx = C D = f(x dx. 2. Mss with vrying density ( The condition y 1 x is the sme s the two conditions y 1 x nd (1 x y. We lso hve the condition x. Let s drw pictures of ech of these three conditions seprtely: x y 1 x (1 x y The region D where ll three conditions hold is the intersection of these three pictures: 2
3 (b The piece of D whose xvlues re between x k 1 nd x k is prllelepiped of width x, with lefthnd edge of length 2(1 x k 1 nd righthnd edge of length 2(1 x k : 1 x k 1 1 x x x To pproximte the re of the prllelepiped by rectngle we cn use either 2(1 x k or 2(1 x k 1 s the side length of the rectngle, so the re of the prllelepiped is pproximtely 2(1 x k x (or 2(1 x k 1 x. (c The density t ny point (x, y is given by ρ(x, y = x 2 ; this is roughly constnt on the strip from prt (b. We cn use either the density when x = x k 1 or x = x k (or nywhere in between in the pproximtion. The totl mss is the re times the density, so possible pproximtions for the mss re 2(1 x k x 2 k x or 2(1 x k 1 x 2 k 1 x (or even 2(1 x k x 2 k 1 x or 2(1 x k 1 x 2 k x. (d The Riemnn sum pproximting the mss of D is simply the sum of ll the pproximtions: n 2(1 x k x 2 k x. k=1 (e As n increses, the sum in prt (d pproches the vlue of the integrl 1 2(1 xx 2 dx; since s n increses the sum in prt (d lso gives better nd better pproximtion to the mss of D, the integrl bove gives the exct mss. 3
4 x (f The exct mss of D is 1 2(1 xx 2 dx = ( 2 3 x3 2 1 ( 2 4 x4 = x= 3 1 ( = Moment of inerti of disk ( The totl mss of D is the re of D times the density, so M = πr 2 ρ Kg. (b The piece of D consisting of ll the points whose distnce is between x k 1 nd x k from the origin is ring of inner rdius x k 1, outer rdius x k, nd width x: x k 1 x k Detil of piece of ring: x x 1 x 2 x k 2 x k+1 (c If we unroll the piece from prt (b we cn pproximte it by rectngle of width x nd height 2πx k (or 2πx k 1. Therefore the pproximte re of the piece is 2πx k x (or 2πx k 1 x. (d The mss is the re times the density. Therefore the pproximte mss of this piece is 2πρ x k x Kg (or 2πρ x k 1 x Kg. (e The points of the piece re pproximtely constnt distnce from the origin (the distnce vries between x k 1 nd x k. The moment of inerti is the mss times the distnce squred. Therefore the moment of inerti of the piece is pproximtely (mss (distnce 2 = (2πρ x k x (x k 2 = 2πρ x 3 k x. (Eqully vlid pproximtions re: 2πρ x 3 k 1 x, or 2πρ x k 1x 2 k x, or 2πρ x kx 2 k 1 x. (f The Riemnn sum pproximting the moment of inerti of D is the sum of the pproximtions of the moment of inerti of the pieces: n 2πρ x 3 k x; k=1 4
5 other sums re lso possible using the different pproximtions from prt (e. (g As we increse n the Riemnn sum from (g pproches the vlue of the integrl r 2πρ x 3 dx; since, s n increses, the Riemnn sum from (g lso gives us better nd better pproximtion for the moment of inerti of the disk D, the integrl bove gives us the exct moment of inerti. (h The exct moment of inerti of D is therefore r 2πρ x 3 dx = 2πρ 4 x4 r x= = ( ( 2πρ 2πρ 4 r4 4 4 = ρπ 2 r4 Kg. (i Since, from prt ( the mss of the disk is M = ρπr 2 Kg, we cn rewrite the moment of inerti of the disk from (h s ρπ 2 r4 Kg = 1 2 (ρπr2 Kg r 2 = Mr2 2. Bonus Problem: By definition, set X hs the sme crdinlity s set Y if there is function f : X Y which is bijection, i.e., function which is both n injection (lso known s onetoone nd surjection (lso known s onto. If X hs the sme crdinlity s Y then there is bijection f : X Y. If Y hs the sme crdinlity s Z then there is bijection g: Y Z. In order to show tht X nd Z hve the sme crdinlity, we need to show tht there exists some bijection from X to Z. Let h = g f; this is function from X to Z. We will show tht h is bijection (nd therefore tht X hs the sme crdinlity s Z by showing tht h is both n injection nd surjection. In doing this we will use the fct tht both f nd g re bijections. h is n injection: Suppose tht there re x 1, x 2 X such tht h(x 1 = h(x 2. We wnt to show tht this implies tht x 1 = x 2. But, if h(x 1 = g(f(x 1 = g(f(x 2 = h(x 2, then since g is n injection this mens tht f(x 1 = f(x 2. But now since f is n injection this mens tht x 1 = x 2. Hence h is n injection (since x 1 nd x 2 were rbitrry in this rgument. h is surjection: Suppose tht we re given some z Z. We wnt to show tht there is n x X so tht h(x = z. Since g is surjection there is y Y so tht g(y = z. Since f is surjection there is n x X so tht f(x = y (the sme y. Therefore h(x = g(f(x = g(y = z, nd so h is surjection. 5
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