palm_ch9

# palm_ch9 - Solutions to Problems in Chapter Nine Solutions...

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Unformatted text preview: Solutions to Problems in Chapter Nine Solutions to Test Your Understanding Problems T9.1-1 The sessions are: (a) syms x E1 = x^3-15*x^2+75*x-125; E2 = (x+5)^2-20*x; S1 = E1*E2; factor(S1) ans = (x-5)^5 S2 = E1/E2; simplify(S2) ans = x-5 S3 = E1+E2; G = sym(subs(S3,x,7.1)) G = 7696088813222736*2^(-49) G = simplify(G) G = 481005550826421/35184372088832 H = double(G) H = 13.6710 T9.2-1 The session is: syms x solve(sqrt(1-x^2)-x) ans = 1/2*2^(1/2) T9.2-2 The session is: S = solve( x+6*y=a , 2*x-3*y=9 ); S.x ans = 9-1 18/5+1/5*a S.y ans =-3/5+2/15*a T9.2-3 After the session shown at the end of Example 9.2-1, type subs(S.y,b,sqrt(3)) ans = 4.7320 3.0180 T9.3-1 The session is: syms x E = diff(sinh(3*x)*cosh(5*x)); subs(E,x,0.2) ans = 9.2288 T9.3-2 The session is: syms x y diff(5*cosh(2*x)*log(4*y),y) ans = 5*cosh(2*x)/y T9.3-3 The session is: syms x int(x*sin(3*x)) ans = 1/9*sin(3*x)-1/3*x*cos(3*x) T9.3-4 The session is: syms x y int(6*y^2*tan(8*x),y) ans = 2*y^3*tan(8*x) 9-2 T9.3-5 The session is: syms x E = int(x*sin(3*x)); subs(E,x,5)-subs(E,x,-2) ans = 0.6672 T9.3-6 The session is: syms x taylor(cos(x),5) ans = 1-1/2*x^2+1/24*x^4 T9.3-7 The session is: syms m symsum(m^3) ans = 1/4*m^4-1/2*m^3+1/4*m^2 T9.3-8 The session is: syms x symsum(cos(pi*x),0,7) ans = T9.3-9 The session is: syms x limit((2*x-10)/(x^3-125),5) ans = 2/75 9-3 T9.4-1 The session is: dsolve( D2y+b^2*y=0 ) ans = C1*cos(b*t)+C2*sin(b*t) T9.4-2 The session is: dsolve( D2y+b^2*y = 0 , y(0) = 1 , Dy(0) = 0 ) ans = cos(b*t)) T9.5-1 The session is: syms a b t laplace(1-exp(-a*t)) ans = 1/s-1/(s+a) ilaplace(ans) ans = 1-exp(-a*t) laplace(cos(b*t)) ans = s/(s^2+b^2) ilaplace(ans) ans = cos(b*t) T9.5-2 Applying the Laplace transform to the equation gives 5[ s 2 Y ( s )- 5 s- 1] + 20[ sY ( s )- 5] + 15 Y ( s ) = 30 U ( s )- 4 sU ( s ) Solve for Y ( s ) using U ( s ) = 1 /s . Y ( s ) = 25 s 2 + 101 s + 30 s (5 s 2 + 20 s + 15) The session is: syms s ilaplace((25*s^2+101*s+30)/(s*(5*s^2+20*s+15)) ans = 2-8/5*exp(-3*t)+23/5*exp(-t) 9-4 T9.6-1 The session is: syms a R = [cos(a),sin(a);-sin(a),cos(a)]; E = subs(R,a,3*a); F = R*R*R; G = F-E; simplify(G) ans = [0,0] [0,0] The variable F represents RRR ; the variable E represents R (3 a ). Because F- E is a zero matrix, F = E , and thus RRR = R (3 a ). T9.6-2 The session is: syms k A = [-2, 1;-3*k, -5]; poly(A); ans = x^2+7*x+10+3*k solve(ans) ans = [ -7/2+1/2*(9-12*k)^(1/2) ] [ -7/2-1/2*(9-12*k)^(1/2) ] T9.6-3 The script file is: syms c A = sym([-4, 6;7, -4]); B = sym([-2*c;23]); x = inv(A)*B x = A\B The two results are equivalent. The solutions are x = 69- 4 c 13 y = 46- 7 c 13 9-5 Solutions to End-of-Chapter Problems 1. The session is: syms x y E = sin(x)^2+cos(x)^2; simplify(E) ans = 1 F = sin(x+y); expand(F) ans = sin(x)*cos(y)+cos(x)*sin(y) G = sin(2*x); expand(G) ans = 2*sin(x)*cos(x) H = cosh(x)^2-sinh(x)^2; simplify(H) ans = 1 2. The session is: syms y E = cos(5*y); expand(E) ans = 16*cos(y)^5-20*cos(y)^3+5*cos(y) Thus the polynomial is 16 x...
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## This note was uploaded on 10/21/2009 for the course ENGR 290 taught by Professor Priritera during the Spring '05 term at S. Alabama.

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palm_ch9 - Solutions to Problems in Chapter Nine Solutions...

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