Math 132 Fall 2007 Exam II22 October 2007Integral Formula: = d⌠⌡( )sec t3t + + 12( )sec t ( )tan t12( )ln + ( )sec t ( )tan t C 1. Suppose that = = = = ( )f x 2x. Calculate ( )( )D f 9 , the derivative of ( )f x evaluated at x = 9.) a) 23 ln(2) b) 29 ln(2) c) 43 ln(2) d) 49 ln(2) e) 427 ln(2) f) 23 ( )ln 2 g) 29 ( )ln 2 h) 43 ( )ln 2 i) 49 ( )ln 20 j) 427 ( )ln 2 Solution ( c )> f := x -> 2^(sqrt(x)); := f → x 2x> D(f)(x);122( )x( )ln 2x> D(f)(9);43( )ln 22. Calculate d⌠⌠⌠⌠⌡⌡⌡⌡13( )log3x x.a) − − − − 21( )ln 3 b) + + + + 21( )ln 3 c) − − − − 2 ( )ln 3 d) + + + + 2 ( )ln 3 e) − − − − 2 ( )ln 3 1f) + + + + 2 ( )ln 3 1 g) − − − − 32( )ln 3 h) + + + + 32( )ln 3 i) − − − − 12( )ln 3 j) + + + + 12( )ln 3 Solution ( g )> Int('log[3](x)',x)= Int(log[3](x),x); = d⌠⌡( )log3x x d⌠⌡( )ln x( )ln 3x> Int(ln(x)/ln(3),x)=student[intparts](Int(ln(x)/ln(3),x),ln(x)); #Integration by parts with u=ln(x) = d⌠⌡( )ln x( )ln 3x − x ( )ln x( )ln 3d⌠⌡1( )ln 3x> Int(ln(x)/ln(3),x)= value(student[intparts](Int(ln(x)/ln(3),x),ln(x))); = d⌠⌡( )ln x( )ln 3x − x ( )ln x( )ln 3x( )ln 3> Answer := Int('log[3](x)',x = 1..3) = subs(x=3, 1/ln(3)*x*ln(x)-1/ln(3)*x) - subs(x=1,1/ln(3)*x*ln(x)-1/ln(3)*x); := Answer = d⌠⌡13( )log3x x − − 32( )ln 3( )ln 1( )ln 3> Answer := value(rhs(Answer)); := Answer − 32( )ln 33. Suppose that = = = = ( )f x xx. Calculate D(f)(4). (D(f)(4) is the derivative of f(x) evaluated at x = 4.) a) + + + + 1 ( )ln 2 b) 2 ( ) + + + + 2 ( )ln 2 c) 4 ( ) + + + + 2 ( )ln 2 d) 2 ( ) + + + + 1 ( )ln 2 e) + + + + 8 ( )ln 2 f) + + + + 2 ( )ln 2 g) + + + + 4 ( )ln 2 h) 4 ( ) + + + + 1 ( )ln 2 i) 2 ( ) + + + + 4 ( )ln 2 j) 8 ( ) + + + + 1 ( )ln 2Solution ( j )> f := x -> x^sqrt(x); := f → x xx> D(f)(x);x( )x + 12( )ln xx1x> D(f)(4); + 4 ( )ln 4 84. A radioactive substance has mass 120g at time = = = = t4 and mass 90g at time = = = = t6. What is the mass at = = = = t 12 ? a) 197932 b) 198532 c) 199132 d) 199732 e) 120321 f) 120932 g) 121532 h) 122132 i) 122732 j) 123332 Solution ( g )> m:= t -> A*exp(-lambda*t); := m → t Ae( )−λ t> eqn1 := m(4) = 120; := eqn1 = A e( )−4 λ120> eqn2 := m(6) = 90; := eqn2 = A e( )−6 λ90> eqn3 := lhs(eqn1)/lhs(eqn2) = rhs(eqn1)/rhs(eqn2); := eqn3 = e( )−4 λe( )−6 λ43> eqn4 := combine(lhs(eqn3), exp) = rhs(eqn3); := eqn4 = e( )2 λ43> eqn5 := lambda = solve(eqn4, lambda); := eqn5 = λ12ln43> eqn6 := subs(eqn5, eqn1); #Substitute this value of lambdainto eqn1 to find A := eqn6 = Ae( )−2 ( )ln / 4 3120> eqn7 := A = solve(eqn6, A); := eqn7 = A6403> Answer := `m(12) = `*m(12) = subs( {eqn5, eqn7, t = 12}, m(t) ); := Answer = m(12) = A e( )−12 λ6403e( )−6 ( )ln / 4 3> answer := simplify(Answer); := answer1215325. The mass of a microbe colony splashing about in a nutrient broth triples every 12 hours. What is the colony's doubling time?a) 8 b) 8 ( )ln 2 c) 8 ( )ln 3 d) 8 ( )ln 2( )ln 3 e) 12 ( )ln 2( )ln 3 f) 12ln32 g) 3 ( )ln 122 h) 3 ( )ln 2 i) 2 ( )ln 3 j) 2 ( )ln 123 Solution ( e )Let T be the doubling time and let = ( )m t A 2tT be the colony's population at time t . > m := t -> A*2^(t/T); := m → t A 2tTThen> eqn := m(12) = 3*m(0); := eqn = A 212T3 A> solve(eqn, T);12 ( )ln 2( )ln 36. Suppose that ( )u t is the unique solution of the initial value problem , = = = = ddt( )u t − − − − M 5 ( )u t = = = = ( )u 0 1 where M is a constant. If lim → → → → t ∞∞∞∞ = = = = ( )u t 12 then what is M? a) 3 b) 4 c) 5 d) 10 e) 12 f) 15 g) 20 h) 30 i) 60 j) 120 Solution ( i )> eqn1 := Int(1/(M-5*u),u) = Int(1,t)+C; := eqn1 = d⌠⌡1 − M 5 uu + d⌠⌡1 t C> eqn2 := map(value, eqn1); := eqn2 = −15( )ln − M 5 u + t C> eqn3 := map( z -> -5*z, eqn2); := eqn3 = ( )ln − M 5 u − − 5 t 5 C> eqn4 := map( z -> exp(z), eqn3); := eqn4 = − M 5 u e( )− − 5 t 5 C> eqn5 := u = solve(eqn4, u); := eqn5 = u − M515e( )− − 5 t 5 C> eqn6 := limit(rhs(eqn5), t = infinity) = 12; := eqn6 = M512> M = solve(eqn6, M); = M 607. Suppose that = = = = ( )fx( )arcsec 5x. Calculate ( )Df −−−−13. (The derivative of f(x) at = = = = x −−−−13 ). a) −−−−94 b) −−−−54 c) −−−−43 d) −−−−53 e) −−−−920 f) 94 g) 54 h) 43 i) 53 j) 920 Solution ( f )> f := x -> arcsec(5*x); := f → x ( )arcsec 5 x> D(f)(x);1x2 − 251x2> D(f)(-1/3);9 16168. Suppose that = = = = ( )f x 120 ( )arctan x . What is ( )( )D f 4 ? (The derivative of f(x) at = = = = x 4 ). a) 2 b) 3 c) 4 d) 5 e) 6 f) 8 g) 10 h) 12 i) 15 j) 20 Solution ( e )> f := x -> 120*arctan(sqrt(x)); := f → x 120 ( )arctan x> D(f)(x);60x ( ) + 1 x> simplify(D(f)(4));6 9. Calculate d⌠⌠⌠⌠⌡⌡⌡⌡012xe( )2 xx. a) 14 b) 12 c) 34 d) 1 e) 2 f) e4 g) e2 h) 3 e4 i) e j) 2 e Solution ( a )> J := Int(x*exp(2*x),x = 0 .. 1/2); := J d⌠⌡0 / 1 2x e( )2 xx> K := student[intparts](J, x); #Integration by parts with u = x := K
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