MATH 122, Final Exam

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1 MATH, Finl Exm Winter Nme: Setion: You must show ll of your work on the exm pper, legily n in etil, to reeive reit. A formul sheet is tthe.. (7 pts eh) Evlute the following integrls. () 3x + x x Solution. Let u = + x. Then u = 4xx, n so 3x + x x = 3 4 ( + x )( 4xx) = 3 4 = u + C = ( + x ) + C 3 uu = 3 4 u u () osx + sin x x Solution. Let u = + sin x. Then u = osx x, u = when x =, u = when x =, n so osx x = ( + sin x + sin x )(osx x) = u u = ln( u) = ln( ) ln() = ln

2 . (7 pts) Use logrithmi ifferentition to fin y x where y = xx. Solution. Sine ln y = ln( x x ) = xln x, it follows tht y y x = x [ln y] = x [x ln x] =(x)( ) + ()(ln x) = + ln x x y x = y(+ ln x) = xx (+ ln x) 3. (7 pts eh) Evlute the following limits. tn x () lim x x Solution. This is n ineterminte form of type. Thus, using L Hopitl s Rule, we hve tn x lim = lim + x x x x = + = () lim (+ x) x x Solution. This is form of type. If L = lim ( + x) x, then x ln L = ln( lim (+ x) x ) = lim (ln( + x) x ) = lim x x x (( x ln( + x) )ln( + x)) = lim x x n the ltter limit is of type. Thus, pplying L Hopitl s Rule, we hve ln( + x) ln L = lim = lim + x x x x = + = lim (+ x) x = L = e ln L = e x

3 4. ( pts) Fin the re of the resent shpe region tht is oune y the prols x = y n x = 3 3y (see figure). Solution. The y-interepts of the two urves re t (,±). Thus, using horizontl re elements, the re of the region is given y A = [(3 3y ) ( y )]y = ( y )y = (y 3 y3 ) = [( 3 ) ( )] = (5 pts) Use Simpson s Rule with n = 4 to otin n pproximtion of the integrl + sin x x. [Leve your nswer in the form of n unevlute sum.] Solution. There re four suintervls, so x = 4 n the points of suivision (inluing enpoints) re, 4,, 3 4,. Thus the Simpson s Rule pproximtion is = + sin x x x 3 {y + 4y + y + 4y 3 + y 4} 4 3 { + sin () sin ( 4 ) + + sin ( ) sin ( 3 4 ) + + sin () } = { + () ( ) + + () ( ) + + () } = { } = { } (ny of these forms ws fully eptle)

4 6. Let R e the region in the first qurnt tht is ove the line y = n uner the irle x + y = 4 (see figure). () ( pts) Use the metho of ross-setions (isks) to fin the volume of the spheril p tht is generte y rotting the region R out the y-xis. Solution. Using horizontl re elements, resulting in irulr ross setions (isks), the volume of the p is given y V= x y = ( 4 y ) y = (4 y )y = (4y y3 3 ) =[(8 8 3 ) (4 3 )] = 5 3 () (5 pts) Use the metho of ylinril shells to set up n integrl tht represents the volume of the spheril p in prt (). DO NOT EVALUATE THE INTEGRAL. Solution. The point of intersetion of the irle n the line is t x = 3, y =. Thus, using vertil re elements (orresponing to ylinril shell volume elements), the volume of the p is given y V = x( 4 x )x 3

5 7. (8 pts) The grvittionl fore exerte on l ojet hel t istne r miles from the enter of the erth is F(r) = k r where k =6 6. How muh work (in ft-ls) is one in lifting l ojet from the surfe of the erth to height miles ove the surfe? Note. The rius of the erth is 4 miles, n there re 58 feet in mile. Solution. The work one in lifting the ojet from the surfe ( r = 4 mi ) to height mi ove the surfe ( r = 5 mi ) is 5 W = F(r)r = 4 =(6 6 )( r ) k r = (6 6 ) r 4 5 r r 4 =(6 6 )( ) = 6 6 = 8 mi - ls, =(8)(58) = 4,4, ft - ls 8. (7 pts) Evlute the integrl 4 x x. Solution. First note tht follows tht x 4 x = ( x ) x. Thus, letting u = x ( u = x), it 4 x x = ( x x = ) ( x ( ) x) = u u = sin (u) = sin ( ) sin () = 6 = 6

6 9. ( pts) A pon is initilly stoke (t time t = ) with 4 fish, n therefter the fish popultion P = P(t) stisfies the ifferentil eqution P = k P where k is onstnt. t There re 6 fish in the pon fter months. Solve the ifferentil eqution to fin n expliit formul for P(t). How long will it e until there re fish in the pon? Solution. The generl solution of the ifferentil eqution is otine y seprting the vriles n then integrting oth sies: P P = kt P = kt P P = kt + C From the initil onition P() = 4 it follows tht 4 = k() + C. Thus C = 4, n so P = kt + 4 Now, from the onition P() =6, it follows tht 6 = k + 4 n so k =. Thus Finlly, there will e fish in then pon when i.e. fter 8 months. P = t + 4 P(t) = P =( t + 4 ) =(t + ) P(t) = (t + ) = t + = or t = 8

7 . ( pts) The riotive ron isotope 4 C hs hlf-life of pproximtely 57 yers. A prhment frgment hs een isovere tht hs 75% s muh 4 C riotivity s oes living plnt mteril. How ol is the prhment? Note. It is reommene tht you use your lultor for this prolem; otherwise you my leve your nswer in ext form (in terms of logrithms). Solution. Riotive sustnes ey t rte tht is proportionl to the mount present (nturl ey). Thus, if A = A(t) is the mount present t time t, then A = ka n t A(t) = A e kt The vlue of k is otine from the hlf-life informtion: A(57) = A A = A e 57k = e 57k 57k = ln( ln( ) ) = ln( ) k = 57.6 We now hve the following formul for A(t): ( ln A(t) = A e 57 )t A e.6 t The mount of 4 C will hve eye to 75% of the initil mount when A(t) =.75A = 3 4 A. This les to: ln ( A e 57 )t = 3 4 A e ( ln 57 )t = 3 4 ( ln 57 )t = ln( 3 4 ) = ln( 4 3 ) t = (ln( 4 57 )( 3 ln( ) ) 365 Thus the prhment is pproximtely 365 yers ol.

8 MISCELLANEOUS FORMULAS f(x)x x (y + y + + y n + y n ) f(x)x x (y 3 + 4y + y + 4y 3 + y 4 + y n + 4y n + y n ) V = y x = [ f (x)] x V = ([ f (x)] [g(x)] )x V = x y = [ f (y)] y V = ([ f (y)] [g(y)] )y V = xyx = xf(x) x V = x( f (x) g(x))x V = yxy = yf(y) y V = y( f (y) g(y))y L = + [ f (x) ]) x = + ( y x L = + [ g (y) ]) y = + ( x x ) y ) y P t = kp P(t) = P e kt A(y) y t = gy W = F(x)x F(x) = kx F(r) = k r W = ρya(y)y x [sin u] = u u x [ln x ] = x x [ex ] = e x u u x = sin u + C x [ln u ] = u u x x [eu ] = e u u x x [tn u u] = + u x u + u = tn u + C u = ln u + C u e u u = e u + C x [se u] = u u u u u = se u + C u x

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