P Calculus BC Chapter 6 - P Eam Problems solutions. E Epand the integrand. ( + ) d = ( + ) 5 4 5 + d = + + + C.. f ( ) = sin+ C, f( ) = +cos+c+k. Option is the only one with this form. sec d = d ( tan ) =tan + C 4. Since f is linear, its second derivative is zero. The integral gives the area of a rectangle with zero height and width ( b a). This area is zero. 5. D d= = 4 = 8. 6. D = + 7 7 ( ) 6 = ( +k) d d k d = 0 + ( ) k = 4k k = 4 7. C = d d = = 8. e e E d d = ln e 0 e = = = e 9 9. D ( ) d = ) ( d) = ( ) ( 8 ) = ( = ( ) 0 0 0 0. E d d lim d lim 0 = L 0 + L = L 0 + = which does not eist. L
. ( + )( d = d = ( )d = ( ) ) = + +. C k 0 8 = k = k k = 7, so k =. k k d= ( k (k ) ) )= 7 0 = ( + = only when k =. 4. The value of this integral is. Option is also and none of the others have a value of. Visualizing the graphs of y = sin and y = cos is a useful approach to the problem. 5. D The graph is a V with verte at =. The integral gives the sum of the areas of the two triangles that the V forms with the horizontal ais for from 0 to. These triangles have areas of / and respectively. 6. C The graph is a V with verte at =. The integral gives the sum of the areas of the two triangles that the V forms with the horizontal ais for from to 4. These triangles have areas of and 0.5 respectively. 7. By the Fundamental Theorem of Calculus c f ( d ) = f( ) = f(c) f(0) 0 0 c
8. B e e f( )d d d= + lne= = + 0 0 9. B 500 500 500 500 ( ) d + ) d = ( ) d ( ( ) d = ( ) d = = = 4.946 ln ln ln ln 0. E Q ( ) = p() degree of Q is n+. Differentiating the epression in (i) gives f ( ) = a + b Let =. Then from (ii), a+ b= f () =6 a+ b= f ( ) = 8 Solving these two equations gives a = and b = 6. Therefore f ( ) = 6 and hence f ( ) = 4 +C Using (iii) gives 8 = (4 +C) d 4 = ( + C) = (6 8 + C) ( + C ) = 8 + C Hence C = 0 and f( ) = 4 + 0.
. D ( d ) +5) ( + 5) ( d= 6 = ( + 5) +C = ( + 5) + C 6 6. D ( ) θ = + sin θ ( cos θdθ ) =( + sin ) ( ) 0 0 4. Let u = +. Then / d = C u / du = u +C = + + + 5. D 9 ( + ) d = ( d)= ( + ) = ( )= ( + ) 6 6 0 0 0 6. C cos() d = cos()(d) = sin( ) + C 7. D = = ( cos cos0)= sin( ) d cos() 0 0 8. C sin( + )d = )(d) = cos( sin( + + ) + C 9. B )= ( cos ) = y ( +C; Let =, 0 cos + C C= 0. y(0) = ( cos0) = 0. a du u =sin u + C, a> 0 a sin ( 0) = 4 d = sin 0 =sin 0
. du sin u d du = d =sin +C a u a 5 5. B f ( ) = e +e = e ( + ); f ( ) < 0 for < < 0. E + e e d = e e ( e d ). This is of the form u edu, u= e, so e + e e e d= +C 4. Use the technique of antiderivatives by parts u = dv= e d du = d v = e e e d= e e +C 4 5. e d (4 d) = 0 4 e = e = ( e 0 4 0 4 4 4 4 ) 6. B sin( ) tan( ) d ln cos( cos( ) d = ) +C = 7. B d d = = + ( ) ln = ( ln0 ln 5 ) + + = ln 8. E du d d ln d ln = = u (ln ) + C.This is udu with u = ln, so the value is
9. C F(9) F() 9 (ln t) = dt =5.87 using a calculator. Since F() = 0, F(9) = 5.87. t 40. B Or solve the differential equation with an initial condition by finding an antiderivative for (ln ). This is of the form udu where u = ln. Hence F( )= 4 (ln) + C and since 4 F() = 0, C = 0. Therefore (ln 9) 4 F(9) = =5.87 4 ( + ) d + = + + + d = ln = (ln 8 ln ) 4. 4. 4. E Since F is an antiderivative of f, f()d= F( ) = F(6) F() u =, du = d; when =, u = and when = 4, u = u u d du = u = u 4 du ( ) B Separate the variables. y dy =d; =+ C; y =. Substitute the point (, ) y + C to find the value of C. Then = C =, so y + C =. When =, y =. 44. C This is the differential equation for eponential growth. t y = y(0)e = e t ; = e t ; t = ln t = ln = ln 45. dy = y d dy = d ln y = + K ; y = Ce and y(0) = 8 so, y= 8e y 46. C dy y t an = sec d ln y = tan + k y= Ce. y(0) = 5 y= 5e tan 47. C,ln y + Ce dy d y = = C,y =. Only C is of this form.
48. (a) dy d = + y : answer dy d = y = 4 = + 4 = 4 8 = (b) y 4 = ( ) f(.) 4 (. ) { : equation of tangent line : uses equation to approimate f(.) (c) f(.) 0. + 4 = 4. y dy = ( + ) d y dy = ( + ) d y = + + C 4 = + + C 4 = C y = + + 4 y = + + 4 is branch with point (, 4) f() = + + 4 : separates variables : antiderivative of dy term : antiderivative of d term 5 : uses y = 4 when = to pick one function out of a family of functions : solves for y 0/ if solving a linear equation in y 0/ if no constant of integration Note: ma 0/5 if no separation of variables Note: ma /5 [-0-0-0-0] if substitutes value(s) for, y, or dy/d before antidifferentiation (d) f(.) =. +. + 4 4.4 : answer, from student s solution to the given differential equation in (c)
49. (a) ln y dy = d y (ln y) = + C or (ln y) = + C or ln y or =± C y = e ± C (b) (ln e ) C = 4 (ln y) = 0 + C = 4 ln y =± 4 But =0, y = e ln y = 4 y = e 4 (c) y If =, then y = and ln y = 0. This causes ln y to be undefined.
50. Å Y E E Y E DY X DX Y X # X # Y LN X # LN Y LN X # ÅÅ# E E ÅSEPRTESÅVRIBLES ÅNTIDERIVTIVEÅOFÅDYÅTERM ÅNTIDERIVTIVEÅOFÅDX ÅTERM ÅCONSTNTÅOFÅINTEGRTION Å ÅUSESÅINITILÅCONDITIONÅF ÅSOLVESÅFORÅY Å.OTEÅÅIFÅY ÅISÅNOTÅÅLOGRITHMICÅFUNCTIONÅOFÅX.OTEÅMXÅÅ;=ÅIFÅNOÅCONSTNTÅOF INTEGRTION.OTEÅÅIFÅNOÅSEPRTIONÅOFÅVRIBLES B $OMINÅX E X E X E E NGEÅd Y d ÅX E ÅDOMIN Å ÅÅÅÅÅÅ.OTEÅÅIFÅÅISÅNOTÅINÅTHEÅDOMIN ÅRNGE.OTEÅÅIFÅYÅISÅNOTÅÅLOGRITHMICÅFUNCTIONÅOFÅX
5. = @O @N MDN! @O Г O ГO=! ГN @N O!Г!Г IBD=I=?=EEK=JJDEIFEJ H *?=KIBEI?JEKKIBH N #JDH EI=EJHL=?J=EECN!MDE?D @O OJDEIEJHL= EIC=JELJ @N @O JDBJBN!=@ EIFIEJELJJD @N HECDJBN!6DHBHBD=I=?= EEK=JN! > O@O O! ГN@N! N Г N + & & Г& + +& O $ N ГN $ O Г $ N ГN $ N!!?=EEK KIJEBE?=JE IF=H=JIL=HE=>I =JE@HEL=JELB@OJH =JE@HEL=JELB@N JH $?IJ=JBEJCH=JE KIIEEJE=?@EJEC$ Г" ILIBHO J=N!$EB?IJ=J BEJCH=JE J$EBIF=H=JEBL=HE=>I
5. (a) dy d dy d dy y (6 Г ) Гy d = y (6 Г) Гy, 4 4 0Г Г 8 : dy : d < Г > product rule or chain rule error : value at, 4 (b) dy (6 )d Г y Г 6 Г C y Г 4 8 Г 9 C 9 C C Г 6 : : separates variables : antiderivative of dy term : antiderivative of d term : constant of integration : uses initial condition f () 4 : solves for y y Г6 Note: ma /6 [---0-0-0] if no constant of integration Note: 0/6 if no separation of variables
5. (a) @ Г f ()d = @ f= ( )d lim f = ()d lim f( ) b@ = lim fb) ( Г f() 0 Г 4 Г4 b@ b b@ b : : use of FTC : answer from limiting process (b) f(.5) N f() f =() (0.5) = 4 Г ()(4)(0.5) Г f() NГ f= (.5)(0.5) NГГ (.5)( Г)(0.5).5 : : Euler's method equations or equivalent table : Euler approimation to f () (not eligible without first point) (c) dy Г d y ln y Г k y Ce Г 4 Ce Г ; C 4e y 4e e Г : separates variables : antiderivatives 5 : : constant of integration : uses initial condition f () 4 : solves for y Note: ma /5 [--0-0-0] if no constant of integration Note: 0/5 if no separation of variables
5. E 4 4 dt = ln t =ln 4 ln = ln 4 t 54. C Use the technique of antiderivatives by parts: u = dv= e d du =d v = e e e d= ( e e ) = e + 0 55. E Use the technique of antiderivatives by parts: Let u = and dv = cos d. ( ) = cos d = sin sin d ( s in +cos ) 0 = 0 0 56. E Use the technique of antiderivatives by parts: Let u = and dv = cos d. ( ) = cos d = sin sin d ( s in +cos ) 0 = 0 0 57 E Use the technique of antiderivatives by parts: u = and dv= sec d sec d= tan tan d= tan +ln cos + C 58. B Use the technique of antiderivative by parts, which is no longer in the B Course Description. The formula is udv= uv vdu. Let u = f( ) and dv = d. This leads to f ( ) d = f ( ) f ( )d.
59. B Use the technique of antiderivatives by parts: = dv =sin d u f ( ) du f ( ) = d v = cos f ( )sin d = f( )cos + f ( )sin d= f ( )cos + f ( )cos d and we are given that cos d f ( ) = f ( ) = 60. Use partial fractions to rewrite ( ) ( + ) as + = d ( ln ln + ) + C = ln ( + d = + C + + ) ( ) 6. D Use partial fractions: d = d = (ln ln + ) = ln ln 5 ln+ ln 4 = ln 8 ( + + 5 ) ( ) 6. Use partial fractions. = = 6+ 8 ( 4)( ) 4 4 d = ( ln 4 ln ) + C = ln + C 6+ 8 6. C ll slopes along the diagonal y = appear to be 0. This is consistent only with option (C). For each of the others you can see why they do not work. Option () does not work because all slopes at points with the same coordinate would have to be equal. Option (B) does not work because all slopes would have to be positive. Option (D) does not work because all slopes in the third quadrant would have to be positive. Option (E) does not work because there would only be slopes for y > 0.
64. Å ÅZEROÅSLOPEÅTÅÅPOINTSÅWITH ÅÅÅÅÅÅY ÅNDÅX Å Å ÅNEGTIVEÅSLOPEÅTÅ ÅNDÅ ÅÅÅÅÅÅPOSITIVEÅSLOPEÅTÅ ÅNDÅ ÅÅÅÅÅÅSTEEPERÅSLOPEÅTÅY ÅTHNÅY Å Å B 4HEÅGRPHÅDOESÅNOTÅHVEÅSLOPEÅÅWHEREÅY nåorån 4HEÅSLOPEÅFIELDÅSHOWNÅSUGGESTSÅTHTÅSOLUTIONS REÅSYMPTOTICÅTOÅY ÅFROMÅBELOWÅBUTÅTHE GRPHÅDOESÅNOTÅEXHIBITÅTHISÅBEHVIOR Å RESON C Y DY X DX Y X # # ÅÅÅ # Y X ÅÅÅY X ÅSEPRTESÅVRIBLES ÅNTIDERIVTIVES ÅCONSTNTÅOFÅINTEGRTION Å ÅUSESÅINITILÅCONDITIONÅF ÅSOLVESÅFORÅY ÅÅÅÅÅÅÅIFÅYÅISÅLINER.OTE MXÅÅ;=ÅIFÅNOÅCONSTNTÅOF INTEGRTION.OTE ÅIFÅNOÅSEPRTIONÅOFÅVRIBLES D ÅRNGEÅISÅ b Y Å NSWER ÅIFÅnÅNOTÅINÅRNGE Copyright 000 by College Entrance Eamination Board and Educational Testing Service. ll rights reserved.
65.
66. = @O @N MDN! @O Г O ГO=! ГN @N O!Г!Г IBD=I=?=EEK=JJDEIFEJ H *?=KIBEI?JEKKIBH N #JDH EI=EJHL=?J=EECN!MDE?D @O OJDEIEJHL= EIC=JELJ @N @O JDBJBN!=@ EIFIEJELJJD @N HECDJBN!6DHBHBD=I=?= EEK=JN! > O@O O! ГN@N! N Г N + & & Г& + +& O $ N ГN $ O Г $ N ГN $ N!!?=EEK KIJEBE?=JE IF=H=JIL=HE=>I =JE@HEL=JELB@OJH =JE@HEL=JELB@N JH $?IJ=JBEJCH=JE KIIEEJE=?@EJEC$ Г" ILIBHO J=N!$EB?IJ=J BEJCH=JE J$EBIF=H=JEBL=HE=>I
67. (a) : : zero slope at each point (, y) where = 0 or y = positive slope at each point (, y) where 0 and y > : negative slope at each point (, y) where 0 and y < (b) Slopes are positive at points (, y) where 0 and y >. : description (c) dy = d y ln y = + C y = 0 y = Ke, K = ±e = Ke = K y = + e e C e C 6 : : separates variables : antiderivatives : constant of integration : uses initial condition : solves for y 0 if y is not eponential Note: ma 6 [--0-0-0] if no constant of integration Note: 0 6 if no separation of variables