SFU UBC UNBC Uvic Calculus Challenge Examination June 5, 2008, 12:00 15:00

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1 SFU UBC UNBC Uvic Calculus Callenge Eamination June 5, 008, :00 5:00 Host: SIMON FRASER UNIVERSITY First Name: Last Name: Scool: Student signature INSTRUCTIONS Sow all your work Full marks are given only wen te answer is correct, and is supported wit a written derivation tat is orderly, logical, and complete Calculators are optional, not required Correct answer tat is calculator ready, like + ln7or e, are preferred Any calculator acceptable for te Provincial Eamination in Principles of Matematics may be used 4 A basic formula seet as been provided No oter notes, books, or aids are allowed In particular, all calculator memories must be empty wen te eam begins 5 If you need more space to solve a problem on page n, work on te back of te page n- 6 CAUTION Candidates guilty of any of te following or similar practices sall be dismissed from te eamination immediately and assigned a grade of 0: (a) Using any books, papers or memoranda (b) Speaking or communicating wit oter candidates (c) Eposing written papers to te view of oter candidates Question Maimum Score Total 00

2 Calculus Callenge Eam, June Page of 8 [9] In eac case eiter compute te limit eplaining briefly ow you obtained te value, or eplain wy te limit does not eist (a) lim 0 4 / + lim = = = and lim / = = = / lim 4 + DNE so 0 / (b) lim ( 5)( ) ( + )( + ) ( )( ) ( 6 + 5) ( ) 4 ( + )( ) lim = lim = lim = lim 4 ( )( ) + = = since te degree of te numerator is te same as te degree of te denominator (c) lim + 6( /) / lim + = lim + = lim + = e / / [6] Find te derivatives of te functions below Do not simplify [6] Given ( ) (a) g( ) = sec 5 tan ( ) ( ) ( ( )) ( ) ( ) ( ) g = sec 5 tan tan 5 tan 0 sec (b) ( ) = ( + ) 5 e 5 / ( ) 0 ( ) = e + + e cos y sin = y use logaritmic differentiation to find an epression for dy and y No need to simplify te epression 6 6 d in terms of

3 Calculus Callenge Eam, June Page of 8 ln cos y = ln y sin cos yln = sin ln y d d cos yln = sin ln y d d dy dy sin y ln + cos y = cos ln y+ sin d y d sin dy cos y + sin yln = cos ln y y d dy cos y sin = cos ln y + sin yln d y [6] 4 Te limit lim 0 9 represents te derivative of some function f at te point = 0 (a) Find one possible function definition for f suc tat f ( 0) lim 0 9 = 9 9 ( 0+ ) 9 ( 0) ( 0) = lim = lim so in general ( ) 9 f any real number c 0 0 (b) Evaluate te limit directly witout using te fact tat it is equal to f ( 0) lim = lim = lim = lim = lim = = = ( 9 + ) ( + ) f = +c for [8] 5 Wat is an equation for te straigt line troug te point (, 0 ) tat is tangent to te grap of y = + at a point in te first quadrant? Let aa, + a be te point of tangency on te grap of y = + Ten te slope m can be a calculated in two ways: () y = and so m = = () Using te two points a a a + (, 0 ) and aa, + a a + m = = Setting te two equations a a a a equal to eac oter we get ( )( ) equation is given by y 0 ( ) te slope formula yields ( ) a a + = a a a ( ) Solving tis equation for a we obtain a+ a =0 Since a must be in te first quadrant a = Terefore, te tangent line = or y = + 6

4 Calculus Callenge Eam, June Page 4 of 8 [5] 6 Recall te definition of te inverse tangent function: θ = tan t t = tanθ and d Sow tat ( tan t) = + t π π < θ < d d dθ () t = ( tanθ) = sec θ Solving for d θ dθ we obtain =, wic is defined sec θ π π for < θ < Now, dθ d = =, so = ( tan t) = sec θ + tan θ + t + t [6] 7 Consider te grap of te function y = f ( ) sown below to te left Estimate te slope of te grap of f at various points and use tese estimates to sketc te grap of y = f ( ) f ( 5) 05, f ( 5) 05, f ( ), f ( ) 0 (, ) = (,6), f ( ) for = for (, ), f ( ) = 05 for, so te grap is sown above to te rigt 4 [6] 8 Use linear approimation (or differentials) to estimate ( 99) 4 Using linear approimation L( ) = f ( a) + f ( a)( a), we coose f ( ) differentiable everywere, and ( ) ( ) ( ) ( 4 ) L( ) ( ) =, wic is = 4 a = Ten f ( ) = = 6, f ( ) 4, and f = 4 =, and so te linearization of f at a is given by L( ) = 6 + ( ) Finally, f 99 = = = 568 [6] 9 Find te largest interval on wic te grap of te function f ( ) ln ln = is concave up + ln Te domain of f is D f = ( 0, ) f ( ) = and f ( ) = f ( ) = 0 / / + ln = 0 ln= / =e Now, on te interval ( ) / and so by te concavity test f is concave down on ( ) f >, and so by te concavity test f is concave up on ( /, ) we ave ( ) 0 <, 0,e we ave f ( ) 0 0,e However, on te interval ( e /, ) e

5 Calculus Callenge Eam, June Page 5 of 8 [8] 0 A cone is to be constructed from a circular piece of paper wit radius centimetres by cutting out a wedge as sown in te diagram below Wat is te maimum volume of te cone? (Hint: Vcone = π r ) r We need to maimize V = π r Te rigt diagram sows us tat r and are related troug Pytagoras Teorem r = = 69 Ten te volume can be epressed in only one 69 variable, namely, as V = π ( 69 ) = π π wit [ 0,] Differentiating te 69 equation wit respect to we get V = π π Solving V = 0 for we obtain te two critical numbers =± However, te negative value must be ecluded as only = [ 0,] Te maimum volume must occur at eiter te critical number or an endpoint 494 of te interval We compare V ( 0) = 0, V = π 69 π =, and V ( ) = 0, and conclude tat te maimum volume is π cubic units 9 [8] A particle is moving along te curve f ( ) = As te particle passes troug te point (, f ( ) ), its -coordinate increases at a rate of 5 cm/s How fast is te distance from te particle to te point 0, f ( 0) ( ) canging at tis instant? Let y = and let s denote te distance from te point s = + y Ten, Using te fact tat ( ) y (, y ) to te point ( 0, f ( 0) ) = ( 0,0) = tis equation simplifies to s = + = + 4 We differentiate tis equation wit respect to time t, and obtain ds d d ( ) d We are given s = + = + fact tat distance is positive we calculate d (,9) = 5 cm/s From s = + 4 and te 4 s = + = 90 = 0 Substituting all tese

6 Calculus Callenge Eam, June Page 6 of 8 quantities back into te equation s ( ) ds 95 = 0 cm/s [9] Sketc te grap of a function f wit te following properties (a) f is continuous on its domain { /, } (b) f ( 0) = and f ( 4) = are inflection points (c) f ( ) = 4, f ( ) = 0, and ( ) f < 0 ds d ds = + we get 0 ( () 5 = + ), and so Tis means tat te grap of f as a relative maimum point (, 4 ) by te second derivative test (d) lim f ( ) = and f ( ) lim = Tis means tat te grap of f as as orizontal asymptote y = (e) lim f, + ( ) = lim f ( ) =, lim f ( ) = and lim f ( ) + Tis means tat te grap of f as vertical asymptotes = and = (f) f ( ) 0 = < for <, < <, and > 4 (sould be ), and < <4 (sould be < < ) > for > f ( ) 0 Tis means tat te grap of f is decreasing for <, < <, and >, and increasing for < <

7 Calculus Callenge Eam, June Page 7 of 8 [9] Te levels of a sedative in a patient s blood were monitored to determine te appropriate time for an operation Every fifteen minutes a blood sample was taken to determine te concentration C of te sedative in milligrams per litre, and ten recorded in te table of data sown below Time (min) Concentration C (mg/l) (a) Estimate te rate of cange of concentration wit respect to time at 0 minutes and 60 minutes Is te rate of cange of concentration wit respect to time t a constant? Tere is a variety of way to estimate te rates asked for dc t= 0 dc t= 60 ( ) C( ) C = = ( ) C( ) C = = Since 05 is not close to 0065 te rate of cange of concentration wit respect to time is not constant (b) Sow tat te rate of cange is rougly proportional to te concentration Write tis relationsip as a differential equation leaving te constant of proportionality, k, undetermined 05 C ( 0) = and 0065 C ( 60) = bot yield quantities close to eac oter, wic indicates tat te rate of cange is rougly proportional to te concentration Terefore, dc = kc for some constant k (c) Solve te differential equation from part (b) and coose te constant of proportionality, k, C 0 = 0 and C 60 = from te table so tat te solution satisfies bot te entries ( ) Write te constant of proportionality accurate to 4 decimal places Te solution to te initial value problem dc kc =, C ( 0) 0 ( ) kt = is C( t) 0e constant of proportionality k, we use te given data ( 60) 60k ln ( / 0) 0454 = e k = Terefore, C( t) 0e = To obtain te 60k C = and solve = 0e for k: t =

8 Calculus Callenge Eam, June Page 8 of 8 [8] 4 Below is te grap of f( ) = + (a) Grap and sade te region enclosed by te curves = ±, y = π, and f( ) = + (b) Find te area of te region described in part (a) (Hint: You may use information from a previous question on tis eam) Using symmetry, te area of te saded region can be calculated wit te following integral to π be π d π tan ( ) π tan ( 0) = = = 0 0 π + or π ( ) square units