CALCULUS MATH*2080 SAMPLE FINAL EXAM
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1 CALCULUS MATH*28 SAMPLE FINAL EXAM Sample Final Exam Page of 2 Prof. R.Gentry Print Your Name Student No. SIGNATURE Mark This exam is worth 45% of your final grade. In Part I In Part II In part III In part IV you will be asked to indicate a choice for the correct procedure. Please indicate only one answer that you believe is correct. If more than one answer choice is indicated the problem will be considered wrong. you need not evaluate the given integral, simply indicate.if it is an improper integral how it should be evaluated. you should give complete answers in the space provided. you should circle the correct multiple choice. If more than one answer is circled the problem will be marked wrong. You may not utilize any calculator or computing device. No cell phones or pagers may be used during the examination. You may utilize a 3" by 5" facts card for formulas, etc., that has your name and ID number and is turned in with the exam. There are pages including this cover page.
2 Sample Final Exam Page 2 of 2 4 pts Part I For each integral indicate in space proceeding the integral which of the following methods should be used first to evaluate the integral. Write only one letter in the space provided. Each problem is worth.5 points. DO NOT EVALUATE THE INTEGRALS, IT WILL NOT INCREASE YOUR MARK. CHOICES are denoted (a), (b),...etc. a. A linear substitution u = ax + b b. A non-linear substitution u = a(x r + b c. The substitution x = a(sin(θ) d. The substitution x = a(tan(θ) e. Integration by Parts f. Partial Fractions g. Division h. Apply a standard integration rule. State the Rule. i. None of the above () B (2) D x(x /4 ) dx x 94x 2 dx (3) C 9x 2 dx (4) E x 2 e -.5x dx x 2 x 3 (5) G dx (6) F 65x x 2 x2 x 2 4x2 dx (7) E cos 2 (x) dx (8) H x 2 (9 - x 3 ) 2 dx
3 Sample Final Exam Page 3 of 2 4 pts Part II In the box below each integral indicate if it is improper? If it is, indicate how it should be evaluated. You do not need to complete the evaluations. Each problem is worth.5 points. II-() x /3 2 dx II-(2) " x 2 e x dx Improper ' Not Improper Improper ' Not Improper II-(3) 5 25x 2 x dx II-(4) " sin(x)e cos(x) dx ' Improper Not Improper Improper ' Not Improper II-(5) dx II-(6) 65x x 2 5 x5 x 2 4x2 dx Improper ' Not Improper Improper ' Not Improper II-(7) e x ln(x) dx II-(8) x 2 e x dx " Improper ' Not Improper Improper ' Not Improper
4 Sample Final Exam Page 4 of 2 27 pts Part III Solve each problem in the space allotted. 2 pts Part III. The concentration of a drug in an individual's blood was monitored following an injection at time t = and the data fit the curve C(t) = 25 t e -.5t µg/l (t in hrs.). What is the individual s average blood concentration of this drug over the 24 hours following the injection? Average blood concentration is A = And to evaluate we use integration by parts. Set u = 25t and v = e -.5t. Then u = 25 and v = -2e -.5t (Note: /-.5 = -2). So, 25t e -.5t dt = -5 t e -.5t - -5 e -.5t dt = -5 t e -.5t - e -.5t = -5[ t + 2] e -.5t = -5[ t + 2] e -.5t t = 24 t = = -5(26 e -2 + = [-3e ] So A = [2-3e -2 ]/24 If you use a calculator you will find that A pts Part III 2. If a sequence {X n } is generated by the rule that the next term is the sum of the present term and twice the previous term, with X = and X =, what is the value of x? (I know, it is a big number!) x = Model: Present term = X n next term = X n+ and previous term = X n-. Dynamical equation: X n+ = X n + 2 X n- or X n+2 - X n+ - 2 X n =. Characteristic equation is λ 2 - λ - 2 =. This factors as (λ -2)(λ +) =, giving λ = - and λ = 2. General form of the solution is thus X n = c (-) n + c 2 2 n. X = c (-) + c 2 2 = c + c 2 and X = c (-) + c 2 2 = -c + 2c 2 Setting X = gives c = - c 2. Substituting this into the equation = X = -c + 2c 2 gives = c 2. Hence, c 2 = 2/3 and c = /3. Thus, X n = /3 (-) n + 2/3 2 n. Finally, X = /3 (-) + 2/3 2 = [ + 2 ]/3
5 Sample Final Exam Page 5 of 2 3 pts Part III 3. Determine the general solution of the given equation: a. y + t y = y(t) = Use the formula for y = a(t) y y = C e a(t) dt a(t) = -t b. y - 4y = y(t) = C e 2t + c 2 e -2t The characteristic equation λ 2-4 GIVES λ = 2 AND λ = -2 c. y + 8y + 2y = y(t) = c e t + c 2 e -2t + c 3 e -6t The characteristic equation is λ λ λ = Which factors as λ(λ 2 + 8λ + 2) = λ(λ + 2)( λ + 6) = 3 pts Part III 4. Consider the equation y + a y + b y = a and b parameters. a. What condition on a and b will ensure that the solution y(t) can be y = c t? For this to happen λ = must be a repeated root. Thus the characteristic equation must be λ 2 = and hence we conclude that a = b =, b. What condition on a and b will ensure that limy(t) = "? t$" For the limit at infinity to equal infinity at least one of the eigenvalues must be positive, as terms such as e -2t approach zero. Furthermore the solution can not oscillate so it can not be complex, As λ =.5[ -a ± (a 2-4b) ½ ] we conclude that a > and b.5 a 2 c. What condition on a and b will ensure that the solution y(t) will be positive for all t if y() = 5? The solution can not have an oscillatory factor since these become negative. Hence b <.25a 2 Beyond this, the problem becomes difficult. The answer depends on the coefficients A function like y = e -2t - 5e -.2t starts at (,5) and then becomes negative before it approaches the t-axis. I will not give this question!!!
6 Sample Final Exam Page 6 of 2 3 pts Part III 5. Compute the average of F(x,y) = x y over the circle of radius centered at the origin. First sketch this region to see how it looks and determine its boundary curves. Show all necessary work. The circular region is R = {(x,y) - x and -( - x 2 ) ½ y ( - x 2 ) ½ } The average A is the integral of F over R divided by the area of R, which is 2π. y = = = = x Consequently the average value is just zero. 3 pts Part III 6. Sketch the region of integration of F(x,y) dy dx and determine the equivalent iterated 2 2x x integral(s) with the order of integration reversed. Note y = - x gives x = -y and y = 2 - x gives x = 2 -y. Easy! The region R = {(x, y) x 2 and -x y 2-x } must be split into THREE regions to integrate first with respect to x: R = {(x,y) y 2 and x 2 -y } plus R 2 = {(x,y) y and - y x 2 - y } R 3 = {(x,y) - y and - y x 2 } Thus the integral is F(x,y) dx dy + F(x,y) dx dy + F(x,y) dx dy y R R R 2 x
7 Sample Final Exam Page 7 of 2 3 pts Part III 7. Consider the surface z = x 3-75x + y 2-4y. a. What are the critical (x, y) points of this surface? z x = 3x 2-75 and z y = 2 y - 4 are defined everywhere so the only critical points are determined from the solutions of z x = z y =. These are y = 2 and x = ±5. Critical points are (5,2) and (-5,2). b. Determine if the stationary points correspond to local maximum or minim values. I.e., Classify the critical points using the Second Partial Derivative Test. As A = z xx = 6x B = z xy = and C = z yy = 2, the discriminate D = B 2 - AC = -2x. At the point (5,2), D = -6 < and A > so the surface has a local minimum. At the point (-5,2), D = 6 > and by the second partial derivative test so the surface has a saddle point and neither a local minimum nor a local maximum. 2pts Part III 8. What is the equation of the tangent plane to the surface z = x sin(πy) at the point (2, )? Standard form of Tangent Plane: z = z + z x (x,y ) (x - x ) + z x (x,y ) (y - y ) z x = sin(πy) and z y = πx cos(πy) z(2,) = 2sin(π) = z x (2,) = sin(π) = and z y (2,) = π2 cos(π) = -2 π So the tangent plane is z = + (x - 2) + -2 π(y - ) or simply z = -2 π(y - )
8 Sample Final Exam Page 8 of 2 6 pts Part III 9. Consider the non-linear dynamical system X = X - XY 2 Y = 3Y - 2XY. a. Determine all equilibriums of this system. Equilibriums are where X = and Y =. Factoring we see that X = X - XY 2 = X( - Y 2 ) = ; X = or Y = or Y = - Y = 3Y - 2XY = Y(3-2X) = ; Y = or X = 3/2. Therefore the equilibrium points are: (,), (3/2,) and (3/2,-). b. Determine the linear system obtained by linearizing these equations about one of the equilibriums with a positive Y-value. Show all necessary work. The point is (3/2, ) Linear Model x = x - 3y y = -2x + y We first linearize F(X,Y) = X - XY 2 and G(X,Y) = 3Y - 2XY about (3/2, ) F X (X,Y) = - Y 2 F Y (X,Y) = - 2XY G X (X,Y) = - 2Y G Y (X,Y) = 3-2X F X (3/2, ) = - 2 = F Y (3/2, ) = - 2 (3/2 ( = -3 G X (3/2, ) = - 2( = -2 G Y (3/2, ) = 3-2(3/2 = c. What is the solution of the linear system you found in part b? Differentiating x = -3y gives x = -3 y Substitute y = -2x to get x = -3((-2)x or x - 6x =. This equation has the characteristic equation λ 2-6 = or λ = ±. Its solution is thus x(t) = c + c 2 Since y = -x/3 this gives y(t) = c /3 - c 2 /3 2 pts Part III. Use the Lagrange method to find the maximum of the function F(x,y) = xy for points lying on the ellipse x 2 + 4y 2 =. The constraint function is g(x) = x 2 + 4y 2 -. The Lagrange equations are F x = λg x, F y = λg y and g =. These become y = 2λx x = 8λy and x 2 + 4y 2 - = The first two equations give y = 6λ 2 y. Hence, either y = or λ = /4 or λ = -/4. Case. If y = the constraint equation gives x = ±. Two critical points (,) and (-,): F(±, ) =. Case 2. If λ = /4 then y = x/2, which when substituted into the constraint equation gives x 2 + 4x 2 /4 = or x 2 = /2. This has two roots, x = +(/2) ½ and x = -(/2) ½. Evaluating the function F we get F((/2) ½, (/2) ½ /2) = (/2) ½ (/2) ½ /2 = /4 and F( -(/2) ½, -(/25) ½ /2) = -(/2) ½ ( (-(/2) ½ /2) = /4. Case 3. If λ = -/4 then y = -x/2 and again substituting this into the constraint equation gives x 2 + 4x 2 /4 = or x 2 = /2. This has two roots, x = +(/2) ½ and x = -(/2) ½, but this time the y-values have the opposite sign and hence evaluating the function we find F((/2) ½, -(/2) ½ /2) = (/2) ½ (-(/2) ½ /2) = -/4 and F( -(/2) ½, (/2) ½ /2) = -(/2) ½ ( (/2) ½ /2 = -/4. So, the maximum value of F(x,y) subject to the constraint is /4, which occurs at two critical points.
9 pts Section IV Multiple choice. Circle on this exam your choice for the correct answer. The problem will be marked wrong if more than one answer is circled. Each problem has only one correct answer and is worth one point. Sample Final Exam Page 9 of 2 Part IV. lim x$ x(ln(x) equals cos(2x) a. b. " c. /2 d. -/2 e. /4 Part IV 2. cos(3x)(e -2x dx = " a. " b. 2/3 c. 5/3 d. -/3 e. /3 Part IV 3. 4 To evaluate dx the correct substitution would be x( 259x2 a. x = 5 sin(3θ) b. x = (3/5) tan(θ) c. x = tan(5θ/3) d. x = (5/3) sin(θ) e. x = (5/3) tan(θ)
10 Sample Final Exam Page of 2 Part IV 4. The solution of the differential equation y = 3y - 6 for which y() = 4 is a. y(t) = 4 e 3t b. y(t) = e 3t + 3 c. y(t) = 2e -3t + 2 d. y(t) = 2e 3t + 2 e. y(t) = 5 e -3t - 9 Part IV 5. a. The general solution of the equation y + 2y + 5y = is y(t) = e -t [C cos(2t) + C 2 sin(2t)] b. y(t) = [C + C 2 t]e -t c. y(t) = C e -2t + C 2 e 2t d. y(t) = C cos(2t) + C 2 sin(2t) e. y(t) = C e -2t + C 2 e t Part IV 6. If F(x,y) = cos(πxy) + x 2 y, then F(,.5) equals a. -/4 b. (.5π +,.5π + ) c. (-π +, -.5π + ) d..5π(x + y)cos(.5πxy) + 2xy + x 2 e. (-.5π +, -π + ) f. [ln(t) - /t]/t - [sin(t) - e -sin(t) ]cos(t)
11 Sample Final Exam Page of 2 Part IV 7. If F = x 2 y + e xy, which of the following is false? a. F x = y(2x + e xy ) b. F y = x(x + e xy ) c. F xy = F yx d. F yy = x 2 e xy e. F xx = y 2 e xy Part IV 8. If u = (3, ) and v = (-3, 2), which of the following is false a. u+v = -7 b. v = SQRT(3) c. u is perpendicular to (, 3) d. v is parallel to (8, -2) e. v + u = (, 3) Part IV 9. If F(x,y) = xe -2y and d = (3,-) then the directional derivative D d F(2,) equals a. (-,4) b. 7 c. d. e. 5
12 Sample Final Exam Page 2 of 2 Part IV. The equation of the plane normal to the vector (3,-,2) that contains the point (,-,3) is a. x - + y+ + z -3 = -6 b. x - - y+ + 2z -3 = c. 3x - y + 2z = -6 d. 3(x - ) - (y+) + 2(z -3) = -6 e. 3(x - ) - (y+) + 2(z -3) =
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