University of Toronto Solutions to MAT187H1S TERM TEST of Thursday, March 20, 2008 Duration: 90 minutes

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University of Toronto Solutions to MAT187H1S TERM TEST of Thursday, March 2, 28 Duration: 9 minutes Only aids permitted: Casio 26, Sharp 52, or Texas Instrument 3 calculator.

1 University of Toronto Solutions to MAT187H1S TERM TEST of Thursday, March 2, 28 Duration: 9 minutes Only aids permitted: Casio 26, Sharp 52, or Texas Instrument 3 calculator. Instructions: Make sure this test has 8 pages. Answer all questions. Present your solutions in the space provided. The value for each question is indicated in parantheses beside the question number. TOTAL MARKS: 6 Comments and Results: The marks on this test were on average much lower than on the first test. Nevertheless, the questions on this test are all very routine; four of them were right out of the homework. Nobody nobody who did their homework! should have failed this test. Concerning Question 5, part b): you can t use the second derivative test to show that x = 2 is a maximum since =.8)x.4)x2 d2 x 2 which means the second derivative test doesn t apply. =.8).8)x d2 x = at the critical value x = 2, 2 Breakdown of Results: 469 students wrote this test. The marks ranged from 7% to 1%, and the average was 63.5%. Some statistics on grade distributions are in the table on the left, and a histogram of the marks by decade) is on the right. Grade % Decade % 9-1% 7.7% A 22.2% 8-89% 14.5% B 15.6% 7-79% 15.6% C 22.2% 6-69% 22.2% D 18.1% 5-59% 18.1% F 21.9% 4-49% 13.% 3-39% 6.% 2-29% 2.1% 1-19%.4% -9%.4 %

2 1. [8 marks] Consider the curve with parametric equations Find both in terms of the parameter t. Solution: x = t 2 + 4t; y = t 3 3t. dy = dy and d2 y 2 dy = 3t2 3 2t + 4 d 2 y 2 = dy = d ) 3t 2 3 2t + 4 2t + 4 = 6t2t + 4) 23t 2 3) 2t + 4) 2 2t + 4 = 6t2 + 24t + 6 2t + 4) 3

3 2. [8 marks] Solve for y as a function of x if y + 4y + 2 y = ; y) = 9, y ) = 1. Solution: the auxiliary quadratic is r 2 + 4r + 2. Solve: Thus r 2 + 4r + 2 = r = 4 ± y = C 1 e 2x cos4x) + C 2 e 2x sin4x). To find C 1 use the initial condition y = 9 when x = : To find C 2 you need to find y : 9 = C 1 e cos + C 2 e sin C 1 = 9. = 2 ± 4i. y = C 1 2e 2x cos4x) 4e 2x sin4x)) + C 2 2e 2x sin4x) + 4e 2x cos4x)). Now substitute x =, y = 1, C 1 = 9 : 1 = 9 2 ) + C 2 + 4) 4C 2 = 28 C 2 = 7. Thus y = 9e 2x cos4x) + 7e 2x sin4x).

4 3. [8 marks] Consider the curve with parametric equations x = sin t, y = sin t cos t, for t π 2. Plot this curve and compute the volume of the solid of revolution produced by revolving the curve around the x-axis. Solution: t = x, y) =, ). t /2 x, y) = 1, ). dy = cos2 t sin 2 t cos t t 4 = t /4 x, y) = 1/ 2, 1/2); which is the maximum point. V = Let u = sin t) πy 2 π/2 π/2 sin 2 t cos 2 t cos t sin 2 t 1 sin 2 t) cos t u 2 1 u 2 ) du u 2 u 4) du [ 1 3 u3 1 5 u ) 5 = 2π 15 Alternate solution: eliminate the parameter: y = x 1 x 2, x 1. So V = πy 2 ] 1 x 2 1 x 2 ) = 2π 15.

5 4. [8 marks] Plot the polar curve with polar equation r 2 = cos θ + sin θ for and find the area within the curve. Solution: θ = π/4 r 2 = θ = r 2 = 1 θ /4 r 2 = 2 θ /2 r 2 = 1 θ = 3π/4 r 2 = π 4 θ 3π 4 r 2 = cos θ + sin θ r = ± cos θ + sin θ r > is the red curve; r < is the green curve. A = 2 = 1 2 3π/4 π/4 3π/4 π/4 r 2 dθ ) cos θ + sin θ) dθ = [sin θ cos θ] 3π/4 π/4 ) 3π = sin cos 4 = 4 2 = 2 2 3π 4 ) sin π ) + cos π ) 4 4

6 5. [1 marks] As the salt KNO 3 dissolves in methanol, the number xt) of grams of the salt in solution after t seconds satisfies the differential equation =.8)x.4)x2. a) [6 marks] If x = 5 when t =, how long will it take an additional 5 g of salt to dissolve? Solution: this is logistic growth. =.8)x.4)x 2 =.8 x 1 x ) 2 2 x =, for some constant A 1 + Ae.8t Use the initial condition t =, x = 5 to find A : 5 = A 1 + A = 4 A = 3. Now let x = 1 and solve for t : 1 = e.8t 3e.8t = 1 e.8t = 3 t = ln So it will take about 1.37 seconds for another 5 grams of KNO 3 dissolve. to b) [4 marks] What is the maximum amount of salt that will ever dissolve in the methanol? Solution: find limit of x as t : lim t 2 2 = 1 + 3e.8t 1 + = 2. Alternate Calculations: if you didn t remember the logistic equation. =.8 x 1 x ) ) 2 =.8 2 x2 x) ) 1 x + 1 =.8t + C ln x ln 2 x =.8t + C 2 x ln x 2 x =.8t + C x 2 x = ±Be.8t 2 x = Ae.8t x 2 x = Ae.8t x = 1 + Ae ; with A =.8t ±B 1 = ±e C Now proceed as above.

7 6. [1 marks] If x is the amount of salt disolved in a saline solution of volume V, at time t, in a large mixing tank, then + r V x = r ic i, where c i is the concentration of salt in a solution entering the mixing tank at rate r i, and r is the rate at which the well-mixed solution is leaving the tank. A 2-liter tank initially contains 1 liters of brine i.e. saline solution) containing 25 kg of salt. Brine containing 1 kg of salt per liter enters the tank at the rate of 5 liters per sec, and the well-mixed brine in the tank flows out at the rate of 3 liters per sec. How much salt will the tank contain when it is full of brine? Solution: V = 1 + r i r o )t = 1 + 2t. The integrating factor of the differential equation t x = 5 is and so ρ = e R 3 1+2t = e 3 ln1+2t)/2 = 1 + 2t) 3/2 x = 5 ρ ρ = 1 + 2t)5/2 + C C = 1 + 2t) t) 3/ t). 3/2 Use the initial condition t =, x = 25 to find C : Thus 25 = 1 + C C = /2 x = 1 + 2t) t) 3/2. The tank is full when V = t = 2 t = 5. At t = 5, x = 1 + 1) = ) 3/ So when the tank is full of brine it contains about kilograms of salt.

8 7. [8 marks] A bomb is dropped initial speed zero) from a helicopter hovering at a height 8 m. A projectile is fired from a gun located on the ground 8 m from the point directly beneath the helicopter. The projectile is supposed to intercept the bomb at a height of exactly 6 m. If the projectile is fired at the same instant that the bomb is dropped, what should be its initial speed and angle of inclination? Use g = 9.8 m/sec 2.) Solution:, 8) Trajectory of the bomb is, 6) y = gt2. So y = 6 2 = 1 2 gt2 v α 8, ) x y, ) t 2 = 4 g t = 2 g The parametric equations of the projectile s trajectory are x = 8 + v cos α)t; y = v sin α)t 1 2 gt2. At t = 2 g, x, y) =, 6) 8 + 2v cos α g = 2 + 2v sin α = 6 g { v cos α = 4 g v sin α = 4 g tan α = 1 α = 45 v = 4 g cos 45 = 4 2g So the projectile should be aimed at an angle of 45 from the horizontal with an initial speed of m/sec. NB: if the projectile is fired from the right side of the helicopter, then α = 135 counter clockwise from the positive x-axis.

University of Toronto FACULTY OF APPLIED SCIENCE AND ENGINEERING FINAL EXAMINATION, JUNE, 2012 First Year - CHE, CIV, IND, LME, MEC, MSE

University of Toronto FACULTY OF APPLIED SCIENCE AND ENGINEERING FINAL EXAMINATION, JUNE, 212 First Year - CHE, CIV, IND, LME, MEC, MSE MAT187H1F - CALCULUS II Exam Type: A Examiner: D. Burbulla INSTRUCTIONS: