math-1004-tests-1-2-3-4-5-6-winter-2019

Studocu is not sponsored or endorsed by any college or university MATH 1004 - Tests 1, 2, 3, 4, 5, 6 Winter 2019 Calculus for Engineering or Physics (Carleton University) Studocu is not sponsored or endorsed by any college or university MATH 1004 - Tests 1, 2, 3, 4, 5, 6 Winter 2019 Calculus for Engineering or Physics (Carleton University) Downloaded by Ammar Alturaihi (ammar.alturaihi@gmail.com) lOMoARcPSD|31437612

1 School of Mathematics and Statistics Carleton University Math. 1004A, Winter 2019 TEST 1 SOLUTIONS Any wireless calculator permitted, 1 or more blank sheets permitted for roughs Print Name : Student Number: Tutorial Section (H1, H2, H3, H4): PART I: Multiple Choice Questions (Choose and CIRCLE only ONE answer - No part marks here.) 1. [2 marks] Evaluate lim x →∞ parenleftbigg 2 x + 10 - 12 parenrightbigg . (a) 1, (b) 0, (c) 10 - 1 , (d) 10 - 12 Answer: (d) 2. [2 marks] Evaluate lim x →∞ x sin x. (a) 0, (b) 1, (c) − 1, (d) The limit does not exist Answer: (d) 3. [2 marks] Evaluate lim x → 0 sin(2 sin x ) sin x . (a) 1 / 2, (b) 2, (c) 0, (d) 1, (e) None of these or the limit does not exist. Answer: (b) 4. [2 marks] Which of the following functions is continuous at x = 0? (a) f ( x ) = 1 x , (b) f ( x ) = | x | , (c) f ( x ) = x, for x negationslash = 0 and f (0) = 1, (d) None of these. Answer: (b) 5. [2 marks] Evaluate lim x → 0 1 − cos( x 2 ) x 2 . (a) − 1, (b) 1 / 2, (c) 0, (d) 1, Answer: (c) PART II: Show all work here and give details. No additional pages will be accepted 6. [5+5 marks] a) A function f has the property that f (0) = 1, f (1) = − 1, f ( − 1) = 2 and f (2) = 0. Evaluate the composition f ( f ( f (0))). Answer: f ( f ( f (0))) = f ( f (1)) = f ( − 1) = 2 . b) Let f be defined by f ( x ) = braceleftbigg x, if x ≥ 0 , 1 , if x< 0 . Find lim x → 0 f ( x ). Give details for your answer. Answer: The limit does not exist because the left and right limits at x = 0 are different. For example, lim x → 0 - f ( x ) = 1 while lim x → 0 + f ( x ) = 0. Downloaded by Ammar Alturaihi (ammar.alturaihi@gmail.com) lOMoARcPSD|31437612

1 School of Mathematics and Statistics Carleton University Math. 1004A, Winter 2019 TEST 2 SOLUTIONS Any wireless calculator permitted, 1 or more blank sheets permitted for roughs Print Name : Student Number: Tutorial Section (H1, H2, H3, H4): PART I: Multiple Choice Questions (Choose and CIRCLE only ONE answer - No part marks here.) 1. [2 marks] Evaluate lim h → 0 f ( h ) − f (0) h where f ( x ) = x | x | , for all x . (a) 2, (b) 0, (c) − 1, (d) The limit does not exist Answer: (b) 2. [2 marks] Let f, g be two differentiable functions such that f ′ (0) = 2, g (1) = 0, g ′ (1) = − 1 and f (1) = 0. Evaluate ( f ◦ g ) ′ (1) . (a) − 2, (b) 0, (c) 2, (d) 1 Answer: (a) 3. [2 marks] Evaluate lim x → 0 sin(Arcsin x ) Arcsin x . (a) π , (b) 0, (c) 1, (d) − 1, (e) None of these or the limit does not exist. Answer: (c) 4. [2 marks] Which of the following functions is differentiable at x = 0? (a) f ( x ) = | x | , (b) f ( x ) = 1 /x , (c) f ( x ) = x 2 | x | , (d) f ( x ) = x | x | for x negationslash = 0 and f (0) = 1. Answer: (c) 5. [2 marks] Let x 2 + 3 xy 2 + sin( y ) = 0 describe a curve in the plane. Assuming that y can be written as a function of x calculate its derivative, y ′ , at the point x = 0 , y = π . (a) π , (b) − 1, (c) − π 2 , (d) 3 π 2 , Answer: (d) PART II: Show all work here and give details. No additional pages will be accepted 6. [5+5 marks] a) Write down the formula for the derivative, F ′ , of an inverse function f in terms of f ′ and F . Answer: F ′ ( x ) = 1 f ′ ( F ( x )) . b) Use the result in (a), above, to show that if a differentiable function f with an inverse F has the property that f ( − 1) = 1, f ′ ( − 1) = 1 / 3, f (1 / 3) = 2 and f (0) = 1. Evaluate the derivative F ′ (1). Answer: Since f ( − 1) = 1 it follows that F (1) = − 1. On the other hand, F ′ (1) = 1 f ′ ( F (1)) = 1 f ′ ( − 1) = 3 . Downloaded by Ammar Alturaihi (ammar.alturaihi@gmail.com) lOMoARcPSD|31437612

2 . 7. [5+5 marks] a) Let f ( x ) = Arctan(sin( x 2 )). Evaluate f ′ ( √ π ). b) Find the derivative of the function g ( x ) = sin(Arcsin ( √ cos x )) at x = 0. Answer: a) f ′ ( x ) = 1 1 + sin 2 ( x 2 ) 2 x cos( x 2 ) = ⇒ f ′ ( √ π ) = − 2 √ π Answer: b) Since sin(Arcsin x ) = x for all x = ⇒ g ( x ) = √ cos x . So, g ′ ( x ) = − 1 2 (cos x ) − 1 / 2 sin x and g ′ (0) = 0. Total: 30 marks Downloaded by Ammar Alturaihi (ammar.alturaihi@gmail.com) lOMoARcPSD|31437612

2 . 7. [2+3 marks] a) Evaluate lim x → 0 + d dx integraldisplay x 2 1 sin t t 3 / 2 dt using Leibniz’s Rule. b) Evaluate integraldisplay 2 0 t 1 + t 4 dt using any method. (Hint: Think arctangent) Answer: (a) Using Leibniz’s Rule, we get, for x > 0, d dx integraldisplay x 2 1 sin t t 3 / 2 dt = sin x 2 x 3 2 x - 0 = 2 sin x 2 x 2 . Since, lim x → 0 sin x 2 x 2 = 1, it follows that lim x → 0 + d dx integraldisplay x 2 1 sin t t 3 / 2 dt = 2 lim x → 0 sin x 2 x 2 = 2 . Answer: (b) Let u = t 2 , du = 2 tdt or tdt = du 2 . For t = 0, u = 0 while for t = 2, u = 4. Hence, integraldisplay 2 0 t 1 + t 4 dt = 1 2 integraldisplay 4 0 du 1 + u 2 = 1 2 Arctan u vextendsingle vextendsingle vextendsingle vextendsingle 4 0 = 1 2 Arctan 4 . Downloaded by Ammar Alturaihi (ammar.alturaihi@gmail.com) lOMoARcPSD|31437612

1 School of Mathematics and Statistics Carleton University Math. 1004 H, Winter 2019 TEST 4 Any wireless calculator permitted, 1 or more blank sheets permitted for roughs Print Name : Student Number: Tutorial Section (H1, H2, H3, H4): PART I: Multiple Choice Questions (Choose and CIRCLE only ONE answer - No part marks here.) 1. [3 marks] A continuous function f has an antiderivative F such that F (0) = 1, F (1) = 2 and F (2) = - 1. Evaluate the definite integral integraldisplay 2 0 f ( x ) dx . (a) 0, (b) - 2, (c) 3, (d) 1. Answer: (b) 2. [3 marks] Find the most general antiderivative of x 3 x 2 , that is, evaluate integraldisplay x 3 x 2 dx . (a) 3 x 2 3 log 2 + C , (b) 3 x 2 − 1 2 log 3 + C , (c) 3 x 2 − 1 log 3 + C , (d) 3 x 2 2 log 3 + C , Answer: (d) 3. [3 marks] Find the antiderivative F of f ( x ) = sin 3 x cos x such that F (0) = - 1. (a) sin 4 x 4 - 1, (b) cos 4 x 4 - 5 4 , (c) sin 4 2 x 4 - 1, (d) sin 3 x 3 - 1. Answer: (a) 4. [3 marks] Evaluate integraldisplay xe 2 x dx . (a) xe 2 x 8 - e 2 x 4 + C , (b) xe 2 x 4 - e 2 x 2 + C , (c) xe 2 x 2 - e 2 x 4 + C , (d) xe 2 x 2 - e 2 x 2 + C , (e) None of these. Answer: (c) 5. [3 marks] Let f be a differentiable function with f (0) = 1 and integraldisplay f ′ ( x ) dx = x 2 + C where C is a constant. Then f ( x ) = x 2 + 1. (a) TRUE, (b) FALSE, Answer: (a) PART II Show all work here and give details. Answers only are insufficient for full credit. No additional pages will be accepted 6. [5+5 marks] a) Evaluate integraldisplay 1 0 xe − x dx . Answer: Use Table method to get integraldisplay xe − x dx = - xe − x - e − x so that integraldisplay 1 0 xe − x dx = - xe − x - e − x vextendsingle vextendsingle vextendsingle vextendsingle 1 0 = 1 - 2 e b) Evaluate integraldisplay 3 x 2 sin xdx . Answer: Table method of integrating by parts is easiest! Downloaded by Ammar Alturaihi (ammar.alturaihi@gmail.com) lOMoARcPSD|31437612

1 School of Mathematics and Statistics Carleton University Math. 1004 H, Winter 2019 TEST 5 SOLUTIONS Any wireless calculator permitted, 1 or more blank sheets permitted for roughs Print Name : Student Number: Tutorial Section (H1, H2, H3, H4): PART I: Multiple Choice Questions (Choose and CIRCLE only ONE answer - No part marks here.) 1. [3 marks] Simplify using long division: 3 x 2 + 2 x + 1 (a) 3 x + 2, (b) 3 x - 1 - 5 x + 1 , (c) 3 x - 3 + 5 x + 1 , (d) 3 x - 1 + 5 x + 1 . Answer: (c) 2. [3 marks] Evaluate integraldisplay 3 sin 2 x cos x dx (a) sin 3 x + C , (b) sin 3 x 3 + C , (c) cos 3 x + C , (d) cos 3 x 3 + C , Answer: (a) 3. [3 marks] Evaluate integraldisplay sec 5 x tan x dx (a) sec 6 x 6 + C - 6, (b) tan 3 x 3 + C , (c) sec 5 x 5 - 1 + C , (d) sec 3 x tan 2 x 3 + C . Answer: (c) 4. [3 marks] Find the form of the partial fraction decomposition of x 2 + 2 x 2 ( x 2 - 1)( x 2 + 1) : (a) A x 2 + B x + 1 + C x - 1 + Dx + E x 2 + 1 , (b) A x + B x + 1 + C x - 1 + Dx + E x 2 + 1 , (c) A x + B x 2 + C x + 1 + D x - 1 + E x 2 + 1 (d) A x + B x 2 + C x + 1 + D x - 1 + Ex + F x 2 + 1 , (e) None of these. Answer: (d) 5. [3 marks] Let F ( x ) = integraldisplay tan 4 x sec 2 x dx with F ( π/ 4) = 3. Then F ( x ) = tan 5 x 5 + 3 (a) TRUE, (b) FALSE, Answer: (b) PART II Show all work here and give details. Answers only are insufficient for full credit. No additional pages will be accepted 6. [4+3 marks] a) Evaluate I = integraldisplay sin 3 x cos 2 x dx Answer: There are many ways of doing this, for example: sin 3 x cos 2 x = sin 2 x cos 2 x sin x = (1 - cos 2 x ) cos 2 x sin x which gives I = integraldisplay (cos 2 x - cos 4 x ) sin x dx = - integraldisplay ( u 2 - u 4 ) du = - cos 3 x 3 + cos 5 x 5 + C Downloaded by Ammar Alturaihi (ammar.alturaihi@gmail.com) lOMoARcPSD|31437612

2 b) Evaluate integraldisplay x + 1 ( x + 2)( x + 3) dx using the method of partial fractions. Answer: x + 1 ( x + 2)( x + 3) = 2 x + 3 - 1 x + 2 so integraldisplay x + 1 ( x + 2)( x + 3) dx = integraldisplay 2 x + 3 dx - integraldisplay 1 x + 2 dx = 2 ln | x +3 |- ln | x +2 | + C . 7. [3 + 3 + 2 marks] Convert and integrate the rational function x 4 + 1 ( x + 2)( x + 1) using partial fractions as follows: a) Use long division to find the remainder b) Find the partial fraction decomposition of the remainder c) Integrate the result in (b). ( NOTE : An answer alone is insufficient and worth only 1 mark) Answer: a) Long division gives x 4 + 1 ( x + 2)( x + 1) = x 2 - 3 x + 7 - 15 x + 13 ( x + 2)( x + 1) . b) 15 x + 13 ( x + 2)( x + 1) = - 2 x + 1 + 17 x + 2 c) integraldisplay x 4 + 1 ( x + 2)( x + 1) dx = integraldisplay ( x 2 - 3 x + 7) dx + integraldisplay 2 x + 1 dx - integraldisplay 17 x + 2 dx = x 3 3 - 3 x 2 2 + 7 x + 2 ln | x + 1 | - 17 ln | x + 2 | . Downloaded by Ammar Alturaihi (ammar.alturaihi@gmail.com) lOMoARcPSD|31437612

2 b) Find an expression for the solid of revolution obtained by rotating the region in the first quadrant bounded by the curves y = 2 x 2 , y = 2 and x = 0, about the y -axis. DO NOT EVALUATE the constants nor the integral. Answer: The two curves intersect when x = 1 and y = 2. Using vertical slices we get r out = x + dx , r in = x and height= 2 - 2 x 2 . When this slice is rotated about the y -axis we get its volume as π ( r 2 out - r 2 in ) dx = 2 π x (2 - 2 x 2 ) dx . Hence the required volume is integraldisplay 1 0 2 π x (2 - 2 x 2 ) dx. Alternately, using horizontal slices r out = radicalbig y/ 2, r in = 0 and height= dy . When this slice is rotated about the y -axis we get its volume as π ( r 2 out - r 2 in ) dy = π ( radicalbig y/ 2) 2 dy . Hence the required volume is also equal to integraldisplay 2 0 π 2 y dy. 7. [5 marks] Evaluate the improper integral integraldisplay 0 - 1 | x | - 1 / 2 dx . (Hint: Remove the absolute value and evaluate the integral using the definition of an improper integral) Answer: Since x < 0, | x | = - x and so, using the substitution - x = t , dx = - dt , and the definition of the improper integral we find that integraldisplay 0 - 1 | x | - 1 / 2 dx = integraldisplay 0 - 1 ( - x ) - 1 / 2 dx = lim ε → 0 - integraldisplay ε - 1 ( - x ) - 1 / 2 dx = lim ε → 0 - - integraldisplay - ε 1 t - 1 / 2 dt = lim ε → 0 - integraldisplay 1 - ε t - 1 / 2 dt = lim ε → 0 - 2 t 1 / 2 vextendsingle vextendsingle vextendsingle vextendsingle 1 - ε = 2 . Downloaded by Ammar Alturaihi (ammar.alturaihi@gmail.com) lOMoARcPSD|31437612