DEPARTMENT OF MATHEMATICS MATH 1510 Applied Calculus I October 12, 2016 8:30 am LAST NAME: FIRST NAME: STUDENT NUMBER: SIGNATURE: (I understand tat ceating is a serious offense DO NOT WRITE IN THIS TABLE Question Points Score 1 10 2 10 3 10 4 7 5 8 Total: 45 INSTRUCTIONS TO STUDENTS: Fill in clearly all te information above. Tis is a 50 minute exam. No calculators, texts, notes, cellpones or oter aids are permitted. Sow your work clearly for full marks. Tis exam as a title page, 5 pages of questions and 1 blank page at te end for roug work. Please ceck tat you ave all pages. Te value of eac question is indicated in te left-and margin beside te statement of te question. Te total value of all questions is 45. Answer all questions on te exam paper in te space provided. If you need more room, you may continue your work on te reverse side of te page, but clearly indicate tat your work is continued tere.
PAGE: 1 of 6 1. Evaluate te following limits, if tey exist. ( 1 [5] (a lim t 1 + t 1 t We see tat lim ( 1 t 1 + t 1 t ( 1 1 + t t 1 + t 1 (1 + t = t 1 + t(1 + 1 + t t t 1 + t(1 + 1 + t 1 1 + t(1 + 1 + t 1 1 + 0(1 + 1 + 0 = 1 2. ( 1 + 1 + t 1 + 1 + t [5] (b lim x 6 x + 6 We first compute te left and limit: lim x 6 x + 6 We also compute te rigt and limit: lim x 6 + x + 6 x 6 2(x + 6 (x + 6 2 x 6 1 = 2. x 6 + 2(x + 6 x + 6 x 6 + 2 = 2. Since te left and rigt and limits are not equal, we see tat lim x 6 x + 6 not exist. does
PAGE: 2 of 6 [10] 2. Find te values of a and b for wic te function ( πx 4 sin x < 2 4 f(x = ax 2 bx + 3 2 x < 3 2x + 3a 3b x 3 is continuous everywere. Since 4 sin( πx 4, ax2 bx + 3, and 2x + 3a 3b are continuous at every point in R, no matter te coice of a and b, we see tat f is continuous at every point in R\{2, 3}. For f to be continuous at x = 2, we need or lim f(x = f(2 f(x x 2 x 2 + ( πx lim 4 sin = 4a 2b + 3 x 2 4 ax2 bx + 3. x 2 + Since bot 4 sin( πx 4 and ax2 bx + 3 are continuous at x = 2, we ave and lim 4 sin(πx x 2 4 = 4 sin(π 2 = 4 lim ax2 bx + 3 = 4a 2b + 3 (= f(2. x 2 + Tus, we need 4 = 4a 2b + 3 or 4a 2b = 1 (. For f to be continuous at x = 3, we need or lim f(x = f(3 f(x x 3 x 3 + lim ax2 bx + 3 = 6 + 3a 3b 2x + 3a 3b. x 3 x 3 + Since bot ax 2 bx + 3 and 2x + 3a 3b are continuous at x = 3, we ave and lim ax2 bx + 3 = 9a 3b + 3 x 3 lim 2x + 3a 3b = 6 + 3a 3b (= f(3. x 3 + Tus, we need 9a 3b + 3 = 6 + 3a 3b or a = 1 2. Plugging a = 1 2 into ( yields 4( 1 2 2b = 1 or b = 1 2. Terefore, f is continuous everywere if a = 1 2 and b = 1 2.
PAGE: 3 of 6 [2] 3. (a Let f(x be a function. Wat is te definition of te derivative of f at te point x = a, f (a? Te derivative of f at te point x = a is f (a f(a + f(a ( f(x f(a or lim, x a x a provided te limit exists. [4] (b Let g(x = x 2 8x + 6. Compute g (4 using te definition. We compute g (4 g(4 + g(4 (4 + 2 8(4 + + 6 (4 2 + 8(4 6 16 + 8 + 2 32 8 + 6 16 + 32 6 2 = 0. [4] (c Let (x = (3x 3 x(5 2x. Compute (x. We use te Product Rule: (x = (9x 2 1(5 2x + (3x 3 x( 2 = 45x 2 18x 3 5 + 2x 6x 3 + 2x = 24x 3 + 45x 2 + 4x 5
PAGE: 4 of 6 [7] 4. Find te equation of te tangent line to te function f(x = 2x x 2 + 1 wen x = 2. Is te tangent line parallel or perpendicular to te line 25x 6y + 12 = 0? We first note tat f(2 = 2(2 (2 2 + 1 = 4 5. To compute f (x we use te Quotient Rule: f (x = (x2 + 1(2 (2x(2x (x 2 + 1 2 = 2x2 + 2 4x 2 (x 2 + 1 2 = 2 2x2 (x 2 + 1 2 So f (2 = 2 2(22 ((2 2 + 1 2 = 6 25. Tus, te tangent line to f(x at te point (2, 4 5 is y = 6 25 (x 2 + 4 5 = 6 25 x + 12 25 + 4 5 = 6 25 x + 32 25. Te line 25x 6y + 12 = 0 can be written as y = 25 6 x + 2. Since ( ( 25 = 1, te 6 25 6 tangent line is perpendicular to te line 25x 6y + 12 = 0.
PAGE: 5 of 6 5. A particle is moving along te x-axis, and its displacement in metres after t seconds is given by x(t = 1 3 t3 11 2 t2 + 28t 34, were t 0. [4] (a Wat is te velocity and speed of te particle at t = 5 seconds? We know tat te velocity v(t is given by Tus, te velocity at t = 5 seconds is v(t = x (t = t 2 11t + 28. v(5 = (5 2 11(5 + 28 = 25 55 + 28 = 2 m/s. We also know tat te speed is given by v(t. So te speed at t = 5 seconds is v(5 = 2 = 2 m/s. [4] (b At wat times is te particle s acceleration 0 m/s 2? We know tat te acceleration a(t is given by a(t = x (t = 2t 11. Tus, te acceleration is 0 m/s 2 wen a(t = 2t 11 = 0 or wen t = 11 2 seconds.
PAGE: 6 of 6 Tis page is intentionally left blank. Use it for roug work.