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Test2practice - MATH 151 SPRING 2006 COMMON EXAM II VERSION A LAST NAME First name(print INSTRUCTOR SECTION NUMBER UIN SEAT NUMBER DIRECTIONS 1 The

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Unformatted text preview: MATH 151, SPRING 2006 COMMON EXAM II - VERSION A LAST NAME, First name (print): INSTRUCTOR: SECTION NUMBER: UIN: SEAT NUMBER: DIRECTIONS: 1. The use of a calculator, laptop or computer is prohibited. 2. In Part 1 (Problems 1-13), mark the correct choice on your ScanTron form No. 815-E using a No. 2 pencil. For your own records, also record your choices on your exam! ScanTrons will be collected from all examinees after 90 minutes and will not be returned. 3. In Part 2 (Problems 14-17), present your solutions in the space provided. Show all your work neatly and concisely and clearly indicate your final answer. You will be graded not merely on the final answer, but also on the quality and correctness of the work leading up to it. 4. Be sure to write your name, section number and version letter of the exam on the ScanTron form. THE AGGIE CODE OF HONOR "An Aggie does not lie, cheat or steal, or tolerate those who do." Signature: DO NOT WRITE BELOW! Question 1-12 13 14 15 16 17 Points Awarded Points 48 9 14 9 10 10 100 1 PART I 1. (4 pts) If g(x) = (x3 + x)5 , then g (-1) = (a) 320 (b) 160 (c) 80 (d) 40 (e) 0 2. (4 pts) 4 ln 2 + ln (a) ln 12 (b) ln 6 (c) ln 25 (d) ln 54 (e) ln 24 3 = 4 Exam continues on next page 2 3. (4 pts) Find the slope of the tangent line to the curve sin(xy) = x2 - 3 at the point (a) -2 (b) -2 - 3 3 (c) -2 + 2 (d) -2 - 3 (e) -2 + 3 3, 3 . 4. (4 pts) If f (x) = (a) x + 2 1 - 2x (b) x 1 + 2x (c) x x (d) 1 + 2x 1 (e) - (x + 2)2 1 , then the inverse function of f (x) is x+2 Exam continues on next page 3 5. (4 pts) lim (a) 0 (b) 1 (c) 2 (d) 4 (e) 8 t0 sin2 (4t) = 2t2 6. (4 pts) If f (x) = x + sin(x) + 2e3x and g(x) = f -1 (x) , then g (2) = (a) 1 1 (b) 2 1 (c) 3 1 (d) 7 1 (e) 8 Exam continues on next page 4 7. (4 pts) If h(x) = sin2 (3x) , then h (x) = (a) 6 cos(3x) (b) 18(cos2 (3x) - sin2 (3x)) (c) 18 (d) 9(cos2 (3x) - 1) (e) 9 sin2 (3x) 8. (4 pts) Find the linear approximation of f (x) = (a) (b) (c) (d) (e) x 1 + 2 16 1 x - 4 8 x 1 - 2 16 1 x + 2 8 1+x 1 at a = 4 . x Exam continues on next page 5 9. (4 pts) The length of leg AB of right triangle ABC increases at a rate of 2 inches per second and the length of leg BC increases at a rate of 6 inches per second. At what rate in inches per second does the hypotenuse increase when AB = 3 and BC = 4 ? (a) 30 (b) 26 (c) 6 26 (d) 5 (e) 4 10. (4 pts) (a) 2 (b) 1 2 - e5x = x- 1 + e2x lim (c) - (d) -1 (e) Exam continues on next page 6 11. (4 pts) A curve C is given by the parametric equations x = 2t3 - 3t2 , y = t2 - t . Find all horizontal and vertical tangents. (a) horizontal tangent at t = 0 and t = 1 , vertical tangent at t = (b) horizontal tangent at t = 1 2 1 2 . , vertical tangent at t = 0 and t = 1 . (c) horizontal tangent at t = -1 and t = 1 , vertical tangent at t = 2 . (d) horizontal tangent at t = 2 , vertical tangent at t = -1 and t = 1 . (e) None of the above 12. (4 pts) Newton's method with the initial guess x1 = -1 is used to find a zero of f (x) = x5 - x + 1 . What value does Newton's method give for x2 ? (a) 5 4 5 4 (b) - (c) 3 2 (d) - 3 2 (e) -5 Exam continues on next page 7 PART II 13. (9 pts) If (x2 + xy)3 = x2 + y 2 - 6 , then find dy when (x, y) = (1, -2) . dx Exam continues on next page 8 14. Find g (x) for the following functions. Don't simplify! (a) (7 pts) g(x) = tan(3x) x2 + sec(x) Exam continues on next page 9 (b) (7 pts) g(x) = xe7x 5 -8x4 Exam continues on next page 10 15. Let h(x) = f (x4 ) where f (-1) = 5 , f (1) = 13 , f (-1) = 3 , f (1) = 2 , f (-1) = -4 and f (1) = -2 . (a) (6 pts) Compute h (-1) . (b) (3 pts) Compute h (-1) . Exam continues on next page 11 16. The position of a particle is given by the vector function r(t) = t - 2 sin t , 2 - 2 cos t . (a) (2 pts) Find the position of the particle at time t = /2 . (b) (3 pts) Find the velocity of the particle at time t = /2 . (c) (2 pts) Find the speed of the particle at time t = /2 . (d) (3 pts) Find the acceleration of the particle at time t = /2 . Exam continues on next page 12 17. (10 pts) A paper cup has the shape of a cone with height 10 cm and radius 3 cm (at the top). If water is poured into the cup at a rate of 2 cm3 /sec, how fast is the water level rising when the water is 5 cm deep? The volume of a cone is V = 1 r2 h . 3 End of exam 13 Spring 2006 Math 151 Exam 2A: Solutions c 2006, Art Belmonte Mon, 27/Mar 1. (a) Now g (x) = 5 x 3 + x 3x 2 + 1 , whence 4 (4) = 320. g (-1) = 5 (-2) 2. (a) We have 4 ln 2 + ln 3 = ln 24 + ln 3 = ln 16 4 4 3 4 4 7. (b) The chain and product rules give h (x) h (x) = = 2 (sin 3x)1 (cos 3x) (3) = 6 sin 3x cos 3x 6 3 cos2 3x - 3 sin2 3x = 18 cos2 3x - sin2 3x . 8. (c) The linearization of f (x) = x -1 at a = 4 is L (x) = f (4) + f (4) (x - 4) . Now f (x) = -x -2 . Thus = ln 12. L (x) = 1 4 - 1 16 (x - 4) = 1 2 - 1 16 x. 3. (d) Implicitly differentiate sin (x y) = x 2 - 3 with respect to x, then substitute the data (x, y) = 3, . 3 9. (c) Let x, y, z be as depicted in the diagram below. Apply the Pythagorean Theorem, then differentiate with respect to t. Finally, plug in the numerical data. z2 dz 2z dt dz dt dz dt A z y cos (x y) y + x y y + xy xy y Substitute data: y y = = = = = = 2x 2x cos (x y) 2x -y cos (x y) y 2 - cos (x y) x 2 3 - 3, at 3 cos 3 -2 - , slope 3 = = = = x 2 + y2 dx dy 2x + 2y dt dt x d x + y dy dt dt z (4) (6) + (3) (2) =6 5 4. (b) Solve y = f (x) for x. y = f (x) = yx + 2y yx x Thus f -1 (x) = 1 - 2x . x sin = 1. 0 (sin 4t)2 (4t)2 sin 4t 4t 2 B x C = = = 1 x +2 1 1 - 2y 1 - 2y y 10. (a) As x - the exponentials decay to zero. 2 - e5x x- 1 + e 2x lim = 2-0 =2 1+0 11. (b) Given x = 2t 3 - 3t 2 and y = t 2 - t, we have d x/dt = 6t 2 - 6t and d y/dt = 2t - 1. A horizontal tangent occurs when dy d y/dt = = 0. dx d x/dt This can occur when d y/dt = 0 and d x/dt = 0. Now d y/dt = 2t - 1 = 0 implies t = 1 , at which 2 1 4 5. (e) Use algebra and the fact that lim sin2 4t lim t0 2t 2 = = = d x/dt = 6 - t0 1 8 t0 lim 1 2 = - 3 = 0. Thus we have a 2 horizontal tangent at t = 1 . 2 A vertical tangent occurs when d y/dt dy = ="". dx d x/dt This can occur when d x/dt = 0 and d y/dt = 0. Now d x/dt = 6t (t - 1) = 0 implies t = 0, 1, at which d y/dt = 1 = 0. Thus we have vertical tangents at t = 0, 1. lim 8 8 (1)2 = 8 6. (e) By inspection, f (0) = 2. Thus g (2) = 0. Accordingly, g (2) = = = 1 f (g (2)) 1 f (0) 1 1 + cos x + 6e3x = x=0 12. (b) Let f (x) = x 5 - x + 1. Then f (x) = 5x 4 - 1. Given x 1 = -1, Newton's method yields x2 1 . 8 = = f (x 1 ) f (x 1 ) 5 1 -1 - = - . 4 4 x1 - 1 13. Implicitly differentiate x 2 + x y = x 2 + y 2 - 6 with respect to x. Solve for y = d y/d x, then substitute the data. 3 x2 + x y 2 3 3 10 r 2x + y + x y 2 = = 2x + 2yy 2x - 3 x 2 + x y 2x - 3 x 2 + x y 2 2 2 3x x 2 + x y - 2y y (2x + y) (2x + y) h y Substitite data: y = = 3x x 2 + x y - 2y 2 - 3 (1) (0) 2 = at (1, -2) . 3 (1) (1) + 4 7 14. (a) The Quotient Rule gives g (x) = x 2 + sec x sec2 3x (3) - (tan 3x) (2x + sec x tan x) x 2 + sec x 2 . (b) Use the Product Rule. g (x) = (1) e7x 15. 5 -8x 4 + xe7x 5 -8x 4 35x 4 - 32x 3 (a) Use the Chain Rule. h (x) = h (x) = h (-1) = h (-1) = f x4 f x 4 4x 3 f (1) (-4) (2) (-4) = -8 (b) Use the chain and product rules. h (x) = f x 4 4x 3 2 + f x4 12x 2 h (-1) = f (1) (16) + f (1) (12) h (-1) = (-2) (16) + (2) (12) = -32 + 24 = -8 16. Given position function r (t) = [t - 2 sin t, 2 - 2 cos t], the velocity is r (t) = [1 - 2 cos t, 2 sin t] and the acceleration is r (t) = [2 sin t, 2 cos t]. (a) At t = (b) At t = (c) At t = (d) At t = , 2 2, 2, , 2 the position is r the velocity is r the speed is r 2 2 2 = = 2 2 - 2, 2 . 12 + 22 = = [2, 0]. 5. = [1, 2]. the acceleration is r 17. The volume of a cone is V = 1 r 2 h, where r is the radius of 3 the cone and h is its height. Similar triangles (see diagram at r 3 3 upper right) give h = 10 , whence r = 10 h. Therefore, V V dV dt dh dt = = = = 1 3 3 10 h 2 h 3 h 3 100 9 dh h 2 100 dt 100 d V 100 (2) 8 dt cm/s. = = 2 2 9 9h 9 (5) 2 Fall 2006 Math 151 Common Exam 2A Thu, 26/Oct/2006 Name (LAST, First): For official use only! QN 114 15 Instructor: 16 17 Section # 18 19 Total Seat # PTS MAX 56 10 8 10 8 8 100 Signature: Instructions 1. In Part 1 (Problems 114), mark the correct choice on your ScanTron form using a No. 2 pencil. For your own record, also mark your choices on your exam! ScanTrons will be collected from all examinees after 90 minutes and will not be returned. 2. Be sure to write your name, section number, and version of the exam (2A or 2B) on your ScanTron. 3. In Part 2 (Problems 1519), present your solutions in the space provided. Show all your work neatly and concisely, and indicate your final answer clearly. You will be graded not merely on the final answer, but also on the quality and correctness of the work leading up to it. 4. Neither calculators nor computers are permitted on this exam. 5. Please turn off all cell phones so as not to interrupt other students. 1 Part 1: Multiple Choice (56 points) Read each question carefully. Each problem in Part 1 is worth 4 points. 1. Evaluate lim (a) 3/2 (b) 2 (c) 3 (d) (e) 2. Find the limit lim cos sin (sin x) . x0 3x cos x . x0 sin 2x (a) (b) 1 (c) cos 1 (d) 0 (e) does not exist 3. For what values of x is the tangent line to the graph of the function f (x) = 2x - sin 2x horizontal? (a) all integer multiples of (b) all odd integer multiples of (c) all integer multiples of (e) there is no such x 4. Suppose that x > 0 and let s denote the distance between the points (x, 0) and (0, 1). If x is changing with time and d x/dt = 2, then ds/dt =? (a) 2x (b) 2 (c) x2 x +1 2x 2 (d) all even integer multiples of (d) x2 + 1 4x 2 (e) 2 x +1 2 5. If log3 (3x + 2) = 2, what is x? (a) 0 (b) 4/3 (c) 3/2 (d) 7/3 (e) does not exist 6. Let f and g be functions defined on R with the following properties: f (3) = 1, f (3) = 3, g (1) = 3, g (1) = 2. Let h = f g be the composition function defined by h (x) = f (g (x)). Compute h (1). (a) 1 (b) 2 (c) 3 (d) 6 (e) We do not have sufficient information to compute h (1). 7. Given that f (x) = (b) [1, ) (a) (-, ) (c) (-, 1] (e) [0, ) 1 - x, find the range of f -1 , the inverse of f . (d) [-1, 1] 8. Find f (x) if f (x) = x cos2 x 3 . (b) -3x 2 cos2 x 3 sin2 x 3 (d) 6x 3 cos x 3 (a) -6x 2 cos x 3 sin x 3 (c) cos2 x 3 - 6x 3 cos x 3 sin x 3 (e) cos2 x 3 - 2x 4 cos x 3 sin x 3 3 9. The position of a particle A is given by r (t) = (t + 1) i + sin Find the tangent vector to the path of the particle at time t = 3. (a) i (b) j (d) i + 1 j 8 (e) 4i + j (c) 4i - j (t + 1) 8 j. 10. Let p be a polynomial of degree 2 such that p (1) = 1, p (0) = 2 and p (1) = 2. What is the value of p (-1)? (b) -1 (c) 0 (e) 3 (d) 1 (a) -3 11. The parametric curve x = t t 2 - 1 , y = t 2 - 1, crosses itself at (0, 0). Find the angle between the two tangent lines at (0, 0). (a) 0 (b) 30 (c) 45 (d) 60 (e) 90 4 12. Let L (x) be the linear approximation of f (x) = (a) 9.3 (b) 9.4 (c) 9.5 (d) 9.6 (e) 9.7 x at a = 100. Compute L (90). 13. Let q (x) be the quadratic approximation of p (x) = x 2 - 2x + 1 about a = 2. Compute q (1). (b) 0 (d) 2 (e) 4 (a) -1 (c) 1 14. Suppose g is the inverse function of f . Given that f (1) = 2, f (1) = 3, and f (2) = 3, we can deduce that g (2) equals (a) 3 (b) 2 (c) 1 (d) 1/2 (e) 1/3 5 Part 2: Work-Out Problems (44 points) Show all your work neatly and concisely, and indicate your final answer clearly. You will be graded not merely on the final answer, but also on the quality and correctness of the work leading up to it. Partial credit is possible. 15. Consider the curve C given by the parametric equations x = te-t , (a) [3 points] Find d x/dt and dy/dt. y = (2t + 1)1/3 , - < t < . (b) [3 points] Find dy/d x in terms of the parameter t. (c) [4 points] Obtain an equation of the tangent line to C at the point (0, 1). 6 16. [8 points] Find an equation of the tangent line to the curve x 2 - x y + y 3 = 25 at (1, 3). 7 17. The position of a particle at time t is given by r (t) = (2 cos 2t) i + (2 sin 2t) j. (a) [3 points] Find the velocity of the particle at time t = /12. (b) [2 points] Find the speed of the particle at a general time t. Simplify your answer. (c) [3 points] Show that at any time t, the particle's velocity and acceleration are orthogonal (perpendicular). (d) [2 points] Sketch the path of the particle on the grid below and indicate its direction of motion. 4 3 2 1 0 -1 -2 -3 -4 -4 -3 -2 -1 0 x 1 2 3 4 y 8 x 18. Let f (x) = . 7 - 3x (a) [2 points] Find f (2). (b) [2 points] Compute f ( f (2)). (c) [2 points] Let h = f f be the composite function defined by h (x) = f ( f (x)). Use the Chain Rule to compute h (2). (d) [2 points] Now write the linear approximation L (x) of h (x) at a = 2. L (x) = Please turn the page over for the LAST problem! 9 19. [8 points] The following rates of change refer to a triangle. The height of the triangle is increasing at a rate of 2 cm/min. The area of the triangle is decreasing at a rate of 4 cm2 /min. At what rate is the base of the triangle changing when the height of the triangle is 10 cm and the area of the triangle is 100 cm2 ? Clearly define all variables used in your solution. Include units with your answer. 10 Fall 2006 Math 151 Exam 2A: Solutions c 2006, Art Belmonte Mon, 30/Oct 1. (a) We have 3x cos x = lim lim x0 x0 sin 2x 3x 2x cos x sin 2x 2x 7. (c) The range of f -1 , the inverse of f (x) = 1 - x, is the domain of f . We require 1 - x 0 or 1 x; i.e., (-, 1]. 8. (c) If f (x) = x cos2 x 3 , the product and chain rules give f (x) = (1) cos2 x 3 + x 2 cos x 3 - sin x 3 3x 2 ; that is, f (x) = cos2 x 3 - 6x 3 cos x 3 sin x 3 . 9. (a) We have v (t) = r (t) = i + whence v (3) = i. 8 = lim 3 cos x 2 x0 sin 2x 2x = 3 2 (1) 3 = . 1 2 cos (t+1) 8 j, 2. (b) The sine and cosine functions are continuous on R. Moreover, compositions of continuous functions are continuous. Therefore, x0 10. (a) Let p (x) = ax 2 + bx + c. Then p (x) = 2ax + b and p (x) = 2a. Now substitute the numerical data. p (1) = a + b + c p (0) = b p (1) = 2a = 1 = 2 lim cos sin (sin x) = = = cos sin (sin 0) cos sin 0 cos 0 = 1. = 2 This gives a = 1, b = 2, and c = 1 - a - b = -2. Hence p (x) = x 2 + 2x - 2 and thus p (-1) = 1 - 2 - 2 = -3. 11. (e) Setting r (t) = t 3 - t, t 2 - 1 = [0, 0], we deduce that t = 1. Now v (t) = r (t) = 3t 2 - 1, 2t . The relevant tangent vectors are v (1) = [2, 2] and v (-1) = [2, -2]. Observe that v1 v2 = 0. This implies that the tangent vectors (and hence the tangent lines) are perpendicular. Hence the angle between them is 90 (graph on reverse). 1 12. (c) With f (x) = x 1/2 , compute f (x) = 2 x -1/2 . Then 3. (a) The slope of a horizontal tangent line is zero. Recall that f (x) = 2x - sin 2x. Thus f (x) = 2 - 2 cos 2x cos 2x 2x x where n is any integer. 4. (d) Use related rates. By the Pythagorean Theorem, we have s 2 = x 2 + y 2 = x 2 + 1, whence 2s ds dt ds dt = = 2x x dx dt dx dt = = = 0 1 2n n, = L (x) L (90) f (100) + f (100) (x - 100) 1 = 10 + 20 (-10) = 9.5. = s = 2x x2 + 1 13. (b) The quadratic approximation of a quadratic function is said function! Now p (x) = (x - 1)2 . So q (1) = p (1) = 0. [Nonbelievers: grind out the details if you must.] 14. (e) Since f (1) = 2 and g = f -1 , we have g (2) = 1. 1 1 1 = = . Thus g (2) = f (g (2)) f (1) 3 15. Recall that r (t) = [x (t) , y (t)] = te-t , (2t + 1)1/3 . . M151 X2A/P4 1 y 0 0 1 s x 2 (a) So r (t) = d x , d y = 1e-t + t -e-t , 1 (2t + 1)-2/3 2 . dt dt 3 5. (d) Use properties of logarithms. log3 (3x + 2) = 3x x 2 7 7/3 32 = 9 3x + 2 = = = (b) Hence m = dy d y/dt = = dx d x/dt 2 3 2 (c) At (x, y) = (0, 1), we have t = 0 and m = 3 . The point-slope formula gives a Cartesian equation of the tangent line. (2t + 1)-2/3 . (1 - t) e-t Checking this answer in the original equation, we verify that log3 9 = 2. 6. (d) We have h (x) = h (1) = h (1) = f (g (x)) g (x) f (g (1)) g (1) f (3) g (1) (3) (2) = 6. 1 y-1 = y = 2 3 (x - 0) 2 3x + 1 h (1) = 2 the tangent line is thus L (t) =[0, 1] + t 1, 3 Alternatively, a direction vector for the line is 2 r (0) = 1, 3 . A parametric representation of 2 or L (t) = t, 3 t + 1 . (See graph on reverse.) 16. Implicitly differentiate x 2 - x y + y 3 = 25 with respect to x, then immediately substitute (x, y) = (1, 3). This let's you do arithmetic instead of algebra (although the latter is fine too). Below, y represents the value of d y/d x at (1, 3). 2x - 1y + x y + 3y 2 y 2 - 3 + y + 27y y (c) Recall that h (x) = f ( f (x)). The Chain Rule gives h (x) h (2) = = f ( f (x)) f (x) f ( f (2)) f (2) = 42 = 16. 26y = = 0 0 1 1/26 (d) Now h (2) = f ( f (2)) = f (2) = 2. Thus L (x) = h (2)+h (2) (x - 2) = 2+16 (x - 2) = 16x-30. 19. Let A, B, and H represent the area, base, and height of the 1 triangle, respectively. Recall that A = 2 B H . At the instant of time t in question, we have A = 100 cm2 , H = 10 cm, and thus B = 2 A/H = 200/10 = 20 cm. Use related rates. Differentiate with respect to t, then substitute the numerical data given in the problem. dA dt = 1 dB dH H+B 2 dt dt 1 10B (t ) + 20 (2) 2 5B (t ) + 20 = = Now use the point-slope formula. y-3 y See graph at lower right. 17. Recall the position function r (t) = (2 cos 2t) i + (2 sin 2t) j. (a) The velocity is v (t) = r (t) = (-4 sin 2t) i + (4 cos 2t) j, whence v (/12) = -2i + 2 3 j. 16 = 4. = = 1 26 (x - 1) 77 1 26 x + 26 -4 = -24 = B (t ) = -4 = 5B (t ) -24/5 = -4.8 cm/min (b) The speed at time t is v (t) = Illustrative graphs M151 X2A/P11 1.5 1 M151 X2A/P15 16 sin2 2t + 16 cos2 2t = (c) The acceleration is a (t) = v (t) = (-8 cos 2t) i + (-8 sin 2t) j. 0 y 1 0.5 y 0 -0.5 -1 0 x 1 -2 -1.5 -1 -0.5 x 0 0.5 1 M151 X2A/P16 (unequal axis scales) The dot product of v (t) and a (t) is 32 sin 2t cos 2t - 32 sin 2t cos 2t = 0. Hence v (t) and a (t) are orthogonal (perpendicular). (d) Now r (t) = [x (t) , y (t)] = [2 cos 2t, 2 sin 2t]. Hence x 2 + y 2 = 4 cos2 2t + 4 sin2 2t = 4 = 22 . As t increases, the circle x 2 + y 2 = 22 is traversed over and over again in a counterclockwise direction. (See graph at bottom right.) 18. With f (x) = y 3.2 -1 M151, X2A/P17 4 3 3.1 2 1 3 0 -1 2.9 y 0 0.5 1 1.5 x 2 2.5 3 -2 -3 , the derivative of f is (7 - 3x)1/2 1 2 x 2.8 -4 -4 -3 -2 -1 0 x 1 2 3 4 f (x) = (7 - 3x)1/2 (1) - x (7 - 3x)-1/2 (-3) 7 - 3x 1 2 . (a) Therefore, f (2) = (1) (1) - 2 1 (1) (-3) = 1+3 = 4. 1 (b) Note that f (2) = 2. Thus f f (2) = f (2) = 4. 2 Spring 2007 Math 151 Common Exam 2 Test Form A PRINT: Last Name First Name: Signature: ID: Instructor's Name: Section # INSTRUCTIONS In Part 1 (Problems 112), mark the correct choice on your ScanTron form using a #2 pencil. For your own records, also record your choices on your exam! The ScanTrons will be collected after 1 hour; they will NOT be returned. In Part 2 (Problems 1318), write all solutions in the space provided. CLEARLY INDICATE YOUR FINAL ANSWERS No Calculators Permitted 1 Math 151 Multiple Choice (4 points each) Part I 1. Find the limit x0 lim 2 cos(x) sin(2x)(x + 1) (a) 0 (b) 1 2 (c) 1 (d) 2 (e) does not exist 2. For f (x) = 2x - 3 , calculate f -1 (x). 7x + 5 7x + 5 5x + 3 (a) f -1 (x) (b) f -1 (x) = 2x - 3 -7x + 2 1 x- 1 2 - 3x 3 (d) f -1 (x) = (e) f -1 (x) = 2 1 7 + 5x x+ 1 7 5 (c) f -1 (x) = 2x - 7 3x + 5 3. Find the derivative of f (x) = -1 - cos x sin2 x 1 (d) f (x) = - 1 + cos x (a) f (x) = 1 + cos x . sin x (c) f (x) = 1 - sin x sin2 x (b) f (x) = - tan x (e) f (x) = csc x + cot x 2 4. Calculate lim (a) -1 esin x - 1 . x0 x (b) 0 (c) 1 (d) 2 (e) does not exist 5. Let f and g be functions with the following properties: f (4) = 2, f (4) = 2, g(1) = 4 and g (1) = 2. Compute h (1), where h(x) = f (g(x)). (a) 2 (b) 4 (c) 6 (d) 12 (e) We do not have sufficient information. 6. Find the derivative of f (x) = xe + ex at x = 1. (a) 0 (b) 1 (c) e (d) 2 (e) 2e 3 7. Find the equation of the line tangent to the curve x4 + y 2 = 25 at the point (2,3). 3 (a) y = x 2 (d) y - 3 = - 16 (x - 2) 3 2x3 (x - 2) y y (e) y - 3 = 3 (x - 2) 2x (b) y - 3 = - 2 13 (c) y = - x + 3 3 8. A ball follows a trajectory given in parametric form by (x, y) = (cos(t) + sin(t), cos2 t + 1). What is the slope of the tangent line at the point (1,2)? (a) 0 (b) cos2 (2)+1 cos(-1)+sin(1) sin(2) cos(2) (c) -2 cos(1)-sin(1) sin(1) cos(1) (d) -2 cos(2)-sin(2) (e) undefined 9. Let Q be a polynomial of degree 2 such that Q(1) = 3, Q (0) = -1 and Q (0) = 2. What is Q(2)? (a) -1 (b) 1 (c) 3 (d) 5 (e) There is not enough information. 4 10. What is the differential of f (x) = sin(cos x)? (a) sin(cos(x))dx (b) - cos(cos(x)) sin(x) (c) - sin(cos(x)) sin(x)dx (d) - cos(cos(x)) sin(x)dx (e) (sin(cos(x)) + sin(sin(x)))dx 11. Approximate the cosine of (a) 1 3 (b) 1 18 (c) 1 radians by using the quadratic approximation of cos x near x = 0. 3 17 18 (d) 2 3 (e) 8 9 12. Suppose g is the inverse function of f . Assuming that f (1) = 2, f (2) = 3, and f (1) = 3 we can deduce that g (2) is (a) 3 (b) 2 (c) 1 (d) 1/2 (e) 1/3 5 Math 151 Work-Out Problems Show your work. No credit will be given to unsupported answers. Part II 13. In the search for a solution of the equation x2 - 3 = 0, consider the initial guess x1 = 2. (a) Use Newton's method with x1 = 2 as input to obtain a better approximation x2 to a solution. (6 points) (b) Use Newton's method with the value of x2 as input to obtain a still better approximation x3 to a solution. (4 points). 6 14. A 13-foot ladder is leaning against the wall. If the foot of the ladder is sliding away from the wall at a constant rate of 7 feet per minute, how fast is the top of the ladder sliding down the wall when the foot of the ladder is 5 feet from the wall? (10 points) 15. Calculate both 1st and 2nd derivatives of f (x) = 3x - 2. (6 points) 7 16. In each case, find dy in terms of x and y. dx (a) ex+y = x2 + y. (6 points) (b) 3x2 + y 3 + xy = 0. (5 points) 17. Calculate the derivative of f (x) = tan(x2 - x + 1). (5 points) 8 18. Let the position vector of a particle be given by r (t) = 2 cos t, 2 sin t at any given time t. (a) Calculate the velocity as a vector function of t. (3 points) (b) Calculate the acceleration as a vector function of t. (3 points) (c) Verify that the velocity is perpendicular to the acceleration at all times. 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Neither calculators nor computers are permitted on this exam. 5. Please turn off all cell phones so as not to interrupt other students. 1 Part 1: Multiple Choice (56 points) Read each question carefully. Each problem in Part 1 is worth 4 points. 1. Find the slope of the tangent line to y = sec x - 2 cos x at x = /3. (a) 3 (b) 2 3 (c) 3 3 (d) 1 (e) 0 2. A table of values for f , g, f , and g is given below. If H (x) = g ( f (x)), find H (1). x 1 2 3 f (x) 3 1 7 g (x) 2 8 2 f (x) 4 5 7 g (x) 6 7 9 (a) 30 (b) 65 (c) 24 (d) 63 (e) 36 3. Determine the limit lim (a) 0 (b) - cos 1 (c) sin 1 (d) cos 1 (e) - sin 1 sin (cos ) . 0 sec 2 4. Let f (x) = (a) 0 (b) 1/4 (c) 1/2 (d) 3/4 (e) 1 1 + xe-2x . Compute f (0). 5. Find a Cartesian equation of the tangent line to the parametric curve x = cos t + cos 2t, y = sin t + sin 2t for t = /2. (a) y = -x (b) y = 2x + 3 (c) y = 3x + 4 (d) y = x (e) y = x + 2 6. The length of a rectangle is increasing at a rate of 8 cm/s and its width is increasing at a rate of 3 cm/s. When the length is 20 cm and the width is 10 cm, how fast (in cm2 /s) is the area of the rectangle increasing? (a) 24 (b) 140 (c) 190 (d) 224 (e) 4800 3 7. A particle moves along the curve y = 1 + x 3 . As it reaches the point (2, 3), the y-coordinate is increasing at a rate of 4 cm/s. How fast (in cm/s) is the x-coordinate of the point changing at that instant? (a) 1 (b) 2 (c) 3 (d) 4 (e) 5 8. Use the linear approximation with a = 64 to estimate the number 3 (a) 4 4 3 70. (b) 4 1 2 (c) 4 1 4 (d) 4 1 8 1 (e) 4 16 9. Determine the quadratic approximation of f (x) = cos (x + ) at a = 0. (a) 1 (b) 1 - 1 x 2 2 (c) -1 + x 2 (d) sin x - (e) -1 + 2 1 2 x 2 4 10. Use Newton's method with initial approximation x1 = 0 to find x2 , the second approximation to a root of the equation e x - 4x - sin x = 0. (a) 1/e (b) 1/3 (c) 1/2 (d) 1/4 (e) 2/e 11. Let p (x) = ax 2 + bx + c, where a, b, c are constants. Given that p (1) = 8, p (1) = 4, and p (1) = 6, 1 what is p 2 ? (a) 3 1 2 (b) 4 1 2 (c) 6 3 4 (d) 8 1 (e) 12 4 12. Let g (x) = f -1 (x) be the inverse of f (x) = 3 + x 2 + tan ( x/2), -1 < x < 1. Find g (3). (a) 0 (b) 4/ (c) 1 (d) 2/ (e) /4 5 13. Evaluate lim (a) 0 (b) 1/2 (c) 1 (d) e/2 (e) x3+ x 1 x -3 . 2 1- x 14. Let y = . Express x in terms of y. 1+ x y-1 (a) x = y+1 (b) x = (1 + y)2 (1 - y)2 1+ y (c) x = 1- y (1 - y)2 (1 + y)2 1 1+ y (e) x = 1 1- y (d) x = 6 Part 2: Work-Out Problems (48 points) Partial credit is possible. Present your solutions in the space provided. Show all your work neatly and concisely, and indicate your final answer clearly. You will be graded not merely on the final answer, but also on the quality and correctness of the work leading up to it! 15. [8 points] Find an equation of the tangent line to the "devil's curve" y 4 - 4y 2 = x 4 - 5x 2 at Devil's Curve 5, 2 . 16. A particle's position in the plane at time t is given by the vector function r (t) = t cos t, t sin t . For time t = , compute the following items regarding the particle. Show all work! (a) [2 points] position: (b) [2 points] velocity: (c) [2 points] speed: (d) [2 points] acceleration: 7 17. [8 points] Find all values of t for which the tangent line to the curve x = 2t 3 + 3t 2 - 12t, y = 2t 3 + 3t 2 + 1 is horizontal or vertical. When you have finished, fill in the following table. For "type," write H for horizontal and V for vertical. (Add additional rows if necessary.) t-value type 18. [8 points] The circumference of a sphere (the length of its "equatorial circle") was measured to be 20 cm with a possible error of 1 cm. Estimate the maximum error in the calculated volume of the sphere using differentials. 4 Recall that the volume of a sphere is V = 3 r 3 and the circumference of a circle is C = 2r . Here r is the radius. 8 19. [8 points] Solve the following two equations. -1 (a) 10 1 + e-x =3 (b) log2 (2x + 1) = 2 - log2 (4x) 20. [8 points] Two French seahorses frolic in the sea at Mediterranean Downs, practicing for the big race on Saturday! Phillipe travels due north of the starting post, while Gigi heads due east. At time t, Phillipe's distance (in cm) from the post is y = 7 + 2t + 1 t 2 , whereas that of Gigi is x = 6 + 4t. Here time is measured in seconds. At what rate is 2 the distance z between Phillipe and Gigi changing after t = 2 seconds? 9 Fall 2007 Math 151 Exam 2A: Solutions c 2007 Art Belmonte Wed, 31/Oct 1. (c) The derivative is y = x tan x + 2 sin x, whence sec y = 2 3 + 3 = 3 3. 3 2. (e) Recall that H (x) = g ( f (x)). Apply the Chain Rule. H (x) H (1) H (1) H (1) 10. (d) With f (x) = e x - 4x - sin x, let g (x) = x - f (x) e x - 4x - sin x =x- x . f (x) e - 4 - cos x 1 1-0-0 = . 1-4-1 4 Then x 2 = g (x 1 ) = g (0) = 0 - 11. (c) Compute the first two derivatives of p = ax 2 + bx + c. p (x) = p (x) = 2ax + b 2a. = = = = g ( f (x)) f (x) g ( f (1)) f (1) g (3) f (1) (9) (4) = 36 The specified data give rise to three equations in the three unknowns a, b, and c. 8 = p (1) 4 = p (1) 6 = p (1) = a+b+c 3. (c) Substitution gives lim sin (cos ) sin (1) = = sin 1. sec 1 0 -1/2 1 4. (c) Now f (x) = 2 1 + xe-2x 1 whence f (0) = 2 . e-2x - 2xe-2x = = 2a + b 2a 5. (b) Recall x = cos t + cos 2t and y = sin t + sin 2t. For t = , we have (x, y) = (-1, 1), a point on the tangent line. 2 The slope of the tangent line for t = /2 is d y/dt cos t + 2 cos 2t -2 dy = = = = 2. dx d x/dt - sin t - 2 sin 2t -1 The point-slope formula then yields y-1 y = 2 (x - (-1)) 2x + 3. The third equation yields a = 3. Accordingly, we have b = 4 - 2a = -2 and c = 8 - a - b = 8 - 3 + 2 = 7. 1 3 Thus p (x) = 3x 2 - 2x + 7. Hence p 2 = 6 4 . 12. (d) Recall f (x) = 3 + x 2 + tan ( x/2), -1 < x < 1, and g (x) = f -1 (x). Since f (0) = 3, we have g (3) = 0. Now f (x) = 2x + sec2 ( x/2) and therefore 2 1 1 1 2 g (3) = = = = . f (g (3)) f (0) /2 3 x + = , whence 13. (a) As x 3+ , we have x -3 0 x 1 x -3 0. 2 1- x 14. (d) Isolate x step-by-step, given that y = . 1+ x y+y x (1 + y) x x x = = = = 1- x 1-y 1-y 1+y 1-y 1+y = 6. (b) The area of a rectangle is A = LW , where L is its length and W its width. At the stated instant we have dL dW dA = W+L = (8) (10) + (20) (3) = 140 cm2 /s. dt dt dt 7. (b) Now y 2 = 1 + x 3 , whence 2y dy dy dx = 3x 2 and thus dt dt 2y dt 2 (3) (4) dx = = = 2 cm/s. (You may alternatively dt 3x 2 3 (2)2 differentiate directly instead of implicitly.) 1 8. (d) Let f (x) = x 1/3 . Then f (x) = 3 x -2/3 . Thus 1 f (64) = 4 and f (64) = 48 . The linear approximation is (64) (x - 64) = 4 + 1 (x - 64). L (x) = f (64) + f 48 Hence 3 70 = f (70) L (70) = 4 1 . 8 2 = (1 + y)2 (1 - y)2 15. Get the slope of the tangent line via implicit differentiation. y 4 - 4y 2 dy dy - 8y 4y 3 dx dx dy dx dy dx 1 5,2 9. (e) With f (x) = cos (x + ), we differentiate to obtain f (x) = - sin (x + ) and f (x) = - cos (x + ). Thus f (0) = -1, f (0) = 0, and f (0) = 1. The quadratic approximation is Q (x) = Q (x) = f (0) + f (0) (x - 0) + -1 + 1 2 2x . 1 2 = = = = x 4 - 5x 2 4x 3 - 10x 2x 3 - 5x 2y 3 - 4y 5 5 [continued] 8 f (0) (x - 0)2 Now use the point-slope formula and finish it off. 5 5 y-2 = x- 5 8 5 5 25 y = x +2- 8 8 9 5 5 x- y = 8 8 16. Velocity is the derivative of position, acceleration is the derivative of velocity, and speed is the magnitude of velocity. r (t) v = r (t) a = v (t) = [t cos t, t sin t] sin t + t cos t] cos t + cos t - t sin t] 18. Now C = 2r implies dC = 2 dr . It was stated that C = 20 and dC = 1, with lengths in centimeters. Thus C 20 10 dC 1 r= = = and dr = = . Now compute 2 2 2 2 the differential d V and plug in the data. V dV dV 19. = = = 4 3 3 r 2 4r dr 10 4 2 1 2 = 200 cm3 2 (a) Isolate x step-by-step. The answer checks out. 10 1 + e-x 1+e -1 = = [cos t - t sin t, [- sin t - sin t - t cos t, At t = , we have position = velocity = speed = r () = [-, 0] v () = [-1, -] a () = [, -2] v () = 1 + 2 -x -1 = 3 = = = 1 + e-x e -x 3 10 10 3 7 3 acceleration = -x x = ln 7 - ln 3 3 = ln 3 - ln 7 = ln 7 17. Recall that x = 2t 3 + 3t 2 - 12t and y = 2t 3 + 3t 2 + 1. The tangent line to the curve will be horizontal where d y/dt dy = = 0. This can occur if dy = 0 & d x = 0 dt dt dx d x/dt dy simultaneously. Now 0 = dt = 6t 2 + 6t = 6t (t + 1) implies t = -1, 0. Observe that for these values of t we have d x = 6t 2 + 6t - 12 = 0. dt (b) Same drill. Check your answers! log2 (2x + 1) = 2 - log2 (4x) 2 2 log2 (2x + 1) + log2 (4x) = = log2 (2x + 1) (4x) 8x + 4x - 4 = 2x 2 + x - 1 = 2 8x 2 + 4x = 22 = 4 0 0 The tangent line to the curve will be vertical where d y/dt dy = = "", colloquially speaking. More dx d x/dt precisely, this can occur where d x = 0 and dy = 0 dt dt simultaneously. Now 0 = d x = 6t 2 + 6t - 12 implies dt 0 = 6 t 2 + t - 2 = 6 (t - 1) (t + 2) and thus t = -2, 1, at which dy = 6t 2 + 6t = 0. dt Here is the requested table and an illustrative plot. Recall that H = horizontal and V = vertical. t -2 -1 0 1 10 (2x - 1) (x + 1) = x = Toss out x = -1. Only x = equation! 1 2 0 1 2 , -1 satisfies the original 20. First note that when t = 2, we have x = 6 + 4t = 14, 1 y = 7 + 2t + 2 t 2 = 13, d x = 4 and dy = 2 + t = 4. dt dt Apply the Pythagorean Theorem, then proceed. z2 dz 2z dt dz dt dz dt = = = = x 2 + y2 dx dy 2x + 2y dt dt x d x + y dy dt dt z (14) (4) + (13) (4) 142 + 132 type V H H V 108 cm/s = 365 5 y 0 Happy Halloween! -5 0 5 x 10 15 20 -5 -10 2 MATH 151, FALL 2008 COMMON EXAM II - VERSION A LAST NAME, First name (print): INSTRUCTOR: SECTION NUMBER: UIN: SEAT NUMBER: DIRECTIONS: 1. The use of a calculator, laptop or computer is prohibited. 2. In Part 1 (Problems 1-10), mark the correct choice on your ScanTron using a No. 2 pencil. For your own records, also record your choices on your exam! 3. In Part 2 (Problems 11-16), present your solutions in the space provided. Show all your work neatly and concisely and clearly indicate your final answer. You will be graded not merely on the final answer, but also on the quality and correctness of the work leading up to it. 4. Be sure to write your name, section number and version letter of the exam on the ScanTron form. THE AGGIE CODE OF HONOR "An Aggie does not lie, cheat or steal, or tolerate those who do." Signature: DO NOT WRITE BELOW! Question 1-10 11 12 13 14 15 16 Points Awarded Points 40 12 10 10 12 8 8 100 1 PART I: Multiple Choice 1. (4 pts) In order to solve the equation x5 - 2x + 5 = 0, we apply Newton's Method with an initial guess x1 = 1. What value does Newton's Method give for x2 , the second approximation? 7 3 1 (b) 4 (a) (c) - (d) 7 4 1 4 1 3 (e) - 2. (4 pts) lim (a) 9 sin2 (3) = 0 2 (b) 3 1 (c) 9 1 (d) 3 (e) The limit does not exist 2 3. (4 pts) Find the tangent vector of unit length for r(t) = e2t , t cos t at t = 0. 1 1 , 2 2 (b) 2, 1 (a) (c) 1, 0 (d) 2 1 , 5 5 (e) 1, 1 4. (4 pts) Solve the equation ln(x + e) + ln(x - e) = 2 + ln 3. (a) x = 3e only (b) x = 1 and x = 3e (c) x = 2e only (d) x = 2e and x = -2e (e) No solution 3 5. (4 pts) If g is the inverse of f , find g (2) if it is known that f (3) = 2, f (3) = 7, f (2) = 11 and f (11) = 8. Assume g to be differentiable. (a) (b) (c) (d) (e) 1 7 1 11 1 8 1 2 1 5 6. (4 pts) If h(x) = f g = f (g(x)), find h (-3) given that g (-3) = 4, f (-3) = 7, g(-3) = -2, f (-2) = 11, and f (4) = -3 (a) 28 (b) 44 (d) - 6 (c) - 14 (e) - 3 4 7. (4 pts) An object is moving with position function f (t) = 2 sin t-3 cos t. Find the velocity, v(t), and the acceleration, a(t), at t = . 6 3 3 (a) v =- 3- = -1 + a 6 2 6 2 3 3 3 = 3- =1- a (b) v 6 2 6 2 3 3 3 (c) v = 3+ = -1 + a 6 2 6 2 3 3 3 = 3- =1- a (d) v 6 2 6 2 3 3 3 =1+ =- 3+ a (e) v 6 2 6 2 8. (4 pts) If Q(x) is the quadratic approximation for f (x) = (a) 3 5 (b) 2 3 (c) 2 7 (d) 2 9 (e) 2 2 at x = 1, then Q x 1 2 = 5 9. (4 pts) Evaluate lim- e1/x x0 (a) 1 (b) 0 (d) - (e) e (c) 10. (4 pts) Find the inverse function of f (x) = (a) f -1 (x) = (b) f -1 (x) = (c) f -1 (x) = (d) f -1 (x) = (e) f -1 (x) = 1 - 3x 4x + 1 3x - 1 4x + 1 4x + 3 1-x 1 - 3x 4x 3x - 1 4x 1-x 4x + 3 6 PART II WORK OUT Directions: Present your solutions in the space provided. Show all your work neatly and concisely and Box your final answer. You will be graded not merely on the final answer, but also on the quality and correctness of the work leading up to it. 11. Find the derivative of: (i) (6 pts) f (x) = tan3 (x) + tan(x3 ) (ii) (6 pts) g(t) = 1+ t. 7 3 12. (10 pts) Water is poured into a conical cup at the rate of cubic inches per second. If the cup is 6 inches tall and 2 the top of the cup has a radius of 2 inches, how fast does the water level rise when the water is 4 inches deep? Be 1 sure to include units with your answer. NOTE: The volume of a cone is V = r2 h. 3 8 13. (10 pts) Find the equation of the tangent line to the curve y 2 sin 2x = 8 - 2y at the point ,2 . 4 9 14. Consider the curve given by parametric equations x = t2 - 10t, y = t3 - 3t2 . (i) (6 pts) Find the equation of the tangent line at t = 1. (ii) (6 pts) Find all points on the curve where the tangent line is: (a) vertical (b) horizontal Exam continues on next page 10 15. (8 pts) Use differentials or a linear approximation to approximate 16.03. 16. (8 pts) Find all value(s) of x, 0 x 2, where f (x) = x + 2 sin x has a horizontal tangent. End of Exam 11 Math 151 Fall 2008 Exam II Solutions-Form A 1. C: Let f (x) = x5 - 2x + 5. Newton's Method says if x1 is the first guess to f (x) = 0, then the second f (x1 ) approximation is given by x2 = x1 - . Thus f (x1 ) 1 f (1) = 1 - 4/3 = - . if x1 = 1, x2 = 1 - f (1) 3 2. A: lim sin2 (3) sin(3) sin(3) = lim 2 0 0 sin(3) sin(3) = 9(1)(1) = 9 = 9 lim 0 3 3 7. C: f (t) = 2 sin t - 3 cos t, v(t) = f (t) = 2 cos t + 3 sin t and a(t) = v (t) = -2 sin t + 3 cos t. Evaluating these at 3 t = yields v = 3 + and 6 6 2 3 3 = -1 + a 6 2 8. D: The quadratic approximation for f (x) at x = 1 f (1) is Q(x) = f (1) + f (1)(x - 1) + (x - 1)2 . 2 f (1) = 2, f (1) = -2 and f (1) = 4. Thus Q(x) = 2 - 2(x - 1) + 2(x - 1)2 , and 7 1 1 1 Q( 2 ) = 2 - 2( 2 - 1) + 2( 2 - 1)2 = . 2 x0 x0 3. D: First we will find the tangent vector at t = 0 and then make it a unit vector by dividing by the magnitude: r(t) = e2t , t cos t thus r (t) = 2e2t , cos t r (0) = 2e0 , cos(0) - (0) sin(0) = 2, 1 . The unit 1 2 2, 1 = , . vector is | 2, 1 | 5 5 4. C: To solve ln(x + e) + ln(x - e) = 2 + ln(3), we will use logarithm properities: ln(x + e) + ln(x - e) - ln(3) = 2 (x + e)(x - e) ln =2 3 (x + e)(x - e) = e2 3 (x + e)(x - e) = 3e2 - t sin t . Therefore 9. B: lim- e1/x = 0 since lim- 1/x = - lim 1/x and lim e1/x = e x0- x0- = lim ey = 0 y- 10. A: Let y = x= 1-x . Interchange x and y: 4x + 3 1-y . Solve for y: x(4y + 3) = 1 - y 4y + 3 1 - 3x . 4xy + y = 1 - 3x, hence y = 4x + 1 11. (i) f (x) = tan3 x + tan(x3 ). By the chain rule, f (x) = 3 tan2 x sec2 x + sec2 (x3 )(3x2 ). (ii) g(t) = 1 + t = (1 + t1/2 )1/2 . By the chain rule, 1 1 g (t) = (1 + t1/2 )-1/2 t-1/2 2 2 1 = 4 t 1+ t 12. We are given 3 dh dV = . We want to find when dt 2 dt 1 h = 4. The volume of a cone is V = r 2 h. To get 3 V in terms of h only, we will use similar triangles. 2 1 r = , thus r = h. Substitute this in for r: h 6 3 1 1 2 3 V = ( h) h = h . Differentiate with respect 3 3 27 dh dV 3 dV = h2 . Now substitute = to time: dt 9 dt dt 2 2 dh dh 27 3 = (4) , thus = and h = 4 yields 2 9 dt dt 32 inches per second. x2 - e2 = 3e2 , yielding x = 2e. Now, the domain of ln(x + e) + ln(x - e) = 2 + ln(3) is x > e, hence x = -2e is extraneous, so the only solution is x = 2e. 5. A: Recall if g is the inverse of f , then 1 1 . Thus g (2) = . g (a) = f (g(a)) f (g(2)) 1 1 Since f (3) = 2, g(2) = 3. g (2) = = f (3) 7 6. B: By the chain rule, if h(x) = f (g(x)), h (x) = f (g(x))g (x). Thus h (-3) = f (g(-3))g (-3) = f (-2)(4) = 44 13. First we will differentiate implicitly with respect to x: y 2 sin 2x = 8-2y. Using the product rule and chain rule, dy dy dy 2y sin 2x + y 2 2 cos 2x = -2 . Solve for . dx dx dx dy (2y sin 2x + 2) = -2y 2 cos 2x, thus dx -2y 2 cos 2x dy = . To find the slope of the tangent dx 2y sin 2x + 2 dy line, we will substitute the point , 2 into . 4 dx -2(2)2 cos 2 = 0. Thus the equation of the m= 2(2) sin( ) + 2 2 line is y - 2 = 0(x - /4), hence y = 2. 14. (i) To find the tangent line, we need a point and the slope. To find the point, substitute t = 1 into the parametric equations: x = t2 - 10t, y = t3 - 3t2 , so when t = 1, x = -9 and y = -2. Thus the point is (-9, -2). Now, the slope of the tangent line is m= dy 3t2 - 6t dy/dt evaluated at t = 1. = , thus dx/dt dx 2t - 10 3 when t = 1, m = . So the equation of the tangent 8 3 line is y + 2 = 8 (x + 9). 15. Using differentials, f (a + dx) f (a) + f (a)dx, where f (x) = x, a = 16 and dx = 0.03. Thus f (16.03) = 16.03 f (16) + f (16)(0.03) 1 = 4 + (0.03) = 4.00375 8 Using Linear Approximation: Find the linear ap proximation for f (x) = x at x = 16. L(x) = f (16) + f (16)(x - 16). 1 1 Now, f (16) = 16 = 4 and f (16) = = . 8 2 16 1 Hence L(x) = 4 + (x - 16). Now, to approximate 8 16.03, substitute x = 16.3 in L(x) for x: 1 1 L(16.3) = 4 + (16.3 - 16) = 4 + (0.03) = 4.00375 8 8 16. To find where the tangent line is horizontal, we need to solve f (x) = 0 for 0 x 2. f (x) = x+2 sin x, thus f (x) = 1 + 2 cos x. Solve f (x) = 0 gives 1 cos x = - . For 0 x 2, this happens when 2 2 4 x= and x = . 3 3 dx = 0 and (ii) The tangent line is vertical where dt dy = 0. This happens when 2t - 10 = 0, thus t = 5. dt Substitute this into x = t2 - 10t, y = t3 - 3t2 , we get the point (-25, 50) dy (iii) The tangent line is horizontal where = 0 dt dx and = 0. This happens when 3t2 - 6t = 0, thus dt t = 0 and t = 2. Substitute this into x = t2 - 10t, y = t3 - 3t2 , we get the points (0, 0) and (-16, -4). ...
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This note was uploaded on 06/25/2009 for the course MATH 151 taught by Professor Artbelmonte during the Spring '06 term at Texas A&M.

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