Math 116 Final Exam Part 1 Fall 2013

Math 116 Final Exam – Part 1 Fall 2013

Name:______This is a closed book exam. For the question in Part 1 you may use the formulas handed out with the exam and a non-graphing calculator, but NOT a graphing calculator. Show all work and explain any reasoning which is not clear from the computations. (This is particularly important if I am to be able to give part credit.) When you have finished this part, turn it in and obtain the remaining questions for this exam for which you can use a graphing calculator. t 1. Consider the curve defined by the parametric equations x = cos t and y = sin(2t) cos t 2t sin(2t) where t varies from 0 to . 0 a. (3 points) Fill in the table of values at the right /6 b. (8 points) Sketch the curve defined by the parametric equations x = cos t and /4 y = sin(2t) where t varies from 0 to .. Indicate by an arrow the direction the /3 curve is traced out as t increases. /2 2/3 3/4 5/6 Math 116 Final Exam – Part 2 Fall 2013 

Name: ______This is a closed book exam. For these questions you may use the formulas handed out with the exam and a graphing calculator. You may find that your calculator can do some of the problems. If this is so, you still need to show how to do the problem by hand. In other words, show all work and explain any reasoning that is not clear from the computations. (This is particularly important if I am to be able to give part credit.) Turn in this exam along with your answers. However, don't write your answers on the exam itself; leave them on the pages with your work. Also turn in the formulas; put them on the formula pile.

2. (11 points) Find if y = tanh(x2) sinh-1(ex + 1) 3. (11 points) Use integration by parts to find 4. (11 points) Use a trig identity and a substitution to find 5. (11 points) Find . Hints: Do some algebra and a substitution to reduce it to something in the formula sheet. Also 2 = 2. 6. (11 points) Find limn 7. a. (6 points) Find the partial fraction decomposition of b. (5 points) Find the sum of the series . Hint: telescoping. n 8. (11 points) Use a well known power series, algebra and integration to find the power series anx of x ln(1 - x). -1 -1 2 3 … 9. (12 points) Suppose the Taylor series of sin x is sin x = a0 + a1x + a2x + a3x + Use Taylor's formula to find a0, a1, a2 and a3. Solutions t cos t 2t sin(2t) 1 a. b. 1.0 0 1 0 0 /6 /2 /3 /2 0.5 /4 /2 /2 1 /3 1/2 2/3 /2

/2 0  0  1.0  0.5 0.5 1.0 2/3 - 1/2 4/3 - /2

3/4 - /2 3/2 - 1  0.5 5/6 - /2 5/3 - /2

 - 1 2 0  1.0

2. [ tanh(x2) sinh-1(ex + 1)] = tanh(x2) [sinh-1(ex + 1)] + sinh-1(ex + 1) [tanh(x2)] } 2 points = tanh(x2) [ex + 1] + sinh-1(ex + 1) sech2(x2) [x2] } 3 + 3 points = + 2x sinh-1(ex + 1) sech2(x2) } 2 + 1 points

3. Let u = ln x and dv = x dx. So du = dx and v = . Therefore = ln x - } 5 + 3 points = - } 3 points 4. = } 2 points Let u = cos x and du = - sin x dx  = - } 4 points = = } 3 points = } 2 points OR = } 2 points Let u = sin x and du = cos x dx  = } 4 points = = } 3 points = } 2 points 5. = } 2 points Let u = and du = dx so du = dx  = } 4 points = tan-1u } 3 points = tan-1 } 2 points 6. Let bn = ln = n ln } 2 points l'Hospital  lim = lim n ln = lim = lim = lim = lim = 5 } 6 points So limn = lim = e5 } 2 points 7. a. = = + } 2 points  1 = A(x + 2) + B(x – 2). x = 2  A = ¼. x = - 2  B = ¼  = - } 4 points b. Using part a, one has = } 1 points = ( - ) + ( - ) + ( - ) + … + ( - ) = ( - ) + ( - ) + ( - ) + ( - ) + ( - ) + ( - ) + … + ( - ) } 1 points the second fraction in every term cancels with the first fraction in the term four places to the right  The sum telescopes to + + + - - - - . Letting n   one obtains = ( + + + ) = = } 3 points 8. = 1 + x + x2 + … + xn + … } 2 points Integrating this we get – ln(1 – x) = C + x + + … + + … . Plugging in x = 0 we get C = 0. } 5 points Multiplying by – x gives x ln(1 – x) = - x2 – - … - - … } 4 points -1 9. Let f(x) = sin x. In general an = } 2 points One has f '(x) = = (1 – x2)-1/2 and f ''(x) = (-1/2)(1 – x2)-3/2(-2x) = x(1 – x2)-3/2 and f '''(x) = (-3/2)x(1 - x2)-5/2(-2x) + (1 - x2)-3/2 = 3x2(1 – x2)-5/2 + (1 – x2)-3/2 } 4 points -1 So a0 = f(0) = sin (0) = 0 and a1 = f '(0) = 1 and a2 = f ''(0)/2 = 0 and a3 = f '''(0)/6 = 1/6 } 4 points So sin-1x = x + x3/6 + … } 2 points