Math 115 - University of Michigan



Math 116 Final Exam – Part 1 Fall 2014

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.

1. Consider the curve defined by the parametric equations x = t3 – 3t and y = 8 – 2t2 where t varies from - 2 to 2.

a. (1.5 points) Find the point(s) where the curve intersects the x axis.

b. (1.5 points) Find the point(s) where the curve intersects the y axis.

c. (1.5 points) Find the point(s) where the tangent line is horizontal.

d. (1.5 points) Find the point(s) where the tangent line is vertical.

e. (5 points) Sketch the curve defined by the parametric equations x = t3 – 3t and y = 8 – 2t2 where t varies from - 2 to 2. Include the points which you found in parts a – d.

Math 116 Final Exam – Part 2 Fall 2014

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. (12 points) Find if y = tan-1 x + ln . Show that can be simplified to where q(x) is a polynomial you are to find.

3. (13 points) Use integration by parts to find

4. (13 points) Find . First make a trig substitution to get rid of the square root. Then use a trig identity on the resulting integral.

5. (11 points) Consider the two curves defined by the polar equations r = 2cos ( and r = 1. Find the area A of the region R that lies inside the first curve and outside the second curve.

6. (16 points) Use the ratio test, the alternating series test and the integral test to determine those x for which the series converges.

7. (12 points) Use a well known power series, algebra and differentiation to find the sum of the power series for – 1 < x < 1.

8. (12 points) Suppose the Taylor series of tan x is tan x = a0 + a1x + a2x2 + a3x3 + … Use Taylor's formula to find a0, a1, a2 and a3.

Solutions

1 a. The curve intersects the x axis when y = 0. Set y equal to 0 and find the corresponding t. y = 0 ( 8 – 2t2 = 0 ( t2 = 4 ( t = ( 2. t = 2 ( x = 23 – (3)(2) = 8 – 6 = 2. t = - 2 ( x = (- 2)3 – (3)(- 2) = - 8 + 6 = - 2. So the curve intersects the x axis at the points = and = . b. The curve intersects the y axis when x = 0. Set x equal to 0 and find the corresponding t. x = 0 ( t3 – 3t = 0 ( t(t2 – 3) = 0 ( t = 0, ( . t = 0 ( y = 8. t = ( ( y = 8 – (2)(( )2 = 8 – 6 = 2. So the curve intersects the y axis at the points = and = . c. The tangent line is horizontal when = 0. = = . = 0 ( t = 0 ( = . d. The tangent line is vertical when = 0. = = . = 0 ( t = ( 1. t = 1 ( x = 13 – (3)(1) = - 2 and y = 8 – (2)(12) = 6. t = - 1 ( x = (- 1)3 – (3)(- 1) = 2 and y = 8 – (2)(- 1)2 = 6. So the tangent line is vertical at the points = and = . e. See right.

2. [tan-1 x + ln ] = + [] (2 + 3 pts) = + (3 pts) = + (1 + 1 pts) = + = + = (2 pts)

3. Let u = x and dv = sec2 x dx. So du = dx and v = tan x. Therefore = x tan x -  ( 6 pts) = . Let y = cos x. Then dy = - sin x dx, so = - = - ln y = - ln(cos x) (4 pts) So = x tan x + ln(cos x) (1 pt)

4. Let x = 2 sin u and dx = 2 cos u du ( = = 4 (4 pts) = 4 = 2(u + ) (4 pts) = 2u + 2sin u cos u = 2 sin-1 + x = 2 sin-1 + x= 2 sin-1 + x (3 pts)

5. The first curve is the circle with center (1, 0) and radius 1. The second curve is the circle with center the origin and radius 1. To find where the curves meet set 2 cos ( = 1. So cos ( = ½ or ( = ( (/3. (1 pt) Since R is symmetrical with respect to the x axis, we can find the area of the part of R in the first quadrant and double it. So A = 2[½ - ½ ] (4 pts) = = = = [sin(2() + (] | (5 pts) = sin + = + ( 1.91. (1 pt)

6. Apply ratio test. = (1 pts) = = = . (3 pts) So series converges if < 1, diverges if > 1 and ratio test is inclusive if = 1. So series converges if – 2 < x < 2, diverges if x > 2 or x < - 2 and ratio test is inclusive if x = 2 or x = - 2. (2 pts) If x = 2 then series becomes = . Use integral test. Series has the form where a(x) = . a(x) is a decreasing function for x > 0. So the series converges if and only if < (. = ln x. So = ln x = ln L - ln 1 = ln L. So = ln L = (. So the series diverges if x = 2. (2 pts) If x = - 2 then the series becomes = Use alternating series test. The series has the form where an = . an decreases as n increases and approaches 0 as n ( (. So the series converges if x = - 2. (3 pts)

7. 1 + x + x2 + … + xn + … = = (1 – x)-1 (1 pts) Differentiating this we get 1 + 2x + 3x2 + … + nxn-1 + … = (1 – x)-2. (3 pts) Multiplying by x gives x + 2x2 + 3x3 + … + nxn + … = x(1 – x)-2. (3 pts) Differentiating again gives 1 + 4x + 9x2 + … + n2xn-1 + … = (1 - x)-2 + 2x(1 - x)-2 (3 pts) Multiplying by x again gives x + 4x2 + 9x3 + … + n2xn + … = x(1 - x)-2 + 2x2(1 - x)-3 = (2 pts)

8. Let f(x) = tan x. In general an = One has f'(x) = sec2x and f''(x) = 2tan x sec2x and f'''(x) = 2 sec4x + 4tan2x sec2x. (6 pts) So a0 = f(0) = tan 0 = 0 and a1 = f'(0) = sec20 = 1 and a2 = f''(0)/2 = (½)(2)tan 0 sec20 = 0 and a3 = f'''(0)/6 = (1/6)(2 sec40 + 4tan20 sec20) = 1/3 (6 pts) So tan x = x + x3/3 + …

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