Math 116Exam 3 – Part 1Fall 2013

Name:______This is a closed book exam. For the questions 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 = 4 + ln t and y = ln(2t3) where t varies from 1 to e.

a. (8 points) Eliminate t so as to describe the curve by an equation of the form y = F(x) where F(x) is a formula involving x that you are to find.

b. (7 points) Sketch the curve defined by the parametric equations x = 4 + ln t and y=ln(2t3) where t varies from 1 to e. Indicate by an arrow the direction the curve is traced out as t increases.

2.Consider the curve defined by the polar equation r= + 1 where  varies from 0 to 2.

a. (3 points) Fill in the table of values at the right

b. (12 points) Sketch this curve.

Math 116Exam 3 – Part 2Fall 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.

3.(20 points) Find . Show all your work. A substitution may be helpful.

4.(20 points) A flat plate with uniform density occupies the region that is above the x axis and below the curve y = sinx for 0 x. Find the center of mass. Use symmetry to make some of your work easier.

5.(15 points) Consider the curve defined by the parametric equations x = t2+ t+ 1 and y = t2– t-1. Find the equation of the tangent line at the point (x, y) corresponding to t = 1.

6.(15 points) Consider the curve defined by the polar equation r = cos3 where  varies from 0 to . Find the length of the curve.

Solutions

1.a.x – 4 = ln t. So y = ln(2t3) = ln(t3) + ln(2) = 3ln(t) + ln(2) = 3(x – 4) + ln(2) = 3x12 + ln(2)  3x – 11.307.

b.The curve is a straight line with slope 3 and y intercept – 12 + ln(2). When t = 1 one has x = 4. When t = e one has x=5. So the curve is the part of the straight line from (x, y) = (4, ln(2))  (4, 0.693) to (5, 3 + ln(2))  (5, 3.693).

2.See right.

3. = . Let u = - . Then - 2du = dx and = 2= - 2eu = - 2. So = - 2|= 2e-1 - 2. Since2e-1 - 2= 2e-1– (2)(0) = 2e-1 one has = 2e-1 0.736.

4.By symmetry = /2. One has = where A is the area and f(x) = sin x and a= 0 and b=. One has A = = ( - cos x)| = (- cos ) – (cos0) = (1) + 1 = 2. Also = = = | = – = – = . So = = .

5.When t = 1 one has (x, y) = ((1)2 + 1 + 1, (1)2 – 1 – 1) = (3, -1). The slope of the tangent line is |x=1. One has = = . So |x=1 = |t=1 = . The equation of the tangent line is (y – (-1)) = (1/3)(x – 3) or y = (1/3)x – 2.

6.L = . One has = - cos2sin. So L ======| = - =  1.5708 + 0.6495 = 2.2203.