Math 103 Midterm Test 1 Feb. 10, min.

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1 Full Name: Student Number: Math 13 Midterm Test 1 Feb. 1, 26 5 min. 1. Ensure that your full name and student number appear on this page. 2. No calculators, books, or notes, or electronic devices of any kind are permitted. 3. Unless otherwise indicated, show all your work. Answers not supported by calculations or reasoning may not receive credit. Correct answers that do not demonstrate an appropriate understanding of the course material might not receive credit. Messy work will not be graded. Read each question carefully to be sure you are answering the question being asked. 4. Exposing your test paper, copying from another student s paper, or sharing information about this test constitutes academic dishonesty. Such behaviour may jeopardize your grade on this test, in this course, and your standing at this university. 5. Five minutes before the end of the test period you will be given a verbal notice. After that time, you must remain seated until all test papers have been collected. 6. When the test period is over, you will be instructed to put away writing implements. Put away all pens and pencils at this point. Continuing to write past this instruction will be considered as cheating. 7. Please remain seated and pass your test paper down the row to the nearest indicated aisle. Once all the test papers have been collected, you are free to leave. Problem # Grade Value I have read and understood the instructions and agree to abide by them. Signed: 4 8 Total 3 Summation Formulae C = NC k = N(N + 1)/2 k 2 = N(N + 1)(2N + 1)/6 k 3 = (N(N + 1)/2) 2 N k= r k = (1 r N+1 )/(1 r)

2 1. Calculate each of the following. (a) 15 k=3 2 k. Solution(2 points) 15 k=3 2k = 15 k= 2k 2 k= 2k = = = (b) 2 e 2x dx. 2 e 2x dx = 1 2 e 2x 2 = 1 2 (1 e 4 ) (c) Find the area under the curve f(x) = x 2 between x = 1 and x = 2. Area = 2 1 x 2 dx = 1 3 x3 2 1 = 8/3 ( 1/3) = 9/3 = 3 (d) The area between the graphs y 1 = 2 x 2 and y 2 = x. These two curves cross when 2 x 2 = x. Because both curves are symmetric about the y-axis, we need only find the intersection point with positive x value so solve 2 x 2 = x. This equation has only one positive solution, x = 1 so y 1 and y 2 cross at x = ±1. Note that y 1 is always higher than y 2 between x = 1 and x = 1 (although we do not actually need to know this) so the area between them is given by 1 1 (y 1 y 2 )dx = 2 1 (2 x x 2 )dx = 2 ) 1 (2x x2 2 x3 = 2(2 1/2 1/3) = 7/3 3 Had we not known which curve was higher and used y 2 y 1, this would have come out as 7/3 and we would simply drop the minus sign so as to give a positive value for area. (e) An approximation for the sum 4 sin ( π 8 k). Is your approximation higher or lower than the actual value? Note that b a f(x)dx f(x k ) x where x = b a N and x k = a + k x for the right-corner sum or x k = a + (k 1) x for the left-corner sum. We must interpret 4 sin ( π 8 k) as an example of the right hand side of this general formula. In this particular case, notice that there is a k but no k 1 2

3 appearing so we are in the right-corner case. Also, based on the upper limit of the sum, N = 4. Moreover, x = 1 as there is no apparent constant term multiplying the sum. Next, we interpret f(x) = sin ( π 8 x) which means x k = k and so the lower limit is a = and the upper limit is b = x 4 = 4. Note that because we are in a right-corner case, the lower limit is not a = x 1 = 1 because 1 is the right edge of the first rectangle (although using a = 1 simply omits one rectangle from the sum which should not introduce a serious error for N large enough - no points were deducted for this error). Finally, we have an integral approximation: 4 ( π ) sin 8 k 4 ( π ) sin 8 x dx = 8 ( π ) 4 π cos 8 x = 8 π Because the function sin ( π 8 x) is increasing between and 4 and the sum is a rightcorner sum, the integral approximation lies entirely below the rectangles. Thus, this approximation is lower than the actual value of the sum. Note that there are several variations that give the same or almost the same approximation. See Problem 3.6 for a similar example. 3

4 2. A cone-shaped coffee filter with a height of 1 cm and an opening radius of 1 cm is used to make coffee. While in use, the filter is full of water and the density of coffee grounds is given by ρ(y) = 1 1y where y = corresponds to the top of the filter. y is measured in cm and ρ is measured in grams/cm 3. 1cm 1cm (a) Use the method of disks to find the volume of water in the filter. Solution (4 points) When sliced into horizontal disks, a disk at height y has a radius given by 1 + y. Therefore, the volume is V = 1 π(1 + y) 2 dy = π (1 + 2y + y 2 )dy = π (1y + 1y ) y3 1 1 = π ( ) 13 = 1 3 π cm3. (b) Calculate the total mass of coffee grounds in the filter. The total mass of coffee grounds is found by adding up the mass in a thin disk for each value of y between -1 and. Each thin disk has ρ(y) V disk (y) cells where ρ(y) = 1 1 is the density in each thin disk and V disk (y) = π(1 + y) 2 dy is the volume of each thin disk. Therefore, the total mass is ( M = π 1 ) (1 + y) 2 dy = π (1y + 2y y3 )dy 1 = π (5y y ) ( 1 y4 = π ) 4 = 25 3 π grams 4

5 3. A sink can hold a maximum of 2 liters before it overflows. Suppose the sink is initially empty. The inflow rate from the tap is given by T (t) = A(1 + e 3t ) and the outflow through the drain is D(t) = A(1 e 4t ). T (t) and D(t) are both measured in liters per minute and t is measured in minutes. (a) How many liters of water are in the sink at time t? Solution (4 points) The total change in water volume in the sink is found by integrating its net rate of change from to t: V (t) = V (t) V () = t (T (s) D(s))ds = t A(e 3s + e 4s )ds ( 1 = A 3 e 3s + 1 ) t ( 1 ( 4 e 4s = A 1 e 3t ) + 1 ( 1 e 4t )) 3 4 (b) What is the maximum value of A for which the sink does not eventually overflow? Notice that the volume of water in the sink V (t) starts at zero and steadily rises to an asymptotic value of 7A/12. The maximum of A such that the sink does not overflowis determined by 7A/12 = 2 or equivalently, A max =

6 4. Below are graphs of the birth rate and death rate of a particular species of wild hare in the arctic over a period of ten years starting in 1995 (x = ). By examining the graphs, answer the following questions. 2 b(t) 1 d(t) t (a) Determine the years in which the population size was increasing and in which years the population size was decreasing. From t = to t = 3 (1995 through 1997 inclusive), the birth rate was higher than the death rate so the population increased. From t = 3 to t = 5 (1998 and 1999) the population decreased. From t = 5 to t = 8 (2 through 22 inclusive) the population increased. From t = 8 to t = 1 (23 and 24) the population decreased. (b) At the start of which year was the population of hares largest and when was it smallest? Briefly explain your answer. The population will be largest at the end of a period of increase so either at t = 3 or t = 8. Between those two values of t, the integral of b(t) d(t) (i.e. the change in population) gains more than it loses (the area between 3 and 5 is smaller than the area between 5 and 8). This means that the population was largest at t = 8 so at the start of 23. By similar reasoning, it was smallest at t = so at the start of

(a) x cos 3x dx We apply integration by parts. Take u = x, so that dv = cos 3x dx, v = 1 sin 3x, du = dx. Thus

(a) x cos 3x dx We apply integration by parts. Take u = x, so that dv = cos 3x dx, v = 1 sin 3x, du = dx. Thus Math 128 Midterm Examination 2 October 21, 28 Name 6 problems, 112 (oops) points. Instructions: Show all work partial credit will be given, and Answers without work are worth credit without points. You