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### Exam 1

Course: M 58365, Spring 2009
School: University of Texas
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Word Count: 2267

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Ilse Garcia, Exam 1 Due: Oct 2 2007, 11:00 pm Inst: Fonken This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Find the value of x lim . x0 16 + 3x 4 1. limit = 2. limit = 3. limit = 4. limit = 3 8 4 3 8 correct 3 1 keywords: limit, evaluate limit analytically,...

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Ilse Garcia, Exam 1 Due: Oct 2 2007, 11:00 pm Inst: Fonken This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Find the value of x lim . x0 16 + 3x 4 1. limit = 2. limit = 3. limit = 4. limit = 3 8 4 3 8 correct 3 1 keywords: limit, evaluate limit analytically, rationalize denominator, 002 (part 1 of 1) 10 points Determine the value of x1 lim f (x) when f satises the inequalities 5x f (x) on [0, 1) (1, 2]. 1. limit does not exist 2. limit = 4 3. limit = 5 correct 4. limit = 3 2 13 x + 4x + 3 3 5. limit = 0 6. limit = 3 4 5. limit = 2 6. limit = 6 Explanation: Set g(x) = 5x, f (x) = h(x) = 13 2 x + 4x + . 3 3 Explanation: After rationalization, Thus x 16 + 3x 4 x 16 + 3x + 4 = , 3x from which it follows that 16 + 3x + 4 f (x) = 3 for x = 0. Now lim 16 + 3x = 4 . x0 (16 + 3x) 16 16 + 3x 4 = . 16 + 3x + 4 Then, by properties of limits, x1 lim g(x) = lim 5x = 5 , x1 while lim h(x) = lim = 2 13 x + 4x + 3 3 x1 x1 2 1 + 4 + = 5. 3 3 Consequently, by properties of limits, x0 By the Squeeze Theorem, therefore, x1 lim f (x) = 8 . 3 lim f (x) = 5 . Garcia, Ilse Exam 1 Due: Oct 2 2007, 11:00 pm Inst: Fonken To see why the Squeeze theorem applies, its a good idea to draw the graphs of g and h using, say, a graphing calculator. They look like 6. limit = 7 Explanation: Since |v| = we see that x 2 + |x 2| = (1, 5) Thus f (x) = g: h: 1x, 1x + 4(x 2) , 2 2 v, v, v < 0, v 0, 2(x 2), 0, x 2, x < 2. x < 2, x 2. In particular, therefore, (not drawn to scale), so the graphs of g and h touch at the point (1, 5) while the graph of f is sandwiched between these two graphs. Thus again we see that x1 h0 lim f (3 + h) f (3) h = d 1x + 4(x 2)2 dx = 1 + 8(x 2) x=3 lim f (x) = 5 . x=3 keywords: limit, squeeze theorem 003 (part 1 of 1) 10 points Let f be the function dened by f (x) = x + (x 2 + |x 2|)2 . Determine if h0 because 3, 3 + h > 2 for all small h. Consequently, limit = 9 . keywords: limit, Newtonian quotient absolute value function 004 (part 1 of 1) 10 points Determine if the limit x0 lim f (3 + h) f (3) h exists, and if it does, nd its value. 1. limit doesnt exist 2. limit = 6 1. limit = 3 3. limit = 10 4. limit = 8 5. limit = 9 correct 2. limit = 2 3. limit = 6 lim sin 2x 6x exists, and if it does, nd its value. Garcia, Ilse Exam 1 Due: Oct 2 2007, 11:00 pm Inst: Fonken 1 4. limit = correct 3 5. limit doesnt exist Explanation: Using the known limit: sin ax lim = a, x0 x we see that lim 1 sin 2x . = 6x 3 3 provided x = 1. On the other hand, by properties of limits, x1 lim x3 2 = . x+2 3 Consequently, the given limit exists, and limit = 2 . 3 keywords: limit, rational function, Properties of Limits, 006 (part 1 of 1) 10 points x0 keywords: limit, trigonometric function 005 (part 1 of 1) 10 points Determine if x1 lim x2 4x + 3 x2 + x 2 Find all the values of a for which the limit xa Consider the function 3 x, x, f (x) = (x 2)2 , x < 1 x 2. 1 x < 2 exists, and if it does, nd its value. 1. limit = 1 2. limit does not exist 3. limit = 1 4. limit = 2 5. limit = 6. limit = 2 3 2 correct 3 1 2 lim f (x) exists, expressing your answer in interval notation. 1. (, 2) (2, ) 2. (, 1) (1, 2) (2, ) correct 3. (, 1) (1, ) 4. (, 1] [2, ) 5. (, ) Explanation: The graph of f is a straight line on (, 1), so xa Explanation: After factoring the numerator and denominator we see that x2 4x + 3 x2 + x 2 = lim f (x) exists (and = f (a)) for all a in (, 1). Similarly, the graph of f on (1, 2) is a straight line, so xa x3 (x 1)(x 3) = (x 1)(x + 2) x+2 lim f (x) Garcia, Ilse Exam 1 Due: Oct 2 2007, 11:00 pm Inst: Fonken exists (and = f (a)) for all a in (1, 2). On (2, ), however, the graph of f is a parabola, so lim f (x) xa 4 Explanation: By denition f (a) = lim When (1 + h)3 1 f (a + h) f (a) = , h h therefore, inspection shows that f (a + h) f (a) . h h0 still exists (and = f (a)) for all a in (2, ). On the other hand, x 1 lim f (x) = 4, x 1+ lim f (x) = 1 , while x 2 lim f (x) = 2, x 2+ lim f (x) = 0 . f (x) = x3 , a=1. Thus neither of the limits x 1 lim f (x), x2 lim f (x) keywords: limit, derivative 008 (part 1 of 1) 10 points Find f (x) when exists. Consequently, the limit exists only for values of a in (, 1) (1, 2) (2, ) . keywords: limit, graph, linear function, piecewise dened function, quadratic function 007 (part 1 of 1) 10 points For which of the following functions f and corresponding numbers a is the limit lim (1 + h)3 1 h f (x) = 1. f (x) = 2. f (x) = x . 1+x 1 correct (1 + x)2 3 (1 + x)2 3 (1 + x)2 3. f (x) = 4. f (x) = h0 2 (1 + x)2 1 (1 + x)2 2 (1 + x)2 the value of f (a)? 1. f (x) = x3 , a = 1 correct a=1 a=1 a=3 5. f (x) = 6. f (x) = 2. f (x) = (x + 1)3 , 3. f (x) = (x 1)3 , 4. f (x) = (x + 1) , 5. f (x) = x3 , 6. f (x) = x , 3 3 Explanation: By the Quotient Rule f (x) = Consequently, f (x) = 1 . (1 + x)2 (1 + x) x . (1 + x)2 a=0 a=3 Garcia, Ilse Exam 1 Due: Oct 2 2007, 11:00 pm Inst: Fonken keywords: derivative, quotient rule 009 (part 1 of 1) 10 points Find the derivative of f when f (x) = 1. f (x) = 2. f (x) = 3. f (x) = 4. f (x) = cos x . sin x 1 010 (part 1 of 1) 10 points Determine the derivative of f when f (x) = x2 sin x + 2x cos x . 1. f (x) = (x2 + 2) cos x 2. f (x) = (x2 2) sin x 3. f (x) = (x2 + 2) sin x 4. f (x) = (x2 + 2) sin x 5. f (x) = (x2 2) cos x 6. f (x) = (x2 + 2) cos x correct Explanation: By the Product Rule, f (x) = 2x sin x + x2 cos x + 2 cos x 2x sin x . Consequently, f (x) = (x2 + 2) cos x . . 5 keywords: dierentiation, trig function, quotient rule 1 1 + cos x 1 correct sin x 1 1 cos x 1 1 1 cos x 5. f (x) = 6. f (x) = 1 sin x + 1 1 cos x + 1 1 7. f (x) = 1 sin x 1 8. f (x) = 1 + sin x Explanation: By the quotient rule, f (x) = = sin x (sin x 1) cos2 x (sin x 1)2 sin2 x + cos2 x + sin x (sin x 1)2 1 + sin x sin x 1 But sin2 x + cos2 x = 1, so f (x) = = Consequently, 1 f (x) = . sin x 1 (sin x 1)2 (sin x 1) 2 keywords: dierentiation, trig function, product rule 011 (part 1 of 1) 10 points . Find an equation for the tangent line to the parabola y = 7x2 4x at the point P (1, y(1)). 1. y = 9x + 6 Garcia, Ilse Exam 1 Due: Oct 2 2007, 11:00 pm Inst: Fonken is horizontal. 2. y = 11x + 8 3. y = 10x + 7 4. y = 10x 7 correct 5. y = 11x 8 6. y = 9x 6 Explanation: When x = 1, a simple calculation shows that y(1) = 3. Thus, using the Newtonian dierence quotient with y = f (x) = 7x2 4x and the point P (1, 3), we see that f (1) = lim = lim f (1 + h) f (1) h 1. x = 2. x = 3. x = 4. x = 5. x = 4 3 8 3 5 3 5 ,0 3 4 ,0 3 6 6. x = 0 8 7. x = , 0 correct 3 Explanation: By the quotient rule, y (x) = 9x2 + 24x 6x(3x + 4) 9x2 = . (3x + 4)2 (3x + 4)2 h0 h0 7(1 + h)2 4(1 + h) 3 h 7h2 + 10h h h0 Now the tangent line is horizontal to the graph when y (x) = 0, i.e., when 9x2 + 24x = 3x(3x + 8) = 0. Hence the tangent line is horizontal when x = 0, 8 3 . = lim h0 = lim (7h + 10) = 10 . By the point-slope formuala, therefore, an equation for the tangent line at P (1, 3) is y 3 = 10(x 1) which after simplication becomes y = 10x 7 . keywords: tangent line, slope equation 012 (part 1 of 1) 10 points Find all values of x at which the tangent line to the graph of 3x2 y= 3x 4 keywords: + derivative, rational function, tangent line horizontal, 013 (part 1 of 1) 10 points The position of a car as a function of time is given by the values in table t (seconds) 0 1 2 3 4 5 s (feet) 0 13 17 33 39 57 Find the average velocity for the time period beginning when t = 3 and lasting 2 seconds. 1. average vel. = 10 feet/sec 2. average vel. = 11 feet/sec Garcia, Ilse Exam 1 Due: Oct 2 2007, 11:00 pm Inst: Fonken 3. average vel. = 8 feet/sec 4. average vel. = 9 feet/sec 5. average vel. = 12 feet/sec correct Explanation: Average velocity between times t = 3 and t = 3 + h is given by s(3 + h) s(3) dist. travelled = . time taken h When h = 2, therefore, s(3 + 2) s(3) s(5) s(3) 57 33 = = . 2 2 2 Consequently, over the time interval from t = 3 to t = 5 the car has average velocity = 12 ft/sec . 7 distance travelled being plotted on the y -axis, time elapsed on the x-axis. Consider the following statements: A. At 12:50pm he was stationary momentarily. B. At 1:00pm he was accelerating. Which statements are true? 1. neither is true 2. B only 3. A only 4. both are true correct Explanation: A. True: tangent line horizontal at that moment. B. True: slope is increasing at that moment. keywords: word problem, speed, distance keywords: average velocity, table of values, velocity, position 014 (part 1 of 1) 10 points Recently, Stewart drove the 120 kms from Austin to San Antonio, leaving at 12noon and arriving in SA at 1:30pm. He stayed briey and then started back. The GPS system in his car recorded his distance from Austin and the time elapsed. It plotted these accurately to scale on the following graph: kms 120 100 80 60 40 20 time 12:30 1:00 1:30 2:00 015 (part 1 of 1) 10 points Below is the graph of a function f . 6 5 4 3 2 1 0 -1 -2 -3 -4 -5 -6 4 2 6 4 2 2 4 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 2 4 6 Use this graph to determine all of the values of x on (7, 7) at which f is discontinuous. 1. None of these 2. x = 5 Garcia, Ilse Exam 1 Due: Oct 2 2007, 11:00 pm Inst: Fonken 3. no values of x 4. x = 5 , 5 correct 5. x = 5 Explanation: Since f (x) is dened everywhere on (7, 7), the function f will be discontinuous at a point x0 in (7, 7) if and only if x x0 8 3. (, 3) (3, 4) (4, ) 4. (, 3) (3, ) 5. (, 3) (3, 4) (4, ) correct Explanation: After factorization the denominator becomes x2 7x + 12 = (x 3)(x 4), so f can be written as lim f (x). f (x) = x5 . (x 3)(x 4) lim f (x) = f (x0 ) or if x x0 lim f (x) = x x0 + As the graph shows, the only possible candidates for x0 are x0 = 5 and x0 = 5. But at x0 = 5, f (5) = 4 = lim f (x) = 0, x5 Being a rational function, it will be continuous everywhere except at the zeros of the denominator since it will not be dened at such points. Thus f is continuous everywhere except at x = 3 and x = 4. Hence it will continuous on (, 3) (3, 4) (4, ). keywords: rational function, continuous 017 (part 1 of 1) 10 points Below is the graph of a function f . 10 9 8 8 7 6 6 5 4 4 3 2 2 1 0 -1 6 4 2 24 6 -2 2 -3 -4 4 -5 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 while at x0 = 5, x 5 lim f (x) = 4 = x 5+ lim f (x) = 0 . Consequently, on (7, 7) the function f is discontinuous at x = 5, 5. keywords: graph, continuous, limit 016 (part 1 of 1) 10 points Find all values of x at which the function f dened by f (x) = x2 x5 7x + 12 is continuous, expressing your answer in interval notation. 1. (, 4) (4, ) 2. (, 4) (4, 3) (3, ) Use the graph to determine all the values of x on (6, 6) at which f is not dierentiable. Garcia, Ilse Exam 1 Due: Oct 2 2007, 11:00 pm Inst: Fonken 1. x = 4, 0 2. x = 4, 3 3. x = 4 4. x = 0, 3 5. x = 4, 0, 3 correct Explanation: The graph shows that f has a removable discontinuity at x = 4 and a jump discontiinuity at x = 3, so f will not be dierentiable at these points. On the other hand, at x = 0 the graph is continuous but has a corner, so it will not be dierentiable at this point also. Thus, on (6, 6) the function f will fail to be dierentiable at the points x = 4, 0, 3, . 6 5 4 1. 3 2 1 0 -1 -2 -3 -4 -5 -6 6 5 4 2. 3 2 1 0 -1 -2 -3 -4 -5 -6 6 5 4 3. 3 2 1 0 -1 -2 -3 -4 -5 -6 6 5 4 4. 3 2 1 0 -1 -2 -3 -4 -5 -6 6 5 4 5. 3 2 1 0 -1 -2 -3 -4 -5 -6 9 4 2 2 2 4 -2 -1 0 1 2 3 4 5 6 7 8 9 10 4 6 8 4 2 2 2 4 -2 -1 0 1 2 3 4 5 6 7 8 9 10 4 6 8 4 2 2 2 4 4 6 8 correct keywords: not dierentiable, graph, discontinuity 018 (part 1 of 1) 10 points If f is a function having 6 5 4 4 3 2 2 1 0 -1 2 46 8 -2 2 -3 -4 4 -5 -6 -2 -1 0 1 2 3 4 5 6 7 8 9 10 -2 -1 0 1 2 3 4 5 6 7 8 9 10 4 2 2 2 4 -2 -1 0 1 2 3 4 5 6 7 8 9 10 4 6 8 4 2 2 2 4 -2 -1 0 1 2 3 4 5 6 7 8 9 10 4 6 8 as its graph, which of the following is the graph of the derivative f of f ? Garcia, Ilse Exam 1 Due: Oct 2 2007, 11:00 pm Inst: Fonken Explanation: The slope of the tangent to the graph is zero at x = 3, is negative to the left of x = 3 and is positive to the right of x = 3. But it is increasing for all x, so the graph of f is 8 7 6 6 5 4 4 3 2 2 1 0 -1 2 46 8 -2 2 -3 -4 -2 -1 0 1 2 3 4 5 6 7 8 9 10 7 6 5 2. 4 3 2 1 0 -1 -2 -3 -4 -5 7 6 5 3. 4 3 2 1 0 -1 -2 -3 -4 -5 7 6 5 4. 4 3 2 1 0 -1 -2 -3 -4 -5 5 4 3 5. 2 1 0 -1 -2 -3 -4 -5 -6 -7 5 4 3 6. 2 1 0 -1 -2 -3 -4 -5 -6 -7 10 6 4 2 6 4 2 2 4 6 4 2 6 4 2 2 4 6 4 2 6 4 2 2 4 4 2 6 4 2 2 4 6 4 2 6 4 2 2 4 6 2 4 6 2 4 6 2 4 6 2 4 6 2 4 6 correct -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 keywords: graph of derivative from graph of function 019 (part 1 of 1) 10 points When 12 11 10 9 8 7 6 5 4 3 2 1 0 -1 -2 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 10 8 6 4 2 6 4 2 2 4 6 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 is the graph of a function f on (7, 7), which of the following is the graph of its derivative f? 5 4 4 3 1. 2 2 1 0 -1 246 -2 6 4 2 2 -3 -4 4 -5 -6 6 -7 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 Explanation: Since the graph of f has corners at two points (x0 , f (x0 )), f (x) will not exist for -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 Garcia, Ilse Exam 1 Due: Oct 2 2007, 11:00 pm Inst: Fonken these two values of x0 . The graph of f will thus have two holes, eliminating immediately one of the possible graphs of f . Now the graph of the derivative of a linear function is a horizontal line which will lie below (resp. above) the x-axis when the slope of the straight line is negative (resp. positive). On the other hand, the derivative of a quadratic function is a straight line whose slope will be positive (resp. negative) if the graph of the quadratic function (a parabola) opens up (resp. down). For the given f , therefore, the graph of f will be 7 6 6 5 4 4 3 2 2 1 0 -1 6 4 2 2 4 6 -2 2 -3 -4 4 -5 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 11 2. 2 1 1 2 3. 2 1 1 2 keywords: graph of derivative, from graph of function 020 (part 1 of 1) 10 points Sketch the graph of a function g for which g(0) = 1, while g (1) = 0, g (2) = 2 . g (0) = 4, 4. 2 1 correct 1 2 1. 2 1 5. 2 1 1 2 1 2 Garcia, Ilse Exam 1 Due: Oct 2 2007, 11:00 pm Inst: Fonken 12 6. 2 1 1 2 Explanation: One possible graph can be eliminated immediately because its y-intercept does not satisfy the condition g(0) = 1. On the other hand, the slope both at x = 0 and x = 2 of the remaining graphs can be estimated using the grid provided. By this method we thus see that the only graph satisfying all the conditions is 2 1 1 2 keywords: graph, derivative, estimate
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University of Texas - M - 58365
Garcia, Ilse Exam 2 Due: Oct 31 2007, 1:00 am Inst: Fonken This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. The due time is Central time. 001 (part 1 of 1)
University of Texas - M - 58365
Garcia, Ilse Exam 3 Due: Dec 5 2007, 1:00 am Inst: Fonken This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. The due time is Central time. 001 (part 1 of 1)
University of Texas - M - 58365
Garcia, Ilse Final 1 Due: Dec 18 2007, 2:00 am Inst: Fonken This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. The due time is Central time. 001 (part 1 of 1
University of Texas - M - 58365
Garcia, Ilse Homework 1 Due: Sep 7 2007, 3:00 am Inst: Fonken This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. The due time is Central time. Yes, Homework
University of Texas - M - 58365
Garcia, Ilse Homework 2 Due: Sep 4 2007, 3:00 am Inst: Fonken This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. The due time is Central time. Yes, Homework
University of Texas - M - 58365
Garcia, Ilse Homework 3 Due: Sep 11 2007, 3:00 am Inst: Fonken This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. The due time is Central time. 001 (part 1 o
University of Texas - M - 58365
Garcia, Ilse Homework 4 Due: Sep 19 2007, 3:00 am Inst: Fonken This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. The due time is Central time. 001 (part 1 o
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COP 3503 CS II CLASS NOTES - DAY #3 Supplemental Asymptotic Notation Big Oh Notation Definition: Let p(n) and q(n) be two nonnegative functions. The function p(n) is asymptotically bigger [p(n) asymptotically dominates q(n)] than the function q(n)
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COP 3503 Computer Science II CLASS NOTES - DAY #11 SupplementElementary Sorting AlgorithmsSelection Sort Selection sort is an attempt to localize the exchanges of array elements by finding a misplaced element first and putting it in its final pl
UCF - COP - 3503c
COP 3503 Computer Science II CLASS NOTES - DAY #12Mergesort The mergesort sorting algorithm uses the divide and conquer strategy in which the original problem is split into two half-size, recursively solved problems. If the overhead of the base c
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UCF - COP - 3503c
COP 3503 Computer Science II CLASS NOTES - DAY #17 PART III - APPLICATIONS Chapter 11 Stacks and Compilers Stacks are a commonly used data structure in compilers. We will examine two of the basic uses of the stack data structure in a compiler, (1)
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COP 3503 Computer Science II CLASS NOTES - DAY #21 Doubly Linked Lists Better than singly linked list as they are bi-directional. Has both a header and a tail node (for the same reason that we added the header node to the singly linked list). nul
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COP 3503 Computer Science II CLASS NOTES - DAY #22Chapter 17 TREES Techniques for defining a tree There are two basic techniques that can be used to define a tree. 1. Recursively: this allows for very simple algorithms to manipulate the tree. 2.
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COP 3503 Computer Science II CLASS NOTES - DAY #23 SupplementHuffman Coding RevisitedExample: Suppose that we have a four letter alphabet consisting of a, b, c, and d, and e only. To encode four letters requires 2 bits. Suppose that these are a
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UCF - COP - 3503c
COP 3503 Final Exam Practice Problems - SOLUTIONS 1. For the binary tree shown below, show the output of a preorder traversal of the tree. Foradditional practice do: inorder, postorder, and level order traversals.ABEC DF G H IPreorder
UCF - COP - 3503c
COP 3503H Mid-term Exam Spring 2001Thursday March 1, 2001 [100 points] NO CALCULATORS MAY BE USED!NAME:KEYStudent ID:1. (15 points Induction Proof) Shown below is a conjecture. Complete an induction proof that proves the conjecture is tru
UCF - COP - 3503c
COP 3503H Spring 2001 Programming Assignment #1Points: This assignment is worth 100 points. [program 60pts write-up 40 pts] Due Date: Thursday February 15, 2001 in class ObjectiveIn class we discussed three different algorithms, each of differen
UCF - COP - 3503c
COP 3503H Spring 2001 Programming Assignment #2Due Date: March 27, 2001 at class time. Points: 100 total program 60 points, write-up 40 points Objective: You will implement the Insertion sort, the Shell sort, and the Quick sort algorithms (all of
UCF - COP - 3503c
COP 3503 Honors OverviewProject #3 PresentationIn this project you will research a data structure which will not be covered in the class. You will prepare a short paper and deliver a presentation to the class on the data structure that you have
UCF - COP - 4710
COP 4710 Fall 2007 Course CalendarAugustSunday MondayTuesday7 14 21 Classes Begin IntroductionWednesday1 2 9 16Thursday3FridaySaturday4 11 18 255 12 196 13 208 15 2210 17 2423 Chapter 1 Notes 30 Chapter 2 Notes262728
UCF - COP - 4710
COP 4710: Database Systems Fall 2007Introduction to Database SystemsInstructor : Dr. Mark Llewellyn markl@cs.ucf.edu HEC 236, 407-823-2790 http:/www.cs.ucf.edu/courses/cop4710/fall2007School of Electrical Engineering and Computer Science Universi
UCF - COP - 4710
COP 4710: Database Systems Fall 2007Introduction to Database SystemsInstructor : Dr. Mark Llewellyn markl@cs.ucf.edu HEC 236, 407-823-2790 http:/www.cs.ucf.edu/courses/cop4710/fall2007School of Electrical Engineering and Computer Science Universi
UCF - COP - 4710
COP 4710: Database Systems Fall 2007Chapter 2 Introduction to Data ModelingInstructor :Mark Llewellyn markl@cs.ucf.edu HEC 236, 823-2790 http:/www.cs.ucf.edu/courses/cop4710/fall2007School of Electrical Engineering and Computer Science Univer
UCF - COP - 4710
COP 4710: Database Systems Fall 2007Chapter 2 Introduction to Data ModelingInstructor :Mark Llewellyn markl@cs.ucf.edu HEC 236, 823-2790 http:/www.cs.ucf.edu/courses/cop4710/fall2007School of Electrical Engineering and Computer Science Univer
UCF - COP - 4710
COP 4710: Database Systems Fall 2007Chapter 2 In Class ExercisesInstructor :Dr. Mark Llewellyn markl@cs.ucf.edu HEC 236, 823-2790 http:/www.cs.ucf.edu/courses/cop4710/fall2007School of Electrical Engineering and Computer Science University of
UCF - COP - 4710
COP 4710: Database Systems Fall 2007Chapter 2 In Class ExercisesInstructor :Dr. Mark Llewellyn markl@cs.ucf.edu HEC 236, 823-2790 http:/www.cs.ucf.edu/courses/cop4710/fall2007School of Electrical Engineering and Computer Science University of
UCF - COP - 4710
COP 4710: Database Systems Fall 2007Chapter 3 The Relational Data ModelInstructor :Dr. Mark Llewellyn markl@cs.ucf.edu HEC 236, 407-823-2790 http:/www.cs.ucf.edu/courses/cop4710/fall2007School of Electrical Engineering and Computer Science Un
UCF - COP - 4710
COP 4710: Database Systems Fall 2007Chapter 3 The Relational Data ModelInstructor :Dr. Mark Llewellyn markl@cs.ucf.edu HEC 236, 407-823-2790 http:/www.cs.ucf.edu/courses/cop4710/fall2007School of Electrical Engineering and Computer Science Un
UCF - COP - 4710
COP 4710: Database Systems Fall 2007Chapter 3 In Class ExercisesInstructor :Dr. Mark Llewellyn markl@cs.ucf.edu HEC 236, 823-2790 http:/www.cs.ucf.edu/courses/cop4710/fall2007School of Electrical Engineering and Computer Science University of
UCF - COP - 4710
COP 4710: Database Systems Fall 2007Chapter 3 In Class ExercisesInstructor :Dr. Mark Llewellyn markl@cs.ucf.edu HEC 236, 823-2790 http:/www.cs.ucf.edu/courses/cop4710/fall2007School of Electrical Engineering and Computer Science University of
UCF - COP - 4710
COP 4710: Database Systems Fall 2007Chapter 4 Relational Query Languages Part 1Instructor :Dr. Mark Llewellyn markl@cs.ucf.edu HEC 236, 823-2790 http:/www.cs.ucf.edu/courses/cop4710/fall2007School of Electrical Engineering and Computer Scien
UCF - COP - 4710
COP 4710: Database Systems Fall 2007Chapter 4 Relational Query Languages Part 1Instructor :Dr. Mark Llewellyn markl@cs.ucf.edu HEC 236, 823-2790 http:/www.cs.ucf.edu/courses/cop4710/fall2007School of Electrical Engineering and Computer Scien
UCF - COP - 4710
COP 4710: Database Systems Fall 2007Chapter 4 Relational Query Languages Part 2Instructor :Dr. Mark Llewellyn markl@cs.ucf.edu HEC 236, 823-2790 http:/www.cs.ucf.edu/courses/cop4710/fall2007School of Electrical Engineering and Computer Scien
UCF - COP - 4710
COP 4710: Database Systems Fall 2007Chapter 4 Relational Query Languages Part 2Instructor :Dr. Mark Llewellyn markl@cs.ucf.edu HEC 236, 823-2790 http:/www.cs.ucf.edu/courses/cop4710/fall2007School of Electrical Engineering and Computer Scien
UCF - COP - 4710
COP 4710: Database Systems Fall 2007Chapter 5 Introduction To SQL Part 1Instructor :Dr. Mark Llewellyn markl@cs.ucf.edu HEC 236, 407-823-2790 http:/www.cs.ucf.edu/courses/cop4710/fall2007School of Electrical Engineering and Computer Science
UCF - COP - 4710
COP 4710: Database Systems Fall 2007Chapter 5 Introduction To SQL Part 1Instructor :Dr. Mark Llewellyn markl@cs.ucf.edu HEC 236, 407-823-2790 http:/www.cs.ucf.edu/courses/cop4710/fall2007School of Electrical Engineering and Computer Science
UCF - COP - 4710
COP 4710: Database Systems Fall 2007Chapter 5 Introduction To SQL Part 2Instructor :Dr. Mark Llewellyn markl@cs.ucf.edu HEC 236, 407-823-2790 http:/www.cs.ucf.edu/courses/ccop4710/fall2007School of Electrical Engineering and Computer Science
UCF - COP - 4710
COP 4710: Database Systems Fall 2007Chapter 5 Introduction To SQL Part 2Instructor :Dr. Mark Llewellyn markl@cs.ucf.edu HEC 236, 407-823-2790 http:/www.cs.ucf.edu/courses/ccop4710/fall2007School of Electrical Engineering and Computer Science
UCF - COP - 4710
COP 4710: Database Systems Fall 2007Chapter 9 Data StorageInstructor :Dr. Mark Llewellyn markl@cs.ucf.edu HEC 236, 407-823-2790 http:/www.cs.ucf.edu/courses/cop4710/fall2007School of Electrical Engineering and Computer Science University of C