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Unformatted text preview: silva (jrs4378) – HW04 – Schultz – (54765) 1 This printout should have 18 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Find the xintercept of the tangent line at the point P ( − 1 , 5) on the graph of f when lim h → f ( − 1 + h ) − 5 h = 4 . 1. xintercept = − 9 4 correct 2. xintercept = − 9 3. xintercept = 1 4. xintercept = 9 4 5. xintercept = 1 4 6. xintercept = 9 Explanation: Since lim h → f ( − 1 + h ) − f ( − 1) h is the slope of the tangent line at P ( − 1 , 5) , an equation for this tangent line is y − 5 = 4( x + 1) , i.e. , y = 4 x + 9 . Consequently, the xintercept = − 9 4 . 002 10.0 points Consider the slope of the given curve at the five points shown. A C E B D List the five slopes in decreasing order. 1. B, D, C, A, E 2. B, D, E, A, C correct 3. D, B, A, E, C 4. D, A, C, E, B 5. C, A, E, D, B 6. C, B, A, D, E Explanation: The order will be the one from most positive slope to most negative slope. Inspection of the graph shows that this is B, D, E, A, C . 003 10.0 points Find the slope of the tangent line to the parabola y = x 2 − 3 x at the point (1, − 2). 1. slope = 1 2. slope = − 2 3. slope = − 1 correct 4. slope = 0 silva (jrs4378) – HW04 – Schultz – (54765) 2 5. slope = 2 Explanation: By definition, the slope, m , of the tangent line at the point P ( a, f ( a )) on the graph of f is the limit m = lim x → a f ( x ) − f ( a ) x − a . For the point (1 , − 2) on the graph of y = f ( x ) = x 2 − 3 x, therefore, m = lim x → 1 f ( x ) − f (1) x − 1 = lim x → 1 ( x 2 − 3 x ) − ( − 2) x − 1 = lim x → 1 ( x − 2)( x − 1) ( x − 1) . Consequently, slope = lim x → 1 ( x − 2) = − 1 . 004 10.0 points A Calculus student leaves the RLM build ing and walks in a straight line to the PCL Library. Her distance from RLM after t min utes is given by the graph 2 4 6 8 10 12 mins yards 100 200 300 400 500 What is her speed after 9 minutes, and in what direction is she heading at that time? 1. away from RLM at 30 yds/min . 2. away from RLM at 25 yds/min . 3. away from RLM at 20 yds/min . 4. towards RLM at 20 yds/min . 5. towards RLM at 30 yds/min . 6. towards RLM at 25 yds/min . correct Explanation: The graph is linear on [8 , 10], so the stu dent’s speed at time t = 9 is the (absolute value of the) slope of this line. Hence slope = 100 − 150 10 − 8 = − 25 . The fact that her distance from RLM is de creasing at t = 9 indicates that she is walking towards RLM at that time. 005 10.0 points Recently, Stewart drove the 120 kms from Austin to San Antonio, leaving at 12noon and arriving in SA at 1:30pm. He stayed briefly and then started back. The GPS system in his car recorded his distance from Austin in terms of the time of day. It plotted these accurately to scale: time kms 20 40 60 80 100 120 12:30 1:00 1:30 2:00 silva (jrs4378) – HW04 – Schultz – (54765)...
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This note was uploaded on 10/23/2010 for the course M 54765 taught by Professor Schultz during the Fall '10 term at University of Texas.
 Fall '10
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