Quest HW 2-solutions

# Quest HW 2-solutions - syed(sms3768 Quest HW 2 seckin(56425...

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syed (sms3768) – Quest HW 2 – seckin – (56425) 1 This print-out should have 28 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points A tank holds 100 gallons oF water, which drains From the bottom oF the tank in 10 minutes. The values in the table t (min) 0 2 4 6 8 10 V (gal) 100 63 40 19 8 0 show the volume, V ( t ), oF water remaining in the tank (in gallons) aFter t minutes. IF P is the point (2 , V (2)) on the graph oF V as a Function oF time t , fnd the slope oF the secant line PQ when Q = (6 , V (6)). 1. slope = 10 2. slope = 11 correct 3. slope = 9 4. slope = 10 5. slope = 11 6. slope = 13 7. slope = 13 8. slope = 9 Explanation: When P = (2 , V (2)) , Q = (6 , V (6)) the slope oF the secant line is given by rise run = V (6) V (2) 6 2 . ±rom the table oF values, thereFore, we see that slope = 19 63 6 2 = 11 . 002 (part 1 oF 5) 10.0 points At which point on the graph F E D C B A (i) is the slope greatest ( i.e. , most positive)? 1. B 2. E 3. D 4. F correct 5. A 6. C Explanation: By inspection the point is F . 003 (part 2 oF 5) 10.0 points (ii) is the slope smallest ( i.e. , most negative)? 1. E 2. F 3. D correct 4. B

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syed (sms3768) – Quest HW 2 – seckin – (56425) 2 5. A 6. C Explanation: By inspection the point is D . 004 (part 3 of 5) 10.0 points (iii) does the slope change from positive to negative? 1. E 2. F 3. B correct 4. C 5. A 6. D Explanation: By inspection the point is B . 005 (part 4 of 5) 10.0 points (iv) does the slope change from negative to positive? 1. B 2. F 3. D 4. C 5. E correct 6. A Explanation: By inspection the point is E . 006 (part 5 of 5) 10.0 points (v) 1. B 2. F 3. A 4. C correct 5. E 6. D Explanation: By inspection the point is C . keywords: slope, graph, change of slope 007 (part 1 of 2) 10.0 points After t seconds the displacement, s ( t ), of a particle moving rightwards along the x -axis is given (in feet) by s ( t ) = 6 t 2 4 t + 7 . (i) Determine the average velocity of the particle over the time interval [1 , 2]. 1. average vel. = 16 ft/sec 2. average vel. = 17 ft/sec 3. average vel. = 15 ft/sec 4. average vel. = 14 ft/sec correct 5. average vel. = 18 ft/sec Explanation: The average velocity over a time interval [ a, b ] is given by dist travelled time taken = s ( b ) s ( a ) b a . For the time interval [1 , 2], therefore, ave. vel. = s (2) s (1) 2 1 ft/sec . Now s (2) = 6 × 4 4 × 2 + 7 = 23 feet ,
syed (sms3768) – Quest HW 2 – seckin – (56425) 3 while s (1) = 6 4 + 7 = 9 feet . Consequently, average vel. = 23 9 = 14 ft/sec . 008 (part 2 of 2) 10.0 points (ii) By determining the average velocity successively over the time intervals [1 , 1 . 1] , [1 , 1 . 01] , [1 , 1 . 001] , [1 , 1 . 0001] , estimate the instantaneous velocity, v (1), of the particle at time t = 1. 1. v (1) = 6 ft/sec 2. v (1) = 7 ft/sec 3. v (1) = 4 ft/sec 4. v (1) = 8 ft/sec correct 5. v (1) = 5 ft/sec Explanation: After calculation, s (1 . 1) s (1) 0 . 1 = 8 . 6 s (1 . 01) s (1) 0 . 01 = 8 . 06 s (1 . 001) s (1) 0 . 001 = 8 . 006 s (1 . 0001) s (1) 0 . 0001 = 8 . 0006 .

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## This note was uploaded on 04/29/2010 for the course MATH 408K taught by Professor Seckin during the Spring '10 term at University of Texas-Tyler.

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Quest HW 2-solutions - syed(sms3768 Quest HW 2 seckin(56425...

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