Explanation Given: • The velocity of a car is recorded at half-second intervals (in feet per second). • Time: First 4 s. • Table: t (min) 0 0.5 1 1.5 2 2.5 3 3.5 4 v ( t ) 0 12 20 29 38 44 32 35 30 Key Concepts Used: Many applications are based on the following principle. The net change in a quantity s ( t ) is equal to the integral of its rate of change: s ( t 2 ) − s ( t 1 ) = ∫ s ' ( t ) ⋅ d t t 1 t 2 . Here s ( t 2 ) − s ( t 1 ) is the net change in s ( t ) over the time interval [ t 1 , t 2 ] and ∫ s ' ( t ) ⋅ d t t 1 t 2 is Integral of rate of change. Here in this problem, velocity of a car in feet per second is recorded at half-second intervals and are tabulated in the given table. The data values are spaced at intervals Δ t = 0.5 seconds. We have to find N t h right-endpoint approximation ( R N ) and N t h left-endpoint approximation ( L N ). Then we have to take the average of N t h right-endpoint approximation ( R N ) and N t h left-endpoint approximation ( L N ) to find the total distance travelled during the first 4 seconds. Formulas Used: N t h Right-endpoint approximation: R N = Δ x ∑ j = 0 N ( f ( a + Δ x ) + f ( a + 2 Δ x ) + ⋅ ⋅ ⋅ ⋅ ⋅ + f ( a + N Δ x ) ) . N t h Left-endpoint approximation: L N = Δ x ∑ j = 0 N − 1 ( f ( a ) + f ( a + Δ x ) + ⋅ ⋅ ⋅ ⋅ ⋅ + f ( a + ( N − 1 ) Δ x ) ) ...

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 5.6, Problem 19E

To determine

Total distance travelled during given time by using the average of left- and right-endpoint approximation

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Chapter 5.6, Problem 18E

Chapter 5.6, Problem 20E

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Total distance travelled during given time by using the average of left- and right-endpoint approximation

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Chapter 5 Solutions