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HW4solution-Part2

HW4solution-Part2 - Math 16A Fall 2007 Professor Stankova...

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Math 16A, Fall 2007 Professor Stankova Félice Lê Homework 4 Solutions – Part 2 Section 1.8 1.8#4 a) The average rate of change is given by 5 2 7 1 2 12 2 ) 2 ( 3 3 12 2 ) 3 ( 3 2 3 ) 2 ( ) 3 ( = - = - + - - + = - - f f . b) The instantaneous rate of change when 2 = t is given by ) 2 ( f . We use the Sum Rule to differentiate each term of ) ( t f separately. - + + = t dt d dt d t dt d t f 12 2 3 ) ( Then we apply the Power Rule to each term. 2 1 12 3 12 0 3 12 2 3 ) ( - - + = - + = - + + = t t dt d t dt d dt d t dt d t f . Then 6 4 12 3 ) 2 ( 12 3 ) 2 ( 2 = + = + = - f . 1.8#6 Note that the t -value in the graph represents the number of years since January 1, 1950. Therefore, t =10 represents January 1, 1910, t =20 represents January 1, 1920, and so forth. a) We are looking at the points (20, 150) and (60, 300) on the curve. The average rate of change is 4 15 20 60 150 300 20 60 ) 20 ( ) 60 ( = - - = - - f f . b) The speed at which the mean farm size was rising on January 1, 1950, is equivalent to the instantaneous rate of change at t = 50, or ) 50 ( f . This in turn is equivalent to the slope of the tangent line to the graph at t = 50. This tangent line is shown on the graph. We can choose any two 1

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points on the line to calculate its slope; using the points (40, 150) and (80, 400), ) 50 ( 4 25 40 80 150 400 f x y m = = - - = = . c) We can see that at t = 60, the graph is steeper than at t = 80. In other words, the slope of the graph (and of the tangent line) is greater at t = 60. Therefore, we conclude that the mean farm size was rising faster on January 1, 1960, than on January 1, 1980. 1.8#10 The rate at which the liquid is flowing into the vat is given by ( 29 t t dt d - 5 . In order to calculate this rate at t = 2, we need to calculate ( 29 t t dt d - 5 2 = t .
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