course MTH271 Monday, July 12th 2010 9:10 AM 005. `query 5
.............................................
Given Solution: `a the specific idea is that ave rate of depth change is [change in depth / change in time] ; rise represents change in depth and run represents change in time so slope = rise/run represents ave rate of depth change. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):okay ------------------------------------------------ Self-critique rating #$&*3 ********************************************* Question: `qexplain why the area of a rate vs. time trapezoid for a given time interval represents the change in the quantity corresponding to that time interval YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The average altitude represents the avg. velocity. The area of a trapezoid involves the altitude, which represents the avg. velocity, and the width, which represents the change in clock time. For any rate vs. clock time graph, average altitude represents approximate average rate, which multiplied by the change in time (not by the time itself) gives you the change in quantity. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a The average altitude represents the avg. velocity. The area of a trapezoid involves the altitude, which represents the avg. velocity, and the width, which represents the change in clock time. When you multiply ave altitude by width you are representing ave vel * change in clock time, which gives change in position. This reasoning isn't confined to velocities. For any rate vs. clock time graph, average altitude represents approximate average rate, which multiplied by the change in time (not by the time itself) gives you the change in quantity ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):okay ------------------------------------------------ Self-critique rating #$&*3 ********************************************* Question: `qtext problem 0.5 #10 add x/(2-x) + 2/(x-2) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x / (2-x) + 2 / (x-2) ( (x-2) / (x-2) ) * ( x / (2-x)) + ( (2-x) / (2-x) )* ( 2 / (x-2) ) x(x-2) / ( (2-x)(x-2) ) + 2 (2-x) / ( (2-x)(x-2) ) (x(x-2) + 2(2-x) ) / ((2-x)(x-2) ) x^2 - 2x + 4 - 2x ) / ( (2-x)(x-2) ) (x^2-4x+4) / ( -x^2+4x-4 ) confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a common denominator could be [ (2-x)(x-2) ]. In this case we have x / (2-x) + 2 / (x-2) = [ (x-2) / (x-2) ] * [ x / (2-x) ] + [ (2-x) / (2-x) ] * [ 2 / (x-2) ] = x(x-2) / [ (2-x)(x-2) ] + 2 (2-x) / [ (2-x)(x-2) ] = [x(x-2) + 2(2-x) ] / [ (2-x)(x-2) ] = [ x^2 - 2x + 4 - 2x ] / [ (2-x)(x-2) ] = (x^2-4x+4) / [ -x^2+4x-4 ] = (x-2)^2 / [-(x-2)^2] = -1. NOTE however that there is a SIMPLER SOLUTION: We can note that x-2 = -(2-x) so that the original problem is -x/(x-2) + 2 /(x-2) = (-x + 2) / (x-2) = -(x-2)/(x-2) = -1. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):okay ------------------------------------------------ Self-critique rating #$&*3 ********************************************* Question: `qtext problem 0.5 #50 cost = 6 x + 900,000 / x, write as single fraction and determine cost to store 240 units YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: [x / x] * 6x + 900,000 / x 6x^2 / x + 900,000 / x (6x^2 + 900,000) / x so cost = (6x^2+900,000)/x x = 240 = (6 * 240^2 + 900000) / 240 = 5190. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a express with common denominator x: [x / x] * 6x + 900,000 / x = 6x^2 / x + 900,000 / x = (6x^2 + 900,000) / x so cost = (6x^2+900,000)/x Evaluating at x = 240 we get cost = (6 * 240^2 + 900000) / 240 = 5190. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):okay ------------------------------------------------ Self-critique rating #$&*3 "