course MTH 163 10/10/09 7:02 p.m. 009. `query 9
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Given Solution: ** For the function y = .1 x^2 - 3 between x = -2 and x = 7 we get: slope = (y2 - y1) / (x2 - x1). For x1 = 2 and x2 = 7 we have y2 = .1 * 7^2 - 3 = 1.9 and y1 = .1 * 2^2 - 3 = -2.6, so slope = (1.9 - (-2.6) ) / ( 7 - 2) = 4.5 / 5 = .9. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I used the -2 in this problem for x I think in your given solution for the x values you have (7-2) should it have been (7--2) or (7+2) ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qproblem 2 symbolic expression for slope, fn depth(t). What is the expression for the slope between the two specified t values? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: function is y=d(t) let x1 = 10 and x2 = 30 let y1 = d(10) and y2 = d(30) rise = (d(30)-d(10)) run = 30-10 = 20 slope = (d(30)-d(10))/20 confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The function is given a name: depth(t). t values are 10 and 30. So rise = depth(30) - depth(10) and run = 30 - 10 = 20. Thus slope = [ depth(30) - depth(10) ] / 20 . ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhat is the rise between the two specified t values? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The rise was (d(30)-d(10)) confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The rise is the change in depth. The two depths are depth(10) and depth(30). The change in depth is final depth - initial depth, which gives us the expression depth(30)-depth(10) ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhat is the run between the two specified t values? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The run was the t values which were 30-10 = 20 confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** run = 30 - 10 = 20 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhat therefore is the slope and what does it mean? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: the slope is (d(30)-d(10))/20 the average rate of depth change for the clock times between 30 and 10 confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** rise = depth(30)-(depth(10) indicates change in depth. run = 30 - 10 = 20 = change in clock time. Slope = rise / run = (depth(30) - depth(10) ) / 20, which is the average rate at which depth changes with respect to clock time between t = 10 and t = 30. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q problem 5 graph points corresponding to load1 and load2 What are the coordinates of the requested graph points? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: sl=spring length let y1 = (sl(load1) and y2 = sl(load2) let x1 = (load1) and x2 = (load2) the coordinates of the graph are therefore (load1,(sl(load1)) and (load2,(sl,(load2)) the rise = {(sl(load2))-(sl(load1)} the run = (load2-load1) the slope = {(sl(load2))-(sl(load1)}/(load2-load1) confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The horizontal axis is the 'load' asix, the vertical axis is the springLength axis. The load axis coordinates are load1 and load2. The corresponding spring lengths are springLength(load1) and springLength(load2). The springLength axis coordinates are springLength(load1) and springLength(load2). The graph points are thereofore (load1, springLength(load1) ) and (load2, springLength(load2) ). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhat is your expression for the average slope of the graph between load1 and load2? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: the slope = {(sl(load2))-(sl(load1)}/(load2-load1) confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** rise = springLength(load2) - springLength(load1) run = load2 - load1 so slope = [ springLength(load2) - springLength(load1)] / (load2 - load1). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q problem 6 symbolic expression for slope of depth function YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Let y1 = d(t1) and y2 = d(t2) let x1 = t1 and x2 = t2 so the function for slope of depth = {d(t2)-d(t1)}/(t2-t1) confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** the name of the function is depth(t). We need the slope between t = t1 and t = t2. The depths are depth(t1) and depth(t2). Thus rise is depth(t2) - depth(t1) and run is t2 - t1. Slope is [ depth(t2) - depth(t1) ] / (t2 - t1). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q problem 8 average rate from formula f(t) = 40 (2^(-.3 t) ) + 25 intervals of partition (10,20,30,40) What average rate do you get from the formula? Show your steps. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: let y1=(40(2^(-.3(10))+25 let y2=(40(2^(-.3(20))+25 let x1=10 let x2 = 20 {[(40(2^(-.3(20))+25]}-((40(2^(-.3(10))+25))/(20-10)=-.4375 let y1=(40(2^(-.3(20))+25) let y2=(40(2^(-.3(30))+25) let x1=20 letx2=30 {(40(2^(-.3(30))+25)- (40(2^(-.3(20))+25)}/(30-20)=-.05469 let y1=(40(2^(-.3(30))+25) let y2=(40(2^(-.3(40))+25 let x1=30 letx2=40 {(40(2^(-.3(40))+25)-(40(2^(-.3(30))+25)}/(40-30)=-.00684 confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** ave rate = change in depth / change in t. For the three intervals we get (f(20)-f(10))/(20-10) = (25.625 - 30 )/(20 - 10) = -4.375 / 10 = -.4375 (f(30)-f(20))/(30-20) = (25.07813 - 25.625)/(30 - 20) = -.5469 / 10 = -.05469. (f(40)-f(30))/(40-30) = (25.00977 - 25.07813)/(40 - 30) = -.0684 / 10 = -.00684. ** Add comments on any surprises or insights you experienced as a result of this assignment. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: STUDENT RESPONSE: Ummm I know the slope formula is (y2-y1)/(x2-x1), but I always just put the number into the expression in the order I see them, but that is ok because I keep the order and get the correct answere because the y2,y1,x2,x1 or all relative. I am correct in doing this? INSTRUCTOR COMMENT: In other words you use (y1 - y2) / (x1 - x2) instead of (y2 - y1) / (x2 - x1). It's more conventional to regard, say, 10 as x1 and 20 as x2, so f(20) is y2 and f(10) is y1. If you start from the lower x number and change to the higher the difference is higher - lower, and this is the way we usually think about changes. According to this convention we calculate change in y as y2 - y1 and change in x as x2 - x1. You are doing (y1 - y2) / (x1 - x2) and you get a negative change in x, a negative denominator, and if you are thinking about change from the first quantity to the second this is backwards. However as you say both numerator and denominator follow the same order so you still get the right answer, since (y1-y2)/(x1-x2)= (y2-y1) / (x2-x1). ** "