course Mth 271
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Given Solution: `aThe steepness of the curve is continually changing. Since it is the slope of the curve then indicates the rate of depth change, the depth vs. clock time curve represents a constantly changing rate of depth change. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ok ********************************************* Question: `q002. As you will see, or perhaps have already seen, it is possible to represent the behavior of the system by a quadratic function of the form y = a t^2 + b t + c, where y is used to represent depth and t represents clock time. If we know the precise depths at three different clock times there is a unique quadratic function that fits those three points, in the sense that the graph of this function passes precisely through the three points. Furthermore if the cylinder and the hole in the bottom are both uniform the quadratic model will predict the depth at all in-between clock times with great accuracy. Suppose that another system of the same type has quadratic model y = y(t) = .01 t^2 - 2 t + 90, where y is the depth in cm and t the clock time in seconds. What are the depths for this system at t = 10, t = 40 and t = 90? ********************************************* Your solution: .01(10 ^2)-2(10)-90= 71 .01(40^2)-2(40)-90= 26 Answer was with 20 but the problem had 40 in it. .01(90^2)-2(90)-90= -9 Confidence Assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aAt t=10 the depth is y(10) = .01(10^2) + 2(10) + 90 = 1 - 20 + 90 = 71, representing a depth of 71 cm. At t=20 the depth is y(20) = .01(20^2) - 2(20) + 90 = 4 - 40 + 90 = 54, representing a depth of 54 cm. At t=90 the depth is y(90) = .01(90^2) - 2(90) + 90 = 81 - 180 + 90 = -9, representing a depth of -9 cm. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): the problem had 40 but the answer had the problem with 20 in it. Self-critique Rating: ok ********************************************* Question: `q003. For the preceding situation, what are the average rates which the depth changes over each of the two-time intervals? ********************************************* Your solution: For the first time interval 10-20 - (71-54)/(10-20)= 1.7 cm/s The second time interval 20-90 = (-9-54)/(90-20)= -.9 For using 40—(26-71)/(40-10)= -1.5 For using 40—(-9-26)/(90-40)= -.7 Confidence Assessment: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aFrom 71 cm to 54 cm is a change of 54 cm - 71 cm = -17 cm; this change takes place between t = 10 sec and t = 20 sec, so the change in clock time is 20 sec - 10 sec = 10 sec. The average rate of change between these to clock times is therefore ave rate = change in depth / change in clock time = -17 cm / 10 sec = -1.7 cm/s. From 54 cm to -9 cm is a change of -9 cm - 54 cm = -63 cm; this change takes place between t = 40 sec and t = 90 sec, so the change in clock time is a9 0 sec - 40 sec = 50 sec. The average rate of change between these to clock times is therefore ave rate = change in depth / change in clock time = -63 cm / 50 sec = -1.26 cm/s. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ok ********************************************* Question: `q004. What is the average rate at which the depth changes between t = 10 and t = 11, and what is the average rate at which the depth changes between t = 10 and t = 10.1? ********************************************* Your solution: For 10 to 11 – (70.82-71)/(10.1-10)= -1.799 For 10 to 10.1—(69.21-71)/(11-10)= -1.79 Confidence Assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aAt t=10 the depth is y(10) = .01(10^2) - 2(10) + 90 = 1 - 20 + 90 = 71, representing a depth of 71 cm. At t=11 the depth is y(11) = .01(11^2) - 2(11) + 90 = 1.21 - 22 + 90 = 69.21, representing a depth of 69.21 cm. The average rate of depth change between t=10 and t = 11 is therefore change in depth / change in clock time = ( 69.21 - 71) cm / [ (11 - 10) sec ] = -1.79 cm/s. At t=10.1 the depth is y(10.1) = .01(10.1^2) - 2(10.1) + 90 = 1.0201 - 20.2 + 90 = 70.8201, representing a depth of 70.8201 cm. The average rate of depth change between t=10 and t = 10.1 is therefore change in depth / change in clock time = ( 70.8201 - 71) cm / [ (10.1 - 10) sec ] = -1.799 cm/s. We see that for the interval from t = 10 sec to t = 20 sec, then t = 10 s to t = 11 s, then from t = 10 s to t = 10.1 s the progression of average rates is -1.7 cm/s, -1.79 cm/s, -1.799 cm/s. It is important to note that rounding off could have hidden this progression. For example if the 70.8201 cm had been rounded off to 70.82 cm, the last result would have been -1.8 cm and the interpretation of the progression would change. When dealing with small differences it is important not around off too soon. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ********************************************* Question: `q005. What do you think is the precise rate at which depth is changing at the instant t = 10? . ********************************************* Your solution: The instant of change at 10 would be -1.79 because at 10.1 and 11 it’s the same. Confidence Assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe progression -1.7 cm/s, -1.79 cm/s, -1.799 cm/s corresponds to time intervals of `dt = 10, 1, and .1 sec, with all intervals starting at the instant t = 10 sec. That is, we have shorter and shorter intervals starting at t = 10 sec. We therefore expect that the progression might well continue with -1.7999 cm/s, -1.79999 cm/s, etc.. We see that these numbers approach more and more closely to -1.8, and that there is no limit to how closely they approach. It therefore makes sense that at the instant t = 10, the rate is exactly -1.8. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ok ********************************************* Question: `q006. In symbols, what are the depths at clock time t = t1 and at clock time t = t1 + `dt, where `dt is the time interval between the two clock times? ********************************************* Your solution: Y(t1)=.01(t1^2)-2(t1)+90 Y(t1+dt)=.01(t1+dt^2)-2(t1)+90 Confidence Assessment: 3
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Given Solution: `aAt clock time t = t1 the depth is y(t1) = .01 t1^2 - 2 t1 + 90 and at clock time t = t1 + `dt the depth is y(t1 + `dt) = .01 (t1 + `dt)^2 - 2 (t1 + `dt) + 90. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ok ********************************************* Question: `q007. What is the change in depth between these clock times? ********************************************* Your solution: change in depth= .01(t1+dt^2)-2(t1)+90 - (01(t1^2)-2(t1)+90) Confidence Assessment: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe change in depth is .01 (t1 + `dt)^2 - 2 (t1 + `dt) + 90 - (.01 t1^2 - 2 t1 + 90) = .01 (t1^2 + 2 t1 `dt + `dt^2) - 2 t1 - 2 `dt + 90 - (.01 t1^2 - 2 t1 + 90) = .01 t1^2 + .02 t1 `dt + .01`dt^2 - 2 t1 - 2 `dt + 90 - .01 t1^2 + 2 t1 - 90) = .02 t1 `dt + - 2 `dt + .01 `dt^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ********************************************* Question: `q008. What is the average rate at which depth changes between these clock time? ********************************************* Your solution: average rate = (.01(t1+dt^2)-2(t1)+90 )- ( 01(t1^2)-2(t1)+90) / ((t1+dt)- (t1)) Confidence Assessment: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe average rate is ave rate = change in depth / change in clock time = ( .02 t1 `dt + - 2 `dt + .01 `dt^2 ) / `dt = .02 t1 - 2 + .01 `dt. Note that as `dt shrinks to 0 this expression approaches .02 t1 - 2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating:ok ********************************************* Question: `q009. What is the value of .02 t1 - 2 at t1 = 10 and how is this consistent with preceding results? ********************************************* Your solution: the value is -1.8 Confidence Assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aAt t1 = 10 we get .02 * 10 - 2 = .2 - 2 = -1.8. This is the rate we conjectured for t = 10. "