#$&*
course Mth 173 9/18 14:00.............................................
Given Solution: `aThe steepness of the curve is continually changing. Since it is the slope of the curve then indicates the rate of depth change, the depth vs. clock time curve represents a constantly changing rate of depth change. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q002. As you will see, or perhaps have already seen, it is possible to represent the behavior of the system by a quadratic function of the form y = a t^2 + b t + c, where y is used to represent depth and t represents clock time. If we know the precise depths at three different clock times there is a unique quadratic function that fits those three points, in the sense that the graph of this function passes precisely through the three points. Furthermore if the cylinder and the hole in the bottom are both uniform the quadratic model will predict the depth at all in-between clock times with great accuracy. Suppose that another system of the same type has quadratic model y = y(t) = .01 t^2 - 2 t + 90, where y is the depth in cm and t the clock time in seconds. What are the depths for this system at t = 10, t = 20 and t = 90? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Given y (t) = 0.01*(t^2) 2t + 90 Thus y (10) = 0.01*(10*10) 2*10 + 90 = 71 cm y (20) = 0.01*(20*20) 2*20 + 90 = 54 cm y (90) = 0.01*(90*90) 2*90 + 90 = - 9 cm thus the depths at t = 10 seconds , t = 20 seconds and t = 90 seconds are 71 cm , 54 cm and 9 cm respectively. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): OK ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q003. For the preceding situation, what are the average rates which the depth changes over each of the two-time intervals? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The first depth of 71cm was at t = 10 seconds and the second depth of 54 cm was at t = 20 seconds. Thus change in depth during the time interval = 54 cm 71 cm = -17 cm Time duration of that interval = 20 seconds 10 seconds = 10 seconds The average rate of change of depth in that duration = -17 cm / 10 seconds = -1.7 cm/seconds The last depth was -9 cm at t = 90 seconds and the second depth to be used to measure the average rate of depth change is 54 cm at t = 20 seconds Thus change in depth during the time interval = -9 cm 54 cm = -63 cm Time duration in that time interval = 90 seconds 20 seconds = 70 seconds The average rate of change of depth in that duration = - 63 cm / 70 seconds = -0.9 cm/second confidence rating #$&*: 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 = 20 sec and t = 90 sec, so the change in clock time is 80 sec - 20 sec = 70 sec. The average rate of change between these to clock times is therefore ave rate = change in depth / change in clock time = -63 cm / 70 sec = -.9 cm/s. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Depth at t = 10 seconds is 71cm, depth at time t = 11 seconds is calculated as y(11) = .01(11^2) - 2(11) + 90 = 1.21 - 22 + 90 = 69.21 cm change in depth in that time interval of t = 10 seconds and t = 11 seconds is 69.21 cm - 71 cm = -1.79 cm time duration in that time interval = 11 seconds 10 seconds = 1 seconds Average rate at which depth changes in that duration = -1.79 cm / 1 second = -1.79 cm / second Depth at t = 10 seconds is 71 cm , depth at time t = 10.1 y(10.1) = 0.01*( 10.1*10.1) 2(10.1) + 90 = 1.0201 20.2 + 90 = 70.8201cm change in depth in that time interval = 70.8201 cm 71 cm = -0.1799 cm time duration in that time interval 10.1 seconds 10 seconds = 0.1 seconds average rate at which depth changes in that duration = - 0.1799 cm / 0.1 seconds = -1.799 cm / seconds confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q005. What do you think is the precise rate at which depth is changing at the instant t = 10? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: From the previous results we see that the average rate of change of depth is 1.7 cm/second when the time duration compared to 10 cm was 10 seconds, an average rate of 1.79 cm / seconds when the time duration was 1 seconds and -1.799 cm/second for time duration of 0.1 seconds. As we notice that as we decrease the time a factor of 10 the average rate of depth change is decreasing and decreasing in way to approach 1.8. Thus when the duration will be even smaller the rate would approach more close to -1.8. Thus we can say that when the duration is such small that both the time intervals are almost on 10 seconds the rate of depth change would reach a value of -1.8 cm / second confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe progression -1.7 cm/s, -1.79 cm/s, -1.799 cm/s corresponds to time intervals of `dt = 10, 1, and .1 sec, with all intervals starting at the instant t = 10 sec. That is, we have shorter and shorter intervals starting at t = 10 sec. We therefore expect that the progression might well continue with -1.7999 cm/s, -1.79999 cm/s, etc.. We see that these numbers approach more and more closely to -1.8, and that there is no limit to how closely they approach. It therefore makes sense that at the instant t = 10, the rate is exactly -1.8. STUDENT COMMENT: I don't really understand this even after reading the solution INSTRUCTOR RESPONSE: You did some rounding in your solutions up to this point (your solutions were otherwise correct), and didn't get all the 9's in some of the numbers. Done without rounding, the rates are -1.7 cm/s, -1.79 cm/s and -1.799 cm/s. These represent average rates over shorter and shorter intervals starting at t = 10 sec. It appears that these average rates are approaching a limit of -1.8 cm/s, which we therefore take to be the instantaneous rate at t = 10 sec. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q006. In symbols, what are the depths at clock time t = t1 and at clock time t = t1 + `dt, where `dt is the time interval between the two clock times? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Given function y(t) = 0.01(t^2) 2t + 90 Thus y(t1) = 0.01(t1*t1) -2*t1 + 90 = 0.01(t1^2) 2t1 + 90 and y( t1 + `dt) = 0.01( [t1 + `dt] * [t1 + dt] ) 2(t1 + `dt) + 90 y(t1 + `dt) = 0.01( t1^2 + 2*t1*`dt + `dt^2 ) 2*t1 2*`dt + 90 = 0.01*(t1^2) + 0.01*(`dt^2) + 0.02*t1*`dt 2*t1 2*`dt + 90 confidence rating #$&*: 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): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q007. What is the change in depth between these clock times? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Change of depth between this 2 points = y ( t + `dt) y(t) = 0.01*(t1^2) + 0.01*(`dt^2) + 0.02*t1*`dt 2*t1 2*`dt + 90 - 0.01(t1^2) + 2t1 - 90 = 0.01*(`dt^2) + 0.02*t1*`dt 2*`dt confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe change in depth is .01 (t1 + `dt)^2 - 2 (t1 + `dt) + 90 - (.01 t1^2 - 2 t1 + 90) = .01 (t1^2 + 2 t1 `dt + `dt^2) - 2 t1 - 2 `dt + 90 - (.01 t1^2 - 2 t1 + 90) = .01 t1^2 + .02 t1 `dt + .01`dt^2 - 2 t1 - 2 `dt + 90 - .01 t1^2 + 2 t1 - 90) = .02 t1 `dt + - 2 `dt + .01 `dt^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q008. What is the average rate at which depth changes between these clock time? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The average rate at which the depth change in that time interval = change in depth / time duration between those clock times Thus average rate = (0.01*(`dt^2) + 0.02*t1*`dt 2*`dt) / `dt = 0.01*`dt + 0.02*t1 - 2 And when `dt tends to zero average rate = 0.02t1 - 2 confidence rating #$&*: 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. STUDENT COMMENT dont understand how that the dt in this equation approaches 0 when .02(t1)-2? INSTRUCTOR RESPONSE If you divide your previous result .02 (t1 dt) + - 2 (dt) + .01 (dt^2) by `dt you get .02 t1 - 2 + .01 * `dt. The shorter the time interval the smaller `dt will be. As `dt gets shorter and shorter it approaches 0. This doesn't affect the terms .02 t1 and -2, but it does affect .01 * `dt. As `dt shrinks to zero, .01 * `dt also shrinks to 0. The limiting value of our expression, as `dt shrinks to 0, is therefore .02 t1 - 2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The value of 0.02t1 2 at t1 = 10 seconds 0.02 (10) 2 = 0.2 2 = -1.8 Thus this is the rate of depth change at time t = 10 seconds. This is the same result as the one we obtained before by calculating average rates for smaller and smaller time intervals. this proves the consistency of the solution. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aAt t1 = 10 we get .02 * 10 - 2 = .2 - 2 = -1.8. This is the rate we conjectured for t = 10. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK "" "
Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ________________________________________ `gr91 #$&*#$&* course Mth 173 9/18 14:00
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Given Solution: `aThe steepness of the curve is continually changing. Since it is the slope of the curve then indicates the rate of depth change, the depth vs. clock time curve represents a constantly changing rate of depth change. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q002. As you will see, or perhaps have already seen, it is possible to represent the behavior of the system by a quadratic function of the form y = a t^2 + b t + c, where y is used to represent depth and t represents clock time. If we know the precise depths at three different clock times there is a unique quadratic function that fits those three points, in the sense that the graph of this function passes precisely through the three points. Furthermore if the cylinder and the hole in the bottom are both uniform the quadratic model will predict the depth at all in-between clock times with great accuracy. Suppose that another system of the same type has quadratic model y = y(t) = .01 t^2 - 2 t + 90, where y is the depth in cm and t the clock time in seconds. What are the depths for this system at t = 10, t = 20 and t = 90? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Given y (t) = 0.01*(t^2) - 2t + 90 Thus y (10) = 0.01*(10*10) - 2*10 + 90 = 71 cm y (20) = 0.01*(20*20) - 2*20 + 90 = 54 cm y (90) = 0.01*(90*90) - 2*90 + 90 = - 9 cm thus the depths at t = 10 seconds , t = 20 seconds and t = 90 seconds are 71 cm , 54 cm and - 9 cm respectively. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): OK ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q003. For the preceding situation, what are the average rates which the depth changes over each of the two-time intervals? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The first depth of 71cm was at t = 10 seconds and the second depth of 54 cm was at t = 20 seconds. Thus change in depth during the time interval = 54 cm - 71 cm = -17 cm Time duration of that interval = 20 seconds - 10 seconds = 10 seconds The average rate of change of depth in that duration = -17 cm / 10 seconds = -1.7 cm/seconds The last depth was -9 cm at t = 90 seconds and the second depth to be used to measure the average rate of depth change is 54 cm at t = 20 seconds Thus change in depth during the time interval = -9 cm - 54 cm = -63 cm Time duration in that time interval = 90 seconds - 20 seconds = 70 seconds The average rate of change of depth in that duration = - 63 cm / 70 seconds = -0.9 cm/second confidence rating #$&*: 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 = 20 sec and t = 90 sec, so the change in clock time is 80 sec - 20 sec = 70 sec. The average rate of change between these to clock times is therefore ave rate = change in depth / change in clock time = -63 cm / 70 sec = -.9 cm/s. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Depth at t = 10 seconds is 71cm, depth at time t = 11 seconds is calculated as y(11) = .01(11^2) - 2(11) + 90 = 1.21 - 22 + 90 = 69.21 cm change in depth in that time interval of t = 10 seconds and t = 11 seconds is 69.21 cm - 71 cm = -1.79 cm time duration in that time interval = 11 seconds - 10 seconds = 1 seconds Average rate at which depth changes in that duration = -1.79 cm / 1 second = -1.79 cm / second Depth at t = 10 seconds is 71 cm , depth at time t = 10.1 y(10.1) = 0.01*( 10.1*10.1) - 2(10.1) + 90 = 1.0201 - 20.2 + 90 = 70.8201cm change in depth in that time interval = 70.8201 cm - 71 cm = -0.1799 cm time duration in that time interval 10.1 seconds - 10 seconds = 0.1 seconds average rate at which depth changes in that duration = - 0.1799 cm / 0.1 seconds = -1.799 cm / seconds confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q005. What do you think is the precise rate at which depth is changing at the instant t = 10? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: From the previous results we see that the average rate of change of depth is - 1.7 cm/second when the time duration compared to 10 cm was 10 seconds, an average rate of - 1.79 cm / seconds when the time duration was 1 seconds and -1.799 cm/second for time duration of 0.1 seconds. As we notice that as we decrease the time a factor of 10 the average rate of depth change is decreasing and decreasing in way to approach - 1.8. Thus when the duration will be even smaller the rate would approach more close to -1.8. Thus we can say that when the duration is such small that both the time intervals are almost on 10 seconds the rate of depth change would reach a value of -1.8 cm / second confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe progression -1.7 cm/s, -1.79 cm/s, -1.799 cm/s corresponds to time intervals of `dt = 10, 1, and .1 sec, with all intervals starting at the instant t = 10 sec. That is, we have shorter and shorter intervals starting at t = 10 sec. We therefore expect that the progression might well continue with -1.7999 cm/s, -1.79999 cm/s, etc.. We see that these numbers approach more and more closely to -1.8, and that there is no limit to how closely they approach. It therefore makes sense that at the instant t = 10, the rate is exactly -1.8. STUDENT COMMENT: I don't really understand this even after reading the solution INSTRUCTOR RESPONSE: You did some rounding in your solutions up to this point (your solutions were otherwise correct), and didn't get all the 9's in some of the numbers. Done without rounding, the rates are -1.7 cm/s, -1.79 cm/s and -1.799 cm/s. These represent average rates over shorter and shorter intervals starting at t = 10 sec. It appears that these average rates are approaching a limit of -1.8 cm/s, which we therefore take to be the instantaneous rate at t = 10 sec. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q006. In symbols, what are the depths at clock time t = t1 and at clock time t = t1 + `dt, where `dt is the time interval between the two clock times? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Given function y(t) = 0.01(t^2) - 2t + 90 Thus y(t1) = 0.01(t1*t1) -2*t1 + 90 = 0.01(t1^2) - 2t1 + 90 and y( t1 + `dt) = 0.01( [t1 + `dt] * [t1 + dt] ) - 2(t1 + `dt) + 90 y(t1 + `dt) = 0.01( t1^2 + 2*t1*`dt + `dt^2 ) - 2*t1 - 2*`dt + 90 = 0.01*(t1^2) + 0.01*(`dt^2) + 0.02*t1*`dt - 2*t1 - 2*`dt + 90 confidence rating #$&*: 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): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q007. What is the change in depth between these clock times? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Change of depth between this 2 points = y ( t + `dt) - y(t) = 0.01*(t1^2) + 0.01*(`dt^2) + 0.02*t1*`dt - 2*t1 - 2*`dt + 90 - 0.01(t1^2) + 2t1 - 90 = 0.01*(`dt^2) + 0.02*t1*`dt - 2*`dt confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe change in depth is .01 (t1 + `dt)^2 - 2 (t1 + `dt) + 90 - (.01 t1^2 - 2 t1 + 90) = .01 (t1^2 + 2 t1 `dt + `dt^2) - 2 t1 - 2 `dt + 90 - (.01 t1^2 - 2 t1 + 90) = .01 t1^2 + .02 t1 `dt + .01`dt^2 - 2 t1 - 2 `dt + 90 - .01 t1^2 + 2 t1 - 90) = .02 t1 `dt + - 2 `dt + .01 `dt^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q008. What is the average rate at which depth changes between these clock time? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The average rate at which the depth change in that time interval = change in depth / time duration between those clock times Thus average rate = (0.01*(`dt^2) + 0.02*t1*`dt - 2*`dt) / `dt = 0.01*`dt + 0.02*t1 - 2 And when `dt tends to zero average rate = 0.02t1 - 2 confidence rating #$&*: 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. STUDENT COMMENT dont understand how that the dt in this equation approaches 0 when .02(t1)-2? INSTRUCTOR RESPONSE If you divide your previous result .02 (t1 dt) + - 2 (dt) + .01 (dt^2) by `dt you get .02 t1 - 2 + .01 * `dt. The shorter the time interval the smaller `dt will be. As `dt gets shorter and shorter it approaches 0. This doesn't affect the terms .02 t1 and -2, but it does affect .01 * `dt. As `dt shrinks to zero, .01 * `dt also shrinks to 0. The limiting value of our expression, as `dt shrinks to 0, is therefore .02 t1 - 2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The value of 0.02t1 - 2 at t1 = 10 seconds 0.02 (10) - 2 = 0.2 - 2 = -1.8 Thus this is the rate of depth change at time t = 10 seconds. This is the same result as the one we obtained before by calculating average rates for smaller and smaller time intervals. this proves the consistency of the solution. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK "" " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: "" " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!