course Mth 173 ??I?a???Tv????????Cv?Student Name: assignment #001
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14:13:47 Note that there are 7 questions in this assignment. `q001. If the water and a certain cylinder is changing depth at a rate of -4 cm/sec at the t = 20 second instant, at which instant the depth is 80 cm, then approximately what do you expect the depth will be at the t = 21 second instant?
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RESPONSE --> 4/20 = .02 = 80/.02 = 4000 seconds -4.2 at t21 seconds
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14:13:50 Note that there are 7 questions in this assignment. `q001. If the water and a certain cylinder is changing depth at a rate of -4 cm/sec at the t = 20 second instant, at which instant the depth is 80 cm, then approximately what do you expect the depth will be at the t = 21 second instant?
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RESPONSE -->
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?d?x??????w??? Student Name: assignment #006 006. goin' the other way
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14:17:53 Note that there are 7 questions in this assignment. `q001. If the water and a certain cylinder is changing depth at a rate of -4 cm/sec at the t = 20 second instant, at which instant the depth is 80 cm, then approximately what do you expect the depth will be at the t = 21 second instant?
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RESPONSE --> -4(20)=80cm per 20s. approx. 84 cm at 21s
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14:18:39 At a rate of -4 cm/s, for the 1-second interval between t = 20 s and t = 21 s the change in depth would be -4 cm/s * 1 sec = -4 cm. If the depth was 80 cm at t = 20 sec, the depth at t = 21 sec would be 80 cm - 4 cm/ = 76 cm.
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RESPONSE --> forgot the ""-"" on the 4cm/sec 80cm-4cm=76cm
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14:18:40 At a rate of -4 cm/s, for the 1-second interval between t = 20 s and t = 21 s the change in depth would be -4 cm/s * 1 sec = -4 cm. If the depth was 80 cm at t = 20 sec, the depth at t = 21 sec would be 80 cm - 4 cm/ = 76 cm.
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RESPONSE -->
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14:20:03 `q002. Using the same information, what the you expect the depth will be depth at the t = 30 sec instant? Do you think this estimate is more or less accurate than the estimate you made for the t = 21 second instant?
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RESPONSE --> less accurate. When we looked at the cylinder depth change we noticed that over time the rate of change in depth changes due to the pressure of water from above. I believe the estimate at 30s would be less accurate because more time has passed.
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14:20:32 At - 4 cm/s, during the 10-second interval between t = 20 sec and t = 30 sec we would expect a depth change of -4 cm/sec * 10 sec = -40 cm, which would result in a t = 30 sec depth of 80 cm - 40 cm = 40 cm.
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RESPONSE --> ok
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14:26:09 `q003. If you know that the depth in the preceding example is changing at the rate of -3 cm/s at the t = 30 sec instant, how will this change your estimate for the depth at t = 30 seconds--i.e., will your estimate be the same as before, will you estimate a greater change in depth or a lesser change in depth?
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RESPONSE --> greater change in depth because the new rate is greater by 1
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h????}??????P???Student Name: assignment #006 006. goin' the other way
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14:27:15 Note that there are 7 questions in this assignment. `q001. If the water and a certain cylinder is changing depth at a rate of -4 cm/sec at the t = 20 second instant, at which instant the depth is 80 cm, then approximately what do you expect the depth will be at the t = 21 second instant?
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RESPONSE --> 80cm - 4cm = 76cm
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14:27:19 At a rate of -4 cm/s, for the 1-second interval between t = 20 s and t = 21 s the change in depth would be -4 cm/s * 1 sec = -4 cm. If the depth was 80 cm at t = 20 sec, the depth at t = 21 sec would be 80 cm - 4 cm/ = 76 cm.
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RESPONSE -->
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14:31:32 `q002. Using the same information, what the you expect the depth will be depth at the t = 30 sec instant? Do you think this estimate is more or less accurate than the estimate you made for the t = 21 second instant?
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RESPONSE --> t30 =40 less accurate because more time has elapsed
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14:31:45 At - 4 cm/s, during the 10-second interval between t = 20 sec and t = 30 sec we would expect a depth change of -4 cm/sec * 10 sec = -40 cm, which would result in a t = 30 sec depth of 80 cm - 40 cm = 40 cm.
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RESPONSE --> ok
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?s???????????? Student Name: assignment #006 006. goin' the other way
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18:37:54 Note that there are 7 questions in this assignment. `q001. If the water and a certain cylinder is changing depth at a rate of -4 cm/sec at the t = 20 second instant, at which instant the depth is 80 cm, then approximately what do you expect the depth will be at the t = 21 second instant?
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RESPONSE --> 80cm-4cm/s=76
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18:38:41 At a rate of -4 cm/s, for the 1-second interval between t = 20 s and t = 21 s the change in depth would be -4 cm/s * 1 sec = -4 cm. If the depth was 80 cm at t = 20 sec, the depth at t = 21 sec would be 80 cm - 4 cm/ = 76 cm.
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RESPONSE --> ok
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18:49:05 `q002. Using the same information, what the you expect the depth will be depth at the t = 30 sec instant? Do you think this estimate is more or less accurate than the estimate you made for the t = 21 second instant?
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RESPONSE --> 76-(9x4)=40 depth = 40cm less accurate due to the amount of time that has elapsed
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18:49:14 At - 4 cm/s, during the 10-second interval between t = 20 sec and t = 30 sec we would expect a depth change of -4 cm/sec * 10 sec = -40 cm, which would result in a t = 30 sec depth of 80 cm - 40 cm = 40 cm.
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RESPONSE --> ok
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18:49:24 At - 4 cm/s, during the 10-second interval between t = 20 sec and t = 30 sec we would expect a depth change of -4 cm/sec * 10 sec = -40 cm, which would result in a t = 30 sec depth of 80 cm - 40 cm = 40 cm.
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RESPONSE -->
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18:53:45 `q003. If you know that the depth in the preceding example is changing at the rate of -3 cm/s at the t = 30 sec instant, how will this change your estimate for the depth at t = 30 seconds--i.e., will your estimate be the same as before, will you estimate a greater change in depth or a lesser change in depth?
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RESPONSE --> 77-(9x3)=50 = 50cm less change in depth
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18:53:56 Since the rate of depth change has changed from -4 cm / s at t = 20 s to -3 cm / s at t = 30 s, we conclude that the depth probably wouldn't change as much has before.
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RESPONSE --> ok
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18:54:30 `q004. What is your specific estimate of the depth at t = 30 seconds?
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RESPONSE --> 80-3=77 77-(9x3)=50 50cm
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18:55:16 Knowing that at t = 20 sec the rate is -4 cm/s, and at t = 30 sec the rate is -3 cm/s, we could reasonably conjecture that the approximate average rate of change between these to clock times must be about -3.5 cm/s. Over the 10-second interval between t = 20 s and t = 30 s, this would result in a depth change of -3.5 cm/s * 10 sec = -35 cm, and a t = 30 s depth of 80 cm - 35 cm = 45 cm.
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RESPONSE --> i should have used the average of -4 and -3 cm/s instead of using simply -3cm/s
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18:56:33 `q005. If we have a uniform cylinder with a uniformly sized hole from which water is leaking, so that the quadratic model is very nearly a precise model of what actually happens, then the prediction that the depth will change and average rate of -3.5 cm/sec is accurate. This is because the rate at which the water depth changes will in this case be a linear function of clock time, and the average value of a linear function between two clock times must be equal to the average of its values at those to clock times. If y is the function that tells us the depth of the water as a function of clock time, then we let y ' stand for the function that tells us the rate at which depth changes as a function of clock time. If the rate at which depth changes is accurately modeled by the linear function y ' = .1 t - 6, with t in sec and y in cm/s, verify that the rates at t = 20 sec and t = 30 sec are indeed -4 cm/s and -3 cm/s.
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RESPONSE --> there is no question here
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18:57:40 At t = 20 sec, we evaluate y ' to obtain y ' = .1 ( 20 sec) - 6 = 2 - 6 = -4, representing -4 cm/s. At t = 30 sec, we evaluate y' to obtain y' = .1 ( 30 sec) - 6 = 3 - 6 = -3, representing -3 cm/s.
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RESPONSE --> i should have answered using the quadratic formula, but the preceding question did not state that.
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19:01:21 `q006. For the rate function y ' = .1 t - 6, at what clock time does the rate of depth change first equal zero?
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RESPONSE --> 6
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19:02:32 The rate of depth change is first equal to zero when y ' = .1 t - 6 = 0. This equation is easily solved to see that t = 60 sec.
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RESPONSE --> ok
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19:06:42 `q007. How much depth change is there between t = 20 sec and the time at which depth stops changing?
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RESPONSE --> at 4cm/s = 76cm at 3.5cm/s =76.5cm at 3cm/s = 77cm
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19:07:27 The rate of depth change at t = 20 sec is - 4 cm/s; at t = 60 sec the rate is 0 cm/s. The average rate at which depth changes during this 40-second interval is therefore the average of -4 cm/s and 0 cm/s, or -2 cm/s. At an average rate of -2 cm/s for 40 s, the depth change will be -80 cm. Starting at 80 cm when t = 20 sec, we see that the depth at t = 60 is therefore 80 cm - 80 cm = 0.
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RESPONSE --> i do not understand what needs to be done here. I will do the query again and figure this one out.
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