Assignment 6

course Mth 173

??I?a???Tv????????Cv?Student Name: assignment #001

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001. Depth vs. Clock Time and Rate of Depth Change

??????U????]?z??Student Name:

assignment #006

006. goin' the other way

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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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"

Take another look on that last one. The key point is that you are given information on rate of depth change and need to find the change in depth, which you do by multiplying ave rate of change by the time interval. Let me know if you have questions.