course 271 ¦‚©ç°„ëüM‚ì¸ímxñýjƒ§û¼ÔÙºnõj–xassignment #006
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13:23:19 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 --> Subtracting 4 you would have 76 cm self critique assessment: 3
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13:25:00 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 --> 80-40 is 40 cm. Yes, because the rate of depth change is changing. self critique assessment: 3
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13:26:01 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 --> There will be less of a change, because the rate of change will be closer to 3 than before self critique assessment: 3
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13:27:08 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 --> Averaging a rate of 4 and 3 would produce a rate of 3.5 over 10 seconds, or 80-35=45 self critique assessment: 3
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13:38:24 `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 --> subtitute 20 and 30 for t to get -4 and -3 set y to 0 and solve, 60 seconds confidence assessment: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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13:41:19 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 --> The depth at 20 seconds is 80, and the depth stops changing at 0, so 80 cm self critique assessment: 3
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