course Mth 271 006. goin' the other way*********************************************
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Given Solution: `aAt 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. Self-critique (if necessary): Self-critique Rating: ok ********************************************* Question: `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? Your solution: 10sec (-4cm/sec) = -40cm 80 cm – 40cm = 40 cm When t=30 sec the depth of the water is 40cm This is less the accurate than the estimate for t=21 sec instant. Confidence Assessment: 3
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Given Solution: `aAt - 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. Self-critique (if necessary): Self-critique Rating: ok ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: -3(10sec) = -30 cm 80-30cm= 50cm No if it was -3 cm/sec we will have a lesser change in depth by 10 cm. Confidence Assessment: 3
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Given Solution: `aSince 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. Self-critique (if necessary): Self-critique Rating: ok ********************************************* Question: `q004. What is your specific estimate of the depth at t = 30 seconds? Your solution: (-4 + -3) / 2 = -3.5 cm/sec -3.5 cm/sec * 10 = -35 cm 80 – 35 cm = 45 cm is the change in depth Confidence Assessment: 3
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Given Solution: `aKnowing 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. Self-critique (if necessary): self-critique Rating: ********************************************* Question: `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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Y’= .1(20)-6 = -4cm/sec Y’ = .1(30) -6 = -3 cm/sec Confidence Assessment: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aAt 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. Self-critique (if necessary): Self-critique Rating:ok ********************************************* Question: `q006. For the rate function y ' = .1 t - 6, at what clock time does the rate of depth change first equal zero? Your solution: 0= .1t-6 6/.1 = .1t/.1 T= 60 sec confidence Assessment: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe 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. Self-critique (if necessary): self-critique Rating: ok ********************************************* Question: `q007. How much depth change is there between t = 20 sec and the time at which depth stops changing? Your solution: T= 20sec, y’= -4cm/sec T= 60 sec, y’= 0 cm/sec (-4+0) / 2 = -2 cm/sec With the average rate of -2cm/sec, the depth will change will be -80cm. When t= 20 sec the depth is 80 Therefore when t= 60 is therefore 80-80= 0 cm Confidence Assessment: 3
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Given Solution: `aThe 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. "