#$&* course Mth 173 9/24 21:40
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Given Solution: `aAt a rate of -4 cm/s, for the 1-second interval between t = 20 s and t = 21 s the change in depth would be -4 cm/s * 1 sec = -4 cm. If the depth was 80 cm at t = 20 sec, the depth at t = 21 sec would be 80 cm - 4 cm/ = 76 cm. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q002. Using the same information, what the you expect the depth will be depth at the t = 30 sec instant? Do you think this estimate is more or less accurate than the estimate you made for the t = 21 second instant? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The rate of changing depth = -4 cm/ sec Initial time = 20 sec Final second = 30 sec Thus the time duration in that interval = 30 sec - 20 sec = 10 sec The rate of change of depth = change in depth / time duration in that interval Thus change in depth = rate of change of depth * time duration in that interval Change in depth = - 4 cm / sec * 10 sec = - 40 cm Initial depth = 80 cm Thus final depth = initial depth + change in depth = 80 cm - 40 cm = 40 cm Thus depth at t = 21 second instant is 40 cm. The estimate is less accurate than the estimate made for the t = 21 second. The smaller the time interval the more accurate the result obtained is. We can see the same through the graph. The smaller the time interval the more closer is the tangent line to the graph to the actual tangent line at that point. Thus the smaller the time interval more accurate is the estimation. confidence rating #$&*: 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): OK ------------------------------------------------ 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: We know that the rate of change of depth at t = 20 sec is -4 cm / sec In the questions we used this rate to be the average rate of change of depth and computed the results. But here it is given that the rate of change of depth change at t = 30 sec is - 3 cm / sec. Thus the average rate of change for that interval thus would be = -3.5 cm / sec Thus the change in depth in this interval will be = average rate of change in depth * time duration = - 3.5 cm/sec * 10 sec = - 35 cm Thus the depth at t = 30 sec will be = depth at time 20 sec + change in depth = 80 cm - 35 cm = 45 cm Initial we estimated the depth at time 30 sec = 40 cm Thus estimate will be a slight greater than above but as compared to the huge values it is almost the same. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aSince the rate of depth change has changed from -4 cm / s at t = 20 s to -3 cm / s at t = 30 s, we conclude that the depth probably wouldn't change as much has before. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q004. What is your specific estimate of the depth at t = 30 seconds? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We know that the rate of change of depth at t = 30 seconds is - 3 cm / sec and rate of change of depth at t = 20 sec is -4 cm / sec. thus the average rate of change of depth between the time interval is ( -3 cm/sec - 4 cm/sec ) / 2 = - 7 / 2 = -3.5 cm/sec. The time duration in this interval = 30 - 20 = 10 seconds Average rate of change of depth = (change in depth) / (time duration in this interval) Thus change in depth = average rate of change of depth * time duration = -3.5 cm / sec * 10 sec = -35 cm Depth at time 20 sec = 80 cm Thus depth at time 30 sec = depth at 20 sec + change in depth = 80 cm - 35 cm = 45 cm confidence rating #$&*: 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): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* 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: The rate of change of depth y’ = 0.1t - 6 where t is in sec and y’ in cm/s Thus we have at time 20 sec y’(20) = 0.1*(20) - 6 = -4 cm/s And time at 30 sec y’(30) = 0.1*(30) - 6 = -3 cm/s Thus rate of change of depth at times t = 20 sec and t = 30 sec are -4 cm/s and -3 cm/s respectively. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aAt t = 20 sec, we evaluate y ' to obtain y ' = .1 ( 20 sec) - 6 = 2 - 6 = -4, representing -4 cm/s. At t = 30 sec, we evaluate y' to obtain y' = .1 ( 30 sec) - 6 = 3 - 6 = -3, representing -3 cm/s. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q006. For the rate function y ' = .1 t - 6, at what clock time does the rate of depth change first equal zero? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Given y’ = 0.1t - 6. Thus the first value of time (t) when the rate of depth changes to zero is 0 = 0.1*t - 6 6 = 0.1 * t t = 6/0.1 = 60 sec thus at time of 60 sec the rate of depth first becomes zero. confidence rating #$&*: 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. STUDENT COMMENT This was pretty straight forward, I could look at it and figure out the time to find zero, but if the times were not spaced out by ten second intervals the finding of zero would be hard to do. INSTRUCTOR RESPONSE If you write down the equation and solve it, it works out easily enough. For example if the equation was .07 t - 12 = 0, you would add 12 to both sides then divide by .07 to get t = 12 / .07, which is approximately 170. Of course if the equation is more difficult (e.g an equation like .02 t^2 + 4.2 t^2 - t + 9 = 0) it gets harder to solve for t, and it doesn't take much to come up the an equation that's impossible to solve. But the linear equation of this problem wouldn't be difficult. You will in any case be expected to be able to solve linear and quadratic equations. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q007. How much depth change is there between t = 20 sec and the time at which depth stops changing? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The rate of change of depth at t = 20 sec will be y’(20) = 0.1*(20) - 6 = 2 - 6 = -4 cm/sec. and as calculated in the previous question the rate of depth change is 0 (time at which depth stops changing) at time 60 sec. the average rate of change of depth between the interval = ( -4 cm/sec + 0 cm/sec ) / 2 = -2 cm/sec The time duration between the interval = 60 sec - 20 sec = 40 sec Thus the average rate of change of depth = ( change in depth ) / ( change in time interval ) Thus change in depth = average rate of change of depth * change in time interval = -2 cm / sec * 40 sec = - 80 cm confidence rating #$&*: 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. STUDENT QUESTION I dont totally understand where the 2 cm/s comes from. INSTRUCTOR RESPONSE The two rates -4 cm/s and 0 cm/s, calculated from the given rate function, are applicable to the interval between t = 20 sec and t = 60 sec. The first is the rate at the beginning of the interval, and the second is the rate at the end of the interval. Without additional information, our first conjecture would be that the average rate is the average of the initial and final rates. For different situations this conjecture might be more or less valid; in this case since the rate function is linear, it turns out that it is completely valid. The average of the two rates -4 cm/s and 0 cm/s is -2 cm/s, and this is the rate we apply to our analysis of this interval. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK "" " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: "" " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!