#$&* course Mth 173 8:54 pm 2/3 005.
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Given Solution: `aThe steepness of the curve is continually changing. Since it is the slope of the curve then indicates the rate of depth change, the depth vs. clock time curve represents a constantly changing rate of depth change. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): It seems like you asked the same question in two different ways, the second part about the what is in a curve confused me. ------------------------------------------------ Self-critique Rating:2 ********************************************* Question: `q002. As you will see, or perhaps have already seen, it is possible to represent the behavior of the system by a quadratic function of the form y = a t^2 + b t + c, where y is used to represent depth and t represents clock time. If we know the precise depths at three different clock times there is a unique quadratic function that fits those three points, in the sense that the graph of this function passes precisely through the three points. Furthermore if the cylinder and the hole in the bottom are both uniform the quadratic model will predict the depth at all in-between clock times with great accuracy. Suppose that another system of the same type has quadratic model y = y(t) = .01 t^2 - 2 t + 90, where y is the depth in cm and t the clock time in seconds. What are the depths for this system at t = 10, t = 20 and t = 90? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: simple plug the numbers in y(10)=.01(10^2)-2(10)+90 y(10)= .01(100)-20+90 y(10)=1-20+90 y(10)=71 y(20)=.01(20^2)-2(20)+90 y(20)=.01(400)-40+90 y(20)=4-40+90 y(20)=54 y(90)=.01(90^2)-2(90)+90 y(90)=.01(8100)-180)+90 y(90)=81-180+90 y(90)= -9 so at t= 10 seconds the depth is 71 cm, at t= 20 seconds the depth is 54cm, and at t= 90 seconds the result would be -9cm, which is not possible, so this is an unrealstic value. Why though? First how is it wrong, how would it react? I think that on a graph, if the hole was made perfect on the bottom of the container, to allow flowing to keep going, it would keep reaching towards zero (I think it would be debatable whether it touches zero or not, in real life because of the little water that would be left over after all the rest flowed out). At any rate it never goes below zero, because tangibly in the real world you cant have less than nothing. This just mean that either before 90 seconds all the water drained out, or that the function doesnt fit well, in this case I think it would be safer to assume that thats just too much time and all the water will have drained.
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Given Solution: `aAt t=10 the depth is y(10) = .01(10^2) + 2(10) + 90 = 1 - 20 + 90 = 71, representing a depth of 71 cm. At t=20 the depth is y(20) = .01(20^2) - 2(20) + 90 = 4 - 40 + 90 = 54, representing a depth of 54 cm. At t=90 the depth is y(90) = .01(90^2) - 2(90) + 90 = 81 - 180 + 90 = -9, representing a depth of -9 cm. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): !!!I believe in the future maybe an explanation should be nice, not just data. I know it is explained elsewhere, but with some repitition, and explanations why, I tend to retain, understand, and am able to use information better. Reason being now I am uncertain if my thoughts of how I explained -9cm are completely correct or not. ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q003. For the preceding situation, what are the average rates which the depth changes over each of the two-time intervals? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (10,71),(20,54),(90,-9) m= (y2-y1)/(x2-x1) (54-71)/20-10 first interval = -17cm/10sec (-9-54)/(90-20) -63/70 second interval = -9cm/10sec ???even with the value being negative, in reality this should still give us a credible average rate, would it not? confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aFrom 71 cm to 54 cm is a change of 54 cm - 71 cm = -17 cm; this change takes place between t = 10 sec and t = 20 sec, so the change in clock time is 20 sec - 10 sec = 10 sec. The average rate of change between these to clock times is therefore ave rate = change in depth / change in clock time = -17 cm / 10 sec = -1.7 cm/s. From 54 cm to -9 cm is a change of -9 cm - 54 cm = -63 cm; this change takes place between t = 20 sec and t = 90 sec, so the change in clock time is 80 sec - 20 sec = 70 sec. The average rate of change between these to clock times is therefore ave rate = change in depth / change in clock time = -63 cm / 70 sec = -.9 cm/s. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): The question didn't ask for per second,so I left it as a fraction, did not put it into a decimal form, even though the decimals were simple ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q004. What is the average rate at which the depth changes between t = 10 and t = 11, and what is the average rate at which the depth changes between t = 10 and t = 10.1? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ???why does this question sound more appropriate for an exponential equation to me y(t)= .01t^2-2t+90 y(10) was already found previously so y(10)= 71 y(11)=.01(11^2)-2(11)+90 y(11)=.01(121)-22+90 y(11)=1.21-22+90 y(11)=69.21 y(10.1)=.01(10.1^2)-2(10.1)+90 y(10.1)=.01(102.01)-20.2+90 y(10.1)=1.0201-20.2+90 y(10.1)=70.8201 (10,71),(10.1,70.8201),(11,69.21) (70.8201-71)/10.1-10 -.1799/.1 *10= -1.799cm/sec for the average rate between 10 and 10.1 (69.21-71)/11-10 -1.79cm/sec for the average rate between 10 and 11 the answers are pretty close together. confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aAt t=10 the depth is y(10) = .01(10^2) - 2(10) + 90 = 1 - 20 + 90 = 71, representing a depth of 71 cm. At t=11 the depth is y(11) = .01(11^2) - 2(11) + 90 = 1.21 - 22 + 90 = 69.21, representing a depth of 69.21 cm. The average rate of depth change between t=10 and t = 11 is therefore change in depth / change in clock time = ( 69.21 - 71) cm / [ (11 - 10) sec ] = -1.79 cm/s. At t=10.1 the depth is y(10.1) = .01(10.1^2) - 2(10.1) + 90 = 1.0201 - 20.2 + 90 = 70.8201, representing a depth of 70.8201 cm. The average rate of depth change between t=10 and t = 10.1 is therefore change in depth / change in clock time = ( 70.8201 - 71) cm / [ (10.1 - 10) sec ] = -1.799 cm/s. We see that for the interval from t = 10 sec to t = 20 sec, then t = 10 s to t = 11 s, then from t = 10 s to t = 10.1 s the progression of average rates is -1.7 cm/s, -1.79 cm/s, -1.799 cm/s. It is important to note that rounding off could have hidden this progression. For example if the 70.8201 cm had been rounded off to 70.82 cm, the last result would have been -1.8 cm and the interpretation of the progression would change. When dealing with small differences it is important not around off too soon. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I don't understand the importance of seeing the progression of average rates from -1.7 to -1.79 to -1.799 ------------------------------------------------ Self-critique Rating:1
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Given Solution: `aThe progression -1.7 cm/s, -1.79 cm/s, -1.799 cm/s corresponds to time intervals of `dt = 10, 1, and .1 sec, with all intervals starting at the instant t = 10 sec. That is, we have shorter and shorter intervals starting at t = 10 sec. We therefore expect that the progression might well continue with -1.7999 cm/s, -1.79999 cm/s, etc.. We see that these numbers approach more and more closely to -1.8, and that there is no limit to how closely they approach. It therefore makes sense that at the instant t = 10, the rate is exactly -1.8. STUDENT COMMENT: I don't really understand this even after reading the solution INSTRUCTOR RESPONSE: You did some rounding in your solutions up to this point (your solutions were otherwise correct), and didn't get all the 9's in some of the numbers. Done without rounding, the rates are -1.7 cm/s, -1.79 cm/s and -1.799 cm/s. These represent average rates over shorter and shorter intervals starting at t = 10 sec. It appears that these average rates are approaching a limit of -1.8 cm/s, which we therefore take to be the instantaneous rate at t = 10 sec. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `q006. In symbols, what are the depths at clock time t = t1 and at clock time t = t1 + `dt, where `dt is the time interval between the two clock times? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ???It says what are the depths at both times, but the way it is phrased it sounds like it wants an interval between the two times. either way it sounds like the derivative. depth=(y(t1+'dt)-y(t1))/'dt ####I forgot I was still using the same fuction, before I thought I had not been given the function for these two times. the function is y(t)=.01t-2t+90. t1 is essentially the same t1+'dt would just be treated as one value in a sence, and be plugged in y(t1+'dt=.01(t1+'dt)-2(t1-'dt)+90 confidence rating #$&*:1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aAt clock time t = t1 the depth is y(t1) = .01 t1^2 - 2 t1 + 90 and at clock time t = t1 + `dt the depth is y(t1 + `dt) = .01 (t1 + `dt)^2 - 2 (t1 + `dt) + 90. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I didn't know I was still using the same information. ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q007. What is the change in depth between these clock times? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I believe this would be what I stated earlier where I would have given the derivative Depth=(y(t1+'dt)-y(t1))/'dt confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe change in depth is .01 (t1 + `dt)^2 - 2 (t1 + `dt) + 90 - (.01 t1^2 - 2 t1 + 90) = .01 (t1^2 + 2 t1 `dt + `dt^2) - 2 t1 - 2 `dt + 90 - (.01 t1^2 - 2 t1 + 90) = .01 t1^2 + .02 t1 `dt + .01`dt^2 - 2 t1 - 2 `dt + 90 - .01 t1^2 + 2 t1 - 90) = .02 t1 `dt + - 2 `dt + .01 `dt^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I forgot to work it out, doing this process still confuses me. on = .01 (t1^2 + 2 t1 `dt + `dt^2) - 2 t1 - 2 `dt + 90 - (.01 t1^2 - 2 t1 + 90) where did the +2 in t1^2+2 t1'dt+'dt^2 come from? ------------------------------------------------ Self-critique Rating:
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Given Solution: `aThe average rate is ave rate = change in depth / change in clock time = ( .02 t1 `dt + - 2 `dt + .01 `dt^2 ) / `dt = .02 t1 - 2 + .01 `dt. Note that as `dt shrinks to 0 this expression approaches .02 t1 - 2. STUDENT COMMENT don’t understand how that the dt in this equation approaches 0 when .02(t1)-2? INSTRUCTOR RESPONSE If you divide your previous result .02 (t1 dt) + - 2 (dt) + .01 (dt^2) by `dt you get .02 t1 - 2 + .01 * `dt. The shorter the time interval the smaller `dt will be. As `dt gets shorter and shorter it approaches 0. This doesn't affect the terms .02 t1 and -2, but it does affect .01 * `dt. As `dt shrinks to zero, .01 * `dt also shrinks to 0. The limiting value of our expression, as `dt shrinks to 0, is therefore .02 t1 - 2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I believe I still struggle in reaching the derivative, I just know what it is, It is difficult for me to obtain it. ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q009. What is the value of .02 t1 - 2 at t1 = 10 and how is this consistent with preceding results? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: it looks exactly as before, it will become -1.8, and in predictions the answers where leading to that going -1.7, -1.79, etc confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aAt t1 = 10 we get .02 * 10 - 2 = .2 - 2 = -1.8. This is the rate we conjectured for t = 10. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!