#$&* course Mth 173 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: `aThe steepness of the curve is continually changing. Since it is the slope of the curve then indicates the rate of depth change, the depth vs. clock time curve represents a constantly changing rate of depth change. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `q002. As you will see, or perhaps have already seen, it is possible to represent the behavior of the system by a quadratic function of the form y = a t^2 + b t + c, where y is used to represent depth and t represents clock time. If we know the precise depths at three different clock times there is a unique quadratic function that fits those three points, in the sense that the graph of this function passes precisely through the three points. Furthermore if the cylinder and the hole in the bottom are both uniform the quadratic model will predict the depth at all in-between clock times with great accuracy. Suppose that another system of the same type has quadratic model y = y(t) = .01 t^2 - 2 t + 90, where y is the depth in cm and t the clock time in seconds. What are the depths for this system at t = 10, t = 20 and t = 90? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: At t=10 the depth is y10 = .01(10^2) + 2(10) + 90 = 1 - 20 + 90 = 71 , depth is 71 cm At t=20 the depth is y20 = .01(20^2) + 2(20) + 90 = 4 - 40 + 90 = 54 , depth is 54 cm At t=90 the depth is y90 = .01(90^2) + 2(90) + 90 = 81 - 180 + 90 = -9 , depth is -9 cm confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique Rating: ok ********************************************* 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: From 71 cm to 54 cm the change is 54 cm - 71 cm = -17 cm, the change takes place from t=10 to t=20 seconds, so the clock time is 20 - 10 = 10 seconds. Average rate = change in depth / change in clock time = -17 cm / 10 seconds = -1.7 cm/sec From 54 cm to -9 cm the change is -9 cm -54 cm = -63 cm, the change takes place from t=20 to t=90 seconds, so the clock time is 90 - 70 = 20 seconds. Average rate = change in depth / change in clock time = -63 cm / 70 seconds = -.9 cm/sec confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aFrom 71 cm to 54 cm is a change of 54 cm - 71 cm = -17 cm; this change takes place between t = 10 sec and t = 20 sec, so the change in clock time is 20 sec - 10 sec = 10 sec. The average rate of change between these to clock times is therefore ave rate = change in depth / change in clock time = -17 cm / 10 sec = -1.7 cm/s. From 54 cm to -9 cm is a change of -9 cm - 54 cm = -63 cm; this change takes place between t = 20 sec and t = 90 sec, so the change in clock time is 80 sec - 20 sec = 70 sec. The average rate of change between these to clock times is therefore ave rate = change in depth / change in clock time = -63 cm / 70 sec = -.9 cm/s. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: I put 90 sec - 20 sec instead of 80 sec - 20 in the second equation. ********************************************* Question: `q004. What is the average rate at which the depth changes between t = 10 and t = 11, and what is the average rate at which the depth changes between t = 10 and t = 10.1? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: At t=10 the depth is y10 = .01(10^2) + 2(10) + 90 = 1 - 20 + 90 = 71 , depth is 71 cm At t=11 the septh is y11 = .01(11^2) + 2(11) + 90 = 1.21 -22 + 90 = 69.21 , depth is 69.21 cm Change in depth / change in clock time = (69.21 - 71) cm / (11 - 10) sec = -1.79 cm/sec At t=10.1 the depth is y10.1 = .01(10.1^2) + 2(10.1) + 90 = 1.0201 - 20.2 + 90 = 70.8201 , depth is 70.8201 cm Change in depth / change in clock time = (70.8201 - 71)cm / (10.1 - 10)sec = -1.799 cm/sec confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `q005. What do you think is the precise rate at which depth is changing at the instant t = 10? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: -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. We have shorter intervals starting at t = 10 seconds. So we expect that the progress might continue with -1.7999 cm/s, -1.79999 cm/s. We see that these numbers approach more closely to -1.8, and that there is no limit to how close they approach. So it makes sense that at the instant t = 10, the rate is exactly -1.8 confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe progression -1.7 cm/s, -1.79 cm/s, -1.799 cm/s corresponds to time intervals of `dt = 10, 1, and .1 sec, with all intervals starting at the instant t = 10 sec. That is, we have shorter and shorter intervals starting at t = 10 sec. We therefore expect that the progression might well continue with -1.7999 cm/s, -1.79999 cm/s, etc.. We see that these numbers approach more and more closely to -1.8, and that there is no limit to how closely they approach. It therefore makes sense that at the instant t = 10, the rate is exactly -1.8. STUDENT COMMENT: I don't really understand this even after reading the solution INSTRUCTOR RESPONSE: You did some rounding in your solutions up to this point (your solutions were otherwise correct), and didn't get all the 9's in some of the numbers. Done without rounding, the rates are -1.7 cm/s, -1.79 cm/s and -1.799 cm/s. These represent average rates over shorter and shorter intervals starting at t = 10 sec. It appears that these average rates are approaching a limit of -1.8 cm/s, which we therefore take to be the instantaneous rate at t = 10 sec. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `q006. In symbols, what are the depths at clock time t = t1 and at clock time t = t1 + `dt, where `dt is the time interval between the two clock times? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: At the clock time t=t1 the depth is y=t1 = .01 t1^2 - 2 t1 + 90 and at clock time t = t1 + dt the depth is y(t1 + dt) = .01 (t1 + dt)^2 - 2 (t1 + dt) + 90 confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `q007. What is the change in depth between these clock times? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The 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 + .01dt^2 - 2 t1 - 2 dt + 90 - .01 t1^2 + 2 t1 - 90) = .02 t1 dt + - 2 dt + .01 dt^2 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe change in depth is .01 (t1 + `dt)^2 - 2 (t1 + `dt) + 90 - (.01 t1^2 - 2 t1 + 90) = .01 (t1^2 + 2 t1 `dt + `dt^2) - 2 t1 - 2 `dt + 90 - (.01 t1^2 - 2 t1 + 90) = .01 t1^2 + .02 t1 `dt + .01`dt^2 - 2 t1 - 2 `dt + 90 - .01 t1^2 + 2 t1 - 90) = .02 t1 `dt + - 2 `dt + .01 `dt^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `q008. What is the average rate at which depth changes between these clock time? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Average rate = change in depth / change in clock time = ( .02 t1 dt + - 2 dt + .01 dt^2 ) / dt = .02 t1 - 2 + .01 dt. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `q009. What is the value of .02 t1 - 2 at t1 = 10 and how is this consistent with preceding results? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: At t1 = 10 we get .02 * 10 - 2 = .2 - 2 = -1.8. Rate conjectured for t = 10. confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique Rating: ok " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!