#$&* course Mth 173 3/18 10:54 PM 014. `query 14
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Given Solution: ** The velocity function is s ' (t) = 10 t and the acceleration function is s '' (t) = 10. *&*& &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I don't know why, but acceleration being a constant confuses me. how is acceleration always 10 for this.
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Given Solution: *&*& The function is increasing so its derivative is positive. The slopes are decreasing, meaning that the rate of increase is decreasing. This means that the first derivative, which is represented by the slope, is decreasing. The second derivative is the rate of change of the derivative. Since the derivative is decreasing its rate of change is negative. Thus the second derivative is negative. *&*& &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `q Query problem 2.5.31 5th; 2.5.23 4th continuous fn increasing, concave down. f(5) = 2, f '(5) = 1/2. Describe your graph.How many zeros does your function have and what are their locations? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: question 31. A function f has f(5)= 20, f'(5)= 2, and f''(x) < 0, for x>= 5. which of the following are possible values for f(7) and which are impossible. (a) 26 (b) 24 and (c) 22. I didnt understand this question or how to solve it. I looked and the book said 22 is the only possible answer. but I don't know why. I notice 20 drop to 2, which seems to be some sort of clue. i guess thinking of what each term means. f(5) is a function at value 5 is 20. at this, the rate of change is 2. and the rate of the rate of change is decreasing. so at f(7) it cant be greater that 2, and most likely under, so f(5) is 20, and +2 is 22. which to me it seems like it has to be lower than 22 as well. but I will go with that. ???I am unsure how to locate the zeros
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Given Solution: ** The average slope between (0,0) and (5, 2) is .4. If the graph goes thru (0,0) then its slope at (5, 2), which is 1/2, is greater than its average slope over the interval, which cannot be the case if the graph is concave down. A constant slope of 1/2, with the graph passing through (5,2), would imply an x-intercept at (1, 0). Since the function is concave down the average slope between the x intercept and (5, 2) must be > 1/2 an x intercept must lie between (1,0) and (2,0). We can't really say what happens x -> infinity, since you don't know how the concavity behaves. It's possible that the function approaches infinity (a square root function, for example, is concave down but still exceeds all bounds as x -> infinity). It can approach an asymptote and 3 or 4 is a perfectly reasonable estimate--anything greater than 2 is possible. However the question asks about the limit at -infinity. As x -> -infinity we move to the left. The slope increases as we move to the left, so the function approaches -infinity as x -> -infinity. f'(1) implies slope 1, which implies that the graph makes an angle of 45 deg with the x axis; it's not horizontal. Because of the downward concavity the ave slope between x = 1 and x = 5 is greater than the slope at x = 5--i.e., greater than 1/2. This is the only restriction. A slope of 1 is therefore indeed possible. A slope of 1/4, or any slope less than 1/2, would be impossible. ** What is the limiting value of the function as x -> -infinity and why must this be the limiting value? STUDENT RESPONSE AND INSTRUCTOR COMMENT: The limiting value is 2, the curve never actually reaches 2 but comes infinitessimally close. INSTRUCTOR COMMENT: The value of the function actually reaches 2 when x = 5, and the function is still increasing at that point. If there is a horizontal asymptote, which might indeed be the case, it would have to be to a value greater than 2, since the function is increasing. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): How did you know the average slope between (0,0) and (5,2) the only data we got was f(5)= 2 and f'(5)= .5 that we know an instantaneous rate at f(5) which is .5 I don't get this problem very well. ------------------------------------------------ Self-critique Rating:1 ********************************************* Question: `q Is it possible that f ' (1) = 1? Is it possible that f ' (1) = 1/4? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: based on your question, f'(5) = 1/2, but the graph is increasing at a decreasing rate, that means before that value, f'(x) is greater, and with an infinite amount of numbers behind it, I believe it would be possible, although I don't know if it is true for that specfic function. but since f'(5)= 1/2, smaller values than 1/2 for f'(x) is not possible for values before f(5). confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** f ' (1) implies slope 1, which implies that if x and y scales are equal the graph makes an angle of 45 deg with the x axis; it's not horizontal. Because of the downward concavity the ave slope between x = 1 and x = 5 is greater than the slope at x = 5--i.e., greater than 1/2. This is the only restriction. A slope of 1 is therefore indeed possible. A slope of 1/4, or any slope less than 1/2, would be impossible. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `q Query problem 5.1.16 was 5.1.3 speed at 15-min intervals is 12, 11, 10, 10, 8, 7, 0 mph YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: problem number 15. Roger runs a marathon. His friend jeff rides behind him on a bicycle and clocks his speed every 15 minutes. Roger starts out strong, but after an hour and a half he is so exhausted that he has to stop. Jeff's data follow: time since start (min) 0 15 30 45 60 75 90 speed (mph) 12 11 10 10 8 7 0 (a) Assuming that Roger's speed is never increasing, give upper and lower estimates for the distance Roger ran during that first half hour. (b) Give upper and lower estimates for the distance roger ran in total during the entire hour and a half (c) How often would Jeff have needed to measure Roger's speed in order to find lower and upper estimates wthin 0.1 mile of the actual distance he ran. I know this is that graph with the curve through data points, and using the upper and lower triangles. its still a little odd to me, and I don't know if I use that summation and i think was called integral. Distance = Velocity * Time. going with his speed = velocty. 12* 15= 180, 11 * 15 = 165, 10 * 15= 150, 10 * 15= 150, 8 * 15= 120, 7 * 15 = 105. = 870 for the left sum 11 * 15 = 165, 10 * 15= 150, 10 * 15= 150, 8 * 15 = 120, 7 * 15 = 105, 0 * 15 = 0. = 690 for the right sum. the units are different though the time is in 15 minute intervals, but we did each velocity as mph. since an hour is 60 minutes, and we want our measurement in minutes, divide by 60. 870 / 60 = 14.5 690 / 60 = 11.5 so at most he went 14.5 miles, at least 11.5 miles ??????I know my division by 60 works, but when I think on how to prove it, it confuses me. how would I prove dividing by 60 gives the right results as changing the data before calculation ?????or maybe what would make it more clear is how i would have to write the units if i kept the values at 870 and 690. for the first half hour 12 * 15 = 180, 11 * 15= 165. = 345 = 5.75miles 11 * 15= 165, 10 * 15= 150 = 315 = 5.25miles to get within .1 mile 12-0 = 0 12 * 'dt <= .1 'dt= .1/12. but that is division by hours, and I wanted minutes, so * 60 and we get .5. so half a minute. that is 30 seconds. 30 seconds intervals to get within a tenth of a mile. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** your upper estimate would assume that the speed on each time interval remained at the maximum value given for that interval. For example on the first interval from 0 to 15 minutes, speeds of 12 mph and 11 mph are given. The upper estimate would assume the higher value, 12 mph, for the 15 minute interval. The estimate would be 12 mi/hr * .25 hr = 3 miles. For the second interval the upper estimate would be 11 mi/hr * .25 hr = 2.75 mi. For the third interval the upper estimate would be 10 mi/hr * .25 hr = 2.5 mi. For the fourth interval the upper estimate would be 10 mi/hr * .25 hr = 2.5 mi. For the fifth interval the upper estimate would be 8 mi/hr * .25 hr = 2.0 mi. For the sixth interval the upper estimate would be 7 mi/hr * .25 hr = 1.75 mi. The upper estimate would therefore be the sum 14.5 mi of these distances. Lower estimates would use speeds 11, 10, 10, 8, 7 and 0 mph in the same calculation, obtaining a lower estimate of the estimate 11.5 mph. ** What time interval would result in upper and lower estimates within .1 mile of the distance? ** The right- and left-hand approximations can differ over an interval (a, b) by at most | f(b) - f(a) | * `dx. We want to find `dx that will make this expression at most .1 mile. Thus we have the equation | f(b) - f(a) | * `dx <= .1 mile. Since f(b) - f(a) = 0 - 12 = -12, we have | -12 mph | * `dx <= .1 mile so `dx <= .1 mile / 12 mph = 1/120 hour = 30 sec. ** STUDENT COMMENT I am lost when it comes to the we have | -12 mph | * `dx <= .1 mile so `dx <= .1 mile / 12 mph = 1/120 hour = 30 sec. ** step. Can you explain why we do this in this way. INSTRUCTOR RESPONSE I'm assuming that you understand the statement 'The right- and left-hand approximations can differ over an interval (a, b) by at most ( f(b) - f(a) ) * `dx.' If not, review it in your text and submit a question specifically on this idea. Here we'll just concentrate on using the expression. Our f(x) function is the speed. The speed f(x) varies from 12 at the beginning of the interval to 0 mph at the end. The interval lasts from clock time x = 0 hr to clock time x = 90 minutes = 1.5 hr. So we can say that f(0) = 12 and f(1.5) = 0, where x is clock time in hours. In terms of the statement 'The right- and left-hand approximations can differ over an interval (a, b) by at most | f(b) - f(a) | * `dx.' a and b are the endpoints of the interval, so b = 1.5 and a = 0 the quantity | f(b) - f(a) | is | f(1.5) - f(0) | = | 0 - 12 mph | = 12 mph `dx is the duration of the time increment into which the interval must be divided, the quantity we wish to find. If the right- and left-hand approximations differ by less than .1, then the desired condition that our approximation be accurate to within .1 mile, is satisfied. So: right- and left-hand approximations differ by at most | f(b) - f(a) | * `dx = 12 mph * `dx our condition is that the approximations differ by at most .1 mile this gives us the equation 12 mph * `dx < .1 mile Solving our equation we get `dx < .1 mile / (12 mph) = .1 mi / (12 mi / hr) = 1 / 120 hr = .0083 hr. 1/120 hr can also be expressed as .5 minute, or 30 seconds. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q Query problem (omitted from 5th edition but work it anyway), previously 5.1.13. Accelerations at clock times 0, 1, 2, 3, 4, 5 seconds are 9.81, 8.03, 6.53, 5.38, 4.41, 3.61, all in meters / second^2. Acceleration is the rate of change of velocity with respect to clock time. Give upper and lower estimates of your t = 5 speed and explain how you obtained your estimates. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: well, this confuses me a bit, but after thinking on it a while. this was my thoughts. !!! I start talking to myself here a little, so if its confusing thats why, trying to work it out in my head. at clock time 0 to 1, the acceleration would have to equal the velocity, if they started out at 0, meaning if at clock time 0 to 1, the acceleration was 9.81, and nothing was traveling till then, the distance went from 0-9.81, and the velocity is also at 9.81. but then i relooked at it. and said wait, okay at clock time 0 there is already acceleration. and at clock time 1 it has changed. so I need an at most, and an at least. but thinking, after I find out the at most and at least values, what can I do with that. and what exactly does that information tell me. maybe if i just take upper and lower estimates of acceleration at the t=5. Well when finding distance we used sums of time and velocity. so that sounds like we took a step down. so maybe the same happens here, and by finding this range with acceleration im actually finding the range of velocity. wait no, from t=0 to t=5 that would mean it would figure out the change to that point. and so for speed at t=5 that would work. seconds 0 1 2 3 4 5 meters 9.81 8.03 6.53 5.38 4.41 3.61 9.81 + 8.03 + 6.53 + 5.38 + 4.41 = 34.16meters/second 8.03 + 6.53 + 5.38 + 4.41 + 3.61 = 27.96meters/second so velocity at t=5 is between 27.92meters/second and 34.16 meters/second, or if an average is wanted. 31.04 meters/second ????could an average be useful at all with something like this? confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Acceleration is the rate of change of velocity with respect to clock time. The average acceleration for an interval is the change in velocity for that interval, divided by the change in clock time: a_Ave = `dv / `dt. It follows that on any interval, `dv = a_Ave * `dt. The total change in velocity is the sum of the changes taken over all five intervals. We expect that on each interval the average acceleration is between the right-hand value of the acceleration and the left-hand value. In this case, for each interval the maximum value of the given values happens to be the left-hand value of the acceleration and the minimum is the right-hand value. So for any interval, right-hand value < a_Ave < left-hand value. Therefore rh value * `dt < a_Ave * `dt < lh value * `dt. So the total change in velocity lies between the right- and left-hand sums. Left-hand values give us the sum 9.81 m/s^2 * 1 sec +8.03 m/s^2 * 1 sec +6.53 m/s^2 * 1 sec +5.38 m/s^2 * 1 sec +4.41 m/s^2 * 1 sec =34.16 m/s. Right-hand values give us the sum 8.03 m/s^2 * 1 sec +6.53 m/s^2 * 1 sec +5.38 m/s^2 * 1 sec +4.41 m/s^2 * 1 sec +3.61 m/s^2 * 1 sec =27.96 m/s. So our lower and upper limits for the change in velocity are 27.96 m/s and 34.16 m/s. *&*& &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): after trying this problem it leads me to question what can using the right and left sums evaluate for us, and what is the inverse of the derivative, etc.
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Given Solution: ** The average of the two estimates is (27.96 m/s + 34.16 m/s) / 2 = 31.06 m/s. The graph of a(t) is decreasing at a decreasing rate. It is concave upward. On every interval a linear approximation (i.e., a trapezoidal approximation) will therefore remain above the graph. So the graph 'dips below' the linear approximation. This means that on each interval the average acceleration will be closer to the right-hand estimate, which is less than the left-hand estimate. Thus the actual change in velocity will probably be closer to the lower estimate than to the upper, and will therefore be less than the average of the two estimates. Another way of saying this: The graph is concave up. That means for each interval the graph dips down between the straight-line approximation of a trapezoidal graph. Your estimate is identical to that of the trapezoidal graph; since the actual graph dips below the trapezoidal approximation the actual velocity will be a bit less than that you have estimated. ** STUDENT QUESTION: I am not sure what I should do after I have established what I did above INSTRUCTOR RESPONSE Your results are done by what amounts to a trapezoidal approximation, approximating the actual function by straight lines over each interval. Thus we can refer to the trapezoidal approximation as the 'broken-line' approximation. The graph is concave up, so on any interval it dips below its broken-line approximation. The area beneath the actual graph is therefore a bit less than the area predicted by the broken-line (i.e., trapezoidal) approximation. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q Query Add comments on any surprises or insights you experienced as a result of this assignment." Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q Query Add comments on any surprises or insights you experienced as a result of this assignment." Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!