course Mth 173 ?I??????????assignment #014s?r??????????????Calculus I
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16:07:40 Query problem 2.6.4. s(t) = 5 t^2 + 3 What are the functions for velocity and acceleration as functions of t?
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RESPONSE --> Velocity = 10t Acceleration = 10 Problem 2.4.13 in 4th edition.
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16:07:45 ** The velocity function is s ' (t) = 10 t and the acceleration function is s '' (t) = 10. *&*&
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RESPONSE --> Ok.
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16:10:29 Query problem 2.6.12. Function negative, increasing at decreasing rate. What are the signs of the first and second derivatives of the function?
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RESPONSE --> The first derivative is positive. The second derivative is negative. Not problem 2.6.12 in 4th edition.
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16:10:44 *&*& 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. *&*&
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RESPONSE --> Ok.
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16:45:57 Query problem 2.6.20 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?
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RESPONSE --> Not problem 2.6.20 in 4th edition. I am not quite sure I have enough information.
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16:56:58 07-12-2007 16:56:58 ** 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. **
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NOTES -------> Ah. Interesting.
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16:58:09 What is the limiting value of the function as x -> -infinity and why must this be the limiting value?
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RESPONSE --> I can't see anything that would indicate there is a horizontal asymptote.
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17:03:06 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. **
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RESPONSE --> Ok.
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17:04:03 Is it possible that f ' (1) = 1? Is it possible that f ' (1) = 1/4?
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RESPONSE --> It is possible that f'(1) = 1 since the curve is concave downward and therefore the slope is decreasing. However, it is not possible that the slope is equal to 1/4, since that would mean the slope is increasing (getting more steep).
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17:04:16 ** 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. **
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RESPONSE --> Ok.
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17:04:29 Query problem 5.1.3 speed at 15-min intervals is 12, 11, 10, 10, 8, 7, 0 mph
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RESPONSE --> Problem 5.1.16 in 4th edition.
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17:18:09 ** 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. **
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RESPONSE --> Hmm. There wasn't anything in prior question asking for anything. Here are my estimates: a) Estimates for the first half hour. Upper: 5.75 miles. Lower: 5.25 miles. b) Estimates for the entire hour and a half. Upper: 14.5. Lower: 11.5. c) h < .00833
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17:19:01 What time interval would result in upper and lower estimates within .1 mile of the distance?
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RESPONSE --> This should be: 12h < 0.1 h < 0.1 / 12 h < .0083
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17:20:09 ** The right- and left-hand limits have to 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. **
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RESPONSE --> Ok. I just didn't run it out to the seconds, leaving it in terms of hours.
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17:36:45 Query problem 5.1.13. Acceleration table for vel, estimate vel (at 1-s intervals 9.81, 8.03, 6.53, 5.38, 4.41, 3.61) Give your upper and lower estimates of your t = 5 speed and explain how you obtained your estimates.
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RESPONSE --> Not problem 5.1.13 in 4th edition. In fact, I can't find this one in the 4th edition. Anyway, here are my answers: Upper estimate: 34.16 I got this estimate by multiplying the velocities by the time interval (1 second). Since the velocity drops, the upper estimate is found by using all of the velocities but the last. Lower estimate: 27.96 Found in the same manner as the upper estimate. However, the lower estimate uses all of the velocities but the first one.
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17:36:54 For each interval we multiply the maximum or minimum value by the time interval. For each interval the maximum value given happens to be the left-hand value of the acceleration and the minimum is the right-hand value. 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. *&*&
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RESPONSE --> Ok.
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17:41:39 What is the average of your estimates, and is the estimate high or low (explain why in terms of concavity)?
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RESPONSE --> The average of the estimate is 31.06. The average of the upper and lower would then be a straight line from the beginning to the end of the interval. If the graph is concave down, then the graph would most likely go outside the average line and therefore the average is most likely lower.
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17:42:13 ** 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. **
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RESPONSE --> I got the graph type wrong, but I think I have the right idea.
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17:42:21 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> Watch the graph type!
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