course mth173 Calculus I10-23-2007
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21:49:23 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 --> If s(t) = 5t^2 +3 represents a Distance vs. Time graph then s'(t) is the function for velocity and s''(t) is the function for acceleration. s'(t) = 5t^2 +3 since +3 is a constant it's derivative is 0 s'(t)= 5 (2t^(2-1)) s'(t)= 10t I'm having trouble finding the s""(t), but it would be acceleration.
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21:51:01 ** The velocity function is s ' (t) = 10 t and the acceleration function is s '' (t) = 10. *&*&
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RESPONSE --> s""(t) = 10, so it is a constant acceleration which makes sense.
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21:55:34 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 --> since the function is INCREASEING the first derivative would be positive indicating the increasing slope of the function but, since it is increasing by a DECRASING RATE the second derivative would be negaitve indicating that the rate (first derivative) is decreasing.
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21:55:54 *&*& 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 --> okay
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22:03:40 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 --> we can get a tangent approximation at point (5,2) 2 = (.5)(5) +b 2= 2.5 +b -.5 = b Ytan = .5 x -.5 0= .5 x -.5 .5 = .5x 1=x so there is a zero somewhere near the x=1 point although it could be greatly off due to the fact that the tangent line is fomr the x=5 point which is somewhat far away.
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22:04:49 ** 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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RESPONSE --> okay
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22:07:55 What is the limiting value of the function as x -> -infinity and why must this be the limiting value?
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RESPONSE --> As x nears infinity the graph will become asymptotic.
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22:08:46 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 --> okay
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22:14:11 Is it possible that f ' (1) = 1? Is it possible that f ' (1) = 1/4?
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RESPONSE --> Assuming since f'(1) =1 then f'(2) = 2 then x= y' so the slope is equal to the x value so you would have an increasing rate, a curve concave up. Again assuming that f'(1) = 1/4 then F(2) = .5 the slope is equal to y'= .25x This would be the same graph except it would take 4 times longer to reach a specified y-unit than the prevous graph.
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22:20:37 ** 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 --> so the prevous question was for a constant slope, If you can't have a slope of less than 1/2 then how can there be a horizontal line whose slope it 0, or is this just for this one instance.
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22:27:47 Query problem 5.1.3 speed at 15-min intervals is 12, 11, 10, 10, 8, 7, 0 mph
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RESPONSE --> The points indicate a slowing down with an avg slope of -.067 then a sudden stop with a slope of -.467
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22:32:12 ** 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 --> I didn' t understand the instructions correctly and didn't relize that you wanted me to do the rectangular approximation to find the integral, but I understand how you've done it here.
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22:46:49 What time interval would result in upper and lower estimates within .1 mile of the distance?
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RESPONSE --> Any intreval that allows the the 12 mph estimant to be less than .1 will get the measurements within .1 mile, such as 15 seconds.
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22:47:24 ** 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 --> okay
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23:04:11 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 --> 9.81 * 2.78*10^(-4)= .002725 8.03 * 2.58*10^(-4)= .002231 6.53 *... = .001814 5.38 * ...= .001494 4.41 * ...= .001225 3.61 *... = .001003 Total high approximation: .010492 Total low approximation: .007767 T= 5 High: .009489 Low: .006764 I obtained the estimation of t=5 by adding the acceleration to that point up.
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23:09:51 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 --> I dind't know that the values were in meters/s, I assumed that they were in miles/hour like the last question. But the as you can see I added up the max and the min sums up to the point of t=5s to get my answer as you have done in the answer window. Thus indicating the maximum and minum limits for velocity change.
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23:26:58 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 all of the values is 6.3 the graph is sloping down and looks fairly linear although if I had to say that there was a concavity I would say Concave down.
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23:36:24 ** 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 --> The average rates for the high and low estimations of t=5 is 31.06 I see now how you got the answer from the trapazoidial approximation because the actual graph dips below the trapazoidial part of the graph and therefore is closer to the lower estimate. This also indicates that the graph is decreasing at a decreasing rate.
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23:45:44 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I think that I understand the process of differential and tangent lines and am close to understanding using the trapazoidial approximation with the rectangle on top, which gives you a maximum and a minimum of what the trapizoidial approximation could be, by further evaluating how the graph lies in relation to the trapazoidial approximation slope lines, will tell you which (min or max) the graph lies closer to.
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