#$&* course Mth 173 6/29 5:35 pm 014. `query 14
.............................................
Given Solution: ** The velocity function is s ' (t) = 10 t and the acceleration function is s '' (t) = 10. *&*& &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q Query problem 2.5.11. Function negative, increasing at decreasing rate. What are the signs of the first and second derivatives of the function? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The first derivative would be positive the second would be negative confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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 zeroes does your function have and what are their locations? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The graph has a point at (5, 2) begins at the origin. The f ‘(x) is a straight line The f’(x) line is the concave down curve Is a zero a point where the line crosses the x axis? This occurs between x= 2 and x= 3 confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q Is it possible that f ' (1) = 1? Is it possible that f ' (1) = 1/4? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Yes F’ (1) = 1 indicates that slope is 1 The slope for f ‘ (5) = ½ So the slope to be any less than ½ is not possible confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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: Lower estimate: (0 min = 0 hours) 15 minutes = .25 hour 30 = .5 hour 45 = .75 60 = 1 75= 1.25 90 = 1.5 (11* .25) + (10 * .5) = 7.75 miles Upper: (12 * 0) + (11* .25) = 2.75 miles Lower: So (11* .25) + (10 * .5) + (10 * .75) + (8 * 1) + (7 * 1.25) + (0 * 1.5) = 32 miles Upper estimate: (12 * 0) + (11* .25) + (10 * .5) + (10 * .75) + (8 * 1) + (7 * 1.25) = 32 miles confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): I thought I did this the correct way and ended up with different results. I multiplied the number of hours rather than the change in hours which was .25 ------------------------------------------------ Self-critique Rating: 2
.............................................
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): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: What is the average of your estimates, and is the estimate high or low (explain why in terms of concavity)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The average would be 27.96 + 37.77 / 2 = 32.865 The estimate is low because it is a line and is below the curve confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): I realize now that because the graph is concave up the line will be above it. And the actual value will be a little less that the average estimated value ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q Query Add comments on any surprises or insights you experienced as a result of this assignment. I now understand the concept of the upper and lower estimates better, I know I need to use the interval not the actual number " 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. I now understand the concept of the upper and lower estimates better, I know I need to use the interval not the actual number " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!