course Mth 173 إwёc۽l}Bassignment #014
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23:01:02 `q001. What are the coordinates of the x = 5 point and the slope at that point of the graph of the function y = .3 x^2? What is the equation of the line through the point and having that slope? Sketch the line and the curve between x = 3 and x = 7 and describe your sketch. How close, in the vertical direction, is the line to the graph of the y = .3 x^2 function when x = 5.5, 6, 6.5 and 7?
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RESPONSE --> x = 5 point has y coordinate y = .3 * 5^2 = 7.5, the point lies at (5, 7.5). The derivative is y ' = .6 x, so the slope is y ' = .6 * 5 = 3. The line passes through the point (5,7.5) and has slope 3. Its equation is y - 7.5 = 3 ( x - 5), y = 3 x - 7.5. When x = 5.5, 6, 6.5 and 7, the y coordinates of the line are y = 3 ( 5.5 ) - 7.5 = 9, y = 3 (6) -7.5 = 10.5, y = 3 (6.5) -7.5 = 12, y = 3 (7) -7.5 = 13.5. At the same x coordinates the function y =.3 x ^ 2 takes values y = .3 (5.5) ^ 2 = 9.075, y =.3 (6) ^ 2 = 10.8, y =.3 (6.5) ^ 2 = 12.675, and y =.3 (7) ^ 2 = 14.7. The straight line is lower than the curve by 9.075 - 9 = .075 units when x = 5.5, 10.8-10.5 = .3 when x = 6, 12.675-12 = .675 when x = 6.5, and 14.7-13.5 = 1.2 when x = 7. We see that as we move away from the common point (5,7.5), the line moves away from the curve more and more rapidly. confidence assessment: 3
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23:01:08 The x = 5 point has y coordinate y = .3 * 5^2 = 7.5, so the point lies at (5, 7.5). The slope that that point is found by evaluating the derivative at that point. The derivative is y ' = .6 x, so the slope is y ' = .6 * 5 = 3. The line therefore passes through the point (5,7.5) and has slope 3. Its equation is therefore y - 7.5 = 3 ( x - 5), which simplifies to y = 3 x - 7.5. When x = 5.5, 6, 6.5 and 7, the y coordinates of the line are y = 3 ( 5.5 ) - 7.5 = 9, y = 3 (6) -7.5 = 10.5, y = 3 (6.5) -7.5 = 12, y = 3 (7) -7.5 = 13.5. At the same x coordinates the function y =.3 x ^ 2 takes values y = .3 (5.5) ^ 2 = 9.075, y =.3 (6) ^ 2 = 10.8, y =.3 (6.5) ^ 2 = 12.675, and y =.3 (7) ^ 2 = 14.7. The straight line is therefore lower than the curve by 9.075 - 9 = .075 units when x = 5.5, 10.8-10.5 = .3 when x = 6, 12.675-12 = .675 when x = 6.5, and 14.7-13.5 = 1.2 when x = 7. We see that as we move away from the common point (5,7.5), the line moves away from the curve more and more rapidly. Your sketch should show a straight line tangent to the parabolic curve y =.3 x^2, with the curve always above the line except that the point tangency, and moving more and more rapidly away from the line the further is removed from the point (5,7.5). The line you have drawn is called the line tangent to the curve at the point (5,7.5).
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RESPONSE --> ok self critique assessment: 3
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23:02:15 `q002. What is the equation of the line tangent to the curve y = 120 e^(-.02 t) at the t = 40 point? If we follow the tangent line instead of the curve, what will be the y coordinate at t = 40.3? How close will this be to the y value predicted by the original function? Will the tangent line be closer than this or further from the original function at t = 41.2?
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RESPONSE --> The tangent line at the x = 40 point will have y coordinate y = 120 e^(-.02 * 40) = 53.9194, accurate to six significant figures. The derivative of the function is y ' = -2.4 e^(-.02 t) and at x = 40 has value y ' = -2.4 e^(-.02 * 40) = -1.07838 The tangent line will therefore pass through the point (40, 53.9194) and will have slope 1.07838. Its equation will be : y - 53.9194 = -1.07838 ( x - 40) y = -1.07838 x + 97.0546. The values of the function at x = 40.3 and x = 41.2 are 53.5969 and 52.6408. The y values corresponding to the tangent line are 53.5958 and 52.6253. We see that the tangent line at x = 40.3 is .0011 units lower than the curve of the function, and at x = 41.2 the tangent line is .0155 units lower--about 14 times as far from the curve as at the x = 40.3 point. confidence assessment:
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23:02:28 The tangent line at the x = 40 point will have y coordinate y = 120 e^(-.02 * 40) = 53.9194, accurate to six significant figures. The derivative of the function is y ' = -2.4 e^(-.02 t) and at x = 40 has value y ' = -2.4 e^(-.02 * 40) = -1.07838, accurate to six significant figures. The tangent line will therefore pass through the point (40, 53.9194) and will have slope 1.07838. Its equation will therefore be y - 53.9194 = -1.07838 ( x - 40), which we can solve for y to obtain y = -1.07838 x + 97.0546. The values of the function at x = 40.3 and x = 41.2 are 53.5969 and 52.6408. The y values corresponding to the tangent line are 53.5958 and 52.6253. We see that the tangent line at x = 40.3 is .0011 units lower than the curve of the function, and at x = 41.2 the tangent line is .0155 units lower--about 14 times as far from the curve as at the x = 40.3 point.
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RESPONSE --> ok self critique assessment: 3
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23:03:20 `q003. Sketch a graph of this curve and the t = 40 tangent line and describe how the closeness of the tangent line to the curve changes as we move away from the t = 40 point.
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RESPONSE --> The tangent line is straight while the slope of the exponential function is increasing. At the t - 40 point the two meet for an instant, but as we move both to the right and to the left the exponential curve moves away from the tangent line, moving away very gradually at first and then with increasing rapidity confidence assessment: 3
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23:03:26 Your sketch should show the gradually curving exponential function very close to the tangent line. The tangent line is straight while the slope of the exponential function is always increasing. At the t - 40 point the two meet for an instant, but as we move both to the right and to the left the exponential curve moves away from the tangent line, moving away very gradually at first and then with increasing rapidity.
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RESPONSE --> ok self critique assessment: 3
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23:04:55 `q004. What is the equation of the line tangent to the curve y = 20 ln( 5 x ) at the x = 90 point?
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RESPONSE --> At x = 90, we have y = 20 ln (5 * 90) = 20 ln(450) = 20 * 6.1 = 126 there is a straight line through the point (90, 126). The slope of this straight line is equal to the derivative of y = 20 ln(5 x) at x = 90. The derivative of ln(5x) is, by the chain rule, (5x) ' * 1 / (5x) = 5 / (5x) = 1/x. y ' = 20 * 1 / x = 20 / x. At x = 90 then we have y ' = 20 / 90 = .22, approx.. Thus the tangent line passes through (90, 126) and has slope .22. The equation of this line is : ( y - 126 ) / (x - 90) = .22 y = .22 x + 106. confidence assessment: 3
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23:05:01 At x = 90, we have y = 20 ln (5 * 90) = 20 ln(450) = 20 * 6.1 = 126, approx.. So we are looking for a straight line through the point (90, 126). The slope of this straight line should be equal to the derivative of y = 20 ln(5 x) at x = 90. The derivative of ln(5x) is, by the chain rule, (5x) ' * 1 / (5x) = 5 / (5x) = 1/x. So y ' = 20 * 1 / x = 20 / x. At x = 90 then we have y ' = 20 / 90 = .22, approx.. Thus the tangent line passes through (90, 126) and has slope .22. The equation of this line is ( y - 126 ) / (x - 90) = .22, which is easily rearranged to the form y = .22 x + 106.
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RESPONSE --> ok self critique assessment: 3
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23:07:07 `q005. Using the tangent line and the value of the function at x = 3, estimate the value of f(x) = x^5 and x = 3.1. Compare with the actual value.
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RESPONSE --> f ' (x) = 5 x^4. When x = 3, f(x) = 3^5 = 243 and f ' (x) = 5 * 3^4 = 405. Thus tangent line passes through (3, 243) and has slope 405. Its equation is: (y - 243) / (x - 3) = 405, y = 405 x - 972. This function could be evaluated at x = 3.1. at x = 3 the derivative, which gives the slope of the tangent line, is 405. This means that between (3, 243) and the x = 3.1 point the rise/run ratio should be close to 405. The run from x = 3 to x = 3.1 is 1, so a slope of 405 therefore implies a rise equal to slope * run = 40.5. A rise of about 40.5 means that y = 3.1^5 should change by about 40.5, from 243 to about 283.5. The actual value of 3.1^5 is about 286.3. The discrepancy between the straight-line approximation and the actual value is due to the fact that y = x^5 is concave up, meaning that the slope is actually increasing. confidence assessment: 3
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23:07:13 f ' (x) = 5 x^4. When x = 3, f(x) = 3^5 = 243 and f ' (x) = 5 * 3^4 = 405. Thus tangent line passes through (3, 243) and has slope 405. Its equation is therefore (y - 243) / (x - 3) = 405, which simplifies to y = 405 x - 972. This function could be evaluated at x = 3.1. However a more direct approach simply uses the differential. In this approach we note that at x = 3 the derivative, which gives the slope of the tangent line, is 405. This means that between (3, 243) and the x = 3.1 point the rise/run ratio should be close to 405. The run from x = 3 to x = 3.1 is 1, so a slope of 405 therefore implies a rise equal to slope * run = 40.5. A rise of about 40.5 means that y = 3.1^5 should change by about 40.5, from 243 to about 283.5. The actual value of 3.1^5 is about 286.3. The discrepancy between the straight-line approximation and the actual value is due to the fact that y = x^5 is concave up, meaning that the slope is actually increasing. The closely-related idea of the differential, which will be developed more fully in the next section, is that if we know the rate at which the function is changing at a given point, we can use this rate to estimate its change as we move to nearby points.
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RESPONSE --> ok self critique assessment: 3
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23:08:44 `q006. Using the differential and the value of the function at x = e, estimate the value of ln(2.8). Compare with the actual value.
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RESPONSE --> At x = e we have ln(x) = ln(e) = 1 The derivative of ln(x) is 1/x, so at x = e rate at which financial log function is changing is 1 / e. Since e = 2.718, approx., between x = e and x = 2.8 x changes by about 2.8 - 2.718 = .082. Since the rate at which x is changing in this vicinity remains close to x = 1/e, `dy = rate * `dx = 1/e * .082 = .082 / 2.718 = .030. Thus at x = 2.8, y = ln(x) takes value approximately 1 + .030 - 1.030. The actual value of ln(2.8) to six significant figures is 1.02961 = 1.030 confidence assessment: 3
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23:08:51 At x = e we have ln(x) = ln(e) = 1 (this is because of the definition of the natural log function as the inverse of the exponential function). The derivative of ln(x) is 1/x, so at x = e the rate at which financial log function is changing is 1 / e. {}Since e = 2.718, approx., between x = e and x = 2.8 x changes by about 2.8 - 2.718 = .082. Since the rate at which x is changing in this vicinity remains close to x = 1/e, the change in y is approximately `dy = rate * `dx = 1/e * .082 = .082 / 2.718 = .030. Thus at x = 2.8, y = ln(x) takes value approximately 1 + .030 - 1.030. The actual value of ln(2.8) to six significant figures is 1.02961, which when rounded off to four significant figures is 1.030, in agreement with our approximation.
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RESPONSE --> ok self critique assessment: 3
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23:09:55 `q007. Using the tangent line verify that the square root of a number close to 1 is twice as close to 1 as the number. Hint: Find the tangent line approximation for the function f(x) = `sqrt(x) at an appropriate point.
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RESPONSE --> The function f(x) = `sqrt(x) takes value 1 when x = 1. The derivative of f(x) = `sqrt(x) is 1 / ( 2 * `sqrt(x) ), which takes value 1/2 when x = 1. The tangent-line approximation to f(x) is therefore a straight line through (1, 1) having slope 1/2. This line has equation (y - 1) / (x - 1) = 1/2, or y = 1/2 ( x - 1) + 1. This number differs from 1 by 1/2 (x-1), which is half the difference between 1 and x. confidence assessment: 3
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23:10:02 The function f(x) = `sqrt(x) takes value 1 when x = 1. The derivative of f(x) = `sqrt(x) is 1 / ( 2 * `sqrt(x) ), which takes value 1/2 when x = 1. The tangent-line approximation to f(x) is therefore a straight line through (1, 1) having slope 1/2. This line has equation (y - 1) / (x - 1) = 1/2, or y = 1/2 ( x - 1) + 1. This number differs from 1 by 1/2 (x-1), which is half the difference between 1 and x. This approximation works for x values near 1. Thus if x is close to 1, then y is approximately twice as close to 1 as is x.
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RESPONSE --> ok self critique assessment: 3
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23:11:00 `q008. Using the tangent line approximaton verify that the square of a number close to 1 is twice as far from 1 as the number.
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RESPONSE --> The tangent-line approximation to f(x) = x^2, near x = 1, is y = 2(x-1) + 1. This number is twice as far from 1 as is x. y = 2(x-1) + 1.{}{}Graph y = 2(x-1) + 1. confidence assessment: 3
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23:11:08 The tangent-line approximation to f(x) = x^2, near x = 1, is y = 2(x-1) + 1. This number is twice as far from 1 as is x.{}{}STUDENT COMMENT: I am not really understanding this part. INSTRUCTOR RESPONSE: *&*& You should understand the part about the tangent-line approximation near x = 1 being y = 2(x-1) + 1.{}{}Graph y = 2(x-1) + 1. If you move horizontally over from the point (1, 1) you are changing x. If you then move vertically up to the graph from this point you will find the change in y. For this function that change will be double the change in x. *&*&.
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RESPONSE --> ok i get it now self critique assessment: 3
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Ά` assignment #014 zKÑQߔ Calculus I 10-12-2008
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23:12:21 Query problem 2.5.13. s(t) = 5 t^2 + 3 What are the functions for velocity and acceleration as functions of t?
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RESPONSE --> The velocity function is s ' (t) = 10 t the acceleration function is s '' (t) = 10
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23:12:25 ** 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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23:13:11 Query problem 2.5.11. 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 function is increasing so its derivative is positive. The slopes are decreasing, so 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. second derivative is negative.
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23:13:14 *&*& 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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23:14:20 Query problem 2.5.23 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 --> 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). 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, 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.
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23:14:23 ** 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 --> ok
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23:15:09 What is the limiting value of the function as x -> -infinity and why must this be the limiting value?
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RESPONSE --> The limiting value is 2, the curve never really reaches 2 but comes very close.
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23:15:17 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 --> oh ok now I see
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23:15:58 Is it possible that f ' (1) = 1? Is it possible that f ' (1) = 1/4?
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RESPONSE --> 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 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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23:16:13 ** 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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23:17:17 Query problem 5.1.3 speed at 15-min intervals is 12, 11, 10, 10, 8, 7, 0 mph
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RESPONSE --> upper estimate is 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.
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23:17:25 ** 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 --> ok
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23:18:11 What time interval would result in upper and lower estimates within .1 mile of the distance?
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RESPONSE --> The right- and left-hand limits have to differ over an interval (a, b) by at most ( f(b) - f(a) ) * `dx. ( f(b) - f(a) ) * `dx <= .1 mile. Since f(b) - f(a) = 0 - 12 = -12, I have | -12 mph | * `dx <= .1 mile so `dx <= .1 mile / 12 mph = 1/120 hour = 30 sec.
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23:18:16 ** 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
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23:19:32 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 --> 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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23:19:37 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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23:20:33 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 --> average of the two estimates is (27.96 m/s + 34.16 m/s) / 2 = 31.06 m/s graph of a(t) is decreasing at a decreasing rate. It is concave upward. On every interval a linear approximation will remain above the graph. So the graph goes below the linear approximation
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23:20:36 ** 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 --> ok
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23:20:43 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> none
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