#$&* course MTH 173 10/23 1:33
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Given Solution: ** Using the binomial Theorem: x^7+7x^6'dx+21x^6'dx^2+35x^4'dx^3+35x^3'dx^4+21x^2'dx^5+7x'dx^6+'dx^7 ** (x + `dx)^7 - x^7 = x^7+7x^6'dx+21x^5'dx^2+35x^4'dx^3+35x^3'dx^4+21x^2'dx^5+7x'dx^6+'dx^7 - x^7 = 7x^6'dx+21x^5'dx^2+35x^4'dx^3+35x^3'dx^4+21x^2'dx^5+7x'dx^6+'dx^7, so [ (x + `dx)^7 - x^7 ] / `dx = (7x^6'dx+21x^5'dx^2+35x^4'dx^3+35x^3'dx^4+21x^2'dx^5+7x'dx^6+'dx^7)/`dx = 7x^6+21x^5'dx+35x^4'dx^2+35x^3'dx^3+21x^2'dx^4+7x'dx^5+'dx^6. As `dx -> 0, every term with factor `dx approaches 0 and the quotient approaches 7 x^6, which is the derivative of y = x^7 with respect to x.** STUDENT QUESTION: why does the x drop off in the last added term and leave just ‘dx^7 INSTRUCTOR RESPONSE: The expansion, written with all the exponents, is x^7 * `dx^0+7x^6'dx+21x^5'dx^2+35x^4'dx^3+35x^3'dx^4+21x^2'dx^5+7x'dx^6+x^0'dx^7. `dx^0 = 1 and x^0 = 1 so the expansion is written simply as x^7+7x^6'dx+21x^5'dx^2+35x^4'dx^3+35x^3'dx^4+21x^2'dx^5+7x'dx^6+'dx^7. STUDENT QUESTION: How does that division by `dx work? INSTRUCTOR RESPONSE: Division by `dx is the same as multiplication by 1 / `dx, so by the distributive law we have (7x^6'dx+21x^5'dx^2+35x^4'dx^3+35x^3'dx^4+21x^2'dx^5+7x'dx^6+'dx^7)/`dx = 7x^6'dx / `dx + 21x^5'dx^2 / `dx + 35x^4'dx^3 / `dx + 35x^3'dx^4 / `dx + 21x^2'dx^5 / `dx + 7x'dx^6 / `dx + 'dx^7 / `dx = = 7x^6+21x^5'dx+35x^4'dx^2+35x^3'dx^3+21x^2'dx^4+7x'dx^5+'dx^6. STUDENT QUESTION I see that this is very wrong and I have worked this multiple times and can not find my mistake, I think it is a possibility that I have my initial formula wrong and that is causeing the problem from the start INSTRUCTOR RESPONSE Your work on this question contains some pretty common errors. You left out several terms and didn't use appropriate signs of grouping You wrote your expression as x^7+7*x^7-1*dx+7(7-1)(7-2)/3*x^7-3*dx^3+dx^7 You neglected some of the necessary signs of grouping. With appropriate signs of grouping your expression would read x^7+7*x^(7-1)*dx+7(7-1)/2*x^7-2*dx^2+7(7-1)(7-2)/3*x^(7-3)*dx^3+dx^7 This contains some of the correct terms so you're on the right track, but you left out a number of terms between your dx^3 and dx^7. You can compare your expression with the expression in the given solution Your lack of some signs of grouping led to some confusion You got some of the terms correct, so again you're on the right track. Your erroneous result 21x+5dx^2 appears to have come from 7(7-1)/2*x^7-2*dx^2 in your preceding line. First I'll note that you should use parentheses to ensure that your expression is valid. Your expression should have been 7(7-1)/2*x^(7-2)*dx^2, which is easily evaluated to give us 21 x^5 dx^2. It should be clear how you got a little careless with the form of your expression and ended up with an erroneous result. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q Query problem 2.1.22 5th; 2.1.16 4th (prev edition 2.1.19 (was 2.1.8)) sketch position fn s=f(t) is vAve between t=2 and t=6 is same as vel at t = 5 Describe your graph and explain how you are sure that the velocity at t = 5 is the same as the average velocity between t=2 and t= 6. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Problem 2.1.22 and 2.1.16 submitted through text_12 assignment. To sketch position time function s = f(t) We know that velocity at t = 5 is the average rate between t = 2 sec and t = 6 sec. Thus from the graph, velocity at t = 5 sec will be the instantaneous slope of the function at t = 5 sec and the average rate of the function between t = 2 sec and t = 6 sec is the slope of the line joining the points on the graph corresponding to t = 2 sec and t = 6 sec. Thus in the graph the instantaneous slope at t = 5 sec that is the slope of the tangent to the function at t = 5 sec should be equal to the slope of line joining the points ( 2,f(2) ) and ( 6 , f(6)). confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The slope of the tangent line at t = 5 is the instantaneous velocity, and the average slope (rise / run between t = 2 and t = 6 points) is the average velocity. The slope of the tangent line at t = 5 should be the same as the slope between the t = 2 and t = 6 points of the graph. If the function has constant curvature then if the function is increasing it must be increasing at a decreasing rate (i.e., increasing and concave down), and if the function is decreasing it must be decreasing at a decreasing rate (i.e., decreasing and concave up). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q What aspect of the graph represents the average velocity? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The average rate of change of the function between 2 clock times represents average velocity. Average rate of change of the function between the 2 clock times is the slope of the line joining the 2 points on the graph. The rise is the distance travelled and the run for the line is the time interval. Thus the slope = distance covered / time interval = average velocity Thus the average rate of change of the function between the 2 clock times represents average velocity. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The straight line through two points has a rise representing the change in position and a run representing the change in clock time, so that the slope represents change in position / change in clock time = average rate of change of position = average velocity ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q What aspect of the graph represents the instantaneous veocity at t = 5? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The instantaneous velocity at any clock time t will be the average velocity such that t tends to 0. Basically the slope of the tangent to the graph at a particular t will represent the instantaneous velocity at that clock time. Thus instantaneous velocity at t = 5 sec is the slope of the tangent to the graph at t = 5 sec. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The slope of the tangent line at the t = 5 graph point represents the instantaneous velocity at t = 5. According to the conditions of the problem this slope must equal the slope between the t = 2 and t = 6 points ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q Query problem 2.1.20 5th; 2.1.14 4th; (3d edition 2.1.16) graph increasing concave down thru origin, A, B, C in order left to right; origin to B on line y = x; put in order slopes at A, B, C, slope of AB, 0 and 1.What is the order of your slopes. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Problem 2.1.20 and 2.1.14 submitted through text_12 assignment. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The graph is increasing so every slope is positive. The downward concavity means that the slopes are decreasing. 0 will be the first of the ordered quantities since all slopes are positive. C is the rightmost point and since the graph is concave down will have the next-smallest slope. The slope of the line from the origin to B is 1. The slope of the tangent line at B is less than the slope of AB and the slope of the tangent line at A is greater than the slope of AB. So slope at A is the greatest of the quantities, 1 is next, followed by slope of AB, then slope at B, then slope at C and finally 0 (in descending order). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q Query problem f(x) = sin(3x)/x. Find the value of f(x) at x = -.1, -.01, -.001, -.0001 and at .1, .01, .001, .0001 and tell what you think the limit of this function, as x approaches zero, should be. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Given f(x) = sin(3x)/x The value of the function at x = -.1, f(-.1) = sin(-.3)/-.1 = -.295520/-.1 = 2.95520 x = -.01, f(-.01) = sin(-.03)/-.01 = -.0299955/-.01 = 2.99955 x = -.001, f(-.001) = sin(-.003)/-.001 = 2.9999955 x = -.0001, f(-.0001) = sin(-.0003)/-.0001 = 2.999999955 x = .1, f(.1) = sin(.3)/.1 = 0.295520/.1 = 2.95520 x = .01, f(.01) = sin(.03)/.01 = 0.0299955/0.01 = 2.99955 x = .001, f(.001) = sin(.003)/.001 = 2.9999955 x = .0001, f(.0001) = sin(.0003)/.0001 = 2.999999955 As we see that as we approach x = 0 from either sides the value of the function tends to 3. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: COMMON ERROR: Here are my values for f(x): -.1, 2.9552 -.01, 2.9996 -.001, 3 -.0001, 3 .1, 2.9552 .01, 2.9996 .001, 3 .0001, 3 . So the limiting value is 3. INSTRUCTOR COMMENT: Good results and your answer is correct. However none the values you quote should be exactly 3. You need to give enough significant figures that you can see the changes in the expressions. The values for .1, .01, .001 and .0001 are 2.955202066, 2.999550020, 2.999995500, 2.999999954. Of course your calculator might not give you that much precision, but you can see the pattern to these values. The limit in any case is indeed 3. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q Describe your graph. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The graph of sin(3x) / x is a curve function with y = 3 at x = 0 which also is basically the limit of the function at that x value. For x = 0 as we move towards more positive x values the function decreases at an increasing rate and reaches the x axis at x = 1.04. Further from x = 1.04 the function further decreases at now at an decreasing rate and reaches a minimum value and then increases at an increasing rate and reaches the x axis again, beyond which it increases at an decreasing rate reaches a point of maxima and then repeats itself along the x axis with one exception that the point of maxima decreases as x moves further and the point of minima increases as x moves further. Since the function is an even function the function has a mirror image for the negative x values confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** the graph passes horizontally through the y axes at (0,3), then as x increases it decreases an increasing rate -- i.e., it is concave downward--for a time, but gradually straightens out then decreases at a decreasing rate as it passes through the x axis, etc.. However the important behavior for this graph is near x = 0, where the graph reaches a maximum of 3 at x = 0, and approaches this value as a limit. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q Find an interval such that the difference between f(x) and your limit is less than .01. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The value of the limit is 3, and thus we want an interval of x for which f(x) is within y = 2.99 and y = 3.01. Using the calculator and the graph we find that The value of the function at x = 0.01 is 2.99955 and that at x = -0.01 will also be the same. Thus both the y values are within the y range but this is a small interval. We also find that the value of the function at x = 0.04 the value of the function 2.9928 And that at x = -0.04 is 2.9928. Since the y value obtained are within the y range of 3.01 to 2.99 this x interval is valid. Thus for 0.04 units on either side of x = 0 the difference between f(x) and my limit will be less than .01. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** As the numbers quoted earlier show, f(x) is within .01 of the limit 3 for -.01 < x < .01. This interval is a good answer to the question. Note that you could find the largest possible interval over which f(x) is within .01 of 3. If you solve f(x) = 3 - .01, i.e., f(x) = 2.99, for x you obtain solutions x = -.047 and x = .047 (approx). The maximum interval is therefore approximately -.047 < x < .047. However in such a situation we usually aren't interested in the maximum interval. We just want to find an interval to show that the function value can indeed be confined to within .01 of the limit. In general we wish to find an interval to show that the function value can be confined to within a number usually symbolized by `delta (Greek lower-case letter) of the limit. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q Query problem 2.2.17 4th (3d edition 2.3.26 was 2.2.10) (5th edition students refer to figure 2.25) In the figure let f(x) represent the cost of manufacturing x kg of the chemical. Then f(x) / x is cost per kg. Describe the line whose slope is f(4) / 4 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Problem 2.2.17 submitted through text_12 assignment. If f(x) represents the cost of manufacturing x kg of chemical, then naturally cost of manufacturing 0 kg of chemical will be 0. Thus f(0) = 0. Now f(4) / 4 can also be written as (f(4) - 0)/(4 - 0) = (f(4) - f(0))/(4 - 0). Thus the ratio basically defines the average rate of change of the function between the points (0,0) and (4,f(4)). confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** A line from (0, 0) to (4, f(4) ) has rise f(4) and run 4 so the slope of this line is rise / run = f(4) / 4. Similarly the slope of the line from the origin to the x=3 point has slope f(3) / 3. If the graph is concave down then the line from the origin to the x = 4 point is less than that of the line to the x = 3 point and we conclude that f(4) / 4 is smaller than f(3) / 3; average cost goes down as the number of units rises. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q Query problem 2.2.27 5th; 2.2.23 4th (3d edition 2.3.32 was 2.2.28) approximate rate of change of ln(cos x) at x = 1 and at x = `pi/4. What is your approximation at x = 1 and how did you obtain it? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Problems submitted via text_12 assignment. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** At this point the text wants you to approximate the value. The values of ln(cos(x)) at x = .99, 1.00 and 1.01 are -0.6002219140, -0.6156264703 and -0.6313736258. The changes in the value of ln(cos(x)) are -.0154 and -.0157, giving average rates of change -.0154 / .01 = -1.54 and -.0157 / .01 = -1.57. The average of these two rates is about -1.56; from the small difference in the two average rates we conclude that this is a pretty good approximation, very likely within .01 of the actual instantaneous rate. The values of ln(cos(x)) at x = pi/4 - .01, pi/4 and pi/4 + .01 are -0.3366729302, -0.3465735902 and -0.3566742636. The changes in the value of ln(cos(x)) are -.009 and 0.0101, giving average rates of change -.0099 / .01 = -.99 and -.0101 / .01 = -1.01. The average of these two rates is about -1; from the small difference in the two average rates we conclude that this is a pretty good approximation, very likely within .01 of the actual instantaneous rate. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q Query Add comments on any surprises or insights you experienced as a result of this assignment. STUDENT QUESTION: I did have another opportunity to go back and look at Pascal`s triangle. I always had a problem with it in earlier calculus courses. I am still uncertain when I use it to get results, but I think it is a matter of becoming more comfortable with the process. INSTRUCTOR RESPONSE: If the first row in Pascal's Triangle is taken to be row number 0, and if the first number in a row is taken to be at the 0th position in the row, then the number in row n, position r represents the number of ways to get r heads on n flips of a fair coin, or equivalently as the number of ways to select a set of r objects from a total of n objects. For example if you have 26 tiles representing the letters of the alphabet then the number of ways to select a set of 6 tiles would be the number in the 26th row at position 6. The six tiles selected would be considered to be dumped into a pile, not arranged into a word. After being selected it turns out there would be 6! = 720 ways to arrange those six tiles into a word, but that has nothing to do with the row 26 position 6 number of the triangle. The number of ways to obtain 4 Heads on 10 flips of a coin is the number in row 10 at position 4. The two interpretations are equivalent. For example you could lay the tiles in a straight line and select 6 of them by flipping a coin once for each tile, pushing a tile slightly forward if the coin comes up 'heads'. If at the end exactly six tiles are pushed forward you select those six and you are done. Otherwise you line the tiles up and try again. So you manage to select 6 tiles exactly when you manage to get six Heads. It should therefore be clear that the number of ways to select 6 tiles from the group is identical to the number of ways to get six Heads. When expanding a binomial like (a + b) ^ 3, we think of writing out (a+b)(a+b)(a+b). When we multiply the first two factors we get a*a + a*b + b*a + b*b. When we then multiply this result by the third (a+b) factor we get a*a*a + a*a*b + a*b*a + a*b*b + b*a*a + b*a*b + b*b*a + b*b*b. Each term is obtained by selecting the letter a or the letter b from each of the three factors in turn, and every possible selection is represented. We could get any one of these 8 terms by flipping a coin for each factor (a+b) to determine whether we choose a or b. We would have 3 flips, and the number of ways of getting, say, two a's and one b would be the same as the number of ways of getting two Heads on three flips. As we can see from Pascal's triangle there are 3 ways to do this. These three ways match the terms a*a*b, a*b*a and b*a*a in the expansion. Since all three terms can be simplified to a^2 b, we have [ 3 * a^2 b ] in our expansion. Using this line of reasoning we see that the expansion a^3 + 3 a^2 b + 3 a b^2 + b^3 of (a+b)^3 has coefficients that match the n=3 row of Pascal's Triangle. This generalizes: the expansion of (a + b) ^ n has as its coefficients the nth row of Pascal's Triangle. The number in position r of row n is designated C(n,r), the number of combinations of r elements chosen from a set of n elements. C(n,r) = n! / [ r! * (n-r)! ]. This formula can to be proven by mathematical induction, or it can be reasoned out as follows: In choosing r elements out of n there are n choices for the first element, n - 1 choices for the second, n-2 for the third, ..., n - r + 1 choices for the rth element, so there are n (n-1)(n-2) ... (n-r+1) ways of choosing r elements in order. There are r! Possible orders for the chosen elements, so the number of combinations, in which order doesn't matter, is n (n-1)(n-2) ... (n-r+1) / r!. This is the same as n! / [ r! (n-r)! ], since n! / (n-r)! = n(n-1) ... (n-r+1). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK "" " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: "" " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!