course Mth 173 ozI۽̸Iassignment #012
......!!!!!!!!...................................
14:11:32 What is the seventh power of (x + `dx) (use the Binomial Theorem)?
......!!!!!!!!...................................
RESPONSE --> (x + 'dx) ^ 7 (7/0) x^7 + (7/1) x^6 'dx^1 + (7/2) x^5 'dx^2 + (7/3) x^4 'dx^3 + (7/4) x^3 'dx^4 + (7/5) x^2 'dx^5 + (7/6) x^1 'dx ^6 + (7/7) 'dx^7 x^7 + 7x^6y + 21x^5y^2 + 35x^4y^3 + 35x^3y^4 + 21x^2y^5 + 7xy^6 + y^7
.................................................
......!!!!!!!!...................................
14:12:06 ** 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
......!!!!!!!!...................................
RESPONSE --> I used x and ys instead of x and dxs.
.................................................
......!!!!!!!!...................................
14:17:50 What is ( (x + `dx)^7 - x^7 ) / `dx and what does the answer have to do with the derivative of x^7?
......!!!!!!!!...................................
RESPONSE --> (x^7 + 7x^6dx + 21x^5dx^2 + 35x^4dx^3 + 35x^3dx^4 + 21x^2dx^5 + 7xdx^6 + dx^7) - x^7 =7x^6dx + 21x^5dx^2 + 35x^4dx^3 + 35x^3dx^4 + 21x^2dx^5 + 7xdx^6 + dx^7 =7x^6dx + 21x^5dx^2 + 35x^4dx^3 + 35x^3dx^4 + 21x^2dx^5 + 7xdx^6 + dx^7/ dx =7x^6 + 21x^5dx + 35x^4dx^2 + 35x^3dx^3 + 21x^2dx^4 + 7xdx^5 + dx^6
.................................................
......!!!!!!!!...................................
14:17:52 ** (x + `dx)^7 - x^7 = 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^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, so [ (x + `dx)^7 - x^7 ] / `dx = (7x^6'dx+21x^6'dx^2+35x^4'dx^3+35x^3'dx^4+21x^2'dx^5+7x'dx^6+'dx^7)/`dx = 7x^6+21x^6'dx+35x^5'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. **
......!!!!!!!!...................................
RESPONSE -->
.................................................
{]zÙC}ȨD assignment #012 ލ܈y[HĒj Calculus I 02-26-2007
......!!!!!!!!...................................
09:13:00 Query problem 2.1.1 (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.
......!!!!!!!!...................................
RESPONSE --> The slope of the tangent line is at 5 and the average slope is between 2 and 6. The graph is increasing at a decreasing rate
.................................................
......!!!!!!!!...................................
09:13:04 ** 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). **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
09:13:40 What aspect of the graph represents the average velocity?
......!!!!!!!!...................................
RESPONSE --> The slope of the tangent represents the average velocity (5).
.................................................
......!!!!!!!!...................................
09:13:42 ** 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 **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
09:14:20 What aspect of the graph represents the instantaneous veocity at t = 5?
......!!!!!!!!...................................
RESPONSE --> The slope of the tangent (5) represents the instantaneous velocity.
.................................................
......!!!!!!!!...................................
09:14:22 ** 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 **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
09:25:13 Query problem 2.1.06 lim (cos h - 1 ) / h, h -> 0. What is the limit and how did you get it?
......!!!!!!!!...................................
RESPONSE --> lim(cos h-1) / h h = .01, .001, and .1 lim(cos (.01) -1.) / .01 lim(.9999500004 - 1.) / .01 lim(-4.99995833) / .01 -.0049999583
.................................................
......!!!!!!!!...................................
09:25:16 ** For h = .1, .01, .001 the values of (cos(h)-1 ) / h are -0.04995834722, -0.004999958472, -0.0005. The limit is zero. **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
09:27:26 Query problem 2.1.15 (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.
......!!!!!!!!...................................
RESPONSE --> The graph is increasing which makes the slopes positive. 0 is first because every slope is positive, from 0 the only way to go is up. C is next, B, and then A
.................................................
......!!!!!!!!...................................
09:27:28 ** 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 at B, then slope of AB, then slope at A and finall 0 (in descending order). **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
09:33:17 Query problem 2.2.8 (was 2.1.16) f(x) = sin(3x)/x. Give your f(x) values at x = -.1, -.01, -.001, -.0001 and at .1, .01, .001, .0001 and tell what you think the desired limit should be.
......!!!!!!!!...................................
RESPONSE --> sin(3(-.1)) / -.1 sin(-.3) / -.1 -.2955202067 / -.1 2.955202067 sin(3(-.01)) / -.01 sin(-.03) / -.01 -.0299955002 / -.01 2.99955002 sin(3(-.001)) / -.001 sin(-.003) / -.001 -.0029999955 / -.001 2.9999955 sin(3(-.0001)) / -.0001 sin(-.0003) / -.0001 -3 / -.0001
.................................................
......!!!!!!!!...................................
09:34:45 ** 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. **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
09:35:53 Find an interval such that the difference between f(x) and your limit is less than .01.
......!!!!!!!!...................................
RESPONSE --> -.01 < x x > .01 with the limit of 3
.................................................
......!!!!!!!!...................................
09:35:55 ** 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. **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
09:37:20 Query problem 2.2.17 (3d edition 2.3.26 was 2.2.10) f(x) is cost so f(x) / x is cost per unit. Describe the line whose slope is f(4) / 4
......!!!!!!!!...................................
RESPONSE --> f(4) is the cost f(4)/4 is the cost per unit which makes it smaller than a small number such as f(2)/2
.................................................
......!!!!!!!!...................................
09:37:21 ** 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. **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
09:41:40 Query problem 2.2.23 (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?
......!!!!!!!!...................................
RESPONSE --> ln(cos(x)) ln(cos(1)) ln(.5403023059) -.6156264703 ln(cos('pi/4)) ln(cos(.7853981634)) ln(.7071067812) -.3465735903
.................................................
......!!!!!!!!...................................
09:41:52 ** 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. **
......!!!!!!!!...................................
RESPONSE --> I suppose I did not understand the question.
.................................................
......!!!!!!!!...................................
09:42:21 Query Add comments on any surprises or insights you experienced as a result of this assignment.
......!!!!!!!!...................................
RESPONSE --> Sometimes it is hard to understand what your questions are asking for.
.................................................
......!!!!!!!!...................................
09:42:23 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). **
......!!!!!!!!...................................
RESPONSE -->
.................................................
"