course Mth 173 |Ln{^_r~|assignment #012 V˪˵}Ӿ Calculus I 02-28-2008
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14:54:56 What is the seventh power of (x + `dx) (use the Binomial Theorem)?
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RESPONSE --> x^7 + 7x^6 'dx + 18x^5 'dx^2 + 12x^4 'dx^3 + 23x^4 'dx^3 + 17x 'dx^4 + 18x^3 'dx^3 + 12x^2 'dx^4 + 9x^2 'dx^5 + 3x 'dx^5 + 4x 'dx^6 + 'dx^7
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14:55:57 ** 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
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RESPONSE --> I used Pascal's Triangle and got that answer, but to double check I worked it out algebraically; I must have messed up somewhere along the way.
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14:58:31 What is ( (x + `dx)^7 - x^7 ) / `dx and what does the answer have to do with the derivative of x^7?
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RESPONSE --> 7x^6+21x^6'dx+35x^4'dx^2+35x^3'dx^3+21x^2'dx^5+7x'dx^5+'dx^6 The derivative of x^7 is 7x^6. This is the first part of the expression above. Other than that, I do not know what it has to do with it.
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15:00:31 ** (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. **
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RESPONSE --> ""every term with factor `dx approaches 0 and the quotient approaches 7 x^6. **"" I understand the fact that it does this, but I do not understand what this can be used for.
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15:19:41 Query problem 2.1.16 (prev edition 2.1.19 (was 2.1.8)) sketch position fn s=f(t) if 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.
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RESPONSE --> I could not think of a way to figure a correct value for velocity without any values for position. I am sure with this question as an example I will be able to answer a similar question later.
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15:23:21 ** 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). **
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RESPONSE --> I know that the two slopes should be equal for the two velocities to be the same, but I do not see how a graph could be drawn with the given information.
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15:26:16 What aspect of the graph represents the average velocity?
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RESPONSE --> I could not make a graph, so I am not sure what my answer would be.
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15:26:56 ** 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 **
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RESPONSE --> Although I could not make a graph, I think that through thinking about I should have been able to come up with that answer.
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15:28:04 What aspect of the graph represents the instantaneous veocity at t = 5?
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RESPONSE --> The instantaneous velocity at t = 5 is the slope of the tangent line at t = 5.
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15:28:12 ** 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 **
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15:37:43 Query problem 2.1.14 (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.
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RESPONSE --> My list (smallest to largest) went as follows: 0, AB, B, C, 1, A I put 0 as the smallest because it is 0/0. The others were estimations based on what I saw in the graph.
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15:39:24 ** 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). **
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RESPONSE --> AB, B, and C were completely out of place and I did not use the same logic in finding the answers. The fact that it is a concave down graph did not come into my thought process.
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15:46:26 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.
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RESPONSE --> x = -.1, -.01, -.001, -.0001 and at .1, .01, .001, .0001 y = 2.955, 2.9995, 2.999995, 2.999999995, 2.955, 2.9995, 2.999995, 2.999999995 I think the limit should be 3
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15:46:40 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.
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15:54:14 Describe your graph.
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RESPONSE --> I will describe the side of the graph has positive x values (the rest of the graph is a reflection voer the x-axis): As the values for x approach, but never reach 0, the y values get closer and closer to 3 but do not reach 3. As the x values increase, the y values fluctuate from postitive to negative but get smaller and smaller at the peaks.
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15:55:07 ** 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. **
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RESPONSE --> I did not think the function would have a point when x = 0.
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16:03:33 Find an interval such that the difference between f(x) and your limit is less than .01.
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RESPONSE --> My first step: f(x) - 3 < .01 f(x) < 3.01 sin(3x)/x < 3.01 Which turned out to be all numbers. Also, the graph f(x) = sin(3x)/x never goes above 3; f(x) is this same graph moved down 3 y-units so all numbers would result in f(x) - 3 being less than .01
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16:05:26 ** 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. **
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RESPONSE --> I thought the question was asking for f(x) - the limit < .01. The limit being 3. I do
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16:13:17 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
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RESPONSE --> The line would be a straight line starting at the origin and the y-values increase based on y*x for every unit of x. Meaning, if x = 4 then the y-value = y*4
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16:13:25 ** 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. **
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16:26:35 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?
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RESPONSE --> I first found the derivative of ln(cos(x)). I found it to be -sin(x) + (1 / cos(x)). At x = 1, I plugged 1 into the derivative and got 1.009344733 At x = 'pi/4, I plugged 'pi/4 into the derivative and got .7071067812
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16:29:15 ** 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. **
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RESPONSE --> The book said that the instantaneous rate of change is called the derivative of f at a, so I found the derivative of f at t.
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16:29:35 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I need to look over instantaneous velocity and rate of change more.
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16:29:53 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). **
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course Mth 173 I do not know why, but I struggled with this assignment. I feel that I do not usually sturggle with derivatives, but my work here is shameful. ҽwb҃gxϜassignment #013 013. Applications of the Chain Rule 02-29-2008
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13:50:59 `qNote that there are 4 questions in this assignment. `q001. The Fahrenheit temperature T of a potato just taken from the oven is given by the function T(t) = 70 + 120 e^(-.1 t), where t is the time in minutes since the potato was removed from the oven. At what rate is the temperature changing at t = 5?
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RESPONSE --> If the temperature function is T(t) = 70 + 120 e^(-.1 t), then the rate of temperature change is the derivative, which is T'(t) = 120e^(-.1 * t) T'(5) = 120 e^(-.1 * 5) = 72.783 degrees F / minute confidence assessment: 1
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13:52:02 The rate of temperature change is given by the derivative function ( T ( t ) ) ', also written T ' (t). Since T(t) is the sum of the constant function 70, whose derivative is zero, and 120 times the composite function e^(-.1 t), whose derivative is -.1 e^(-.1 t), we see that T ' (t) = 120 * ( -.1 e^(-.1 t) ) = -12 e^(-.1 t). Note that e^(-.1 t) is the composite of f(z) = e^z and g(t) = -.1 t, and that its derivative is therefore found using the chain rule. When t = 5, we have T ' (5) = -12 e^(-.1 * 5 ) = -12 e^-.5 = -7.3, approx.. This represents rate = change in T / change in t in units of degrees / minute, so at t = 5 minutes the temperature is changing by -7.3 degrees/minute.
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RESPONSE --> I thought the derivative of e^x = e^x, not x * e^x self critique assessment:
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13:54:53 `q002. The weight in grams of a growing plant is closely modeled by the function W(t) = .01 e^(.3 t ), where t is the number of days since the seed germinated. At what rate is the weight of the plant changing when t = 10?
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RESPONSE --> If the function for weight is W(t) = .01 e^(.3 t ), then the function for the rate of weight change is the derivative, which is W'(t) = (.01)(.3) e^(.3 t) W'(10) = (.01)(.3) e^(.3 * 10) = .0602566 grams / day confidence assessment: 2
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13:55:11 The rate of change of weight is given by the derivative function ( W ( t ) ) ', also written W ' (t). Since W(t) is .01 times the composite function e^(.3 t), whose derivative is .3 e^(.3 t), we see that W ' (t) = .01 * ( .3 e^(.3 t) ) = .003 e^(.3 t). Note that e^(.3 t) is the composite of f(z) = e^z and g(t) = .3 t, and that its derivative is therefore found using the chain rule. When t = 10 we have W ' (10) = .003 e^(.3 * 10) = .03 e^(3) = .06. Since W is given in grams and t in days, W ' will represent change in weight / change in clock time, measured in grams / day. Thus at t = 10 days the weight is changing by .06 grams / day.
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RESPONSE --> self critique assessment:
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14:00:56 `q003. The height above the ground, in feet, of a child in a Ferris wheel is given by y(t) = 6 + 40 sin ( .2 t - 1.6 ), where t is clock time in seconds. At what rate is the child's height changing at the instant t = 10?
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RESPONSE --> If the function for the child's height above ground is y(t) = 6 + 40 sin ( .2 t - 1.6 ), then the rate of change in the child's height is the derivative. The derivative has to be found using the g'(x) + f'(g(x)) f(x) = 6 + 40sin(g(x)) g(x) = .2t - 1.6 f'(x) = 40cos(g(x)) g'(x) = .2 Derivative: .2 + 40cos(.2t - 1.6) y'(t) = .2 + 40cos(.2 * 10 - 1.6) = 37.0424 feet/second confidence assessment: 2
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14:03:21 The rate of change of altitude is given by the derivative function ( y ( t ) ) ', also written y ' (t). Since y(t) is the sum of the constant term 6, with derivative zero, and 40 times the composite function sin (.2 t - 1.6), whose derivative is .2 cos(.2 t - 1.6), we see that y ' (t) = 40 * ( .2 cos(.2 t - 1.6) ) = 8 cos(.2 t - 1.6). Note that sin(.2t - 1.6) is the composite of f(z) = sin(z) and g(t) = .2 t - 1.6, and that its derivative is therefore found using the chain rule. Thus at t = 10 seconds we have rate y ' (10) = 8 cos( .2 * 10 - 1.6) = 8 cos( .4) = 7.4, approx.. Since y represents altitude in feet and t represents clock time in seconds, this represents 7.4 feet per second. The child is rising at 7.4 feet per second when t = 10 sec.
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RESPONSE --> Again I did not bring the power coefficient down. Instead of 40(.2 cos(.2t - 1.6), I left out the .2 self critique assessment: 3
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14:18:00 `q004. The grade point average of a certain group of students seems to be modeled as a function of weekly study time by G(t) = ( 10 + 3t ) / (20 + t ) + `sqrt( t / 60 ). At what rate does the grade point average go up as study time is added for a typical student who spends 40 hours per week studying? Without calculating G(40.5), estimate how much the grade point average for this student would go up if she spend another 1/2 hour per week studying.
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RESPONSE --> I feel like I am off on my derivative on this one because I forgot the process in finding the derivative of f(x)/g(x). The derivative I worked out was: (3/1) + (1 / (2 'sqrt (t / 60))) When t = 40, G'(40) = 3.6123 If this is right, my estimation for G(40.5) is 3.0283 confidence assessment: 0
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14:20:28 The rate of change of grade point average is given by the derivative function ( G ( t ) ) ', also written G ' (t). Since G(t) is the sum of the quotient function (10 + 3 t ) / ( 20 + t), with derivative 50 / ( 20 + t ) ^ 2, and the composite function `sqrt( t / 60) , whose derivative is 1 / (120 `sqrt( t / 60) ), we see that G ' (t) = 50 / ( 20 + t ) ^ 2 + 1 / (120 `sqrt( t / 60) ). Note that `sqrt(t / 60) is the composite of f(z) = `sqrt(z) and g(t) = t / 60, and that its derivative is therefore found using the chain rule. Thus if t = 40 we have rate G ' (40) = 50 / ( 20 + 40 ) ^ 2 + 1 / (120 `sqrt( 40 / 60) ) = .024, approx.. Since G represents grade point and t represents weekly study time in hours, this represents .024 grade points per hour of weekly study time. The grade point is rising by .024 per additional hour of study. To estimate G(40.5) we assume that the .024 grade point rise per additional hour of study time remains valid as we increase study time from 40 to 40.5 hours. This is in increase of .5 hours in weekly study time so we would expect the grade point to go up by grade point change = .5 hours * .024 points / hour = .012 points. Since G(40) = ( 10 + 3 * 40) / (20 + 40) + `sqrt( 40 / 60) = 2.97 approx, the additional half-hour per week will tend to raise this by .012 to around 2.97. If the student is aiming for a 3-point, a couple more hours would do but the .5 hours won't.
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RESPONSE --> In these four question I missed 3/4 derivatives. That is not normal for me. Were these four derivatives found with the same rules as before? self critique assessment: 3
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