course mth271 `questionNumber 60000 Aug 99 to Nov 99
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RESPONSE --> we have 10600 and we subtract from itself and get 0 thus 0% and then the value changed in may to sept so we subtract 10800 from 10600 and get a -200 and then divide that from 10600 and get .019 or 1.9% self critique assessment: 2
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???????Oz??r?o??? assignment #007 007. `query 7 Applied Calculus I 08-03-2007
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20:58:14 `questionNumber 70000 **** Query class notes #07 **** Explain how we obtain the tangent line to a y = k x^3 function at a point on its graph, and explain why this tangent line gives a good approximation to the function near that point.
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RESPONSE --> we plug in the point for x and y confidence assessment: 0
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20:59:29 `questionNumber 70000 If we know that y=kx^3, as in the sandpile model, we can find the derivative as y = 3kx^2. This derivative will tell us the rate at which the volume changes with respect to the diameter of the pile. On a graph of the y = k x^3 curve the slope of the tangent line is equal to the derivative. Through the given point we can sketch a line with the calculated slope; this will be the tangent line. Knowing the slope and the change in x we easily find the corresponding rise of the tangent line, which is the approximate change in the y = k x^3 function. In short you use y' = 3 k x^2 to calculate the slope, which you combine with the change `dx in x to get a good estimate of the change `dy in y. **
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RESPONSE --> we find the derivative and get y=3kx^2 and the curve of the slope of the tangent line is equal to the derivative and we use this equation to find the slope self critique assessment: 2
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21:00:24 `questionNumber 70000 Query class notes #08 **** What equation do we get from the statement 'the rate of temperature change is proportional to the difference between the temperature and the 20 degree room temperature'? What sort of graph do we get from this equation and why?
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RESPONSE --> dy/dt-20 we get a parabola confidence assessment: 1
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21:01:29 `questionNumber 70000 y proportional to x means that for some k we have y = k x. The rate of change of the temperature is the derivative dT/dt. The difference between temp and room temp is T ?20. So the statement says that dT/dt = k (T ?20). Whenever the rate dT/dt is proportional to a quantity like T - Troom, which is a linear function of T, the result is that T-Troom is an exponential function of clock time. **
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RESPONSE --> it should have been dT/dt and the room temp would be T-20 and so we get dT/dt=k(T-20) its a linear function self critique assessment: 2
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21:02:39 `questionNumber 70000 1.2.10 (was 1.2.08 graph matching y = `sqrt(9-x^2) Describe the graph that matches this function and explain how you know this is the graph
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RESPONSE --> it is half of a circle with closed points on 3 and -3. confidence assessment: 1
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21:03:42 `questionNumber 70000 It turns out that this graph is in fact the upper half of a circle of radius 3 centered at the origin. We can show that this graph is part of a circle: If y = `sqrt(9 - x^2) then y^2 = 9 - x^2 so x^2 + y^2 = 9. This is the Pythagorean Theorem for a right triangle defined by center (0,0) and legs x and y; we see that the square of the hypotenuse is 9 so the hypotenuse is 3. The hypotenuse represents the radius of the circle. **
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RESPONSE --> we work out the equation and get x^2+y^2=9 and we can use the pythagorean theorem where the hypotenuse represents the r of the circle self critique assessment: 2
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21:04:28 `questionNumber 70000 Extra problem: (was problem 1.2.10) If air freshener initially contains 30 grams, what is the formula for the number of grams present if 12% of the amount present evaporate per day?
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RESPONSE --> not sure confidence assessment: 0
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21:05:24 `questionNumber 70000 If 12% evaporates per day then 88% remains at the end of each day. That is, the growth rate is -.12 so the growth factor would be 1 + (-.12) = .88 and the function would be Q = 30 gram * .88^t. **
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RESPONSE --> 100%-12%= 88% that remains and the growth rate is -.12 and the growth factor is 1+(-.12) which gives us .88 so Q=30gram*.88^t self critique assessment: 2
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21:06:23 `questionNumber 70000 Extra Problem (was problem 1.2.18): What is the formula for exponential function whose graph passes through the points (1,6) and (2,18)?
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RESPONSE --> not sure confidence assessment: 0
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21:08:08 `questionNumber 70000 STUDENT ANSWER AND INSTRUCTOR COMMENT: I did not use simultaneous equation to solve. I just tried different number for the given original and the number which would be raised to the x power. I then plugged in the two points and found an equation. }INSTRUCTOR COMMENT: Trial and error might work for this problem but only simultaneous equations will work if the numbers are less obvious, so you need to understand that procedure. Using P = P0 * a^t we plug in the coordinates of the first point to get 6 = P0 * a^1. For the second point we get 18 = P0 * a^2. Dividing the second equation by the first we obtain 18/6 = (P0 a^2) / (P0 a^1) or 3 = a, so we know that a = 3. Substituting this into the first equation we find that 6 = P0 * 3^1, which we easily solve for P0 to obtain P0 = 2. So our model P = P0 a^t becomes P = 2 * 3^t. **
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RESPONSE --> so we us P=P0*a^t and then put in 6 for P and 1 for t and get 6=P0*a^1 and the second pt we get 18=P0*a^2 and then divide and get 3=a and then put 3 in for a in the first equation and get 6=P0*3^1 and we get P0=2 self critique assessment: 2
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21:09:12 `questionNumber 70000 1.2.32 (was 1.2.28 graph of y=`sqrt(x+1) Describe your graph, including coordinates of intercepts, whether increasing or decreasing (if both, where it does each), and concavity.
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RESPONSE --> it starts at -1 and then goes towards the right in a curvature and it is increasing and is a upward concavity i think. confidence assessment: 2
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21:09:56 `questionNumber 70000 This graph intercepts the x axis where y = 0, which occurs when x+1 = 0 or x = -1. As x increases the square roots increase, but more and more slowly (just consider the square roots for x = 0, 1, 2, 3 and you'll see how the values increase by less and less each time). So the graph will be increasing at a decreasing rate, which means it is concave downward. The square root of a negative number is not a real number, so this function is undefined when x + 1 < 0, which happens when x < -1. So the function is undefined, and there is no graph at all, for x < -1. **
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RESPONSE --> ok, so it is increasing but at a decreasing rate because it is slowly slowing down and thsi means that its a concave downward. self critique assessment: 2
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21:10:59 `questionNumber 70000 1.2.56 (was 1.2.52 pts of intersection of x+y=7 and 3x-2y=11 What are the point(s) of intersection?
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RESPONSE --> i got the point of intersection as (5,2) confidence assessment: 2
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21:11:24 `questionNumber 70000 This system could be solved by elimination but that solution is confined to linear equations (for which it is very appropriate) and won't be demonstrated here. The methods used here can be used with nonlinear equations. We can solve both equations for y and then set the two results equal: The first equation is x + y = 7. Subtract x from both sides to get y = 7 - x. The second equation is 3x - 2y = 11. Subtract 3x from both sides: -2y = 11 - 3 x. Divide both sides by -2: y = (11 - 3x) / (-2) so y = -11/2 + 3 x / 2. Now set both expressions for y equal to one another: 7 - x = -11/2 + 3 x / 2. Add x and 11 /2 to both sides: 7 - x + x + 11/2 = -11/2 + 3 x / 2 + x + 11/2. 7 + 11/2 = 3 x / 2 + x. Put each side over common denominator: 14 / 2 + 11 / 2 = 3 x / 2 + 2 x / 2. Add: 25 / 2 = 5 x / 2. Multiply both sides by 2/5: 2/5 * 25 / 2 = 2/5 * 5x / 2. Simplify 5 = x. So at the point of intersection x = 5. Thus, substituting this result into the first equation, y = 7 - x = 7 - 5 = 2. Alternatively we could have substituted into the second equation to get y = -11/2 + 3 x / 2 = -11/ 2 + 3 * 5 / 2 = -11/2 + 16 / 2 = 4 / 2 = 2. We get the same y value either way, which must be the case at a point of intersection. So the intersection point is at x = 5, y = 2, i.e., the point (5, 2). ALTERNATIVE SOLUTION: Solve first equation for y then substitute into the second. You could have solved your first equation for y, obtaining y = 7 - x. Substituting into the second equation you would have obtained 3x - 2(7-x) = 11 so 5x - 14 = 11 and x = 5. Then substituting this value in either equation would have given you y = 2. **
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RESPONSE --> i believe that is the point i got self critique assessment: 3
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21:11:50 `questionNumber 70000 **** Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> not really. self critique assessment: 2
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`questionNumber 70000 This system could be solved by elimination but that solution is confined to linear equations (for which it is very appropriate) and won't be demonstrated here. The methods used here can be used with nonlinear equations. We can solve both equations for y and then set the two results equal: The first equation is x + y = 7. Subtract x from both sides to get y = 7 - x. The second equation is 3x - 2y = 11. Subtract 3x from both sides: -2y = 11 - 3 x. Divide both sides by -2: y = (11 - 3x) / (-2) so y = -11/2 + 3 x / 2. Now set both expressions for y equal to one another: 7 - x = -11/2 + 3 x / 2. Add x and 11 /2 to both sides: 7 - x + x + 11/2 = -11/2 + 3 x / 2 + x + 11/2. 7 + 11/2 = 3 x / 2 + x. Put each side over common denominator: 14 / 2 + 11 / 2 = 3 x / 2 + 2 x / 2. Add: 25 / 2 = 5 x / 2. Multiply both sides by 2/5: 2/5 * 25 / 2 = 2/5 * 5x / 2. Simplify 5 = x. So at the point of intersection x = 5. Thus, substituting this result into the first equation, y = 7 - x = 7 - 5 = 2. Alternatively we could have substituted into the second equation to get y = -11/2 + 3 x / 2 = -11/ 2 + 3 * 5 / 2 = -11/2 + 16 / 2 = 4 / 2 = 2. We get the same y value either way, which must be the case at a point of intersection. So the intersection point is at x = 5, y = 2, i.e., the point (5, 2). ALTERNATIVE SOLUTION: Solve first equation for y then substitute into the second. You could have solved your first equation for y, obtaining y = 7 - x. Substituting into the second equation you would have obtained 3x - 2(7-x) = 11 so 5x - 14 = 11 and x = 5. Then substituting this value in either equation would have given you y = 2. **
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RESPONSE --> i believe that is the point i got self critique assessment: 3
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21:11:50 `questionNumber 70000 **** Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> not really. self critique assessment: 2
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l?????n|???????assignment #008 008. `query 8 Applied Calculus I 08-03-2007
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21:19:10 `questionNumber 80000 What functions f(z) and g(t) express the function 2^(3t-5) as a composite f(g(t))?
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RESPONSE --> you put in the equation for g(t) and then multiply it with f(z) and f(z) and g(t) are what you can use to do that confidence assessment: 2
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21:20:05 `questionNumber 80000 CORRECT STUDENT RESPONSE: f(z)=2^z and g(t)=3t-5, so that f(g(t)) = 2^g(t) = 2^(3t-5).
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RESPONSE --> thus you get f(z)= 2^z and g(t)=3t-5 and get for f(g(t))= 2^g(t)=2^(3t-5) self critique assessment: 2
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21:21:20 `questionNumber 80000 1.3.81 (was 1.3.66 temperature conversion. What linear equation relates Celsius to Fahrenheit?
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RESPONSE --> F=9/5C+32 or C=5/9F- 160/9 confidence assessment: 2
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21:21:37 `questionNumber 80000 CORRECT STUDENT RESPONSE: degrees Fahrenheit=1.8(degrees Celsius)+32, or F = 1.8 C + 32. INSTRUCTOR COMMENT: Since each Fahrenheit degree is 1.8 Celsius degrees a graph of F vs. C will have slope 1.8. Since F = 32 when C = 0 the graph will have vertical intercept at (0, 32) so the y = m x + b form will be F = 1.8 C + 32.
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RESPONSE --> i think thats what i got self critique assessment: 3
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21:22:19 `questionNumber 80000 How did you use the boiling and freezing point temperatures to get your relationship?
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RESPONSE --> you plug in the boiling pt or the freezing pt for F or C confidence assessment: 2
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21:27:04 `questionNumber 80000 A graph of Fahrenheit vs. Celsius temperatures gives us the two (x,y) points (o,32) and (100,212). We use these two points to find the slope m=(y2-y1)/(x2-x1) = (212 - 32) / 100 = 1.8. Now we insert the coordinates of the point (0,32) and into the point-slope form y = 1.8 x + b of a line to get 32 = 1.8 * 0 + b. We easily solve to get b = 32. {So the equation is y = 1.8 x + 32, or F = 1.8 C + 32. **
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RESPONSE --> we use the Fv.C and get points of (o,32) (100,212) and then plug those in to find the slope m= (212-32)/100=1.8 then we get y=1.8x+b and then solve for b and get 32 so we have y=1.8x+32 self critique assessment: 2
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?????V?”?????assignment #009 009. `query 9 Applied Calculus I 08-03-2007
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21:29:47 `questionNumber 90000 **** Query problem 1.4.06 diff quotient for x^2-x+1 **** What is the simplified form of the difference quotient for x^2-x+1?
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RESPONSE --> 2x-1 confidence assessment: 2
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21:30:03 `questionNumber 90000 The difference quotient would be [ f(x+`dx) - f(x) ] / `dx = [ (x+`dx)^2 - (x+`dx) + 1 - (x^2 - x + 1) ] / `dx. Expanding the squared term, etc., this is [ x^2 + 2 x `dx + `dx^2 - x - `dx + 1 - x^2 + x - 1 ] / `dx, which simplifies further to }[ 2 x `dx - `dx + `dx^2 ] / `dx, then dividing by the `dx we get 2 x - 1 + `dx. For x = 2 this simplifies to 2 * 2 - 1 + `dx = 3 + `dx. **
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RESPONSE --> i forgot to add the 'dx self critique assessment: 3
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21:31:59 `questionNumber 90000 1.4.40 (was 1.4.34 f+g, f*g, f/g, f(g), g(f) for f=x/(x+1) and g=x^3 the requested functions and the domain and range of each.
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RESPONSE --> x^3+ x/(x+1)= x/(x-1)*x^3= confidence assessment: 0
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21:35:12 `questionNumber 90000 (f+g)(x) = x / (x + 1) + x^3 = (x^4 + x^3 + x) / (x + 1). Domain: x can be any real number except -1. (f * g)(x) = x^3 * x / (x+1) = x^4 / (x+1). Domain: x can be any real number except -1. (f / g)(x) = [ x / (x+1) ] / x^3 = 1 / [x^2(x+1)] = 1 / (x^3 + x), Domain: x can be any real number except -1 or 0 f(g(x)) = g(x) / (g(x) + 1) = x^3 / (x^3 + 1). Domain: x can be any real number except -1 g(f(x)) = (f(x))^3 = (x / (x+1) )^3 = x^3 / (x+1)^3. Domain: x can be any real number except -1 **
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RESPONSE --> f+g= (x^4+x^3+x)/(x+1)= x can be any real number except -1 f*g= x^4/(x+1)= x can be any real # except -1 f/g= 1/[x^2(x+1)]=1/(x^3+x)= x can be any real # but not -1, 0 f(g)= x^3/(x^3+1)= x can be any but -1 g(f)= (x/(x+1))^3= x^3/(x+1)^3 so x cant be -1 self critique assessment: 2
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21:36:44 `questionNumber 90000 1.4.66 (was 1.4.60 graphs of |x|+3, -.5|x|, |x-2|, |x+1|-1, 2|x| from graph of |x|
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RESPONSE --> first graph goes up 3 on the y axis , 2nd graph is fatter, 3rd graph is moved to the right on the x axis 2, graph 4 moves 1 to the left and down 1 and the final graph is skinnier confidence assessment: 2
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21:37:18 `questionNumber 90000 The graph of y = | x | exists in quadrants 1 and 2 and has a 'v' shape with the point of the v at the origin. It follows that: The graph of | x |+3, which is shifted 3 units vertically from that of | x |, has a 'v' shape with the point of the v at (0,3). The graph of -.5 | x |, which is stretched by factor -.5 relative to that of | x |, has an inverted 'v' shape with the point of the v at (0,0), with the 'v' extending downward and having half the (negative) slope of the graph of | x |. The graph of | x-2 |, which is shifted 2 units horizontally from that of |x |, has a 'v' shape with the point of the v at (2, 0). The graph of | x+1 |-1, which is shifted -1 unit vertically and -1 unit horizontally from that of | x |, has a 'v' shape with the point of the v at (-1, -1). The graph of 2 |x |, which is stretched by factor 2 relative to that of | x |, has a 'v' shape with the point of the v at (0,0), with the 'v' extending upward and having double slope of the graph of | x |. |x-2| shifts by +2 units because x has to be 2 greater to give you the same results for |x-2| as you got for |x|. This also makes sense because if you make a table of y vs. x you find that the y values for |x| must be shifted +2 units in the positive direction to get the y values for |x-2|; this occurs for the same reason given above For y = |x+1| - 1 the leftward 1-unit shift is because you need to use a lesser value of x to get the same thing for |x+1| that you got for |x|. The vertical -1 is because subtracting 1 shifts y downward by 1 unit **
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RESPONSE --> I believe that is what i said self critique assessment: 3
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21:38:51 `questionNumber 90000 1.4.71 (was 1.4.64 find x(p) from p(x) = 14.75/(1+.01x)
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RESPONSE --> p/14.75= 1/1+.01x * 1+.01/1 = x/1*p1+.01/14.75 = x= 14.75/p(1+.01) confidence assessment: 2
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21:41:28 `questionNumber 90000 p = 14.75 / (1 + .01 x). Multiply both sides by 1 + .01 x to get (1 + .01 x) * p = 14.75. Divide both sides by p to get 1 + .01 x = 14.75 / p. Subtract 1 from both sides to get 1 x = 14.75 / p - 1. Multiply both sides by 100 to get = 1475 / p - 100. Put the right-hand side over common denominator p: = (1475 - 100 p) / p. If p = 10 then x = (1475 - 100 p) / p = (1475 - 100 * 10) / 10 = 475 / 10 = 47.5 **
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RESPONSE --> I think i did it wrong, so i should have multiplied both sides with 1+.01 and get (1+.01x)*p= 14.75 and then divide by p and get 1+.01x= 14.75/p and then subtract 1 from both sides and get 1x=14.75/p-1 and then * it with 100 and get 1474/p-100 and then that = (1475-100p)/p and if p=10 then that equals to 47.5 by just plugging 10 in for p self critique assessment: 3
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21:41:58 `questionNumber 90000 What is the x as a function of p, and how many units are sold when the price is $10?
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RESPONSE --> x would be the cost confidence assessment: 1
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21:42:27 `questionNumber 90000 If p = 10 then x = (1475 - 100 p) / p = (1475 - 100 * 10) / 10 = 475 / 10 = 47.5 **
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RESPONSE --> it would be the same thing as the previous problem just plug in 10 for p and get 47.5 self critique assessment: 3
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??????????????assignment #010 010. `query 10 Applied Calculus I 08-03-2007
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21:45:28 `questionNumber 100000 What is a polynomial with zeros at -3, 4 and 9? Describe the graph of your polynomial.
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RESPONSE --> y=(x+3)(x-4)(x-9) the graph would look like a wavy line confidence assessment: 2
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21:46:04 `questionNumber 100000 A polynomial with zeros at -3, 4 and 9 must have factors (x + 3), ( x - 4) and (x - 9), and so must contain (x+3) (x-4) (x-9) as factors. These factors can be multiplied by any constant. For example 8 (x+3) (x-4) (x-9), -2(x+3) (x-4) (x-9) and (x+3) (x-4) (x-9) / 1872 are all polynomizls with zeros at -3, 4 and 9. If -3, 4 and 9 are the only zeros then (x+3), (x-4) and (x-9) are the only possible linear factors. It is possible that the polynomial also has irreducible quadratic factors. For example x^2 + 3x + 10 has no zeros and is hence irreducible, so (x+3) (x-4) (x-9) (x^3 + 3x + 10) would also be a polynomial with its only zeros at -3, 4 and 9. The polynomial could have any number of irreducible quadratic factors. **
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RESPONSE --> i believe that is what i got self critique assessment: 3
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21:47:10 `questionNumber 100000 1.5.18 (was 1.5.16 right-, left-hand limits and limit (sin fn)
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RESPONSE --> -2, -2, -2 confidence assessment: 1
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21:48:15 `questionNumber 100000 What are the three limits for your function (if a limit doesn't exist say so and tell why)?
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RESPONSE --> lim f(x) as x approaches c to the right and limf(x) as x approaches c to the left and limf(x) as x approaches c confidence assessment: 2
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21:48:35 `questionNumber 100000 Imagine walking along the graph from the right. As you approach the limiting x value, your y 'altitude' gets steadier and steadier, approaching closer and closer what value? You don't care what the function actually does at the limiting value of x, just how it behaves as you approach that limiting value. The same thing happens if you walk along the graph from the left. What does you y value approach? Is this the value that you approach as you 'walk in' from the right, as well as from the left? If so then it's your limit. **
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RESPONSE --> i think thats what i said self critique assessment: 3
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21:49:22 `questionNumber 100000 1.5.22 (was 1.5.20 right-, left-hand limits and limit (discont at pt) What are the three limits for your function (if a limit doesn't exist say so and tell why)?
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RESPONSE --> 0, 2,-1 three limits are lim f(x) as x approaches c from the left and right and just as it approaches c confidence assessment: 2
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21:49:44 `questionNumber 100000 STUDENT RESPONSE: The three limits are a. 0 b. 2 c. no limit The limit doesn't exist because while the function approaches left and right-handed limits, those limite are different. INSTRUCTOR COMMENT: That is correct. ADVICE TO ALL STUDENTS: Remember that the limit of a function at a point depends only on what happens near that point. What happens at the point iself is irrelevant to the limiting behavior of the function as we approach that point.
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RESPONSE --> there should be no limit for c. self critique assessment: 3
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21:50:50 `questionNumber 100000 1.5.30 (was 1.5.26 lim of (x+4)^(1/3) as x -> 4 What is the desired limit and why?
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RESPONSE --> lim (x+4)^(1/3) = (4+4)^(1/3) = 8^(1/3)= 2 confidence assessment: 2
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21:50:58 `questionNumber 100000 The function is continuous, so f(c) = c and in this case the limit is 8^(1/3), which is 2. **
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RESPONSE --> i got that self critique assessment: 3
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21:52:00 `questionNumber 100000 1.5.48 (was 1.5.38 lim of (x^3-1)/(x-1) as x -> 1
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RESPONSE --> that = x^4+x^3-x+1/x-1 = x^4+x^3-1=2-1=1 confidence assessment: 2
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21:52:38 `questionNumber 100000 As x -> 0 both numerator and denominator approach 0, so you can't tell just by plugging in numbers what the limit will be. If you factor x^3-1 into (x-1)(x^2+x+1) you can reduce the fraction to (x-1)(x^2+x+1) / (x-1) = x^2 + x + 1, which is equal to the original function for all x except 1 (we can't reduce for x = 1 because x-1 would be zero, and we can't divide by zero). As x -> 1, x^2 + x + 1 -> 3. It doesn't matter at all what the function does at x = 1, because the limiting value of x never occurs when you take the limit--only x values approaching the limit count. 3 is therefore the correct limit. **
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RESPONSE --> 3 is the correct limit self critique assessment: 2
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21:53:22 `questionNumber 100000 1.5.70 (was 1.5.56 lim of 1000(1+r/40)^40 as r -> 6%
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RESPONSE --> 1000(1+.06/4)^40= 1000(1+.015)^40= 1814.02 confidence assessment: 2
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21:53:55 `questionNumber 100000 $1000 *( 1+.06 / 40)^40 = 1061.788812. Since this function changes smoothly as you move through r = .06--i.e., since the function is continuous at r = .06--this value will be the limit. **
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RESPONSE --> oh i should have made the 4 40 when i was dividing .06 with 40. self critique assessment: 2
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21:54:05 `questionNumber 100000 Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> dont have any self critique assessment: 3
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x???{W??????assignment #011 011. `query 11 Applied Calculus I 08-03-2007
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21:59:26 `questionNumber 110000 1.6.16 (was 1.6.14 intervals of cont for (x-3)/(x^2-9) What are the intervals of continuity for the given function?
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RESPONSE --> continuous, polynomial, (-infinity,-infinity)(infinity, -infinity) confidence assessment: 1
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22:01:10 `questionNumber 110000 The function is undefined where x^2 - 9 = 0, since division by zero is undefined. x^2 - 9 = 0 when x^2 = 9, i.e., when x = +-3. So the function is continuous on the intervals (-infinity, -3), (-3, 3) and (3, infinity). The expression (x - 3) / (x^2 - 9) can be simplified. Factoring the denominator we get (x - 3) / [ (x - 3) ( x + 3) ] = 1 / (x + 3). This 'removes' the discontinuity at x = +3. However in the given fom (x-3) / (x^2 + 9) there is a discontinuity at x = -3. **
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RESPONSE --> i should have x^2-9=0 then bring over 9 and get x^2=9 and then take the sqrt of 9 and get x=+-3. and then its continuous at (-infinity,-3)(-3,3) and (3, infinity) self critique assessment: 2
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22:01:54 `questionNumber 110000 1.6.24 (was 1.6.22 intervals of cont for |x-2|+3, x<0; x+5, x>=0 What are the intervals of continuity for the given function?
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RESPONSE --> i wasnt sure about this one confidence assessment: 0
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22:04:12 `questionNumber 110000 The graph of y = |x-2|+3 is translated 2 units in the x direction and 3 in the y direction from the graph of y = |x|. It forms a V with vertex at (2, 3). The given function follows this graph up to x = 0. It has slope -1 and y-intercept at y = | 0 - 2 | + 3 = 5. The graph then follows y = x + 5 for all positive x. y = x + 5 has y-intercept at y = 5. From that point the graph increases along a straight line with slope 1. So the graph of the given function also forms a V with vertex at (0, 5). Both functions are continuous up to that point, and both continuously approach that point. So the function is everywhere continuous. **
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RESPONSE --> the vertex of the equation is at (2,3). the slope is at -1 and the y= abs0-2 +3=5 so y=x+5 and the slope is 1 and then get the vertex at (0,5) and both are continuous up to that point and so its continuous everywhere self critique assessment: 2
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22:05:00 `questionNumber 110000 1.6.66 (was 1.6.54 lin model of revenue for franchise Is your model continuous? Is actual revenue continuous?
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RESPONSE --> no it isnt continuous but the actual revenue is confidence assessment: 1
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22:07:09 `questionNumber 110000 revenue comes in 'chunks'; everytime someone pays. So the actual revenue 'jumps' with every payment and isn't continuous. However for a franchise the jumps are small compared to the total revenue and occur often so that a continuous model isn't inappropriate for most purposes. **
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RESPONSE --> oops, the actual revenue is not continuous and the the model is self critique assessment: 2
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22:07:16 `questionNumber 110000 Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> none self critique assessment: 3
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}????l????Y??{? assignment #012 012. `query 12 Applied Calculus I 08-03-2007
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22:11:31 `questionNumber 120000 Class Notes #13 Explain how we obtain algebraically, starting from the difference quotient, the expression for the derivative of the y = x^2 function.
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RESPONSE --> y'=2x confidence assessment: 2
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22:13:48 `questionNumber 120000 The difference quotient is [ f(x + `dx ) - f(x) ] / `dx. In this case we get [ (x+`dx)^2 - x^2 ] / `dx = [ x^2 + 2 x `dx + `dx^2 - x^2 ] / `dx = [ 2 x `dx - `dx^2 ] / `dx = 2 x - `dx. Taking the limit as `dx -> 0 this gives us just 2 x. y ' = 2 x is the derivative of y = x^2. **
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RESPONSE --> we get this from [f(x+dx)-f(x)/dx and so we replace x with [(x+dx)^2-x^2]/dx=[x^2+2xdx+dx^2-x^2]/dx= [2xdx-dx^2]/dx= 2x-dx and this just gives us 2x which is the derivative of y=x^2 self critique assessment: 3
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22:15:27 `questionNumber 120000 **** Explain how the binomial formula is used to obtain the derivative of y = x^n.
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RESPONSE --> dy/dx= n[x]^n-1 so we get y'= nx^n-1 confidence assessment: 3
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22:15:48 `questionNumber 120000 The key is that (x + `dx)^n = x^n + n x^(n-1)*`dx + n(n-1)/2 x^(n-2) * `dx^2 + ... + `dx^n. When we form the difference quotient the numerator is therefore f(x+`dx) - f(x) = (x + `dx)^n - x^n = (x^n + n x^(n-1)*`dx + n(n-1)/2 x^(n-2) * `dx^2 + ... + `dx^n) - x^n = n x^(n-1)*`dx + n(n-1)/2 x^(n-2) * `dx^2 + ... + `dx^n. The difference quotient therefore becomes (n x^(n-1)*`dx + n(n-1)/2 x^(n-2) * `dx^2 + ... + `dx^n) / `dx = n x^(n-1) + n (n-1) / 2 * x^(n-2) `dx + ... +`dx^(n-1). After the first term n x^(n-1) every term has some positive power of `dx as a factor. Therefore as `dx -> 0 these terms all disappear and the limiting result is n x^(n-1). **
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RESPONSE --> i think that is what i said but not in as many words self critique assessment: 3
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22:17:09 `questionNumber 120000 **** Explain how the derivative of y = x^3 is used in finding the equation of a tangent line to that graph.
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RESPONSE --> we do the derivative first and we get y'=3x^2 confidence assessment: 2
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22:18:05 `questionNumber 120000 The derivative is the slope of the tangent line. If we know the value of x at which we want to find the tangent line then we can find the coordinates of the point of tangency. We evaluate the derivative to find the slope of the tangent line. Know the point and the slope we use the point-slope form to get the equation of the tangent line. **
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RESPONSE --> we need a value of x to actually find the coordinates, and we take the derivative to find the slope of the tangent line, we need the point and the slope to use the point-slope form self critique assessment: 3
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22:18:31 `questionNumber 120000 2.1.9 estimate slope of graph.................................................
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RESPONSE --> 1/2 confidence assessment: 0
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22:18:58 `questionNumber 120000 You can use any two points on the graph to estimate the slope. The slope of a straight line is the same no matter what two points you use. Of course estimates can vary; the common approach of moving over 1 unit and seeing how many units you go up is a good method when the scale of the graph makes it possible to accurately estimate the distances involved. The rise and the run should be big enough that you can obtain good estimates. One person's estimate: my estimate is -1/3. I obtained this by seeing that for every 3 units of run, the tangent line fell by one unit. Therefore rise/run = -1/3. **
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RESPONSE --> it looks like 2 self critique assessment: 2
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22:20:56 `questionNumber 120000 2.1.24 limit def to get y' for y = t^3+t^2
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RESPONSE --> f(x+dx)-f(x)/dx = t^3+2tdx+dx^2-t^2-t^3+t^2/dx= 2tdx+dx^2/dx = 2t+dx confidence assessment: 2
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22:21:59 `questionNumber 120000 f(t+`dt) = (t+'dt)^3+(t+'dt)^2. f(t) = t^3 + t^2. So [f(t+`dt) - f(t) ] / `dt = [ (t+'dt)^3+(t+'dt)^2 - (t^3 + t^2) ] / `dt. Expanding the square and the cube we get [t^3+3t^2'dt+3t('dt)^2+'dt^3]+[t^2+2t'dt+'dt^2] - (t^3 - t^2) } / `dt. } We have t^3 - t^3 and t^2 - t^2 in the numerator, so these terms subtract out, leaving [3t^2'dt+3t('dt)^2+'dt^3+2t'dt+'dt^2] / `dt. Dividing thru by `dt you are left with 3t^2+3t('dt)+'dt^2+2t+'dt. As `dt -> 0 you are left with just 3 t^2 + 2 t. **
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RESPONSE --> I think i did something funky in the multiplication and division near the end, the answer is 3t^2+2t self critique assessment: 2
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22:22:58 `questionNumber 120000 2.1.32 tan line to y = x^2+2x+1 at (-3,4) What is the equation of your tangent line and how did you get it?
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RESPONSE --> y= -x-4/3 confidence assessment: 1
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22:23:41 `questionNumber 120000 STUDENT SOLUTION WITH INSTRUCTOR COMMENT: f ' (x)=2x+2. I got this using the method of finding the derivative that we learned in the modeling projects The equation of the tangent line is 2x+2. I obtained this equation by using the differential equation. the slope is -4...i got it by plugging the given x value into the equation of the tan line. INSTRUCTOR COMMENT: If slope is -4 then the tangent line can't be y = 2x + 2--that line has slope 2. y = 2x + 2 is the derivative function. You evaluate it to find the slope of the tangent line at the given point. You have correctly found that the derivative is -4. Now the graph point is (-3,4) and the slope is -4. You need to find the line with those properties--just use point-slope form. You get y - 4 = -4(x - -3) or y = -4 x - 8. **
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RESPONSE --> I messed up doing the derivative otherwise i would have gotten y=-4x-8 self critique assessment: 2
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22:24:52 `questionNumber 120000 2.1.52 at what pts is y=x^2/(x^2-4) differentiable (graph shown) At what points is the function differentiable, and why?
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RESPONSE --> it is differentiable at 4x^2 confidence assessment: 1
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22:25:49 `questionNumber 120000 At x = -2 and x = +2 the function approaches a vertical asymptote. When the tangent line approaches or for an instant becomes vertical, the derivative cannot exist. The reason the derivative doesn't exist for this function this is that the function isn't even defined at x = +- 2. So there if x = +- 2 there is no f(2) to use when defining the derivative as lim{`dx -> 0} [ f(2+`dx) - f(2) ] / `dx. f(2+`dx) is fine, but f(2) just does not exist as a real number. **
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RESPONSE --> it should have been x=-2 and a +2 but 2 doesnt exist as a real # self critique assessment: 2
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22:26:27 `questionNumber 120000 **** Query 2.1.52 at what pts is y=x^2/(x^2-4) differentiable (graph shown)
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RESPONSE --> x=-2 confidence assessment: 0
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22:27:00 `questionNumber 120000 The derivative is defined on(-infinity,-2)u(-2,2)u(2,infinity). The reason the derivative doesn't exist at x = +-2 is that the function isn't even defined at x = +- 2. The derivative at 2, for example, is defined as lim{`dx -> 0} [ f(2+`dx) - f(2) ] / `dx. If f(2) is not defined then this expression is not defined. The derivative therefore does not exist. At x = -2 and x = +2 the function approaches a vertical asymptote. When the tangent line approaches or for an instant becomes vertical, the derivative cannot exist.**
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RESPONSE --> oh yeah, it should be (-infinity, -2)u(-2,2)u(2,infinity) self critique assessment: 2
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22:27:35 `questionNumber 120000 If x is close to but not equal to 2, what makes you think that the function is differentiable at x?
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RESPONSE --> it's differentiable at x because it's not equal to 2 but it's close, there may be an asymptote at 2 confidence assessment: 2
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22:28:27 `questionNumber 120000 If x is close to 2 you have a nice smooth curve close to the corresponding point (x, f(x) ), so as long as `dx is small enough you can define the difference quotient and the limit will exist. **
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RESPONSE --> x is close to 2 which gives a smooth curve and dx is small enough that it can be defined as the difference quotient and the limit will exist self critique assessment: 2
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22:28:42 `questionNumber 120000 If x is equal to 2, is the function differentiable? Explain why or why not.
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RESPONSE --> no because it has gotten bigger confidence assessment: 0
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22:28:59 `questionNumber 120000 GOOD ANSWER FROM STUDENT: if the function does not have limits at that point then it is not differentiableat at that point.
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RESPONSE --> it doesnt have limits there self critique assessment: 3
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22:29:08 `questionNumber 120000 Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> none self critique assessment: 3
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?????d???H???????assignment #013 013. `query 13 Applied Calculus I 08-03-2007
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22:33:18 `questionNumber 130000 2.2.20 der of 4 t^-1 + 1. Explain in detail how you used the rules of differentiation to obtain the derivative of the given function, and give your final result.
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RESPONSE --> I did s(t)=4t^-1 +1 = -1(4)t^-1-1 +1 = -4t^-2+1 = -4/x^2 confidence assessment: 2
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22:34:03 `questionNumber 130000 STUDENT SOLUTION: To solve this using the rules of differentiation, I used the power and constant multiple rules. In dealing with t^-1, I applied the power rule and that gave me derivative -1t^-2. By the constant multiple we multiply this result by the constant 4 to get - 4 t^-2. To deal with 1, I used the constant rule which states that the derivative of a constant is 0. My final result was thus s'(t)=-4t^-2 + 0 = - 4 t^-2. **
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RESPONSE --> I brought down the t^-2 to make it positive but otherwise i got the same answer self critique assessment: 3
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22:36:21 `questionNumber 130000 22.2.30 der of 3x(x^2-2/x) at (2,18) What is the derivative of the function at the given point?
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RESPONSE --> y=9x-160 confidence assessment: 1
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22:38:14 `questionNumber 130000 You could use the product rule with f(x) = 3x and g(x) = x^2 - 2 / x. Since f ' = 3 and g ' = 2 x + 2 / x^2 we have (f g) ' = f ' g + f g ' = 3 (x^2 - 2 / x) + 3x ( 2x + 2 / x^2), which expands to (f g ) ' = 3 x^2 - 6 / x + 6 x^2 + 6 / x. This simplifies to give us just (f g ) ' = 9 x^2. It's easier, though, to just expand the original expression and take the derivative of the result: 3x ( x^2 - 2 / x ) = 3 x^3 - 6. The derivative, using the power-function rule, constant multiple rule and constant rule is thus y ' = 9x^2. At x = 2 we get derivative 9 * ( -2)^2 = 36. Note that (2, 18) is indeed on the graph because 3x ( x^2 - 2/x) evaluated at x = 2 gives us y = 3 * 2 ( 2^2 - 2 / 2) = 6 * 3 = 18. **
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RESPONSE --> I think i did the wrong problem. but we get f g) ' = f ' g + f g ' = 3 (x^2 - 2 / x) + 3x ( 2x + 2 / x^2)= (f g ) ' = 3 x^2 - 6 / x + 6 x^2 + 6 / x = 9 x^2 and then plug in the (2,18) self critique assessment: 2
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22:39:10 `questionNumber 130000 Query 22.2.38 f'(x) for f(x) = (x^2+2x)(x+1) What is f'(x) and how did you get it?
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RESPONSE --> = (2x+2x)(x+1) = 4x(x+1) = 4x^2 +4 confidence assessment: 1
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22:40:27 `questionNumber 130000 You could use the product rule, which would give you (x^2 + 2x) ' ( x + 1) + (x^2 + 2x) ( x + 1) ' = (2x + 2) ( x + 1) + (x^2 + 2 x ) ( 1) = 2 x^2 + 4 x + 2 + x^2 + 2 x = 3 x^2 + 6 x + 2. An easier alternative: If you multiply the expressions out you get x^3+3x^2+2x. Then applying the constant multiple rule and the simple power rule to the function you get f ' (x) = 3 x^2 + 6 x + 2 . **
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RESPONSE --> I should have done the product rule so i would have gotten (x^2 + 2x) ' ( x + 1) + (x^2 + 2x) ( x + 1) ' =2x + 2) ( x + 1) + (x^2 + 2 x ) ( 1) = 2 x^2 + 4 x + 2 + x^2 + 2 x = and get 3 x^2 + 6 x + 2 self critique assessment: 2
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22:41:00 `questionNumber 130000 22.2.66 vbl cost 7.75/unit; fixed cost 500 What is the cost function, and what is its derivative?
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RESPONSE --> not sure confidence assessment: 0
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22:42:34 `questionNumber 130000 The terminology means that it costs 7.75 per unit to manufacture the item, and 500 to run the plant or whatever. So if you produce x units it's going to cost 7.75 * x, plus the 500. The cost function is therefore 7.75 x + 500. If you take the derivative of the cost function you are looking at the slope of a graph of cost vs. number produced. The rise between two points of this graph is the difference in cost and the run is the difference in the number produced. When you divide rise by run you are therefore getting the average change in cost, per unit produced, between those two points. That quantity is interpreted as the average cost per additional unit, which is the average variable cost. The derivative is the limiting value of the slope when you let the two graph points get closer and closer together, and so gives the instantaneous rate at which cost increases per additional unit. Note that the fixed cost doesn't influence this rate. Changing the fixed cost can raise or lower the graph but it can't change the slope. **
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RESPONSE --> it should have been 7.75*x +500 which is 7.75x+500 and then take the derivative to find the slope. Then divide rise over run and you get hte average change in cost self critique assessment: 2
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22:42:54 `questionNumber 130000 Why should the derivative of a cost function equal the variable cost?
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RESPONSE --> because it needs the variable cost confidence assessment: 1
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22:43:18 `questionNumber 130000 The variable cost is defined as the rate at which the cost changes with repect to the number of units produced. That's the meaning of variable cost. That rate is therefore the derivative of the cost function. **
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RESPONSE --> the variable cost is the derivative of the cost function self critique assessment: 2
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??M??????vz????assignment #014 014. `query 14 Applied Calculus I 08-03-2007
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22:48:15 `questionNumber 140000 **** Query 2.3.8 ave rate compared with inst rates at endpts on [1,4] for x^-.5 **** What is the average rate of change over the interval and how did you get it?
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RESPONSE --> not sure confidence assessment: 0
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22:49:39 `questionNumber 140000 STUDENT SOLUTION: The average rate of change over the interval is -1/6. I got this answer by taking the difference of the numbers obtained when you plug both 1 and 4 into the function and then dividing that difference by the difference in 1 and 4. f(b)-f(a)/b-a
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RESPONSE --> ok so take the difference and plug both 1 and 4 into the equation f(b)-f(b)/a-b and get -1/6 self critique assessment: 2
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22:49:49 `questionNumber 140000 **** How does the average compare to the instantaneous rates at the endpoints of the interval?
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RESPONSE --> it is close confidence assessment:
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22:53:47 `questionNumber 140000 The average rate of change is change in y / change in x. For x = 1 we have y = 1^-.5 = 1. For x = 4 we have y = 4^-.5 = 1 / (4^.5) = 1 / 2. So `dx = 1/2 - 1 = -1/2 and `dy = 4 - 1 = 3. `dy / `dx = (.5-1) / (4-1) = -.5 / 3 = -1/6 = -.166... . To find rates of change at endpoints we have to use the instantaneous rate of change: The instantaneous rate of change is given by the derivative function y ' = (x^-.5) ' = -.5 x^-1.5. The endpoints are x=1 and x=4. There is a rate of change at each endpoint. The rate of change at x = 1 is y ' = -.5 * 1^-1.5 = -.5. The rate of change at x = 4 is y ' = -.5 * 4^-1.5 = -.0625. The average of these two rates is (-.5 -.0625) / 2 = -.281 approx, which is not equal to the average rate -.166... . Your graph should show the curve for y = x^-.5 decreasing at a decreasing rate from (1, 1) to (4, .5). The slope at (1, 1) is -.5, the slope at (4, .5) is -.0625. and the average slope is -.166... . The average slope is greater than the left-hand slope and less than the right-hand slope. That is, the graph shows how the average slope between (1,1) and (4,.5), represented by the straight line segment between those points, lies between steeper negative slope at x=1 and the less steep slope at x = 4. If your graph does not clearly show all of these characteristics you should redraw the graph so that it does. **
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RESPONSE --> y/change in x when we plug in 1 for x we get 1^-.5 and get 1 and for 4 we get 4^-.5 and get 1/2 so the dx is 1/2-1 and get -1/2 and the dy is 4-1=3 and then divide -1/2 /3 and get -.166 then we do rate pf change at x=1 and get y'=-.5*1^-1.5 and get -.5 and do the same for 4 and get -.0625 then add those together and divide by 2 and get -.281. the graph shows a curve at a decreasing decreasing rate. the slope at 1,1 is -.5 and the slope at 4,-.5 is -.0625 self critique assessment: 2
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22:55:40 `questionNumber 140000 Query 2.3.14 H = 33(10`sqrt(v) - v + 10.45): wind chill; find dH/dv, interpret; rod when v=2 and when v=5 What is dH/dt and what is its meaning?
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RESPONSE --> dH/dt is the derivative of H as change in t i got 459.523 when v=2 and 418.640 when v=5 confidence assessment: 2
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22:57:30 `questionNumber 140000 ERRONEOUS STUDENT SOLUTION: dH/dt is equal to [33(10 sqrt v+x - v+x + 10.45) + (33(10 sqrt v - v + 10/45))] / [(v+x)-x]. dH/dt represents the average heat loss from a person's body between two difference wind speeds; v+x and v INSTRUCTOR COMMENT: You give the difference quotient, which in the limit will equal the rate of change or the derivative, which is dH / dv = 33 * 10 * .5 * v^-.5 + 33 * -1 = 165 v^-.5 - 33. When v = 2, dH / dv is about 85 and when v = 5, dH / dv is about 40. Check my mental approximations to be sure I'm right (plug 2 and 5 into dH/dv = 165 v^-.5 - 33). H is the heat loss and v is the wind velocity. On a graph of H vs. v, the rise measures the change in heat loss and the run measures the change in wind velocity. So the slope measures change in heat loss / change in wind velocity, which is the change in heat loss per unit change in wind velocity. We call this the rate of change of heat loss with respect to wind velocity. dH / dv therefore measures the instantaneous rate of change of heat loss with respect to wind velocity. **
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RESPONSE --> dH/dt represents the ave heat loss of a person's body b/t 2 different wind speeds. I should have done 33*10*.5*v^-.5+33*-1= 165v^-.5-33 and then plug in 2 and 5 and get 85 and 40 self critique assessment: 2
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22:58:27 `questionNumber 140000 Query 2.3.20 C = 100(9+3`sqrt(x)); marginal cost **** What is the marginal cost for producing x units?
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RESPONSE --> I didnt get the right answer for this one confidence assessment: 1
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22:59:55 `questionNumber 140000 STUDENT SOLUTION: To get the marginal cost for producing x units, I think you take the first derivative of the cost function. If this is true, the marginal cost for 100(9+3sqrt(x)) is 100 * 3 * .5 x^-.5 = 150 x^-.5. The marginal cost is the rate at which cost changes with respect to the number of units produced. For this problem x is the number produced and C = 100 ( 9 + 3 sqrt(x) ). Marginal cost is therefore dC/dx = 100 * 3 / (2 sqrt(x)) = 150 / sqrt(x). **
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??P??????????assignment #015 015. `query 15 Applied Calculus I 08-03-2007
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23:05:37 `questionNumber 150000 2.3.32 P=22t^2+52t+10000, t from 1970; find P at t=0,10,20,25 and explain; find dP/dt; evaluate at given t and explain your results.
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RESPONSE --> dP/dt= 44t+52 and t=0 then it is 52, t=10 then it is492, t=20 it is 932, t= 25 it is 602 confidence assessment: 2
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23:06:54 `questionNumber 150000 dP/dt=44t + 52 (power function rule on each nonconstant term) When t = 0, 10, 20 and 25 you would have P = 10,000, 12,700, 20,000, 25,000 approx. At these values of t we have dP / dt = 52, 492, 932 and 1152 (these are my mental calculations--check them). dP / dt is the rate of change of the population with respect to time t **
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RESPONSE --> i forgot to plug in t=0,10,20,25 in the original equation and that dP/dt is the rate of change of a population with respect to t self critique assessment: 2
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23:08:22 `questionNumber 150000 2.3.48 demand fn p = 50/`sqrt(x), cost .5x+500. Find marginal profit for x=900,1600,2500,3600 Explain how you found the marginal profit, and give your results.
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RESPONSE --> R=xp so x* 50/sqrt(x) - .5x+500 and plug in the numbers confidence assessment: 1
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23:09:47 `questionNumber 150000 x represents the number of items sold. If x items are sold at price p = 50 / `sqrt(x), then revenue is price of item * number sold = 50 / `sqrt(x) * x = 50 `sqrt(x). The profit is revenue - cost = 50 `sqrt(x) - .5 x - 500. The marginal profit is the derivative of the profit function, which is (50 `sqrt(x) - .5 x - 500 ) ' = 25 / `sqrt(x) - .5. Evaluating the marginal profit at x = 900, 1600, 2500 and 3600 we get values .33..., .125, 0 and -.0833... . This shows us that the marginal profit, which is the limiting value of the increase in profit per additional item manufactured, is positive until x = 2500. This means that it is to the advantage of the producer to produce new items when x = 900 and when x = 1600, but that the advantage disappears as soon as x reaches 2500. So 2500 is the best selling price. When x = 3600 production of additional items reduces profits. **
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RESPONSE --> (50 `sqrt(x) - .5 x - 500 ) ' = 25 / `sqrt(x) - .5 when we plug in the numbers we get .33..., .125, 0 and -.0833 and it is positive until x=2500. self critique assessment: 2
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???????a????}???assignment #016 016. `query 16 Applied Calculus I 08-03-2007
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23:15:03 `questionNumber 160000 2.4.12 der of f(x) = (x+1)/(x-1) at (2,3) What is the derivative of f(x) at the given point?
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RESPONSE --> it equals 0 confidence assessment: 2
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23:15:42 `questionNumber 160000 f ' (x) = [ (x+1)'(x-1) - (x+1)(x-1)'] / (x-1)^2 = [ (x-1) - (x+1) ] / (x-1)^2 = -2 / (x-1)^2. When x = 2 we get f ' (x) = f ' (2) = -2 / (2-1)^2 = -2. **
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RESPONSE --> oh i messed up at (x-1) - (x+1) ] / (x-1)^2 = -2 / (x-1)^2. and then plug in 2 and get a -2 self critique assessment: 3
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23:17:27 `questionNumber 160000 2.4.30 der of (t+2)/(t^2+5t+6) What is the derivative of the given function and how did you get it?
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RESPONSE --> i used the quotient rule confidence assessment: 2
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23:17:59 `questionNumber 160000 we get (f ' g - g ' f) / g^2 = [ 1 ( t^2 + 5 t + 6) - (2t + 5)(t + 2) ] / (t^2 + 5 t + 6 )^2 = [ t^2 + 5 t + 6 - ( 2t^2 + 9 t + 10) ] / (t^2 + 5 t + 6)^2 = (-t^2 - 4 t - 4) / (t^2 + 5 t + 6)^2 ) = - (t+2)^2 / [ (t + 2) ( t + 3) ]^2 = - 1 / (t + 3)^2. DER**
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RESPONSE --> [ 1 ( t^2 + 5 t + 6) - (2t + 5)(t + 2) ] / (t^2 + 5 t + 6 )^2 = [ t^2 + 5 t + 6 - ( 2t^2 + 9 t + 10) ] / (t^2 + 5 t + 6)^2 = (-t^2 - 4 t - 4) / (t^2 + 5 t + 6)^2 ) = - (t+2)^2 / [ (t + 2) ( t + 3) ]^2 = - 1 / (t + 3)^2. self critique assessment: 3
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23:19:25 `questionNumber 160000 2.4.48 What are the points of horizontal tangency for(x^4+3)/(x^2+1)?
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RESPONSE --> i used the quotient rule to get the answer confidence assessment: 1
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23:20:36 `questionNumber 160000 the derivative is ( f ' g - g ' f) / g^2 = (4 x^3 ( x^2 + 1) - (2x) (x^4 + 3) ) / (x^2+1)^2 = [ 4 x^5 + 4 x^3 - 8 x^5 - 6 x ] / (x^2 + 1)^2 = -(2 x^5 + 4x^3 - 6x ) / (x^2 + 1)^2 = 2x (x^4 + 2 x^2 - 3) / (x^2+1)^2. The tangent line is horizontal when the derivative is zero. The derivative is zero when the numerator is zero. The numerator is 2x ( x^4 + 2 x^2 - 3), which factors to give 2x ( x^2 + 3) ( x^2 - 1). 2x ( x^2 + 3) ( x^2 - 1) = 0 when 2x = 0, x^2 + 3 = 0 and x^2 - 1 = 0. }2x = 0 when x = 0; x^2 + 3 cannot equal zero; and x^2 - 1 = 0 when x = 1 or x = -1. Thus the function has a horizontal tangent when x = -1, 0 or 1. **
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RESPONSE --> (4 x^3 ( x^2 + 1) - (2x) (x^4 + 3) ) / (x^2+1)^2 = [ 4 x^5 + 4 x^3 - 8 x^5 - 6 x ] / (x^2 + 1)^2 = -(2 x^5 + 4x^3 - 6x ) / (x^2 + 1)^2 = 2x (x^4 + 2 x^2 - 3) / (x^2+1)^2. and the tangent line is horizontal when the d=0 and that is 0 when the numerator is 0 self critique assessment: 3
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23:21:41 `questionNumber 160000 What would the graph of the function look like at and near a point where it has a horizontal tangent?
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RESPONSE --> it would look like a parabola only the tip of it is pushed upward so it creates a dip and the tangent line appears to be at 2 confidence assessment: 2
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23:21:54 `questionNumber 160000 At or near a point of horizontal tangency the graph would become at least for an instant horizontal. This could occur at a peak (like a hilltop, which is level at the very top point) or a valley (level at the very bottom). It could also occur if an increasing function levels off for an instant then keeps on increasing; or if a decreasing function levels off for and instant then keeps decreasing. **
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RESPONSE --> i believe that is what i said self critique assessment: 3
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23:23:32 `questionNumber 160000 2.4.58 defective parts P = (t+1750)/[50(t+2)] t days after employment What is the rate of change of P after 1 day, and after 10 days?
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RESPONSE --> 11.673 and 2.93 and the rate of change is 8.74 confidence assessment: 1
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23:24:46 `questionNumber 160000 It doesn't look like you evaluated the rate of change function to get your result. You have to use the rate of change function to find the rate of change. The rate of change function is the derivative. The derivative is ( f ' g - g ' f) / g^2 = ( 1 * 50(t+2) - 50(t + 1750) ) / ( 50(t+2) ) ^ 2 = -50 (1748) / ( 2500 ( t^2)^2 ) = - 874 / ( 25 ( t + 2) ^ 2 ). Evaluating when t = 1 and t = 10 we get -3.88 and -.243. **
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RESPONSE --> oh i needed to use the quotient function and get ( 1 * 50(t+2) - 50(t + 1750) ) / ( 50(t+2) ) ^ 2 =-50 (1748) / ( 2500 ( t^2)^2 ) =- 874 / ( 25 ( t + 2) ^ 2 ). and we get -3.88 and -.243 when t=1 and 10 self critique assessment: 2
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RESPONSE --> d'= 100*3*.5x^-.5= 150x^-.5 dC/dx= 100*3/(2sqrt(x))= 150/sqrt(x). self critique assessment: 3
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?}???s????????h? assignment #017 017. `query 17 Applied Calculus I 08-03-2007
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23:31:34 `questionNumber 170000 2.5.2 inner, outer fns for (x^2-3x+3)^3
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RESPONSE --> inner is x^2-3x+3 which can be 2x-3 and the outer is u^3 confidence assessment: 2
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23:31:53 `questionNumber 170000 The first function you evaluate is x^2 - 3x + 3. You then cube this result. So the breakdown to get f(g(x)) form is f(z) = z^3 g(x) = x^2 - 3x + 3. **
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RESPONSE --> i believe i did that self critique assessment: 3
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23:32:19 `questionNumber 170000 2.5.8 inner, outer fns for (x+1)^-.5
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RESPONSE --> f(z)= x+1 and the g(z)= z^-.5 confidence assessment: 3
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23:32:28 `questionNumber 170000 The first function you evaluate is x+1. You then take this result to the -5 power. So the breakdown to get f(g(x)) form is f(z) = z^-.5 g(x) = x+1. **
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RESPONSE --> that's what i got self critique assessment: 3
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23:33:14 `questionNumber 170000 2.5.24 der of f(t) = (9t+2)^(2/3) by gen power rule
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RESPONSE --> (2/3)(9t+2)^(-1/3) confidence assessment: 3
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23:34:22 `questionNumber 170000 This function is of the form u^(2/3), where u = 9 t + 2. The derivative of f(t) = u^p is f ' (t) = p * u^(p-1) * u ', by the general power rule. Here p = 2/3, and u ' = (9t + 2)' = 9 so we have f ' (t) = p u^(p-1) * u ' = 2/3 * u^(2/3 - 1) * 9 = 6 * u^(-1/3), or since u = 9 t + 2 f ' (t) = 6 ( 9 t + 2)^(-1/3). **
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RESPONSE --> I messed up hereu^(2/3 - 1) * 9 = 6 * u^(-1/3) or u = 9 t + 2 and get f ' (t) = 6 ( 9 t + 2)^(-1/3) self critique assessment: 2
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23:35:47 `questionNumber 170000 2.5.32 der of f(x) = (25+x^2)^(-1/2) by gen power rule
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RESPONSE --> (-1/2)[2x]^(-3/2) confidence assessment: 1
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23:36:38 `questionNumber 170000 Using inner function u = 25+x^2 and outer function = (u)^(-1/2) we get n u^(n-1) du/dx is -1/2 * u^(-3/2) * 2x = - x ( 25+x^2)^(-3/2). **
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RESPONSE --> oh i should have put the 2x for du/dx and then i would have gotten - x ( 25+x^2)^(-3/2) self critique assessment: 2
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?????????_??? assignment #018 018. `query 18 Applied Calculus I 08-03-2007
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23:39:01 `questionNumber 180000 ** Query problem 2.5.48 der of 3/(x^3-4) **** What is your result?
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RESPONSE --> I got -18x^3/(x^3-4)^3 self critique assessment: 1
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23:39:47 `questionNumber 180000 This function can be expressed as f(g(x)) for g(x) = x^3-4 and f(z) = 3 / z. The 'inner' function is x^3 - 4, the 'outer' function is 1 / z. So f'(z) = -3 / z^2 and g'(x) = 3x^2. Thus f'(g(x)) = -3/(x^3-4)^2 so the derivative of the whole function is [3 / (x^3 - 4) ] ' = g'(x) * f'(g(x)) = 3x^2 * (-3/(x^3-4)^2) = -9 x^2 / (x^3 - 4)^2. DER**
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RESPONSE --> i should have done it as f(g(x)) so i would have gotten [3 / (x^3 - 4) ] ' = g'(x) * f'(g(x)) = 3x^2 * (-3/(x^3-4)^2) = -9 x^2 / (x^3 - 4)^2 self critique assessment: 2
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23:40:09 `questionNumber 180000 **** Query problem 2.5.66 tan line to 1/`sqrt(x^2-3x+4) at (3,1/2) **** What is the equation of the tangent line?
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RESPONSE --> I messed this one up when I tried it confidence assessment: 0
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23:41:21 `questionNumber 180000 The derivative is (2x - 3) * -1/2 * (x^2 - 3x + 4) ^(-3/2) . At (3, 1/2) we get -1/2 (2*3-3)(3^2- 3*3 + 4)^(-3/2) = -1/2 * 3 (4)^-(3/2) = -3/16. The equation is thus ( y - 1/2) = -3/16 * (x - 3), or y = -3/16 x + 9/16 + 1/2, or y = -3/16 x + 17/16. DER**
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RESPONSE --> this is what i should have done 2x - 3) * -1/2 * (x^2 - 3x + 4) ^(-3/2) and then plug in (3,1/2) and get a -1/2 and a -3/16 and then the equation is y=-3/16x+17/16 self critique assessment: 2
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23:42:00 `questionNumber 180000 **** Query problem 2.5.72 rate of change of pollution P = .25 `sqrt(.5n^2+5n+25) when pop n in thousands is 12 **** At what rate is the pollution changing at the given population level?
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RESPONSE --> not sure confidence assessment: 0
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23:43:03 `questionNumber 180000 The derivative is .25 [ (n + 5) * 1/2 * (.5 n^2 + 5 n + 25) ^(-1/2) ) = (n+5) / [ 8 `sqrt(.5n^2 + 5n + 25) ] When n = 12 we get (12+5) / ( 8 `sqrt(.5*12^2 + 5 * 12 + 25) ) = 17 / 100 = .17, approx. DER**
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RESPONSE --> i should do .25 [ (n + 5) * 1/2 * (.5 n^2 + 5 n + 25) ^(-1/2) ) = (n+5) / [ 8 `sqrt(.5n^2 + 5n + 25) ] and then when n=12 then we get .17 self critique assessment: 2
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????????????? assignment #019 019. `query 19 Applied Calculus I 08-03-2007
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23:51:44 `questionNumber 190000 2.6.12 2d der of -4/(t+2)^2 What is the second derivative of your function and how did you get it?
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RESPONSE --> the second d is -32/(t+2)^2 confidence assessment: 1
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23:55:39 `questionNumber 190000 You need to use the Constant Rule. [ -4(t+2)^-2 ] ' = -4 [ (t+2)^-2 ] ' By the chain rule, with g(z) = z^-2 and h(t) = t + 1 this gives us -4 [ h '(t) * g ' ( h(t) ] = -4 [ (t+2) ' * -2(t+2)^-3 ] = -8 ( t+2)^-3. So g ' (t) = -8 ( t+2)^-3. Using the same procedure on g ' (t) we obtain g '' (t) = 24 ( t + 2)^-4. **
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RESPONSE --> I should use the constant rule and get [ -4(t+2)^-2 ] ' = -4 [ (t+2)^-2 ] ' and then get g(z)= z^-2 and h(t)= t+1 and that gives us -4 [ h '(t) * g ' ( h(t) ] = -4 [ (t+2) ' * -2(t+2)^-3 ] = -8 ( t+2)^-3. so we get g'(t) = -8 ( t+2)^-3 and thus we get g''(t)= 24(t+2)^-4 self critique assessment: 2
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23:57:48 `questionNumber 190000 2.6.30 f'''' if f'''=2`sqrt(x-1)
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RESPONSE --> use the chain rule to get the f''''and i believe you get x-1 confidence assessment: 0
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23:58:33 `questionNumber 190000 The fourth derivative f '''' is equal to the derivative of the third derivative. So we have f '''' = (f ''') ' = [ 2 sqrt(x-1) ] ' = 2 [ sqrt(x-1) ] '. Using the Chain Rule (noting that our function is sqrt(z) with z = x - 1, and that sqrt(z) ' = 1 / (2 sqrt(z) ) we get 2 [ (x-1)' * 1/(2 sqrt(x-1) ) ] = 2 [ 1 * 1/(2 sqrt(x-1) ) ] = 1 / sqrt(x-1). **
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RESPONSE --> so the f'''' =f''' and if we use the chain rule we get 2 [ (x-1)' * 1/(2 sqrt(x-1) ) ] = 2 [ 1 * 1/(2 sqrt(x-1) ) ] = 1 / sqrt(x-1). self critique assessment: 2
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23:58:49 `questionNumber 190000 2.6.42 brick from 1250 ft
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RESPONSE --> what confidence assessment: 0
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23:59:22 `questionNumber 190000 The detailed analysis is as follows: The equation comes from a = -32; since a = v' we have v' = -32 so that v = -32 t + c. Then s' = v so s' = -32 t + c and s = -16 t^2 + c t + k. c and k are constants. If s(t) = -16 t^2 + c t + k and at t = 0 we have s = 1250, then s(0) = -16 * 0^2 + c * 0 + k so k = 1250 and s(t) = -16 t^2 + c t + 1250. If the ball is dropped from rest then the initial velocity is v(0) = 0 so v(0) = -32 t + c = 0 and -32 * 0 + c = 0 so c = 0. So s(t) = -16 t^2 + c t + k becomes s(t) = -16 t^2 + 1250. To find how long it takes to hit the sidewalk: Position function, which gives altitude, is y = -16 t^2 + 1250. When the brick hits the sidewalk its altitude is zero. So -16 t^2 + 1250 = 0, and t = + - `sqrt(1250 / 16) = + - 8.8, approx. The negative value makes no sense, so t = 8.8 seconds. To find how fast the brick was moving when it hit the sidewalk: velocity = -32 t so when t = 8.8 we have velocity = -32 * 8.8 = -280 approx. That is, when t = 8.8 sec, v = -280 ft/sec. **
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RESPONSE --> ok, i should do all of that equation comes from a = -32; since a = v' we have v' = -32 so that v = -32 t + c. Then s' = v so s' = -32 t + c and s = -16 t^2 + c t + k. c and k are constants. If s(t) = -16 t^2 + c t + k and at t = 0 we have s = 1250, then s(0) = -16 * 0^2 + c * 0 + k so k = 1250 and s(t) = -16 t^2 + c t + 1250. If the ball is dropped from rest then the initial velocity is v(0) = 0 so v(0) = -32 t + c = 0 and -32 * 0 + c = 0 so c = 0. So s(t) = -16 t^2 + c t + k becomes s(t) = -16 t^2 + 1250. To find how long it takes to hit the sidewalk: Position function, which gives altitude, is y = -16 t^2 + 1250. When the brick hits the sidewalk its altitude is zero. So -16 t^2 + 1250 = 0, and t = + - `sqrt(1250 / 16) = + - 8.8, approx. The negative value makes no sense, so t = 8.8 seconds. To find how fast the brick was moving when it hit the sidewalk: velocity = -32 t so when t = 8.8 we have velocity = -32 * 8.8 = -280 approx. That is, when t = 8.8 sec, v = -280 ft/sec. ** self critique assessment: 2
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