course Mth 173 Fa\|assignment #007
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17:44:19 Query class notes #07Explain 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 --> The tangent line to y = k x^3 is its derivative, y = 3 k x^2. I do not know why it gives a good approximation to the function near that point.
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17:45:00 ** 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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18:30:37 Query class notes #08What 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 --> dT/dt = k(T - roomT) This graph would be a curve that starts with a high 'y' and slopes downward slower and slower.
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18:30:59 STUDENT SOLUTION AND INSTRUCTOR COMMENT: Would it be y = x-20 degrees., with y being the rate of temperature change and x being the temperature?You get a graph with a straight line and a slope of -20? INSTRUCTOR COMMENT: Not a bad attempt. However, you wouldn't use y to represent a rate, but rather dy /dt or y'. An in this example I'm going to use T for temperature, t for clock time. Read further. We need a graph of temperature vs. clock time, not rate of change of temperature vs. clock time. The difference between temperature and room temperature is indeed (T - 20). The rate of change of the temperature would be dT / dt. To say that these to our proportional is to say that dT / dt = k ( T - 20). To solve the situation we would need the proportionality constant k, just as with sandpiles and other examples you have encountered. Thus the relationship is dT / dt = k ( T - 20). Since dT / dt is the rate of change of T with respect to t, it tells you the slope of the graph of T vs. t. So the equation tells you that the slope of the graph is proportional to T - 20. Thus, for example, if T starts high, T - 20 will be a relatively large positive number. We might therefore expect k ( T - 20) to be a relatively large positive number, depending on what k is. For positive k this would give our graph a positive slope, and the temperature would move away from room temperature. If we are talking about something taken from the oven, this wouldn't happen--the temperature would move closer to room temperature. This could be accomplished using a negative value of k. As the temperature moves closer to room temperature, T - 20 becomes smaller, and the steepness of the graph will decrease--i.e., as temperature approaches room temperature, it will do so more and more slowly. So the graph approaches the T = 20 value more and more slowly, approaching as an asymptote. **
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18:55:45 Query Inverse Functions and Logarithms, Problem 7. Construct table for the squaring function f(x) = x^2, using x values between 0 and 2 with a step of .5. Reverse the columns of this table to form a partial table for the inverse function.
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RESPONSE --> Column 'x': 0, .5, 1, 1.5, 2 Column 'y': 0, .25, 1, 2.25, 4 The Inverse function is x = `sqrty Column 'x': 0, .5, 1, 1.5, 2 Column 'y': 0, .25, 1, 2.25, 4
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18:55:54 STUDENT SOLUTION: We get the following ordered pairs: Table 1-- (0,0),(.5,.25),(1,1),(1.5,2.25),(2,4) Table2--(0,0),(.25,.5),(1,1),(2.25,1.5),(4,2).
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19:07:03 Plot the points corresponding to the table of the squaring function, and plot the points corresponding to the table of its inverse. Sketch a smooth curve corresponding to each function. The diagonal line on the graph is the line y = x. Connect each point on the graph of the squaring function to the corresponding point on the graph of its inverse function. How are these pairs of points positioned with respect to the y = x line?
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RESPONSE --> They are perpindicular to the line y = x.
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19:07:08 ** The segments connecting the graph points for function and for its inverse will cross the y = x line at a right angle, and the graph points for the function and for the inverse will lie and equal distances on either side of this line. The graph of the inverse is therefore a reflection of the graph of the original function through the line y = x. **
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19:17:04 **** 8. If we reversed the columns of the 'complete' table of the squaring function from 0 to 12, precisely what table would we get?
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RESPONSE --> Column 'x': 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 Column 'y': 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144 I am not sure if this is what the question is asking for.
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19:17:14 ** We would get a table of the square root function with the first column running from 0 to 144, the second column consisting of the square roots of these numbers, which run from 0 to 12. **
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19:21:37 Sketch the graphs of the functions described by both tables. 9. If we could construct the 'complete' table of the squaring function from 0 to infinity, listing all possible positive numbers in the x column, then why would we be certain that every possible positive number would appear exactly one time in the second column?
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RESPONSE --> Because every number can be squared; therefore, every number is the square root of that number's square.
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19:21:52 ** The table you constructed had only some of the possible x and y values. A complete table, which couldn't actually be written down but can to an extent be imagined, would contain all possible x values. We could be sure because every number is the square of some other number. If the function was, for example, x / (x^2 + 1) there would be a great many positive numbers that wouldn't appear in the second column. But this is not the case for the squaring function. **
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19:22:49 What number would appear in the second column next to the number 4.31 in the first column?
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RESPONSE --> `sqrt(4.31) = 2.07605
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19:23:15 ** In the original table the second column would read 18.57, approx.. This is the square of 4.31. **
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RESPONSE --> I did `sqrt(4.31), not 4.31^2
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19:23:35 What number would appear in the second column next to the number `sqrt(18) in the first column?
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RESPONSE --> `sqrt(18) ^ 2 = 18
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19:23:36 ** 18 would appear in the second column because the square of sqrt(18) is 18. **
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19:24:25 What number would appear in the second column next to the number `pi in the first column?
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RESPONSE --> `pi ^ 2 = `pi^2 or 9.869604
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19:24:27 ** The number would be `pi^2 **
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19:24:45 What would we obtain if we reversed the columns of this table?
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RESPONSE --> The inverse function.
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19:24:50 STUDENT ANSWER: We would obtain the inverse, the square roots of the squares being in the y colume and the squared numbers being in the x column.
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19:25:28 What number would appear in the second column next to the number 4.31 in the first column of this table?
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RESPONSE --> sqrt(4.31) = 2.07605
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19:25:50 This number would be 4.31 squared,18.5761.
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RESPONSE --> I thought we switched the columns and the equation.
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19:26:27 What number would appear in the second column next to the number
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RESPONSE --> ""What number would appear in the second column next to the number "" This is a pasted copy of the question, it is incomplete.
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19:26:29 `pi^2 in the first column of this table?
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19:26:31 STUDENT ANSWER: This number would be the square root, 'pi
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19:26:43 What number would appear in the second column next to the number -3 in the first column of this table?
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RESPONSE --> -3 ^2 = 9
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19:27:02 ** The number -3 doesn't appear in the second column of the original table so it won't appear in the first column of the inverted table. Note that sqrt(-3) is not a real number, since the square of a real number must be positive. **
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RESPONSE --> Oo, tricky tricky
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19:36:45 13. Translate each of the following exponential equations into equations involving logarithms, and solve where possible: 2 ^ x = 18 2 ^ (4x) = 12 5 * 2^x = 52 2^(3x - 4) = 9.
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RESPONSE --> 2^x = 18 xlog2 = log18 x = log18/log2 x = appr. 4.1699 2^(4x) = 12 4xlog2 = log12 x = log12/4log2 x = appr. .89624 5 * 2^x = 52 2^x = 10.4 xlog2 = log10.4 x = log10.4/log2 x = appr. 3.378511623 2^(3x - 4) = 9 (3x - 4)log2 = log9 3xlog2 - 4log2 = log9 3xlog2 = log9 + 4log2 x = (log9 + 4log2) / 3log2 x = appr. 2.389975
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19:37:11 2^(3x-5) + 4 = 0 rearranges to 2^(3x-5) =-4, which we translate as 3x-5 = log {base 2}(-4) = log(-4) / log (2). However log(-4) doesn't exist. When you invert the 10^x table you don't end up with any negative x values. So there is no solution to this problem. Be sure that you thoroughly understand the following rules: 10^x = b translates to x = log(b), where log is understood to be the base-10 log. e^x = b translates to x = ln(b), where ln is the natural log. a^x = b translates to x = log{base a} (b), where log{base a} would be written in your text as log with subscript a. log{base a}(b) = log(b) / log(a), where log is the base-10 log. It also works with the natural log: log{base a}(b) = ln(b) / ln(a).
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RESPONSE --> ok
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19:39:53 Solve 2^(1/x) - 3 = 0
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RESPONSE --> 2^(1/x) - 3 = 0 2^(1/x) = 3 (1/x)log2 = log3 log2/x = log3 log2 = xlog3 x = log2/log3 x = appr. .630929
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19:39:57 ** Rearrange to 2^(1/x) = 3. Then take log of both sides: log(2^(1/x) ) = log(3). Use properties of logs: (1/x) log(2) = log(3). Solve for x: x = log(2) / log(3). **
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19:56:35 Solve 2^x * 2^(1/x) = 15
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RESPONSE --> 2^x * 2^(1/x) = 15 4^(1/x + x) = 15 (1/x + x)log4 = log15 (1/x + x) = log15/log4 I tried on paper and could not solve
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19:57:49 ** 2^x * 2^(1/x) = 15. By the laws of exponents we get 2^(x +1/x) = 15 so that x + 1/x = log {base2}(15) or x + 1/x =log(15) / log(2). Multiply both sides by x to get x^2 + 1 = [log(15) / log(2) ] * x. This is a quadratic equation. }Rearrange to get x^2 - [ log(15) / log(2) ] * x + 1 = 0 or x^2 - 3.91 * x + 1 = 0. Solve using the quadratic fomula. **
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RESPONSE --> I put the quadratic in my calculator and graphed it. I used calc: zero to try and find the zero. I was just lazy and didn't put it in the quadratic formula.
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20:02:42 Solve (2^x)^4 = 5
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RESPONSE --> (2^x)^4 = 5 2^x = 5^(1/4) xlog2 = (1/4)log5 x = (1/4)log5 / log2 x = appr. .5804820237
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20:03:09 ** log( (2^x)^4 ) = log(5). Using laws of logarithms 4 log(2^x) = log(5) 4 * x log(2) = log(5) 4x = log(5) / log(2) etc.**
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20:17:02 problem 1.3.22. C=f(A) = cost for A sq ft. What do f(10k) and f^-1(20k) represent?
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RESPONSE --> f(10k) represents the cost of a building with 10k square feet. f^-1(20k) represents the area of a building with the cost 20k
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20:17:22 ** f(10,000) is the cost of 10,000 sq ft. f^-1(20,000) is the number of square feet you can cover for $20,000. **
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20:19:53 problem 1.3.38. vert stretch y = x^2 by factor 2 then vert shift 1.
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RESPONSE --> y = 2x^2 + 1
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20:19:57 ** Vertically stretching y = x^2 we get y = 2 x^2. The vertical shift adds 1 to all y values, giving us the function y = 2 x^2 + 1. **
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20:21:36 Give the equation of the function.Describe your sketch in detail.
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RESPONSE --> If this is refferring to the previous question, then the function is y = 2x^2 + 1 (0, 1) (-2, 9) (2, 9)
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20:21:41 ** The function would be y = f(x) = 2 x^2 + 1. The factor 2 stretches the y = x^2 parabola verticall and +1 shifts every point of the stretched parabola 1 unit higher. The result is a parabola which is concave up with vertex at point (0,1). The parabola has been stretched by a factor of 2 as compared to a x^2 parabola. If the transformations are reversed the the graph is shifted downward 1 unit then stretched vertically by factor 2. The vertex, for example, shifts to (0, -1) then when stretched shifts to (0, -2). The points (-1, 1) and (1, 1) shift to (-l, 1) and (1, 0) and the stretch leaves them there. The shift would transform y = x^2 to y = x^2 - 1. The subsequent stretch would then transform this function to y = 2 ( x^2 - 1) = 2 x^2 - 2. The reversed pair of transformations results in a parabola with its vertex at (0, -2), as opposed to (0, -1) for the original pair of transformations. **
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20:24:28 problem 1.3.43 (was 1.8.30) estimate f(g(1))what is your estimate of f(g(1))?Explain how you look at the graphs of f and g to get this result
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RESPONSE --> I estimated g(1) to be 2.2 I then estimated f(2.2) to be .65
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20:24:33 *&*& right problem? *&*&
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20:24:39 ** You have to first find g(1), then apply f to that value. To find g(1), you note that this is g(x) for x = 1. So you look on the x-axis where x = 1. Then you move up or down to find the point on the graph where x = 1 and determine the corresponding y value. On this graph, the x = 1 point lies at about y = 2. Then you look at the graph of f(x). You are trying to find f(g(1)), which we now see is f(2). So we look at the x = 2 point on the x-axis and then look up or down until we find the graph, which for x = 2 lies just a little bit above the x axis. Looking over to the y-axis we see that at this point y is about .1. **
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20:34:45 problem 1.5.12 graph, decide if inverse, approximate inverse at x = 20 for f(x) = x^2+e^x and g(x) = x^3 + 3^x
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20:34:47 ** The inverse of a function at a certain value is the x that would give you that value when plugged into the function. At x = 20 for g(x) = x^2 + e^x is the x value for which x^3 + 3^x = 20. The double use of x is confusing and way the problem is stated in the text isn't as clear as we might wish, but what you have to do is estimate the required value of x. It would be helpful to sketch the graph of the inverse function by reflecting the graph of the original function through the line y = x, or alternatively and equivalently by making an extensive table for the function, then reversing the columns. **
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20:36:44 query text problem 1.3 #13 temperature function H = f(t), meaning of H(30)=10, interpret vertical and horizontal intercepts
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RESPONSE --> In the textbook, sectioin 1.3 #13 asks to graph f(5x)
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20:36:53 the verticle ** vertical ** intercept is the temperature of the object when it is placed outside
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20:36:56 the horizontal intercept is the time when the object became the same temperature as the outside *&*& H is the temperature, t is the clock time. H(30) is the temperature at clock time t = 30, so H(30) = 10 tells us that a clock time t = 30 the temperature was 10 degrees. The vertical coordinate is the temperature, and the vertical intercept of the graph occurs when t = 0 so the vertical intercept gives us the temperature at clock time 0. The vertical coordinate is the clock time, and the horizontal intercept occurs when H = 0, so the horizontal intercept gives us the clock time when temperature is 0. *&*&
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