course Mth 173 쳤͢|XEԼizìCassignment #007
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20:50:02 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 --> y=kx^3 y ' = 3kx^2.
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20:50:10 ** 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 --> ok
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20:52:16 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 --> dy / dt = x-20 degrees dy/dt is the rate of temperature change and x is the temperature I get a graph with a straight line and a slope of -20
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20:52:20 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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RESPONSE --> ok
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20:53:24 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 --> 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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20:53:28 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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RESPONSE --> ok
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20:54:25 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 --> The graph of the inverse is a reflection of the graph of the original function through the line y = x
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20:54:30 ** 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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RESPONSE --> ok
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20:55:49 **** 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 --> I get a table of the square root function with the first column 0 to 144, and the second column 0 to 12
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20:55:54 ** 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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RESPONSE --> ok
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20:57:01 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 --> The table only has some of the x and y vlaues, a real table would have all values of x and y
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20:57:05 ** 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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RESPONSE --> ok
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20:57:47 What number would appear in the second column next to the number 4.31 in the first column?
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RESPONSE --> The second column would read: 18.57, 4.31
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20:57:51 ** In the original table the second column would read 18.57, approx.. This is the square of 4.31. **
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RESPONSE --> ok
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20:59:15 What number would appear in the second column next to the number `sqrt(18) in the first column?
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RESPONSE --> 18 because the square of sqrt(18) is 18
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20:59:19 ** 18 would appear in the second column because the square of sqrt(18) is 18. **
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RESPONSE --> ok
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20:59:45 What number would appear in the second column next to the number `pi in the first column?
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RESPONSE --> 'pi ^2
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20:59:49 ** The number would be `pi^2 **
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RESPONSE --> ok
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21:00:25 What would we obtain if we reversed the columns of this table?
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RESPONSE --> the square roots of the squares being in the y column and the squared numbers in the x column
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21:00:31 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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RESPONSE --> ok
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21:01:09 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 --> 4.31 squared which = 18.5761
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21:01:14 This number would be 4.31 squared,18.5761.
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RESPONSE --> ok
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21:01:36 What number would appear in the second column next to the number
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RESPONSE --> `pi^2
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21:01:43 `pi^2 in the first column of this table?
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RESPONSE --> ok
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21:01:50 STUDENT ANSWER: This number would be the square root, 'pi
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RESPONSE --> ok
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21:02:33 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 doesn't appear in the second column of the table so it won't appear in the first column of the inverted table
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21:02:43 ** 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 --> ok
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21:04:59 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 = log{bas3 2}(18) = x. 2^x = 10.4. 2x = log{base 2}(10.4) x = 1/2 log{base 2}(10.4) 2^(3x-4) = 9 = log{base 2}(9) = 3x - 4
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21:05:02 b^x = c is translated into logarithmic notation as log{base b}(c) = x. So: 2^x = 18 translates directly to log{bas3 2}(18) = x. For 5 * 2^x = 52, divide both sides by 5 to get 2^x = 10.4. Now take logs: 2x = log{base 2}(10.4) so x = 1/2 log{base 2}(10.4). Evaluate on your calculator. 2^(3x-4) = 9 translates to log{base 2}(9) = 3x - 4.
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RESPONSE --> ok
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21:05:41 14. Solve 2^(3x-5) + 4 = 0
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RESPONSE --> 2^(3x-5) + 4 = 0 2^(3x-5) =-4, 3x-5 = log {base 2}(-4) = log(-4) / log (2)
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21:06:01 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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21:06:57 Solve 2^(1/x) - 3 = 0
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RESPONSE --> 2^(1/x) = 3 log(2^(1/x) ) = log(3) (1/x) log(2) = log(3) x = log(2) / log(3)
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21:07:01 ** 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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RESPONSE --> ok
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21:07:58 Solve 2^x * 2^(1/x) = 15
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RESPONSE --> 2^x * 2^(1/x) = 15 2^(x +1/x) = 15 x + 1/x = log {base2}(15) x + 1/x =log(15) / log(2) x^2 + 1 = [log(15) / log(2) ] * x x^2 - 3.91 * x + 1 = 0
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21:08:09 ** 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 --> ok
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21:08:39 Solve (2^x)^4 = 5
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RESPONSE --> log( (2^x)^4 ) = log(5) 4 log(2^x) = log(5) 4 * x log(2) = log(5) 4x = log(5) / log(2)
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21:08:43 ** 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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RESPONSE --> ok
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21:09:19 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(10,000) = cost of 10,000 sq ft. f^-1(20,000) = number of square feet you can cover for $20,000
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21:09:22 ** 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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RESPONSE --> ok
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21:10:25 problem 1.3.38. vert stretch y = x^2 by factor 2 then vert shift 1.
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RESPONSE --> y = x^2 y = 2 x^2 +1 = y = 2 x^2 + 1
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21:10:29 ** 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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RESPONSE --> ok
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21:12:04 Give the equation of the function.Describe your sketch in detail.
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RESPONSE --> y = f(x) = 2 x^2 + 1 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. y = 2 ( x^2 - 1) = 2 x^2 - 2 vertex = (0, -2)
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21:12:07 ** 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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RESPONSE --> ok
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21:12:34 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 couldnt find this problem
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21:12:38 *&*& right problem? *&*&
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RESPONSE --> ?
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21:12:58 ** 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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RESPONSE --> ok
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21:14:39 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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RESPONSE --> At x = 20 for g(x) = x^2 + e^x is the x value for which x^3 + 3^x = 20
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21:14:44 ** 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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RESPONSE --> ok
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21:15:14 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 --> the vertical intercept is the temperature of the object when it is placed outside
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21:15:18 the verticle ** vertical ** intercept is the temperature of the object when it is placed outside
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RESPONSE --> ok
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21:15:35 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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RESPONSE --> ok
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~OCy̯r assignment #007 007. Depth functions and rate functions. 09-24-2008
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15:08:36 `qNote that there are 9 questions in this assignment. `q001. The function y = .05 t^2 - 6 t + 100 is related to the rate function y ' = .1 t - 6 in that if y = .05 t^2 - 6 t + 100 represents the depth, then the depth change between any two clock times t is the same as that predicted by the rate function y ' = .1 t - 6. We saw before that for y ' = .1 t - 6, the depth change between t = 20 and t = 30 had to be 35 cm. Show that for the depth function y = .05 t^2 - 6t + 100, the change in depth between t = 20 and t = 30 is indeed 35 cm.
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RESPONSE --> depth at t=20 is: .05(20^2) - 6(20) + 100 = 20 - 120 + 100 = 0. depth at t = 30 is: .05(30^2) - 6(30) + 100 = 45 - 180 + 100 = -35 ave rate = change in depth / change in clock time = -35 cm / (10 sec) = -3.5 cm/s confidence assessment: 3
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15:08:45 The depth at t = 20 will be .05(20^2) - 6(20) + 100 = 20 - 120 + 100 = 0. The depth at t = 30 will be .05(30^2) - 6(30) + 100 = 45 - 180 + 100 = -35. Thus the depth changes from 0 cm to -35 cm during the 10-second time interval between t = 20 s and t = 30 s. This gives us and average rate of ave rate = change in depth / change in clock time = -35 cm / (10 sec) = -3.5 cm/s.
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RESPONSE --> ok self critique assessment: 3
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15:11:48 `q002. What depth change is predicted by the rate function y ' = .1 t - 6 between t = 30 and t = 40? What is the change in the depth function y = .05 t^2 - 6 t + 100 between t = 30 and t = 40? How does this confirm the relationship between the rate function y ' and the depth function y?
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RESPONSE --> t = 30 gives us y ' = .1 * 30 - 6 = -3 t = 40 gives us y ' = .1 * 40 - 6 = -2. average rate = -2.5 cm/s. predicted depth change = -2.5 cm/s * 10 s = -25 cm. confidence assessment: 3
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15:11:55 At t = 30 and t = 40 we have y ' = .1 * 30 - 6 = -3 and y ' = .1 * 40 - 6 = -2. The average of the two corresponding rates is therefore -2.5 cm/s. During the 10-second interval between t = 30 and t = 40 we therefore predict the depth change of predicted depth change based on rate function = -2.5 cm/s * 10 s = -25 cm. At t = 30 the depth function was previously seen to have value -35, representing -35 cm. At t = 40 sec we evaluate the depth function and find that the depth is -60 cm. The change in depth is therefore depth change has predicted by depth function = -60 cm - (-35 cm) = -25 cm. The relationship between the rate function and the depth function is that both should predict a same change in depth between the same two clock times. This is the case in this example.
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RESPONSE --> ok self critique assessment: 3
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15:16:33 `q003. Show that the change in the depth function y = .05 t^2 - 6 t + 30 between t = 20 and t = 30 is the same as that predicted by the rate function y ' = .1 t - 6. Show the same for the time interval between t = 30 and t = 40. Note that the predictions for the y ' function have already been made.
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RESPONSE --> t = 20 gives y = .05(20^2) - 6(20) + 30 = -70 t = 30 gives y = .05(30^2) - 6(30) + 30 = -105 depth change = -105 cm - (-70 cm) = -35 cm t = 40 gives y = .05(40^2) - 6(40) + 30 = -130 depth change = -130 cm - (-105 cm) = -25 cm confidence assessment: 3
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15:16:38 The prediction from the rate function is a depth change of -35 cm, and has already been made in a previous problem. Evaluating the new depth function at t = 20 we get y = .05(20^2) - 6(20) + 30 = -70, representing -70 cm. Evaluating the same function at t = 30 we get y = -105 cm. This implies the depth change of -105 cm - (-70 cm) = -35 cm. Evaluating the new depth function at t = 40 sec we get y = depth = -130 cm. Thus the change from t = 30 to t = 40 is -130 cm - (-105 cm) = -25 cm. This is identical to the change predicted in the preceding problem for the given depth function.
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RESPONSE --> ok self critique assessment: 3
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20:26:57 `q004. Why is it that the depth functions y = .05 t^2 - 6 t + 30 and y = .05 t^2 - 6 t + 100 give the same change in depth between two given clock times?
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RESPONSE --> the only difference between these two equations is the constant C--one is 30 and the other is 100...so there will always be a 70% difference between these two equations confidence assessment: 3
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20:27:05 The only difference between the two functions is a constant number at the end. One function and with +30 and the other with +100. The first depth function will therefore always be 70 units greater than the other. If one changes by a certain amount between two clock times, the other, always being exactly 70 units greater, must also change by the same amount between those two clock times.
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RESPONSE --> ok self critique assessment: 3
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20:30:34 `q005. We saw earlier that if y = a t^2 + b t + c, then the average rate of depth change between t = t1 and t = t1 + `dt is 2 a t1 + b + a `dt. If `dt is a very short time, then the rate becomes very clost to 2 a t1 + b. This can happen for any t1, so we might as well just say t instead of t1, so the rate at any instant is y ' = 2 a t + b. So the functions y = a t^2 + b t + c and y ' = 2 a t + b are related by the fact that if the function y represents the depth, then the function y ' represents the rate at which depth changes. If y = .05 t^2 - 6 t + 100, then what are the values of a, b and c in the form y = a t^2 + b t + c? What therefore is the function y ' = 2 a t + b?
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RESPONSE --> y = .05 t^2 - 6 t + 100 a = .05 b = -6 c = 100. y ' = 2 a t + b. y ' = 2 ( .05) t + (-6), y ' = .1 t - 6 confidence assessment: 3
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20:30:41 If y = .05 t^2 - 6 t + 100 is of form y = a t^2 + b t + c, then a = .05, b = -6 and c = 100. The function y ' is 2 a t + b. With the given values of a and b we see that y ' = 2 ( .05) t + (-6), which simplifies to y ' = .1 t - 6
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RESPONSE --> ok self critique assessment: 3
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20:32:40 `q006. For the function y = .05 t^2 - 6 t + 30, what is the function y ' ?
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RESPONSE --> y = .05 t^2 - 6 t + 30 a = .05 b = -6 c = 30 y ' = 2 a t + b y ' = 2(.05) t + (-6) y ' = .1 t - 6. confidence assessment: 3
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20:32:51 The values of a, b and c are respectively .05, -6 and 30. Thus y ' = 2 a t + b = 2(.05) t + (-6) = .1 t - 6. This is identical to the y ' function in the preceding example. The only difference between the present y function and the last is the constant term c at the end, 30 in this example and 100 in the preceding. This constant difference has no effect on the derivative, which is related to the fact that it has no effect on the slope of the graph at a point.
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RESPONSE --> ok self critique assessment: 3
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20:35:35 `q007. For some functions y we can find the rate function y ' using rules which we will develop later in the course. We have already found the rule for a quadratic function of the form y = a t^2 + b t + c. The y ' function is called the derivative of the y function, and the y function is called an antiderivative of the y ' function. What is the derivative of the function y = .05 t^2 - 6 t + 130? Give at least two new antiderivative functions for the rate function y ' = .1 t - 6.
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RESPONSE --> y = .05 t^2 - 6 t + 130 y ' = .1 t - 6 a = .05 b = -6 c = 130 y ' = .1 t - 6. c = 17 y = .05 t^2 - 6 t + 17 c = -54 y = .05 t^2 - 6 t - 54. confidence assessment: 3
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20:35:41 The derivative of y = .05 t^2 - 6 t + 130 is .1 t - 6; as in the preceding problem this function has a = .05 and b = -6, and differs from the preceding two y functions only by the value of c. Since c has no effect on the derivative, the derivative is the same as before. If y ' = .1 t - 6, then a = 1 / 2 ( .1) = .05 and b = -6 and we see that the function y is y = .05 t^2 - 6 t + c, where c can be any constant. We could choose any two different values of c and obtain a function which is an antiderivative of y ' = .1 t - 6. Let's use c = 17 and c = -54 to get the functions y = .05 t^2 - 6 t + 17 and y = .05 t^2 - 6 t - 54.
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RESPONSE --> ok self critique assessment: 3
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20:38:58 `q008. For a given function y, there is only one derivative function y '. For a given rate function y ', there is more than one antiderivative function. Explain how these statements are illustrated by the preceding example.
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RESPONSE --> y = .05 t^2 - 6 t +25 y = .05 t^2 - 6 t + 50 y = .05 t^2 - 6 t + 100 All three of these functions are antiderivatives of y ' = .1 t - 6. confidence assessment: 3
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20:39:03 The derivative function give the rate at which the original function changes for every value of t, and there can be only one rate for a given t. Thus the values of the derivative function are completely determined by the original function. In the previous examples we saw several different functions with the same derivative function. This occurred when the derivative functions differed only by the constant number at the end. However, for a given derivative function, if we get one antiderivative, we can add any constant number to get another antiderivative. y = .05 t^2 - 6 t +17, y = .05 t^2 - 6 t + 30, and y = .05 t^2 - 6 t + 100, etc. are all antiderivatives of y ' = .1 t - 6.
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RESPONSE --> ok self critique assessment: 3
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20:40:39 `q009. What do all the antiderivative functions of the rate function y ' = .1 t - 6 have in common? How do they differ? How many antiderivative functions do you think there could be for the given rate function?
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RESPONSE --> antidervitave .05 t^2 - 6 t. y = .05 t^2 - 6 t - 4. You can use any number for the last constant. confidence assessment: 3
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20:40:44 All antiderivatives must contain .05 t^2 - 6 t. They may also contain a nonzero constant term, such as -4, which would give us y = .05 t^2 - 6 t - 4. We could have used any number for this last constant.
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RESPONSE --> ok self critique assessment: 3
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