course mth 173 10/5 1pm Question: `q001. Note that there are 5 questions in thie assignment.
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Given Solution: `aThe graph of this function has an intecept on the y' axis at (0,-6) and an intercept on the x axis at (60,0). The graph is a straight line with slope .1. At t = 100 the graph point is (100,4). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q002. Now sketch the graph of a function y which starts at the point (0, 100), and whose slope at every t is equal to the value of y ' at t. Thus, for example, at t = 0 we have y ' = .1 * 0 - 6 = -6, so our graph of y will start off a the t = 0 point (0,100) with a slope of -6, and the graph begins by decreasing rather rapidly. But the slope won't remain at -6. By the time we get to t = 10 the slope will be y ' = .1 * 10 - 6 = -5, and the graph is decreasing less rapidly than before. Then by the time we get to t = 20 the slope will be y ' = . 1 * 20 - 6 = -4, etc.. If you sketch a graph of y vs. t with these calculated slopes at the corresponding t values, what does the graph look like? Will it be increasing or decreasing, and will it be doing so at an increasing, decreasing or constant rate? Is the answer to this question different for different parts of the graph? If so over what intervals of the graph do the different answers apply? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Like in your example, the slopes will be 6 at t = 0, -5 at t = 10, -4 at t = 9, etc. since the numbers are negative, the slope will be negative and will be decreasing. Since the slopes are getting less steep, it will be decreasing at a decreasing rate. When it finally reaches 0(t=60) the slope will be zero and will be leveled off. When the slopes are greater than 0, which means they are positive, the slopes will be positive, eventually increasing at an increasing rate. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe graph will have slopes -6 at t = 0, -5 at t = 10, -4 at t = 9, etc.. At least for awhile, the slope will remain negative so the graph will be decreasing. The negative slopes will however become less and less steep. So the graph be decreasing at a decreasing rate. It's obvious that the slopes will eventually reach 0, and since y' is the slope of the y vs. t graph it's clear that this will happen when y' = .1 t - 6 becomes 0. Setting .1 t - 6 = 0 we get t = 60. Note, and think about about the fact, that this coincides with the x-intercept of the graph of y' vs. t. At this point the slope will be 0 and the graph will have leveled off at least for an instant. Past t = 60 the values of y' will be positive, so the slope of the y vs. t graph will be positive and graph of y vs. t will therefore be increasing. The values of y' will also be increasing, so that the slopes of the y vs. t graph will be increasing, and we can say that the graph will be increasing at an increasing rate. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q003. The graph of y vs. t corresponding to the given rate function y ' = .1 t - 6 has slope -6 at the point (0,100). This slope immediately begins changing, and becomes -5 by the time t = 10. However, let us assume that the slope doesn't change until we get to the t = 10 point. This assumption isn't completely accurate, but we're going to see how it works out. If the slope remained -6 for t = 0 to t = 10, then starting at (0, 100) what point would we reach when t = 10? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Slope=rise/run we are given the slope(-6) and the run(10) so the equation has to be rearranged so the rise can be figured out. Slope*run=rise so -6*10=rise so -60=rise. Then take 100-60=40, so the two points are our original 10 and now 40 so (10,40) confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe slope of the graph is the ratio slope = rise / run. If the slope remains at -6 from t = 0 to t = 10, then the difference between 10 is the run. Thus the run is 10 and the slope is -6, so the rise is rise = slope * run = -6 * 10 = -60. The y coordinate of the graph therefore changes by -60, from y = 100 to y = 100 + (-60) = 40. The corresponding point is (10, 40). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q004. We see that we reach the point (10, 40) by assuming a slope of -6 from t = 0 to t = 10. We have seen that at t = 10 the slope will be y ' = .1 * 10 - 6 = -5. If we maintain this slope for the interval t = 10 to t = 20, what will be the coordinates of the t = 20 point? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 20-10=10(run) and our original -5, so we set up our slope formula, rearranging it again so that slope*run=rise so -5*10=-50 so we are looking at the point (20, -10) confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe run from t = 10 to t = 20 is 10. With a slope of -5 this implies a rise of rise = slope * run = -5 * 10 = -50. Starting from point (10,40), a rise of -50 and a run of 10 takes the graph to the point (20, -10). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q005. Continue this process up to the t = 70 point, using a 10-unit t interval for each approximation. Describe the graph that results. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The slope at t = 20 is y ' = .1 * 20 - 6 = -4. From t = 20 to t = 30 the run is 10, so the rise is rise = slope * run = -4 * 10 = -40. Starting from (20,-10) a rise of -40 and a run of 10 takes us to (30, -50). The slope at t = 30 is y ' = .1 * 30 - 6 = -3. From t = 30 to t = 40 the run is 10, so the rise is rise = slope * run = -3 * 10 = -30. Starting from (30,-50) a rise of -30 and a run of 10 takes us to (40, -80). The slope at t = 40 is y ' = .1 * 40 - 6 = -2. From t = 40 to t = 50 the run is 10, so the rise is rise = slope * run = -2 * 10 = -20. Starting from (40,-80) a rise of -20 and a run of 10 takes us to (50, -100). The slope at t = 50 is y ' = .1 * 50 - 6 = -1. From t = 50 to t = 60 the run is 10, so the rise is rise = slope * run = -1 * 10 = -10. Starting from (50,-100) a rise of -10 and a run of 10 takes us to (60, -110). The slope at t = 60 is y ' = .1 * 70 - 6 = -0. From t = 60 to t = 70 the run is 10, so the rise is rise = slope * run = -0 * 10 = 0. Starting from (60,-110) a rise of and a run of 10 takes us to (70, -110). confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe slope at t = 20 is y ' = .1 * 20 - 6 = -4. From t = 20 to t = 30 the run is 10, so the rise is rise = slope * run = -4 * 10 = -40. Starting from (20,-10) a rise of -40 and a run of 10 takes us to (30, -50). The slope at t = 30 is y ' = .1 * 30 - 6 = -3. From t = 30 to t = 40 the run is 10, so the rise is rise = slope * run = -3 * 10 = -30. Starting from (30,-50) a rise of -30 and a run of 10 takes us to (40, -80). The slope at t = 40 is y ' = .1 * 40 - 6 = -2. From t = 40 to t = 50 the run is 10, so the rise is rise = slope * run = -2 * 10 = -20. Starting from (40,-80) a rise of -20 and a run of 10 takes us to (50, -100). The slope at t = 50 is y ' = .1 * 50 - 6 = -1. From t = 50 to t = 60 the run is 10, so the rise is rise = slope * run = -1 * 10 = -10. Starting from (50,-100) a rise of -10 and a run of 10 takes us to (60, -110). The slope at t = 60 is y ' = .1 * 70 - 6 = -0. From t = 60 to t = 70 the run is 10, so the rise is rise = slope * run = -0 * 10 = 0. Starting from (60,-110) a rise of and a run of 10 takes us to (70, -110). " ********************************************* Question: `q What number would appear in the second column next to the number 4.31 in the first column? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 18.5761 confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** In the original table the second column would read 18.57, approx.. This is the square of 4.31. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q What number would appear in the second column next to the number `sqrt(18) in the first column? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 18 confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 18 would appear in the second column because the square of sqrt(18) is 18. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q What number would appear in the second column next to the number `pi in the first column? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Pi^2 confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The number would be `pi^2 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: ********************************************* Question: `q What would we obtain if we reversed the columns of this table? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The inverse of what we had, with the square roots and squares being on opposite sides of the table confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q What number would appear in the second column next to the number 4.31 in the first column of this table? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 18.5761 confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: This number would be 4.31 squared,18.5761. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: What number would appear in the second column next to the number -3 in the first column of this table? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): There isn’t a -3 Self-critique Rating: ********************************************* Question: `q ** 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Note that sqrt(-3) is not a real number, since the square of a real number must be positive. ** 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. 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q 14. Solve 2^(3x-5) + 4 = 0 ** YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2^(3x-5)=-4 then 3x-5=log(base2)(-4)=log -4/log 2. There isnt a log for -4 so theres no solution confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q Solve 2^(1/x) - 3 = 0 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2^(1/x)=3 Log (2^(1/x))=log 3 (1/x) log(2) = log(3). Solve for x: x = log(2) / log(3). Solve 2^x * 2^(1/x) = 15 ** 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) x^2 + 1 = [log(15) / log(2) ] * x. confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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). ** Solve 2^x * 2^(1/x) = 15 ** 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. ** Solve (2^x)^4 = 5 ** 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.** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I forgot to go past the quadratic equation. ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q problem 1.3.22. C=f(A) = cost for A sq ft. What do f(10k) and f^-1(20k) represent? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: F(10k)= 10000 sq ft and f^-1(20k)= how many square feet can be covered for $20000 confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q problem 1.3.38. vert stretch y = x^2 by factor 2 then vert shift 1. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Y=x^2 ---> 2x^2 then with the vertical shift you just add one to it so y=2x^2 +1 confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q Give the equation of the function.Describe your sketch in detail. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y = 2 x^2 + 1. A squared function is a parabola(concave up), if its a vertical shift of +1 then you add one unit to the top confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q 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 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Plug in the values and graph. confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q query text problem 1.3 #13 temperature function H = f(t), meaning of H(30)=10, interpret vertical and horizontal intercepts YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Since H is temperature, and the f(t) is clock time, the H(30)=10 means that the temperature was 10 degrees when t=30. Vertical is temperature, and the vertical intercept is at t=30, so it gives us temp at t=0. Vertical clock time and horizontal intercept is when H=0, gives us clock time when H=0 confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: the verticle ** vertical ** intercept is the temperature of the object when it is placed outside 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. *&*& &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:3 "