#$&* course Mth 163 Question: `qquery the family of linear functions, Problem 2.
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Given Solution: ** For the basic linear function f(x) = x the A = -.3 graph is obtained by vertically stretching the y = x function by factor -.3, resulting in a straight line thru the origin with slope -.3, basic points (0,0) and (1, -.3), and the A = 1.3 graph is obtained by vertically stretching the y = x function by factor 1.3, resulting in is a straight line thru the origin with slope 1.3, basic points (0,0) and (1, 1.3). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):
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Given Solution: ** The graphs will have slopes identical to that of the original function, but their y intercepts will vary from -2.7 to .3. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qquery problem 4. linear function y = f(x) = -1.77 x - 3.87 What are your symbolic expressions, using x1 and x2, for the corresponding y coordinates y1 and y2. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Y1= F(x1) = -1.77(x1x) - 3.87 Y2= F(x2) = -1.77(x2) - 3.87 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** y1 = f(x1) = -1.77 x1 - 3.87 y2 = f(x2) = -1.77 x2 - 3.87. `dy = y2 - y1 = -1.77 x2 - 3.87 - ( -1.77 x1 - 3.87) = -1.77 x2 + 1.77 x1 - 3.87 + 3.87 = -1.77 ( x2 = x1). Thus slope = `dy / `dx = -1.77 (x2 - x1) / (x2 - x1) = -1.77. This is the slope of the straight line, showing that these symbolic calculations are consistent. ** STUDENT QUESTION My question is how did you take -1.77 x2 + 1.77 x1 and get -1.77(x2 - x1)? I understand the x2-x1 but what happened to the 1.77? INSTRUCTOR RESPONSE This may be clearer if we work backwards: -1.77 * (x2 - x1) = -1.77 * x2 - (-1.77 * x1) = -1.77 x2 + 1.77 x1, which is the same thing as 1.77 x1 - 1.77 x2. -1.77 * (x2 - x1) was chosen as the form for the numerator, so we could easily divide it by x2 - x1. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qquery problem 5. graphs of families for y = mx + b. Describe your graph of the family: m = 2, b varies from -3 to 3 by step 1. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The graph passes through the points -3 and 3 and has the slope of 2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The graphs will all have slope 2 and will pass thru the y axis between y = -3 and y = 3. The family will consist of all such graphs. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):
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Given Solution: ** This is of the form y = mx + b with b= 1. So the y intercept is (0, 1). The point 1 unit to the right is (1, 1.5). The x-intercept occurs when y = 0, which implies .5 x + 1 = 0 with solution x = -2, so the x-intercept is (-2, 0). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qquery problem 6. three basic points graph of y = .5 x + 1 What are your three basic points? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (0,1) (1,1.5) (-2,0) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The y intercept occurs when x = 0, at which point we have y = .5 * 0 + 1 = 1. So one basic point is (0, 1). The point 1 unit to the right of the y axis occurs at x = 1, where we get y = .5 * 1 + 1 = 1.5 to give us the second basic point (1, 1.5) }The third point, which is not really necessary, is the x intercept, which occurs when y = 0. This gives us the equation 0 = .5 x + 1, with solution x = -2. So the third basic point is (-2, 0). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qquery problem 7. simple pendulum force vs. displacement What are your two points and what line do you get from the two points? The two points are (1.1, .21) and ( 2.0, .54) Once we plug these points into the equation y=mx+b we get .21= m(1.1) + b .54= m(2.0) + b &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qSTUDENT RESPONSE: The two points are (1.1, .21) and (2.0, .54). These points give us the two simultaneous equations .21- m(1.1) + b .54= m(2.0) +b. If we solve for m and b we will get our y = mx + b form. INSTRUCTOR COMMENT: I believe those are data points. I doubt if the best-fit line goes exactly through two data points. In the future you should use points on your sketched line, not data points. However, we'll see how the rest of your solution goes based on these points. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qwhat equation do you get from the slope and y-intercept? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: From the 2 equations from above and solving for m and b we get the equation y= .376x - .193. this gives us the slope which is m= .376 and y intercept which is b= -.193 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: STUDENT RESPONSE: b= .21 m=.19 INSTRUCTOR COMMENT: ** b would be the y intercept, which is not .21 since y = .21 when x = 1.1 and the slope is nonzero. If you solve the two equations above for m and b you obtain m = .367 and b = -.193. This gives you equation y = mx + b or y = .367 x - .193. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qwhat is your linear regression model? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The linear regression model I got that the line has a slope of .367 and passes through the y axis at point -.973 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Your linear regression model would be obtained using a graphing calculator or DERIVE. As a distance student you are not required to use these tools but you should be aware that they exist and you may need to use them in other courses. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhat force would be required to hold the pendulum 47 centimeters from its equilibrium position? what equation did you solve to obtain this result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The equation is the y= .367x - .193 when x=47 we get y=.367*47 - .193 y=17.249 - .193 y=17.056 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** If your model is y = .367 x - .193, with y = force and x= number of cm from equilibrium, then we have x = 47 and we get force = y = .367 * 47 - .193 = 17 approx. The force would be 17 force units. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhy would it not make sense to ask what force would be necessary to hold the pendulum 80 meters from its equilibrium position? what equation did you solve to obtain this result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Using the equation y= .10*47+.21 = 15.41 which is way to long because the pendulum can not exceed the length of the equilibrium. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: STUDENT RESPONSE: I used the equation f= .10*47+.21 and got the answer 15.41 which would be to much force to push or pull INSTRUCTOR COMMENT: ** The problem is that you can't hold a pendulum further at a distance greater than its length from its equilibrium point--the string isn't long enough. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qHow far could you hold the pendulum from its equilibrium position using a string with a breaking strength of 25 pounds? what equation did you solve to obtain this result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Using the equation y= .367x - .193 we get 25 = .376x - .193 x= 67.002 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Using the model y = .367 * x - .193 with y = force = 25 lbs we get the equation 25 = .367 x - .193, which we solve to obtain x = 69 (approx.). Note that this displacement is also unrealistic for this pendulum. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhat is the average rate of change associated with this model? Explain this average rate in common-sense terms. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Using the equation y = .367 x - .193, y = force and x = equilibrium from the points (x,y). The change in y = the change in the force and the change in x = the change in position. And the rate of change tell us how much the force changes per unit of position. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Using the model y = .367 * x - .193 with y = force and x = displacement from equilibrium we can use any two (x, y) points to get the rate of change. In all cases we will get rate of change = change in y / change in x = .367. The change in y is the change in the force, while the change in x is the change in position. The rate of change therefore tells us how much the force changes per unit of change in position (e.g., the force increases by 15 pounds for every inch of displacement). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhat is the average slope associated with this model? Explain this average slope in common-sense terms. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The average slope is determined by using the equation y = .367x - .193 with the slope being .367. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Using the model y = .367 * x - .193 with y = force and x = displacement from equilibrium the average slope is .367. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qAs you gradually pull the pendulum from a point 30 centimeters from its equilibrium position to a point 80 centimeters from its equilibrium position, what average force must you exert? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To pull the pendulum 30 cm we get y = .367*30 - .193 = 10.817 To pull the pendulum 80 cm we get y= .367*80 - .193 = 29.167 To find the average between the two we get (29.167 + 10.817) / 2 =19.992 or approx.. 20 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** if it was possible to pull the pendulum back this far and if the model applies you will get Force at 30 cm: y = .367 * 30 - .193 = 10.8 approx. and Force at 80 cm: y = .367 * 80 - .193 = 29 approx. so that ave force between 30 cm and 80 cm is therefore (10.8 + 29) / 2 = 20 approx.. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qquery problem 8. flow range What is the linear function range(time)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT RESPONSE: I obtained model one by drawing a line through the data points and picking two points on the line and finding the slope between them. I then substituted this value for m and used one of my data points on my line for the x and y value and solved for b. the line I got was range(t) = -.95t + 112.38. y = -16/15x + 98 INSTRUCTOR COMMENT: This looks like a good model. According to the instructions it should however be expressed as range(time) = -16/15 * time + 98. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhat is the significance of the average rate of change? Explain this average rate in common-sense terms. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The average rate of change is change in range / change in clock time. So once this is defined we will know on average how much the range changes. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** the average rate of change is change in range / change in clock time. The average rate of change indicates the average rate at which range in cm is changing with respect to clock time in sec, i.e., the average number of cm / sec at which the range changes. Thus the average rate tells us how fast, on the average, the range changes. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhat is the average slope associated with this model? Explain this average slope in common-sense terms. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: It is the average rate at which the range changes. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** it's the average rate at which the range of the flow changes--the average rate at which the position of the end of the stream changes. It's the speed with which the point where the stream reaches the ground moves across the ground. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qquery problem 9. If your total wealth at clock time t = 0 hours is $3956, and you earn $8/hour for the next 10 hours, then what is your total wealth function totalWealth( t ), where t is time in hours? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: A graph of wealth vs. time has a slope of 8 and has a y intercept of 3956, giving us the equation totalWealth(t)= 8*t + 3956 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Total wealth has to be expressed in terms of t. A graph of total wealth vs. t would have y intercept 3956, since that is the t = 0 value, and slope 8, since slope represents change in total wealth / change in t, i.e., the number of dollars per hour. A graph with y-intercept b and slope m has equation y = m t + b. Thus we have totalWealth(t) = 8 * t + 3956 . ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qAt what clock time will your total wealth reach $4000? what equation did you solve to obtain this result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: With a total wealth of 4000$ we get the equation 4000 = 8t + 3956 ; -3956 from both sides 44= 8t ; divide both sides by 8 5.5 = t confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: STUDENT RESPONSE: To find the clock time when my total wealth will reach 4000 I solved the equation totalWealth(t) = 4000. The value I got when I solved for t was t = 5.5 hours. 4.4 hours needed to reach 4000 4000 = 10x + 3956 INSTRUCTOR COMMENT: Almost right. You should solve 4000 = 8 x + 3956, obtaining 5.5 hours. This is equivalent to solving totalWealth(t) = 4000 = 8 t + 3956, which is the more meaningful form of the relationship. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhat is the meaning of the slope of your graph? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The slope shows us how much money is being made per hour confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: GOOD STUDENT RESPONSE: The slope of the graph shows the steady rate at which money is earned on an hourly basis. It shows a steady increase in wealth. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qquery problem 10. Experience shows that when a certain widget is sold for $30, a certain store can expect to sell 200 widgets per week, while a selling price of $28 increases the number sold to 300. What linear function numberSold(price) describes this situation? A graph of numbersold vs. price we get the points (30, 200) and (28, 300). To find the slope of this line we have to use the formula y2-y1 / x2-x1 = 300 - 200 / 28 - 30 = 100 / -2 = -50 Now we have to plug in our numbers into the equation y = mx+b y = -50x + 1700 numbersold(price) = -50*price + 1700 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qIf you make a graph of y = numberSold vs. x = price you have graph points (30, 200) and (28, 300). You need the equation of the straight line through these points. You plug these coordinates into the form y = m x + b and solve for m and b. Or you can use another method. Whatever method you use you get y = -50 x + 1700. Then to put this into the notation of the problem you write numberSold(price) instead of y and price instead of x. You end up with the equation numberSold(price) = -50 * price + 1700. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qIf the store must meet a quota by selling 220 units per week, what price should they set? what equation did you solve to obtain this result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To find the price we need to use the equation numbersold(price) = -50t + 1700 So now we have to plug in the number 220 into the equation 220 = -50t + 1700 and solve for t. -1480 = -50t 29.60 = t confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** If the variables are y and x, you know y so you can solve for x. For the function numberSold(price) = -50 * price + 1700 you substitute 220 for numbersold(price) and solve for price. You get the equation 220 = -50 * price + 1700 which you can solve to get price = 30, approx. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qIf each widget costs the store $25, then how much total profit will be expected from selling prices of $28, $29 and $30? what equation did you solve to obtain this result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If each widget costs $25 then the expected profit for $28 is $300, the profit for $29 is 250, the profit for $30 is 200. TO obtain these results I used the equation numbersold(price) = -50t + 1700 and I plugged in 28, 29, and 30 into the equation for t. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: STUDENT RESPONSE: If each widget costs the store $25, then they should expect to earn a profit of 300 dollars from a selling price of $28, 250 dollars from a price of $29 and 200 dollars from a price of $30. To find this I solved the equations numberSold(28); numberSold(29), and numberSold(30). Solving for y after putting the price values in for p. They will sell 300, 250 and 200 widgets, respectively (found by solving the given equation). To get the total profit you have to multiply the number of widgets by the profit per widget. At $28 the profit per widgit is $3 and the total profit is $3 * 300 = $900; at $30 the profit per widgit is $5 and 200 are sold for profit $1000; at $29 what happens? ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qquery problem 11. quadratic function depth(t) = .01 t^2 - 2t + 100 representing water depth vs. What is the equation of the straight line connecting the t = 20 point of the graph to the t = 60 point? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The equation for t=20 is depth(t) = .01*20^2 - 2*20 + 100 = 64 so the point is (20, 64). The equation for t=60 is depth(t) = .01*60^2 - 2*60 + 100 = 16 so the point is (60, 16). Now we have to find the slope and y intercept to put into the equation y = mx+b The slope is -1.2 the y intercept is 88 so the equation is Y=-1.2x + 88 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The t = 20 point is (20,64) and the t = 60 point is (60, 16), so the slope is (-48 / 20) = -1.2. This can be plugged into the form y = m t + b to get y = -1.2 t + b. Then plugging in the x and y coordinates of either point you get b = 88. y = -1.2 t + 88 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qquery problem 13. quadratic depth function y = depth(t) = .01 t^2 - 2t + 100. What is `dy / `dt based on the two time values t = 30 sec and t = 40 sec. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: T=30 Y=.01*30^2 - 2*30 + 100 Y=49 (30, 49) t=40 y=.01*40^2 - 2*40 +100 y=36 (40, 36) To find the slope we take 36 - 49 / 40-30 = -1.3 SO we know that the depth is changing -1.3cm / second confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** For t = 30 we have y = 49 and for t = 40 we have y = 36. The slope between (30, 49) and (40,36) is (36 - 49) / (40 - 30) = -1.3. This tells you that the depth is changing at an average rate of -1.3 cm / sec. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qwhat is `dy / `dt based on t = 30 sec and t = 31 sec. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: From the same steps above we get that the depth is changing -1.39 cm / second. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Based on t = 30 and t = 31 the value for `dy / `dt is -1.39, following the same steps as before ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qwhat is `dy / `dt based on t = 30 sec and t = 30.1 sec. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Following the same steps before I got that the depth changes -1.399 cm / second. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT RESPONSE: The value for 'dy / `dt based on t = 30 sec and t = 30.1 sec is -1.4 INSTRUCTOR COMMENT: ** Right if you round off the answer. However the answer shouldn't be rounded off. Since you are looking at a progression of numbers (-1.3, -1.39, and this one) and the differences in these numbers get smaller and smaller, you have to use a precision that will always show you the difference. Exact values are feasible here and shoud be used. I believe that this one comes out to -1.399. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhat do you think you would get for `dy / `dt if you continued this process? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The slope would keep getting smaller and more precise. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: STUDENT RESPONSE: An even more and more accurate slope value. I don't think it would have to continue to decrease. INSTRUCTOR COMMENT **If you look at the sequence -1.3, -1.39, -1.399, ..., what do you think happens? It should be apparent that the limiting value is -1.4 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhat does the linear function tell you? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The function tells us that the clock times change at a given rate of depth by the function .02t - 2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The function tells you that at any clock time t the rate of depth change is given by the function .02 t - 2. For t = 30, for example, this gives us .02 * 30 - 2 = -1.4, which is the rate we obtained from the sequence of calculations above. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qquery problem 14. linear function y = f(x) = .37 x + 8.09 . &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qwhat are the first five terms of the basic sequence {f(n), n = 1, 2, 3, ...} for this function. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The first five terms are 8.46, 8.83, 9.2, 9.57, 9.94 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The first five terms are 8.46, 8.83, 9.2, 9.57, and 9.94 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhat is the pattern of these numbers? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We are adding .37 each time. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** These numbers increase by .37 at each interval. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qIf you didn't know the equation for the function, how would you go about finding the 100th member of the sequence? How can you tell your method is valid? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Between 1-100 we are counting by ones, so all we have to do is take 99 and multiply it by .37 and add the first value 8.46. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** You could find the 100th member by noting that you have 99 jumps between the first number and the 100 th, each jump being .37. Multiplying 99 times .37 and then adding the result to the 'starting value' (8.46). STUDENT RESPONSE: simply put 100 as the x in the formula .37x +8.09 INSTRUCTOR COMMENT: That's what you do if you have the equation. Given just the numbers you could find the 100th member by multiplying 99 times .37 and then adding the result to the first value 8.46. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qfor quadratic function y = g(x) = .01 x^2 - 2x + 100 what are the first five terms of the basic sequence {g(n), n = 1, 2, 3, ...}? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The first five terms are 98.01, 96.04, 94.09, 92.16, 90.25. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** We have g(1) = .01 * 1^2 - 2 * 1 + 100 = 98.01 g(2) = .01 * 2^2 - 2 * 2 + 100 = 96.04, etc. The first 5 terms are therefore {98.01, 96.04, 94.09, 92.16, 90.25} &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhat is the pattern of these numbers? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The pattern is 1.97, 1.95, 1.93, 1.91. .02 is being subtracted each time. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The changes in these numbers are -1.97, -1.95, -1.93, -1.91. With each interval of x, the change in y is .02 greater than for the previous interval. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qIf you didn't know the equation for the function, how would you go about finding the next three members of the sequence? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Knowing that .02 is being subtracted each time we just subtract .02 from the numbers three different times. 1.91-.02 = 1.89 1.89-.02 = 1.87 1.87-.02 = 1.85 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** According to the pattern established above, the next three changes are -1.89, -1.87, -1.85. This gives us g(6) = g(5) - 1.89, g(7) = g(6) - 1.87, g(7) = g(6) - 1.85. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qHow can you verify that your method is valid? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Using the formula from above we can simply just plug 6,7,8 into the equation and solve to see if our answers are correct. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** You can verify the result using the original formula; if you evaluate it at 5, 6 and 7 it should confirm your results. That's the best answer that can be given at this point. You should understand, though that even if you verified it for the first million terms, that wouldn't really prove it (who knows what might happen at the ten millionth term, or whatever). It turns out that to prove it would require calculus or the equivalent. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qquery problem 15. The difference equation a(n+1) = a(n) + .4, a(1) = 5 If you substitute n = 1 into a(n+1) = a(n) + .4, how do you determine a(2)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Once we substitute n=1 into the equation we get A(1+1) = a(1) + .4 = a(2) = a(1)+.4 Or A(2) = a(1) + .4 Knowing that a(1) = 5 we get the new equation A(2) = 5+ .4 = A(2) = 5.4 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** You get a(1+1) = a(1) + .4, or a(2) = a(1) + .4. Knowing a(1) = 5 you get a(2) = 5.4. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q If you substitute n = 2 into a(n+1) = a(n) + .4 how do you determine a(3)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: N=2 A(2+1) = a(2) + .4 A(3) = a(2) + .4 Knowing that a(2) = 5.4 we get A(3) = 5.4 + .4 A(3) = 5.8 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** You have to do the substitution. You get a(2+1) = a(2) + .4, or since 2 + 1 = 3, a(3) = a(2) + .4 Then knowing a(2) = 5.4 you get a(3) = 5.4 + .4 = 5.8. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qIf you substitute n = 3 into a(n+1) = a(n) + .4, how do you determine a(4)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: N=3 A(3+1) = a(3) + .4 A(4) = a(3) + .4 Knowing that a(3) = 5.8 we get A(4) = 5.8 + .4 A(4) = 6.2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** We get a(4) = a(3) +.4 = 5.8 + .4 = 6.2 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhat is a(100)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: we know that n always equals a number below what a equals. So using this information we know that n = 99 and that a(1) = 5 so we get 5 + 99*.4 = a(100) = 44.6 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** a(100) would be equal to a(1) plus 99 jumps of .4, or 5 + 99*.4 = 44.6. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qquery problem 17. difference equation a(n+1) = a(n) + 2 n, a(1) = 4. What is the pattern of the sequence? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: A(n+1) = a(n) + 2n A(1+1) = a(1) + 2*1 A(2) = 4 + 2 A(2) = 6 A(2+1) = a(2) + 2*2 A(3) = 6 + 4 A(3) = 10 A(3+1) = a(3) +2*3 A(4) = 10 + 6 A(4) = 16 The pattern is 4, 6, 10, 16. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** For n = 1 we have n+1 = 2 so that the equation a(n+1) = a(n) + 2 n becomes a(2) = a(1) + 2 * 1 Since a(1) = 4 (this was given) we have a(2) = a(1) + 2 * 1 = 4 + 2 = 6. Reasoning similarly, n = 2 gives us a(3) = a(2) + 2 * 2 = 6 + 4 = 10. n = 3 gives us a(4) = a(3) + 2 * 3 = 10 + 6 = 16; etc. The sequence is 4, 6, 10, 16, 24, 34, ... . ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhat kind of function do you think a(n) is (e.g., linear, quadratic, exponential, etc.)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Quadratic because the sequences are changing by the same number each time. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The differences of the sequence are 4, 6, 8, 10, 12, . . .. The difference change by the same amount each time, which is a property of quadratic functions. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qquery the slope = slope equation Explain the logic of the slope = slope equation (your may take a little time on this one) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The slope = slope equation sets the slope between two given points equal to the slope between one of those points and the variable point (x, y). Since all three points lie on the same straight line, the slope between any two of the three points must be equal to the slope between any other pair. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I dont understand the slope= slope formulation. ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qquery problem 7. streamRange(t), 50 centimeters at t = 20 seconds, range changes by -10 centimeters over 5 seconds. what is your function? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution:
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Given Solution: ** The rate at which streamRange changes is change in streamRange / change in t = -10 cm / (5 sec) = -2 cm/s. This will be the slope m of the graph. Since streamRange is 50 cm when t = 20 sec the point (20, 50) lies on the graph. So the graph passes through (20, 50) and has slope -2. The function is therefore of the form y = m t + b with m = -2, and b such that 50 = -2 * 20 + b. Thus b = 90. The function is therefore y = -2 t + 90, or using the meaningful name of the function steamRange(t) = -2t + 90 You need to use function notation. y = f(x) = -2x + 90 would be OK, or just f(x) = -2x + 90. The point is that you need to give the funcion a name. Another idea here is that we can use the 'word' streamRange to stand for the function. If you had 50 different functions and, for example, called them f1, f2, f3, ..., f50 you wouldn't remember which one was which so none of the function names would mean anything. If you call the function streamRange it has a meaning. Of course shorter words are sometimes preferable; just understand that function don't have to be confined to single letters and sometimes it's not a bad idea to make the names easily recognizable. STUDENT RESPONSE: y = -2x + 50 INSTRUCTOR COMMENT: ** At t = 20 sec this would give you y = -2 * 20 + 50 = 10. But y = 50 cm when t = 20 sec. Slope is -10 cm / (5 sec) = -2 cm/s, so you have y = -2 t + b. Plug in y = 50 cm and t = 20 sec and solve for b. You get b = 90 cm. The equation is y = -2 t + 90, or streamRange(t) = -2t + 90. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qwhat is the clock time at which the stream range first falls to 12 centimeters? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Using the correct equation streamRange(t) = -2t + 90, you would set streamRange(t) = 12 and solve 12 = -2t + 90, obtaining t = 39 sec. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qquery problem 9. equation of the straight line through t = 5 sec and the t = 7 sec points of the quadratic function depth(t) = .01 t^2 - 2t + 100 What is the slope and what does it tell you about the depth function? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** You have to get the whole equation. y = m t + b is now y = -1.88 t + b. You have to solve for b. Plug in the coordinates of the t = 7 point and find b. You get 90.9 = -1.88 * 7 + b so b = 104, approximately. Find the correct value. The equation will end up something like y = depth(t) = -1.88 t + 104. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qThe slope of the linear function is -1.88. This tells me that the depth is decreasing as the time is increasing at a rate of 1.88 cm per sec. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qHow closely does the linear function approximate the quadratic function at each of the given times? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The deviations are for t = 3, 4, 5, 6, 7, & 8 as follows: .08, .03. 0. -.01, 0, .03. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qat what t value do we obtain the closest values? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Not counting t= 5 and t = 7, which are 0, the next closest t value is t = 6, the deviation for this is -.01. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qOn which side of the t = 5 and t = 7 points is the linear approximation closer to the quadratic function? On which side does the quadratic function 'curve away' from the linear most rapidly? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** On the t = 4 side the approximation is closer. The quadratic function curves away on the positive x side. ** Query Add comments on any surprises or insights you experienced as a result of this assignment. The slope = slope helped me out a lot. Learning that I can solve a linear in different ways was helpful. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: STUDENT COMMENT I found the difference equation to be a bit challenge to comprehend (it seems it can get pretty complicated) but very exciting as well. I'm still not entirely sure what uses it will have in the future, but it seems like an important concept to have for future reference. INSTRUCTOR RESPONSE The difference equation is a way of specifying how a quantity changes, step by step. There are numerous situations in which all was know is the initial value of a quantity and the rules for how it changes. For example when water flows from a hole in the bottom of a uniform cylinder, it is the depth of water that determines how fast it comes out. All we know, then, is the initial depth of the water and the rule for how quickly the depth changes. It turns out that we can approximate the behavior of the depth function using the difference equation y(n+1) = y(n) - k * sqrt(y(n)), where k is a constant number determined by the diameter of the cylinder. If you continue your study of mathematics you will eventually get to the fourth semester of the standard calculus sequence, a course entitled 'Introduction to Ordinary Differential Equations'. Most second-semester calculus courses also include a briefer introduction to the subject. Your exposure to difference equations in this course will be usefu helpful to you when you reach that point. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: STUDENT COMMENT I found the difference equation to be a bit challenge to comprehend (it seems it can get pretty complicated) but very exciting as well. I'm still not entirely sure what uses it will have in the future, but it seems like an important concept to have for future reference. INSTRUCTOR RESPONSE The difference equation is a way of specifying how a quantity changes, step by step. There are numerous situations in which all was know is the initial value of a quantity and the rules for how it changes. For example when water flows from a hole in the bottom of a uniform cylinder, it is the depth of water that determines how fast it comes out. All we know, then, is the initial depth of the water and the rule for how quickly the depth changes. It turns out that we can approximate the behavior of the depth function using the difference equation y(n+1) = y(n) - k * sqrt(y(n)), where k is a constant number determined by the diameter of the cylinder. If you continue your study of mathematics you will eventually get to the fourth semester of the standard calculus sequence, a course entitled 'Introduction to Ordinary Differential Equations'. Most second-semester calculus courses also include a briefer introduction to the subject. Your exposure to difference equations in this course will be usefu helpful to you when you reach that point. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!