course Mth163 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
.............................................
Given Solution: f(x) = x^2 + 4. To find f(3) we replace x by 3 to obtain f(3) = 3^2 + 4 = 9 + 4 = 13. Similarly we have f(7) = 7^2 + 4 = 49 + 4 = 53 and f(-5) = (-5)^2 + 9 = 25 + 4 = 29. Graphing f(x) vs. x we will plot the points (3, 13), (7, 53), (-5, 29). The graph of f(x) vs. x will be a parabola passing through these points, since f(x) is seen to be a quadratic function, with a = 1, b = 0 and c = 4. The x coordinate of the vertex is seen to be -b/(2 a) = -0/(2*1) = 0. The y coordinate of the vertex will therefore be f(0) = 0 ^ 2 + 4 = 0 + 4 = 4. Moving along the graph one unit to the right or left of the vertex (0,4) we arrive at the points (1,5) and (-1,5) on the way to the three points we just graphed. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): add more description to graph, like passing through these points and arriving at these points on the way to the three points we just arrived. Self-critique Rating: ********************************************* Question: `q002. If f(x) = x^2 + 4, then give the symbolic expression for each of the following: f(a), f(x+2), f(x+h), f(x+h)-f(x) and [ f(x+h) - f(x) ] / h. Expand and/or simplify these expressions as appropriate. ********************************************* Your solution: If f(x) = x^2 + 4 then any expression we put in place of (x) will also replace x in the equation. They would look as following: f(a) = a^2 + 4 f(x+2) = (x+2)^2 + 4 or if you simplify it would look like this: (x+2)(x+2)+4 = x^2 + 4x + 4 + 4 = x^2 + 4x + 8 f(x + h) = (x + h)^2 + 4 or simplified to (x + h)(x + h) + 4 = x^2 + 2xh + h^2 + 4 f(x + h) – f(x) = [(x + h)^2 + 4] – [x^2 + 4] simplify this to: (x^2 + 2xh + h^2 + 4) – (x^2 + 4) = 2xh + h^2 [ f(x + h) – f(x)] / h = [( x^2 + 2xh + h^2 + 4) –( x^2 +4)] / h = (2xh + h^2) / h = 2x + h Confidence Assessment: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: If f(x) = x^2 + 4, then the expression f(a) is obtained by replacing x with a: f(a) = a^2 + 4. Similarly to find f(x+2) we replace x with x + 2: f(x+2) = (x + 2)^2 + 4, which we might expand to get (x^2 + 4 x + 4) + 4 or x^2 + 4 x + 8. To find f(x+h) we replace x with x + h to obtain f(x+h) = (x + h)^2 + 4 = x^2 + 2 h x + h^2 + 4. To find f(x+h) - f(x) we use the expressions we found for f(x) and f(x+h): f(x+h) - f(x) = [ x^2 + 2 h x + h^2 + 4 ] - [ x^2 + 4 ] = x^2 + 2 h x + 4 + h^2 - x^2 - 4 = 2 h x + h^2. To find [ f(x+h) - f(x) ] / h we can use the expressions we just obtained to see that [ f(x+h) - f(x) ] / h = [ x^2 + 2 h x + h^2 + 4 - ( x^2 + 4) ] / h = (2 h x + h^2) / h = 2 x + h. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: ********************************************* Question: `q003. If f(x) = 5x + 7, then give the symbolic expression for each of the following: f(x1), f(x2), [ f(x2) - f(x1) ] / ( x2 - x1 ). Note that x1 and x2 stand for subscripted variables (x with subscript 1 and x with subscript 2), not for x * 1 and x * 2. x1 and x2 are simply names for two different values of x. If you aren't clear on what this means please ask the instructor. ********************************************* Your solution: Same as the problems above, we will substitute the given values of x into the equations and then simplify. f(x1) = 5 * x1 + 7 f(x2) = 5 * x2 + 7 [f(x2) –f(x1)] / (x2 – x1) = (5*x2 + 7) – (5*x1 – 7) / (x2 – x1) = (5*x2 – 5*x1) / (x2 – x1) we can simplify by taking the 5 out to make an original form:[ 5(x2 - x1) ] / (x2 – x1) = 5 Confidence Assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: Replacing x by the specified quantities we obtain the following: f(x1) = 5 * x1 + 7, f(x2) = 5 * x2 + 7, [ f(x2) - f(x1) ] / ( x2 - x1) = [ 5 * x2 + 7 - ( 5 * x1 + 7) ] / ( x2 - x1) = [ 5 x2 + 7 - 5 x1 - 7 ] / (x2 - x1) = (5 x2 - 5 x1) / ( x2 - x1). We can factor 5 out of the numerator to obtain 5 ( x2 - x1 ) / ( x2 - x1 ) = 5. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): word it as factoring out the 5 from the numerator Self-critique Rating: ********************************************* Question: `q004. If f(x) = 5x + 7, then for what value of x is f(x) equal to -3? ********************************************* Your solution: If f(x) = 5x + 7 + -3 to solve for x you need to first take away 7 from both sides 5x = -10 then divide by 5 x = -2 then check equation by substituting x value into equation 5(-2) + 7 = -3 -10 + 7 = -3 -3 = -3 Confidence Assessment: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: If f(x) is equal to -3 then we right f(x) = -3, which we translate into the equation 5x + 7 = -3. We easily solve this equation (subtract 7 from both sides then divide both sides by 5) to obtain x = -2. 004. `query 4 ********************************************* Question: `qWhere f(x) = x^3, what are f(-2), f(-a), f(x-4) and f(x) - 4? ********************************************* Your solution: Again we are substituting x values into equation and then simplifying. 1.f(-2) = (-2)^3 = (-2)(-2)(-2) = -8 2. f(-a) = (-a)^3 = -a^3 3. f(x – 4) =(x-4)^3 = (x-4)(x-4)(x-4) = (x^2 -8x – 16)(x-4) = x^3 – 12x^2 + 48x - 64 4. f(x) -4 = x^3 – 4 because the only substituted value of x is x Confidence Assessment: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** COMMON ERROR WITH COMMENT: Where f(x) = x^3, f(-2) = -2^3 INSTRUCTOR CRITIQUE: Write that f(-2) = (-2)^3. By order of operations -2^3 = -(2^3) (you exponentiate first, before you apply the negative, which is effectively a multiplication by -1). For an odd power like 3 it makes no difference in the end---in both cases the result is -8 COMMON ERROR WITH COMMENT: f(-a) = -a^3. INSTRUCTOR COMMENT:f(-a) = (-a)^3, which because of the odd power is the same as -a^3. If g(x) = x^2, g(-a) would be (-a)^2 = a^2. ANSWERS TO THE REMAINING TWO QUESTIONS: f(x-4) = (x-4)^3. If you expand that you get (x^2 - 8 x + 16) ( x - 4) = x^3 - 12 x^4 + 48 x - 64. In more detail the expansion is as follows: (x-4)^3 = (x-4)(x-4)(x-4) = [ (x-4)(x-4) ] * (x-4) = [ x ( x - 4) - 4 ( x - 4) ] ( x - 4) = (x^2 - 4 x - 4 x + 16) ( x - 4) = (x^2 - 8x + 16) ( x - 4) = (x^2 - 8x + 16) * x - (x^2 - 8x + 16) * 4 = x^2 * x - 8 x * x + 16 * x - x^2 * 4 - (-8x) * 4 - 16 * 4 = x^3 - 8 x^2 + 16 x - 4 x^2 + 32 x - 64 = x^3 - 12 x^2 + 48 x - 64. f(x) - 4 = x^3 - 4. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): In your solution you reverted from (-a)^3 to (-a)^2. I understand that you do exponent first before the – but you will still get –a^3 why did you revert to ^2 In you second to the last problem you answered x^3 -12x^4 + 48x – 64 in one solution and x^3 – 12x^2 +48 – 64 in the other. This was probably a typing error. I was just wandering if you were referring to adding exponents, I didn’t think you could and wanted to make sure I was right. Self-critique Rating: ********************************************* Question: `qWhere f(x) = 2^x, find f(2), f(-a), f(x+3) and f(x) + 3? ********************************************* Your solution: If f(x) = 2^x then the above values of x would be simplified to the follow solutions: f(2) = 2^2 = 4 f(-a) = 2^-a = 2^a f(x+3) = 2^(x+3) f(x) +3 = 2^x + 3 Confidence Assessment: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** Where f(x) = 2^x we have: f(2)= 2^2 or 4; f(a) = 2^a; f(x+3) = 2^(x+3); and f(x) + 3 = 2^x + 3. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: ********************************************* Question: `qquery functions given by meaningful names. What are some of the advantages of using meaninful names for functions? ********************************************* Your solution: They make it easier and help us keep track of what things mean. Confidence Assessment: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** TWO STUDENT RESPONSES: Using the function notation gives more meaning to our equations. Example... 'depth(t) = ' is a lot more understandable than ' y = ' I read of the pros and cons in using names. I think using names helps to give meaning to the equation, especially if it's one you haven't looked at for a couple days, you know right away what you were doing by reading the 'meaningful' notation.** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): It also reminds us of what we were doing if we go back and look at in a few days. Self-critique Rating: ********************************************* Question: `qWhat were your results for value(0), value(2), value(t+3) and value(t+3)/value(t)? ********************************************* Your solution: Where value(t) = $1000(1.07)^t substitute value(0), value(2), value(t+3),value(t+3) /value(t) into the place of t. value (0) = $1000(1.07)^0 = $1000(1) = $1000 value(2) = $1000(1.07)^2 =1000(1.1449) = $1144.90 value(t+3) = $1000(1.07)^(t + 3) or $1000(1.07^t * 1.07^3) value(t + 3) / value(t) = $1000(1.07)^(t +3) / [$1000(1.07)^t] = (1.07)^(t + 3) / (1.07^t) We can simplify 1.07^t * 1.07^3 / (1.07^t) = 1.07^3 Confidence Assessment: mostly ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** Substitute very carefully and show your steps: value(0) = $1000(1.07)^0 = $ 1000 value(2) = $1000(1.07)^2 = $1144.90 value(t + 3) = $1000(1.07)^(t + 3) value(t + 3) / value (t) = [$1000(1.07)^(t + 3)] / [ $1000(1.07)^t] , which we simplify. The $1000 in the numerator can be divided by the $1000 in the denominator to give us value(t+3) / value(t) = 1.07^(t+3) / [ 1.07^t]. By the laws of exponents 1.07^(t+3) = 1.07^t * 1.07^3 so we get value(t+3) / value(t) = 1.07^t * 1.07^3 / [ 1.07^t]. The 1.07^t divides out and we end up with value(t+3) / value(t) = 1.07^3. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Refer to as law of exponents when breaking apart exponents Self-critique Rating: ********************************************* Question: `qWhat did you get for illumination(distance)/illumination(2*distance)? Show your work on this one. ********************************************* Your solution: illumination(distance) / illumination(2*distance) where illumination(distance) = 50 / distance^2 Then you will have: 50 / distance^2 / [50/ (2*distance)^2] = We can invert denominator in order to multiply: (50 / distance ^2) * [(2* distance)^2 / 50] The 50 cancels out the other 50 so we have: (2*distance)^2 / distance^2 Simplify the exponents and get: 2^2 * distance^2 / distance^2 or 4*distance / distance^2 Your answer will be 4 Confidence Assessment: mostly ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** We substitute carefully and literally to get illumination (distance) / illumination (2*distance) = [50 / distance^2] / [50 / (2*distance)^2] which is a complex fraction and needs to be simplified. You invert the denominator and multiply to get [ 50 / distance^2 ] * [ (2 * distance)^2 / 50 ] = (2 * distance)^2 / distance^2 = 4 * distance^2 / distance^2 = 4. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: ********************************************* Question: `qquery #3. Sketch a reasonable graph of y = f(x), if it is known that f(2) = 80, f(5) = 40 and f(10) = 25. Explain how you constructed your graph. ********************************************* Your solution: I drew the graph on graph paper using each block as two units on the x and y axis, I plotted out each point (2,80), (5,40), (10,25) then drew the line. Confidence Assessment: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** STUDENT ANSWER WITH INSTRUCTOR COMMENT: I used graph paper counted each small square as 2 units in each direction, I put a dot for the points (2,80), (5,40), & (10,25) and connected the dots with lines. INSTRUCTOR COMMENT: The points could be connected with straight lines, but you might also have used a smooth curve. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): explain my line as a curved line Self-critique Rating: ********************************************* Question: `qwhat is your estimate of value of x for which f(x) = 60? ********************************************* Your solution: Looking at my graph and the line I drew, finding 60 on the y axes and moving over to the line, x = approximately 3.5 because the line is a little over halfway between 2 and 4 on the x axes. Confidence Assessment: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: **If your graph was linear from the point (2, 80) to the point (5, 40) you would estimate x = 3.5, since 3.5 is halfway from 2 to 5 and 60 is halfway from 80 to 40. However with f(10) = 25 a straightforward smooth curve would be decreasing at a decreasing rate and the y = 60 value would probably occur just a bit earlier, perhaps somewhere around x = 3.3** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): use my points that this (?,60) value comes between to estimate my x value because my line might be straight or curved. Self-critique Rating: ********************************************* Question: `qwhat is your estimate of the value f(7)? ********************************************* Your solution: 7 falls between the points (5,40) and (10,25) the is a difference of 3 between 5&7, and 7&10 the point would rest on the y value of approximately 34 Confidence Assessment: mostly ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** STUDENT RESPONSE WITH INSTRUCTOR COMMENT: My estimation for the value of f(7) was f(7) = 34. A linear graph between (5, 40) and (10, 25) would have slope -3 and would therefore decline by 6 units between x = 5 and x = 7, giving your estimate of 34. However the graph is probably decreasing at a decreasing rate, implying a somewhat greater decline between 5 and 7 that would be implied by a straight-line approximation. A better estimate might be f(7) = 32 or 33. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): remember the decreasing rate changes your decline Self-critique Rating: ********************************************* Question: `qwhat is your estimate of the difference between f(7) and f(9)? ********************************************* Your solution: If f(7) estimated value of x is 32 – 34 then f(9) value would be approximately around 27 Confidence Assessment: unsure ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** If f(7) = 32 and f(9) = 27, not unreasonable estimates, then the difference is 5. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: ********************************************* Question: `qwhat is your estimate of the difference in x values between the points where f(x) = 70 and where f(x) = 30? ********************************************* Your solution: plotting out the y values of 70 and 30 I perceive to get the values ox as 3 and 8 which is a difference of 5 Confidence Assessment: mostly ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** f(x) will be 70 somewhere around x = 3 and f(x) will be 30 somewhere around x = 8 or 9. The difference in these x values is about 5 or 6. On your graph you could draw horizontal lines at y = 70 and at y = 30. You could then project these lines down to the x axis. The distance between these projection lines, which again should probably be around 5 or 6, can be estimated by the scale of the graph. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: ********************************************* Question: `qquery #4. temperature vs. clock time function y = temperature = T(t), what is the symbolic expression for ... ********************************************* Your solution: If you have t = 3 then you have t(3) If you have t = 5 then you have t(5) The difference between the two is t(3) = t(5) = 2 that is your change The average between the two would be [t(3) + t(5)] / 2 = t(4) average Confidence Assessment: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** STUDENT SOLUTION WITH INSTRUCTOR COMMENT: The temperature at time t = 3; T(3)The temperature at time t = 5; T(5) The change in temperature between t = 3 and t = 5; T(3) - T(5) The order of the expressions is important. For example the change between 30 degrees and 80 degrees is 80 deg - 30 deg = 50 degrees; the change between 90 degrees and 20 degrees is 20 deg - 90 deg = -70 deg. The change between T(3) and T(5) is T(5) - T(3). When we specify the change in a quantity we subtract the earlier value from the later. INSTRUCTOR COMMENT: To average two numbers you add them and divide by 2. The average of the temperatures at t = 3 and t = 5 is therefore [T(3) + T(5)] /2 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: ********************************************* Question: `qWhat equation would we solve to find the clock time when the model predicts a temperature of 150? How would we find the length of time required for the temperature to fall from 80 to 30? ********************************************* Your solution: In order to solve the temperature prediction of 150 we must solve the temperatures at 80 and 30, which would be T(t) = 80 and T(t) = 30. To find the time required for the temperature to fall we must subtract the two. Confidence Assessment: unsure ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** GOOD STUDENT SOLUTION: To find the clock time when the model predicts a temperature of 150 we would solve the equation T(t) = 150. To find the length of time required for the temperature to fall from 80 to 30 we would first solve the equation T(t) = 80 to get the clock time at 80 degrees then find the t value or clok time when T(t) = 30. Then we could subtract the smaller t value from the larger t value to get the answer. [ value of t at T(t) = 30] - [ value of t at T(t) = 80)] ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Instead of subtracting the difference between the two9 we must subtract the smaller t value from the larger t value Self-critique Rating: ********************************************* Question: `qquery. use the f(x) notation at every opportunity:For how long was the depth between 34 and 47 centimeters? ********************************************* Your solution: To find the two times by using the equation f(t) = 34 and f(t) = 47 then subtract the two to find how long. Confidence Assessment: unsure ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** If depth = f(t) then you would solve f(t) = 34 to obtain t1 and f(t) = 47 to obtain t2. Then you would find abs(t2-t1). We use the absolute value because we don't know which is greater, t1 or t2, and the questions was 'how long' rather than 'what is the change in clock time ... ' ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): remember that we didn’t know which # was greater so we are not really sure which # went first. Self-critique Rating: ********************************************* Question: `qBy how much did the depth change between t = 23 seconds and t = 34 seconds? ********************************************* Your solution: The f(t) would be f(23) and f(34) to find how much it changed you must subtract the two: f(34) – f(23) Confidence Assessment: mostly ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** This would be f(34) - f(23). If for example if your model was f(t) = .01 t^2 - 1.5 t + 90, you would find that f(34) = 50.6 and f(23) = 60.8 so f(34) - f(23) = 50.6 - 60.8 = -10.2. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): you could give an example using the quadratic function f(t) = at^2 + bt + c and filled in values for a,b,&c to come up with solutions for both f(43) & f(23), subtracted and gave solution. Self-critique Rating: ********************************************* Question: `qOn the average, how many seconds did it take for the depth to change by 1 centimeter between t = 23 seconds and t = 34 seconds? ********************************************* Your solution: First you must establish that the depth would be f(34) – f(23) and the time would be 34sec. – 23 sec. = 11 sec. You would need to divide the time by depth to establish how many seconds for depth to change. 11 sec. / f(34) – f(23).Since it is given in centimeters the formula would be seconds / centimeters. Confidence Assessment: mostly ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** Change in depth would be f(34) - f(23). Change in t would be 34 s - 23 s = 11 s. Thus change in t / change in depth = 11 s / [ f(34) - f(23) ]. If f(t) gives depth in cm this result will be in seconds / centimeter, the ave number of seconds required for depth to change by 1 cm. If for example if your model was f(t) = .01 t^2 - 1.5 t + 90, f(t) in cm, you would find that f(34) = 50.6 cm and f(23) = 60.8 cm so f(34) - f(23) = 50.6 cm - 60.8 cm = -10.2 cm so that 11 s / [ f(34) - f(23) ] = 11 s / (-10.2 cm) = -1.08 sec / cm, indicating that depth decreases by 1 cm in 1.08 seconds. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: ********************************************* Question: `qOn the average, how by many centimeters did the depth change per second between t = 23 seconds and t = 34 seconds? ********************************************* Your solution: To establish did the depth change per second you would have to calculate cm / sec Which would be the reverse of the problem before giving you the problem: [f(34) – f(23)] / 11 sec. ( 11 sec. found by 34sec. – 23sec.) Since it was asked in centimeters you would calculate centimeters / seconds. Confidence Assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** Change in depth would be f(34) - f(23). Change in t would be 34 s - 23 s = 11 s. Thus change in depth / change in t = [ f(34) - f(23) ] / (11 s). If f(t) gives depth in cm this result will be in centimeters / second, the ave number of centimeters of depth change during each second. If for example if your model was f(t) = .01 t^2 - 1.5 t + 90, f(t) in cm, you would find that f(34) = 50.6 cm and f(23) = 60.8 cm so f(34) - f(23) = 50.6 cm - 60.8 cm = -10.2 cm so that [ f(34) - f(23) ] / (11 s) = (-10.2 cm) / (11 s) = -.92 cm / sec, indicating that between t = 23 sec and t = 34 sec depth decreases by -.92 cm in 1 sec, on the average. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: ********************************************* Question: `qquery. A hypothetical depth vs. time model based on three points, none of which are actual data points. Describe how you constructed your graph. ********************************************* Your solution: I would take data from t and depth on a bar graph and plot the points for constructing my graph. Confidence Assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** STUDENT RESPONSE: I sketched a graph using the depth vs. time data:(0, 96), (10, 89), (20, 68), (30, 65), (40, 48), (50, 49), (60, 36), & (70, 41). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: ********************************************* Question: `qWhat 3 data point did you use as a basis for your model? ********************************************* Your solution: After drawing the lines on this graph I chose three different x values and went up the y axis to find the corresponding y value , the points are ( 5,92), ( 35,57), and ( 65,37) Confidence Assessment: unsure ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** STUDENT RESPONSE: After drawing a curved line through the scatter data I placed 3 dots on the graph at aproximately equal intervals. I obtained the x and y locations form the axis, thier locations were at points (4,93), (24, 68) & (60, 41). I used these 3 data points as a basis for obtaining the model equation.** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): need to describe line since it is curved but the points are all over, you would say curved line with scattered points. Self-critique Rating: ********************************************* Question: `qWhat was your function model? ********************************************* Your solution: using the points I came by in my graph: (5,92), (35,57), and (65,37) and using the quadratic function y = ax^2 + bx + c, then solving for values of a, b, & c my function model is .00833x^2 – 14.998x + 541.732 = depth(t) Confidence Assessment: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** STUDENT RESPONSE CONTINUED: The function model obtained from points (4, 93), (24, 68), & (60, 41) is depth(t) = .0089x^2 - 1.4992x + 98.8544. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok. I used my poi8nts which is slightly different data but if no calculation errors, derived equation in the correct manner using the elimination process. Self-critique Rating: ********************************************* Question: `qWhat is the average deviation for your model? ********************************************* Your solution: I took our points from the points of our original data a few problems before and subtracted the a value from the y and found our model numbers then I subtracted the model data from the y7 data and formed my deviation numbers. The average of deviations was found from adding the deviation data together and dividing them by 8 ( the # of deviations we have) and came up with an average of 0.009 Confidence Assessment: unsure ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** STUDENT RESPONSE CONTINUED: I added a column 2 columns to the chart given in our assignment one labeled 'Model data' and the other 'Deviation'. I then subtracted the model data readings from the corresponding given data to get the individual deviations. I then averaged out the numbers in the deviation column. The average deviation turned out to be 3.880975.** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I went back to my notes from quadratic model of depth vs. clock and found the way I did this as to what I understood you explain. I did not receive the same data. Self-critique Rating: ********************************************* Question: `qHow close is your model to the curve you sketched earlier? ********************************************* Your solution: The model graphed is extremely close to the one I sketched before. Confidence Assessment: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** STUDENT RESPONSE CONTINUED: I was really suprised at how close the model curve matched the curve that I had sketched earlier. It was very close.** Query Add comments on any surprises or insights you experienced as a result of this assignment. ********************************************* Your solution: ok, I was surprised on how close they really were and how different points picked from different people can give different results. Confidence Assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** STUDENT RESPONSE: I was surprized that the scattered data points on the hypothetical depth vs. time model would fit a curve drawn out through it. The thing that sank in for me with this was to not expect all data to be in neat patterns, but to look for a possible pattern in the data. INSTRUCTOR COMMENT: Excellent observation **