course mth 163 ??????????????assignment #009009. `query 9
......!!!!!!!!...................................
19:56:38 Symbolic calculation of slope, preliminary exercise What was the function, between which two points were you to calculate the average slope and how did you get this slope?
......!!!!!!!!...................................
RESPONSE --> The function was y = .1x^2 - 3 The two points were x = -2 and x=7 or the coordinates: (-2, -2.6) and (7, 1.9) I used the slope formula : (y2-y1) / (x2-x1) since I had points to work with. confidence assessment: 3
.................................................
......!!!!!!!!...................................
19:59:30 ** For the function y = .1 x^2 - 3 between x = -2 and x = 7 we get: slope = (y2 - y1) / (x2 - x1). For x1 = 2 and x2 = 7 we have y2 = .1 * 7^2 - 3 = 1.9 and y1 = .1 * 2^2 - 3 = -2.6, so slope = (1.9 - (-2.6) ) / ( 7 - 2) = 4.5 / 5 = .9. **
......!!!!!!!!...................................
RESPONSE --> in this problem x1 = -2 not 2, therefore the correct answer is: slope = (1.9 - (-2.6) ) / ( 7 -(- 2)) = 4.5 / 9 = .5. self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:02:00 problem 2 symbolic expression for slope, fn depth(t). What is the expression for the slope between the two specified t values?
......!!!!!!!!...................................
RESPONSE --> the expression would be: slope: [f(x2) - f(x1)] / 30-10 Simplified: [f(x2) - f(x1)] / 20 confidence assessment: 1
.................................................
......!!!!!!!!...................................
20:03:00 ** The function is given a name: depth(t). t values are 10 and 30. So rise = depth(30) - depth(10) and run = 30 - 10 = 20. Thus slope = [ depth(30) - depth(10) ] / 20 . **
......!!!!!!!!...................................
RESPONSE --> I should have put the t value in for the x2 and x1 values. self critique assessment: 2
.................................................
......!!!!!!!!...................................
20:05:33 ** The rise is the change in depth. The two depths are depth(10) and depth(30). The change in depth is final depth - initial depth, which gives us the expression depth(30)-depth(10) **
......!!!!!!!!...................................
RESPONSE --> This program is touchy...I went to type the rise which is f(30) - f(10), but it went ahead and gave me the answer..... self critique assessment: 2
.................................................
......!!!!!!!!...................................
20:05:48 What is the run between the two specified t values?
......!!!!!!!!...................................
RESPONSE --> the run would be 30-10 or 20 confidence assessment: 2
.................................................
......!!!!!!!!...................................
20:06:24 ** run = 30 - 10 = 20 **
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:07:15 What therefore is the slope and what does it mean?
......!!!!!!!!...................................
RESPONSE --> the slope is [f(30) - f(10)] / 20 It means the change in depth vs the change in time. confidence assessment: 2
.................................................
......!!!!!!!!...................................
20:08:36 ** rise = depth(30)-(depth(10) indicates change in depth. run = 30 - 10 = 20 = change in clock time. Slope = rise / run = (depth(30) - depth(10) ) / 20, which is the average rate at which depth changes with respect to clock time between t = 10 and t = 30. **
......!!!!!!!!...................................
RESPONSE --> I just mentioned the change, not the average rate of change.(I do know that this is what slope is.) self critique assessment: 2
.................................................
......!!!!!!!!...................................
20:11:20 problem 5 graph points corresponding to load1 and load2 What are the coordinates of the requested graph points?
......!!!!!!!!...................................
RESPONSE --> (load1, springlenth(load1)) (load2, sprinlength(load2)) confidence assessment: 2
.................................................
......!!!!!!!!...................................
20:12:50 ** The horizontal axis is the 'load' asix, the vertical axis is the springLength axis. The load axis coordinates are load1 and load2. The corresponding spring lengths are springLength(load1) and springLength(load2). The springLength axis coordinates are springLength(load1) and springLength(load2). The graph points are thereofore (load1, springLength(load1) ) and (load2, springLength(load2) ). **
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:15:24 What is your expression for the average slope of the graph between load1 and load2?
......!!!!!!!!...................................
RESPONSE --> [(sprinlength(load2) - springlenth(load1)] / (load2 - load1) confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:15:31 ** rise = springLength(load2) - springLength(load1) run = load2 - load1 so slope = [ springLength(load2) - springLength(load1)] / (load2 - load1). **
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:18:16 problem 6 symbolic expression for slope of depth function
......!!!!!!!!...................................
RESPONSE --> [(depth(t2) - depth(t1)] (t2-t1) confidence assessment: 2
.................................................
......!!!!!!!!...................................
20:18:21 ** the name of the function is depth(t). We need the slope between t = t1 and t = t2. The depths are depth(t1) and depth(t2). Thus rise is depth(t2) - depth(t1) and run is t2 - t1. Slope is [ depth(t2) - depth(t1) ] / (t2 - t1). **
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:37:45 problem 8 average rate from formula f(t) = 40 (2^(-.3 t) ) + 25 intervals of partition (10,20,30,40) What average rate do you get from the formula? Show your steps.
......!!!!!!!!...................................
RESPONSE --> For function we substitute in: f(t) = 40 (2^(-.3 t) ) + 25 f(10) = 40 (2^(-.3 * 10) ) + 25 = 40 (2^(-3)) + 25 = 40(.125) + 25 = 5 + 25 =30 (10, 30) f(20) = 40 (2^(-.3*20) ) + 25 = 40 (2^-6) + 25 = 40(.015625) + 25 = .625 + 25 = 25.625 (20, 25.625) f(30) = 40 (2^(-.3 * 30) ) + 25 = 40 (2(-9)) + 25 = 40 (.001953125) + 25 = .078 + 25 = 25.078 (30, 25.078) f(40) = 40 (2^(-.3 *40) ) + 25 = 40 (2^-12) + 25 = 40 (.000244140625) + 25 = .0098 + 25 = 25.0098 (40, 25.0098) Now that we have y values for each one, we can use the slope formula: 10-20: (30 - 25.625) / (10-20) = 4.38/-10 = -.44 20-30: (25.625-25.078) / 20-30 = .55/-10 = -.06 30-40: (25.078-25.0098) / 30-40 = .07 = -.01 confidence assessment: 2
.................................................
......!!!!!!!!...................................
20:38:21 ** ave rate = change in depth / change in t. For the three intervals we get (f(20)-f(10))/(20-10) = (25.625 - 30 )/(20 - 10) = -4.375 / 10 = -.4375 (f(30)-f(20))/(30-20) = (25.07813 - 25.625)/(30 - 20) = -.5469 / 10 = -.05469. (f(40)-f(30))/(40-30) = (25.00977 - 25.07813)/(40 - 30) = -.0684 / 10 = -.00684. **
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 2
.................................................
......!!!!!!!!...................................
20:38:26 Add comments on any surprises or insights you experienced as a result of this assignment.
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:38:32 STUDENT RESPONSE: Ummm I know the slope formula is (y2-y1)/(x2-x1), but I always just put the number into the expression in the order I see them, but that is ok because I keep the order and get the correct answere because the y2,y1,x2,x1 or all relative. I am correct in doing this? INSTRUCTOR COMMENT: In other words you use (y1 - y2) / (x1 - x2) instead of (y2 - y1) / (x2 - x1). It's more conventional to regard, say, 10 as x1 and 20 as x2, so f(20) is y2 and f(10) is y1. If you start from the lower x number and change to the higher the difference is higher - lower, and this is the way we usually think about changes. According to this convention we calculate change in y as y2 - y1 and change in x as x2 - x1. You are doing (y1 - y2) / (x1 - x2) and you get a negative change in x, a negative denominator, and if you are thinking about change from the first quantity to the second this is backwards. However as you say both numerator and denominator follow the same order so you still get the right answer, since (y1-y2)/(x1-x2)= (y2-y1) / (x2-x1). **
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
"