course Mth271 ???????????????Student Name:
......!!!!!!!!...................................
13:15:36 Note that there are 9 questions in this assignment. `q001. The function y = .05 t^2 - 6 t + 100 is related to the rate function y ' = .1 t - 6 in that if y = .05 t^2 - 6 t + 100 represents the depth, then the depth change between any two clock times t is the same as that predicted by the rate function y ' = .1 t - 6. We saw before that for y ' = .1 t - 6, the depth change between t = 20 and t = 30 had to be 35 cm. Show that for the depth function y = .05 t^2 - 6t + 100, the change in depth between t = 20 and t = 30 is indeed 35 cm.
......!!!!!!!!...................................
RESPONSE --> y(20)= .05 (20)^2 -6(20) + 100= 0 y(30)= .05(30)^2-6(30)+100= -35 0- -35 = ch in depth = 53 -35 ch in depth/ 10 ch in t = -3.5
.................................................
......!!!!!!!!...................................
13:16:07 The depth at t = 20 will be .05(20^2) - 6(20) + 100 = 20 - 120 + 100 = 0. The depth at t = 30 will be .05(30^2) - 6(30) + 100 = 45 - 180 + 100 = -35. Thus the depth changes from 0 cm to -35 cm during the 10-second time interval between t = 20 s and t = 30 s. This gives us and average rate of ave rate = change in depth / change in clock time = -35 cm / (10 sec) = -3.5 cm/s.
......!!!!!!!!...................................
RESPONSE --> I think that I got his one
.................................................
......!!!!!!!!...................................
13:17:17 `q002. What depth change is predicted by the rate function y ' = .1 t - 6 between t = 30 and t = 40? What is the change in the depth function y = .05 t^2 - 6 t + 100 between t = 30 and t = 40? How does this confirm the relationship between the rate function y ' and the depth function y?
......!!!!!!!!...................................
RESPONSE --> y'(30)= .1 (30) -6= -3 y'(40)= .1 (40) -6 =-2 -2.5
.................................................
......!!!!!!!!...................................
13:18:37 At t = 30 and t = 40 we have y ' = .1 * 30 - 6 = -3 and y ' = .1 * 40 - 6 = -2. The average of the two corresponding rates is therefore -2.5 cm/s. During the 10-second interval between t = 30 and t = 40 we therefore predict the depth change of predicted depth change based on rate function = -2.5 cm/s * 10 s = -25 cm. At t = 30 the depth function was previously seen to have value -35, representing -35 cm. At t = 40 sec we evaluate the depth function and find that the depth is -60 cm. The change in depth is therefore depth change has predicted by depth function = -60 cm - (-35 cm) = -25 cm. The relationship between the rate function and the depth function is that both should predict a same change in depth between the same two clock times. This is the case in this example.
......!!!!!!!!...................................
RESPONSE --> I am not sure as to where the -60 came from ......
.................................................
......!!!!!!!!...................................
13:19:53 `q003. Show that the change in the depth function y = .05 t^2 - 6 t + 30 between t = 20 and t = 30 is the same as that predicted by the rate function y ' = .1 t - 6. Show the same for the time interval between t = 30 and t = 40. Note that the predictions for the y ' function have already been made.
......!!!!!!!!...................................
RESPONSE --> y(20)= .05 (20)^2 -6 (20) + 30= -70 y(30)= .05 (30)^2 -6 (30) + 30= -105
.................................................
......!!!!!!!!...................................
13:21:32 The prediction from the rate function is a depth change of -35 cm, and has already been made in a previous problem. Evaluating the new depth function at t = 20 we get y = .05(20^2) - 6(20) + 30 = -70, representing -70 cm. Evaluating the same function at t = 30 we get y = -105 cm. This implies the depth change of -105 cm - (-70 cm) = -35 cm. Evaluating the new depth function at t = 40 sec we get y = depth = -130 cm. Thus the change from t = 30 to t = 40 is -130 cm - (-105 cm) = -25 cm. This is identical to the change predicted in the preceding problem for the given depth function.
......!!!!!!!!...................................
RESPONSE --> why is depth -130......I think that you just lost me
.................................................
......!!!!!!!!...................................
13:21:32 The prediction from the rate function is a depth change of -35 cm, and has already been made in a previous problem. Evaluating the new depth function at t = 20 we get y = .05(20^2) - 6(20) + 30 = -70, representing -70 cm. Evaluating the same function at t = 30 we get y = -105 cm. This implies the depth change of -105 cm - (-70 cm) = -35 cm. Evaluating the new depth function at t = 40 sec we get y = depth = -130 cm. Thus the change from t = 30 to t = 40 is -130 cm - (-105 cm) = -25 cm. This is identical to the change predicted in the preceding problem for the given depth function.
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
13:22:32 `q004. Why is it that the depth functions y = .05 t^2 - 6 t + 30 and y = .05 t^2 - 6 t + 100 give the same change in depth between two given clock times?
......!!!!!!!!...................................
RESPONSE --> Because c doesnt really matter....it is just a constand....when you take the deriv. of both of these it is the same
.................................................
......!!!!!!!!...................................
13:23:31 The only difference between the two functions is a constant number at the end. One function and with +30 and the other with +100. The first depth function will therefore always be 70 units greater than the other. If one changes by a certain amount between two clock times, the other, always being exactly 70 units greater, must also change by the same amount between those two clock times.
......!!!!!!!!...................................
RESPONSE --> Well in this case I guess it does matter....I understand the effect that it has it is going to make one 70 units greater than the other
.................................................
......!!!!!!!!...................................
13:25:24 `q005. We saw earlier that if y = a t^2 + b t + c, then the average rate of depth change between t = t1 and t = t1 + `dt is 2 a t1 + b + a `dt. If `dt is a very short time, then the rate becomes very clost to 2 a t1 + b. This can happen for any t1, so we might as well just say t instead of t1, so the rate at any instant is y ' = 2 a t + b. So the functions y = a t^2 + b t + c and y ' = 2 a t + b are related by the fact that if the function y represents the depth, then the function y ' represents the rate at which depth changes. If y = .05 t^2 - 6 t + 100, then what are the values of a, b and c in the form y = a t^2 + b t + c? What therefore is the function y ' = 2 a t + b?
......!!!!!!!!...................................
RESPONSE --> y= .05t^2-6t+100 ...... = ........y= at^2 + bt+c a= .05 b= -6 c= 100 y'=.1t+-6 ....=......y'= 2at+b a= .05 b= -6
.................................................
......!!!!!!!!...................................
13:25:37 If y = .05 t^2 - 6 t + 100 is of form y = a t^2 + b t + c, then a = .05, b = -6 and c = 100. The function y ' is 2 a t + b. With the given values of a and b we see that y ' = 2 ( .05) t + (-6), which simplifies to y ' = .1 t - 6
......!!!!!!!!...................................
RESPONSE --> I understand this part
.................................................
......!!!!!!!!...................................
13:26:28 `q006. For the function y = .05 t^2 - 6 t + 30, what is the function y ' ?
......!!!!!!!!...................................
RESPONSE --> y'= .1t+ -6 ....it is the same as the last because the c doesnt have an affect with it comes to this
.................................................
......!!!!!!!!...................................
13:26:34 The values of a, b and c are respectively .05, -6 and 30. Thus y ' = 2 a t + b = 2(.05) t + (-6) = .1 t - 6. This is identical to the y ' function in the preceding example. The only difference between the present y function and the last is the constant term c at the end, 30 in this example and 100 in the preceding. This constant difference has no effect on the derivative, which is related to the fact that it has no effect on the slope of the graph at a point.
......!!!!!!!!...................................
RESPONSE --> I understnat
.................................................
......!!!!!!!!...................................
13:29:11 `q007. For some functions y we can find the rate function y ' using rules which we will develop later in the course. We have already found the rule for a quadratic function of the form y = a t^2 + b t + c. The y ' function is called the derivative of the y function, and the y function is called an antiderivative of the y ' function. What is the derivative of the function y = .05 t^2 - 6 t + 130? Give at least two new antiderivative functions for the rate function y ' = .1 t - 6.
......!!!!!!!!...................................
RESPONSE --> It is going to be the same thing because the c doesnt matter........y'= .1(t) + -6.....I am not sure what you are asking for on this last part......I think that what you are trying to show us that is c doesnt matter so I will guess that this is an antiderivative of this .... y=.05t^2- 6t + 445 y=.05t^2- 6t + 5
.................................................
......!!!!!!!!...................................
13:29:41 The derivative of y = .05 t^2 - 6 t + 130 is .1 t - 6; as in the preceding problem this function has a = .05 and b = -6, and differs from the preceding two y functions only by the value of c. Since c has no effect on the derivative, the derivative is the same as before. If y ' = .1 t - 6, then a = 1 / 2 ( .1) = .05 and b = -6 and we see that the function y is y = .05 t^2 - 6 t + c, where c can be any constant. We could choose any two different values of c and obtain a function which is an antiderivative of y ' = .1 t - 6. Let's use c = 17 and c = -54 to get the functions y = .05 t^2 - 6 t + 17 and y = .05 t^2 - 6 t - 54.
......!!!!!!!!...................................
RESPONSE --> I think that I am on the right track
.................................................
......!!!!!!!!...................................
13:30:44 `q008. For a given function y, there is only one derivative function y '. For a given rate function y ', there is more than one antiderivative function. Explain how these statements are illustrated by the preceding example.
......!!!!!!!!...................................
RESPONSE --> Basicly it is just saying that since c is a constant that it really does not have an affect when it comes to finding the derviative.
.................................................
......!!!!!!!!...................................
13:31:10 The derivative function give the rate at which the original function changes for every value of t, and there can be only one rate for a given t. Thus the values of the derivative function are completely determined by the original function. In the previous examples we saw several different functions with the same derivative function. This occurred when the derivative functions differed only by the constant number at the end. However, for a given derivative function, if we get one antiderivative, we can add any constant number to get another antiderivative. y = .05 t^2 - 6 t +17, y = .05 t^2 - 6 t + 30, and y = .05 t^2 - 6 t + 100, etc. are all antiderivatives of y ' = .1 t - 6.
......!!!!!!!!...................................
RESPONSE --> I think that I explained this correctly...I just shortened it a lot!
.................................................
......!!!!!!!!...................................
13:32:35 `q009. What do all the antiderivative functions of the rate function y ' = .1 t - 6 have in common? How do they differ? How many antiderivative functions do you think there could be for the given rate function?
......!!!!!!!!...................................
RESPONSE --> I think there could be as many antiderviatives as you could think of for this problem....the c is just going to keep going away so it doesnt matter what it is
.................................................
......!!!!!!!!...................................
13:32:46 All antiderivatives must contain .05 t^2 - 6 t. They may also contain a nonzero constant term, such as -4, which would give us y = .05 t^2 - 6 t - 4. We could have used any number for this last constant.
......!!!!!!!!...................................
RESPONSE --> I understnad
.................................................
???|??????? Student Name: assignment #008 008. Approximate depth graph from the rate function
......!!!!!!!!...................................
15:13:29 `q001. Note that there are 5 questions in thie assignment. Sketch a graph of the function y ' = .1 t - 6 for t = 0 to t = 100. Describe your graph.
......!!!!!!!!...................................
RESPONSE --> y= .1(0) -6 y=-6 0=.1(t) -6 t= 60 60,0 and 100, 4
.................................................
......!!!!!!!!...................................
15:14:03 The 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).
......!!!!!!!!...................................
RESPONSE --> I just forgot to say that it was a straight line.....opps
.................................................
......!!!!!!!!...................................
15:17:47 `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?
......!!!!!!!!...................................
RESPONSE --> y'= .1(0) -6= -6 y'= .1(10) -6= -5 y'= .1(20) -6=-4 y'= .1(30) -6=-3 y'= .1(40) -6=-2 y'= .1(50) -6=-1 y'= .1(60) -6=0 y'= .1(70) -6=1 y'= .1(80) -6=2 y'= .1(90) -6=3 y'= .1(100) -6=4 it starts out at (0, -6) and goes to (100, 4) so the slope is positive and the grapph is increasing
.................................................
......!!!!!!!!...................................
15:18:33 The 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.
......!!!!!!!!...................................
RESPONSE --> I think I did that one ok....I could have went into more detail but all in all I think it ok
.................................................
......!!!!!!!!...................................
15:19:46 `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?
......!!!!!!!!...................................
RESPONSE --> -6 * 10 = -60 100+ -60 = 40 so then it would be (10, 40)
.................................................
......!!!!!!!!...................................
15:19:54 The 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).
......!!!!!!!!...................................
RESPONSE --> I got this on right
.................................................
......!!!!!!!!...................................
15:22:43 `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?
......!!!!!!!!...................................
RESPONSE --> -5* 10= -50 10+10= 20 40+-50= -10 (20, -10)
.................................................
......!!!!!!!!...................................
15:22:56 The 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).
......!!!!!!!!...................................
RESPONSE --> I understand this stuff I think
.................................................
......!!!!!!!!...................................
15:27:28 `q005. Continue this process up to the t = 70 point, using a 10-unit t interval for each approximation. Describe the graph that results.
......!!!!!!!!...................................
RESPONSE --> t= 20 ...= -4 * 10= -40 + -10= -50 and 20+10= 30 so (30, -50) t= 30 ....= -3 so (40, -80) t= 40 ....= -2 so (50, -100) t= 50 ...= -1 so (60, -110) t= 60....= 0 so (70, -110) t= 70 ....= 1 so (80,-100) As the x increases the y decreases until you get to x= 70 then it starts increasing
.................................................
......!!!!!!!!...................................
15:28:26 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).
......!!!!!!!!...................................
RESPONSE --> At t= 70 it should have stayed at -110 but somehow I got mine to go up to -100....I dont know how I did that but I think I understand
.................................................
"