course Mth 173 p먩h]ɗєassignment #007 007. Depth functions and rate functions. 03-05-2008
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09:15:26 `qNote 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.
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RESPONSE --> t = 20 y = .05 (20)^2 - 6 (20) + 100 = 0 t = 30 y = .05 (30)^2 - 6 (30) + 100 = -35 confidence assessment: 3
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09:16:09 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.
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RESPONSE --> ok self critique assessment: 3
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09:24:53 `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?
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RESPONSE --> y ' = .1 (30) - 6 = -3 y ' = .1 (40) - 6 = -2 y = .05 (30)^2 - 6 (30) + 100 = -35 y = .05 (40)^2 - 6 (40) + 100 = -60 confidence assessment: 2
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09:26:55 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.
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RESPONSE --> ok self critique assessment: 3
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09:35:54 `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.
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RESPONSE --> y = .05 (20)^2 - 6 (20) + 30 = -70 y = .05 (30)^2 - 6 (30) + 30= -105 average = -43.75 -43.75 * 10 = 437.5 y' = .1 (30) - 6 = -3 y' = .1 (40) - 6 = -2 average = -2.5 -2.5 * 10 = -25 confidence assessment: 2
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09:37:19 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.
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RESPONSE --> ok self critique assessment: 3
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09:40:12 `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?
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RESPONSE --> because when you take the derivitive of either y it gives you the same thing. y = .05 t^2 - 6 t + 30 y' = .1 t - 6 y = .05 t^2 - 6 t + 100 y' = .1 t -6 confidence assessment: 2
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09:40:48 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.
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RESPONSE --> ok self critique assessment: 2
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09:43:51 `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?
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RESPONSE --> confidence assessment:
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09:44:51 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
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RESPONSE --> ok self critique assessment: 3
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09:45:34 `q006. For the function y = .05 t^2 - 6 t + 30, what is the function y ' ?
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RESPONSE --> y' = .1 t - 6 confidence assessment: 3
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09:45:59 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.
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RESPONSE --> ok self critique assessment: 3
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09:50:13 `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.
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RESPONSE --> y = .05 t^2 - 6 t + 130 y' = .1 t -6 y' = .1 t - 6 y = .05 t^2 - 6 t + 90 y = .05 t^2 - 6 t + 150 confidence assessment: 3
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09:51:59 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.
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RESPONSE --> ok self critique assessment: 3
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09:56:33 `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.
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RESPONSE --> when you take the derivative of a function you have all the information required so you should have 1 sure answer. when you take the antiderivative you do not have enough information to get 1 sure answer. therefore you can use multiple different numbers for your answer. confidence assessment: 3
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09:57:03 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.
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RESPONSE --> ok self critique assessment: 3
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10:00:23 `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?
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RESPONSE --> What they have in common. All the antiderivative functions of y' will begin with .05 t^2 - 6 t. How they differ. All the antiderivative functions of y' can have a different c. There could be infinite antiderivative functions for y'. confidence assessment: 3
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10:00:47 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.
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RESPONSE --> ok self critique assessment: 3
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