course mth271 9/29,1400 assignment #007
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12:10:14 `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 --> y=.05t^2 - 6t + 100 a=.05 b=-6 Rate: y'=2(.05)t - 6 y'=.1t - 6 y'(20)=.1(20) -6 y'(20)= -4cm/sec y'(30)=.1(30) - 6 y'(30)= -3cm/sec Average of rate change between t=20 and t=30: (-4cm/sec + -3cm/sec) / 2 = -3.5 cm/sec t=30sec - t=20 sec =10 sec (-3.5cm/sec) * 10sec= -35cm confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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12:18:13 `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'=.1t - 6 y'(30)= .1(30) - 6 y'(30)= -3cm/sec y'(40)= .1(40) - 6 y'(40)= -2cm/sec Avg. Rate: (-3cm/sec + -2cm/sec) / 2 = -2.5cm/sec (-2.5cm/sec) * 10 sec= -25cm y=.05t^2 -6t + 100 y(30)=.05(30^2) -6(30) + 100 y(30)=-35cm y(40)=.05(40^2) - 6(40) + 100 y(40)=-60cm -60cm - (-35cm) = -25cm Both the rate function and the depth function give the same depth change. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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13:33:34 `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=.05t^2 - 6t + 30 y(20)=.05(20^2) - 6(20) + 30 y(20)=-70 cm y(30)=.05(30^2) - 6(30) + 30 y(30)=-105 cm Depth change: -105 cm - -70 cm= -35cm y'=.1t - 6 y'(20)=.1(20) - 6 y'(20)=-4cm/sec y'(30)=.1(30) - 6 y'(30)=-3cm/sec Average rate between t=20 and t=30: (-4cm/sec + -3cm/sec) / 2=-3.5cm/sec (-3.5cm/sec) * 10 sec=-35cm Depth function and rate function predict same change in depth. y=.05t^2 - 6t + 30 y(40)=.05(40^2) - 6(40) + 30 y(40)=-130cm y(30)=-105cm Depth change: -130cm - -105cm= -25cm y'=.1t - 6 y'(40)=.1(40) - 6 y'(40)=-2cm/sec y'(30)=-3cm/sec Average rate between t=30 and t=40: (-3cm/sec + -2cm/sec) / 2 = -2.5cm/sec (-2.5cm/sec) * 10 sec =-25cm Rate function and depth function predict same depth change. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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13:35:09 `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 --> The two depth functions give the same depth change because the only difference in the depth functions is the c value. The c value is of no consequence when calculating the rate of change function between two clock times. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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13:36:49 `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 --> y=.05t^2 - 6t + 100 a=.05, b=-6, c=100 y'=2(.05)t -6 y'=.1t - 6 confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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13:38:29 `q006. For the function y = .05 t^2 - 6 t + 30, what is the function y ' ?
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RESPONSE --> y=.05t^2 - 6t + 30 a=.05, b=-6, c=30 y'=2(.05)t - 6 y'=.1t - 6 The y' value is the same here as in the previous problem, despite the differing c value in the depth functions. This further exemplifies that the c value is of little consequence. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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13:41:20 `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=.05t^2 - 6t + 130 a=.05, b=-6, c=130 y'=2(.05) - 6 y'=.1 t - 6 y'=.1 t - 6 2a=.1 a=.05 b=-6 Two new antiderivatives: y=.05t^2 -6t + 32 y=.05t^2 -6t + 200 confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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13:49:35 `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 --> The function y can only have one derivative. This is because the a and b values of the y are used to find y'. For instance, if y=.05t^2 - 6t + 100, a=.05 and b=-6. These values are transferred into the y' function of 2at + b. In this case, 2(.05)t - 6. Any variance in the a or b value in the y' function would make that function not a derivative of y. The function y' can have multiple antiderivatives because there is no given c value in this function. Therefore, as you place the a and b of y' into an antiderivative, you can have any value for c. For example, in the function y'=.1t - 6, we solve for a and b. a=.05 and b=-6. These values are the a and b values used in the y function. But, the y function also has a c value. Since the c value is not included in the y' function, it can vary: c=799-->y=.05t^2 - 6t +799 or c=2-->y=.05t^2 -6t + 2. As long as the a and b values remain the same, the y function will be an antiderivative of y'. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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13:51:13 `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 --> All of the antiderivative functions of y'=.1t - 6 have an a value of .05 and a b value of -6. They differ in the value of the constant, c. There could be limitless possibilities for the antiderivatives of a rate function. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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