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course mth 173
7/13 12
If the function y = .015 t2 + -1.7 t + 93 represents depth y vs. clock time t, then what is the average rate of depth change between clock time t = 13.9 and clock time t = 27.8? What is the rate of depth change at the clock time halfway between t = 13.9 and t = 27.8?Solution:
y(13.9) = .015*13.9^2 - 1.7*13.9 + 93 = 72.27
y(27.8) = .015*27.8^2 - 1.7*27.8 + 93 = 57.33
average rate of depth change = (57.33-72.27)/(27.8-13.9) = -1.07
average rate halfway:
27.8-13.9 = 13.9
13.9/2 = 6.95
13.9 + 6.95 = 20.85
y(20.85) = .015(20.85^2) -1.7(20.85) + 93 = 64.08
average rate = (64.08-72.27)/(20.85-13.9) = -1.18
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You were asked for the rate at the halfway point, not for an average rate.
What is the rate function, and what is its value at the halfway point?
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You answered this question correctly, below.
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What function represents the rate r of depth change at clock time t?
y'(t) = 2at + b
a = .015
b = -1.7
y(t) = .03t -1.7
What is the clock time halfway between t = 13.9 and t = 27.8, and what is the rate of depth change at this instant?
27.8-13.9 = 13.9
13.9/2 = 6.95
13.9 + 6.95 = 20.85
y'(20.85) = .03(20.85) - 1.7 = -1.07
If the function r(t) = .193 t + -2.1 represents the rate at which depth is changing at clock time t, then how much depth change will there be between clock times t = 13.9 and t = 27.8?
r(20.85)= 1.924
1.924*(27.8-13.9)=26.744
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You can't assume that the average rate occurs at the halfway point. That is the case for linear rate functions, so it works here, but it isn't almost never the case for nonlinear rate functions.
To get the result, you would get the change-in-depth function, which can be any antiderivative of the rate-of-depth-change function and find the change in its value between the two given clock times.
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What function represents the depth?
y(t) = .0965t^2 - 2.1t
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This is a good change-in-depth function, and could be used to answer the preceding question.
However it's not quite correct as an answer to the question of the depth function.
y(t) = .0965t^2 - 2.1t + c
where c is an arbitrarty constant, would be a good answer to this question.
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What would this function be if it was known that at clock time t = 0 the depth is 130 ?
y(t) = .0965t^2 - 2.1t + 130
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Well done, but there are some details I'll want you to attend to.
Mainly you should just check my notes and be sure you understand.
I do want you to use the rate-of-depth-change function to answer the question about the change in depth. Your result should confirm your previous answer, but in a way that would work for a nonlinear rate function.
&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes. &#
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