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mth 173
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Question Form_labelMessages **
7/15 7
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You made some comments on my work submitted for week 2 quiz 1. Here is what you said.
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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Was I correct in the work but just mislabeled it as the average rate halfway? Or is the work wrong too? I have put the specific work below.
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 had a correct answer to the question in your original document.
It appears to be missing from this document.
Remember to always include all information.
In any case, here is your correct solution for that question:
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
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