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At the site 08-50-041 I see the following menu:
06-13-2013_____orientation_part_ii
06-13-2013_____orientation_part_ii_b--understanding_learning_objectives
06-13-2013_____orientation_part_iv--more_about_communication
06-13-2013_____orientation_part_v--how_to_succeed_in_your_course
06-14-2013_____Establishing_Communication_Amanda_Karstetter
06-14-2013_____First_Two_Questions
06-14-2013_____First_Twelve_Questions
06-14-2013_____Typewriter_Notation
06-14-2013_____Describing_Graphs
06-15-2013_____Precalculus_Initial_Questions
06-15-2013_____Calculus_Initial_Questions
06-15-2013_____Rates
06-15-2013_____Volumes
06-15-2013_____Surface_Areas_
06-15-2013_____QA1_Depth_vs_Clock_Time_and_Rate_of_Depth_Change
06-17-2013_____question_form
06-17-2013_____question_form
06-17-2013_____question_form
06-19-2013_____question_form
06-19-2013_____question_form
06-20-2013_____question_form
06-20-2013_____question_form
06-20-2013_____QA_2
06-21-2013_____QA_3
06-22-2013_____question_form
06-22-2013_____Query_1
06-24-2013_____question_form
06-24-2013_____Week_2_Quiz_1
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Does this agree with what you see?
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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 **
DVDs
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How do you access Modeling Project #2? I can only find the solutions on Notes #4.
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That assignment is also a link on your Assignments Page. Just click on the link.
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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 **
Week 2 Quiz 1 Question
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In your notes from Week 2 Quiz 1, underneath the question, 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?What function represents the rate r of depth change at clock time t? 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?
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?
• What function represents the depth?
• What would this function be if it was known that at clock time t = 0 the depth is 130 ?
I had:What is the rate of depth change at the clock time halfway between t = 13.9 and t = 27.8?:
clock time halfway=t=20.85
y2-y1=.015(20.85+¬`dt)^2-1.7(20.85+`dt)+93
y2-y1=.015(434.723+41.7`dt+`dt^2)-1.7(20.85+`dt)+93
y2-y1=6.521+.626`dt+`dt^2-35.445-1.7`dt+93
y2-y1=64.076-1.074`dt+`dt^2
y2-y1-64.076=(-1074`dt+`dt^2)/`dt
y2-y1=-1.074+`dt
y2-y1=-1.074+20.85
y2-y1=19.776
You said:dt is the duration of the interval over which the rate is approximated.
An instantaneous rate corresponds to an interval that shrinks toward zero duration. In the limit, then `dt approaches zero and the rate would be -1.074.
The result you obtained would be the average rate for an interval of length 20.85, starting at the instant t = 20.85; that is, this would be the average rate for an interval running from t = 20.85 to t = 41.7.
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I am unclear on if you meant that the result I obtained should be the average rate or if it was the average rate? I am just confused if I did this problem wrong or not. Thanks!
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The question asked for the rate halfway through the interval, which would be the rate at t = 20.85.
Had you let `dt approach zero rather than substituting 20.85 for `dt, you would have obtained the correct rate -1.074.
However you substituted 20.85 for `dt, which resulted in the average rate for the interval from t = 20.85 to t = 41.7.
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