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course Mth 271
7/20 1:20pm
Revised!Week 2 Quiz asmt 5
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course Mth 271
7/19 6:20Do these quizes count as just part of our homework and participation grade?
Right, but they anticipate quite a few of the upcoming questions on the Major Quiz.
If the function y = .023 t2 + -1.4 t + 67 represents depth y vs. clock time t, then what is the average rate of depth change between clock time t = 8.3 and clock time t = 16.6? What is the rate of depth change at the clock time halfway between t = 8.3 and t = 16.6?
At 8.3 the depth is 56.96447
At 16.6, 47.09788
The avg rate of change between these two points is -9.86659/8.3= - 1.18875
The avg distance between 8.3 and 16.6 is 12.45
At 12.45 the depth is 53.135
The rate from 8.3 to 12.45 would be -3.82947/4.15= -0.8509
The rate from 12.45 to 16.6 would be -6.03712/ 4.15=-1.4547
The avg of these two numbers 1.1528 for the avg rate of depth change at the halfway point
Not bad, but at this point you should be using the rate-of-change function y ' to calculate the midpoint rate.
What function represents the rate r of depth change at clock time t? What is the clock time halfway between t = 8.3 and t = 16.6, and what is the rate of depth change at this instant?
The clock time half way between was 12.45 with the depth being 53.135. and as I stated above the rate of depth change avgs 1.1528
Y= 2at+b is the function that represents the rate of depth change at clock time t
In this example that would be 0.046- 1.4
good, but of course you mean 0.046 t - 1.4. You left out the t.
Now what would this be at the midpoint t = 12.45, and how does this compare to the average rate of change for the interval?
&&&& the rate of depth change at clock time t should be 0.045t-1.4
Substituting the 12.45 into this problem we would find the rate of depth change to be -0.8273&&&&&
If the function r(t) = .024 t + -2.3 represents the rate at which depth is changing at clock time t, then how much depth change will there be between clock times t = 8.3 and t = 16.6?
0.012t^2-2.3t+C where C can be any number
• What function represents the depth?
function y = .023 t2 + -1.4 t + 67
• What would this function be if it was known that at clock time t = 0 the depth is 50 ?
0.012t^2-2.3t+C must equal 50 and t equals 0
So 0.012*0^2-2.3*0+C=50
C=50
The depth function would then be 0.012t^2-2.3t+50
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Good answers to nearly all the questions. Just one needs revision, but it's a very important question with a very important idea. And it usually comes up on the test.
&#Please see my notes and submit a copy of this document with revisions 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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The rate you obtained from the rate function and the average rate should be identical. They are close, but not identical. I believe it's because you used .045 t in your rate function rather than .046 t, which would have corresponded to the .023 t^2.
In any case, good work.
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