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course mth 173
Version 5If the function y = .028 t2 + -1.3 t + 61 represents depth y vs. clock time t, then what is the average rate of depth change between clock time t = 6.7 and clock time t = 13.4?
Y = .028(6.7^2) - 1.3(6.7) + 61 = 1.25692 - 8.71 + 61 = 53.54692 @ 6.7
Y = .028(13.4^2) - 1.3(13.4) + 61 = 5.02768 - 17.42 + 61 = 48.60768 @ 13.4
`dy/`dt = ( 53.54692 - 48.60768) / (13.4 - 6.7) = 4.93924/6.7 = .7372
What is the rate of depth change at the clock time halfway between t = 6.7 and t = 13.4?
(13.4 + 6.7) / 2 = 20.1/2 = 10.05
Y = 2at + b = 2(.028)(10.05) + -1.3 = -0.7372
What function represents the rate r of depth change at clock time t?
R(t) = .056t + -1.3
What is the clock time halfway between t = 6.7 and t = 13.4, and what is the rate of depth change at this instant?
(13.4 + 6.7) / 2 = 20.1/2 = 10.05
Y` = .056(10.05) - 1.3 = .5628 - 1.3 = -.7372
If the function r(t) = .289 t + -2.7 represents the rate at which depth is changing at clock time t, then how much depth change will there be between clock times t = 6.7 and t = 13.4?
R = .289(6.7) - 2.7 = -.7637
R = .289(13.4) - 2.7 = 1.1726
1.1726 - .7637 = .4089
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This is the change in the rate of depth change, not the change in the depth.
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• What function represents the depth?
o Y = .1445(t^2) + -2.7t + c
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What is the change in the value of this function between t = 6.7 and t = 13.4?
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• What would this function be if it was known that at clock time t = 0 the depth is 200 ?
o Y = .1445(t^2) + -2.7t + 200
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Good on the first part.
Check my last two notes and see if you can modify your solution accordingly.
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