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course Mth 271
6/21/13 12:00 pm
If the function y = .021 t2 + -1.3 t + 87 represents depth y vs. clock time t, then what is the average rate of depth change between clock time t = 7.5 and clock time t = 15? What is the rate of depth change at the clock time halfway between t = 7.5 and t = 15?What function represents the rate r of depth change at clock time t? What is the clock time halfway between t = 7.5 and t = 15, and what is the rate of depth change at this instant?
If the function r(t) = .276 t + -1.2 represents the rate at which depth is changing at clock time t, then how much depth change will there be between clock times t = 7.5 and t = 15?
What function represents the depth?
What would this function be if it was known that at clock time t = 0 the depth is 130 ?
'A:The average rate of depth change between clock time (t=7.5) and (t=15) is:
y=.021t^2 + -1.3t + 87
y=.021(7.5)^2 + -1.3(7.5) + 87
y=78.43
y=.021(15)^2 + -1.3(15) + 87
y=72.23
The two points are:
(7.5, 78.43)
(15, 72.23)
(72.23-78.43)/(15-7.5)= -6.2/7.5
The rate of depth change halfway between t=7.5 and t=15 would be a point at t=11.25, because 15-7.5= 7.5/2= 3.25 so 7.5+3.25= 11.25 marking the distance halfway between the two points.
Now, plug in t=11.25, y= .021(11.25)^2+ -1.3(11.25) +87
y=75.03 adding a halfway point at (11.25, 75.03).
The function that represents the rate r of depth change at clock time t is y'= 2(.021t) + -1.3.
The rate of depth change at this instant is y= 2(.021)(11.25) + -1.3 = -0.83.
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Good.
How does this compare with the average rate you found previously?
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r(t)= .276t + -1.2
r(7.5)= .276(7.5)+ -1.2
r(7.5)= 0.87
r(15)= .276(15)+ -1.2
r(15)= 2.94
The depth change from t=15 to t=7.5 is:
2.94cm- 0.87cm= 2.07 cm.
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r(t) doesn't give you depths. r(t) gives you the rate at which depth is changing at a given clock time t.
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The function that represents depth is y= at^2 +bt + c.
Since we are given the rate of depth change at clock time, r(t)= .276t + -1.2, we have to work backwards from the form, y'= 2at +b.
while knowing that when t=0 sec, the depth (y) is equal to 130.
First we have to divide.276 by 2 because in the rate of change function ""a"" is multiplied by 2. Then, ""t"" gets multiplied back into ""a"" and ""b"" parts of the function.
y=.138t^2+ -1.2t + c.
If t=0 means depth is 130, then you know that ""c"" is 130. But you can also work out the problem by pluggin in 0 for t.
130=.138(0)^2+ -1.2(0)+c
130= 0+0+c
130= c
y= .138t^2+ -1.2t+ 130 is the depth function.
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Good.
How much does this function change between t = 7.5 and t = 15?
How does this compare with the result you should have obtained previously without having found this function?
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Good, but you've got a couple of things to reconcile in order to completely understand this.
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