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course Mth 173
2/16 4pm
If the function y = .029 t2 + -1.4 t + 88 represents depth y vs. clock time t, then what is the average rate of depth change between clock time t = 6.4 and clock time t = 12.8?
We have y = f(t) = .029 t2 + -1.4 t + 88 so f(6.4) = .029*6.42 + -1.4*6.4 + 88 = 80.22784 and f(12.8) = .029*12.82 + -1.4*12.8 + 88 = 74.83136 therefore the average rate of depth change between clock time t = 6.4 and clock time t = 12.8 is (74.83136 - 80.22784) / (12.8 - 6.4) = -5.39648 / 6.4 = -.8432. Using dy/dx, we have ((2*.029*6.4 - 1.4) + (2*.029*12.8 - 1.4)) / 2 = -.8432.
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minor note on notation: dy/dx indicates an instantaneous rate. `d stands for the capital Greek letter Delta, which as I'm sure you know stands of 'change in', indicating a finite change.
Thus `dy / `dx would be more appropriate here.
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????Would answer using significant figures = -8.4 E-1?
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Yes, if the given quantities aren't regarded as exact. The minimum number of significant figures would then be 2, and a 2-figure approximation would be -.84.
Note that when in doubt, and when it makes a difference, we will be assuming exact values. In a physics class we would worry more about significant figures. In this class it's pretty much a judgement call.
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What is the rate of depth change at the clock time halfway between t = 6.4 and t = 12.8?
(12.8 + 6.4) / 2 = 9.6 therefore the rate of depth change at the clock time halfway between t = 6.4 and t = 12.8, if the function y = .029 t2 + -1.4 t + 88 represents depth y vs. clock time t, y` = 2*.029*9.6 - 1.4 = -8342.
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Do you mean -.8432?
This being a quadratic function, as it turns out its rate of change is linear, so for any interval the midpoint rate is equal to the average rate. This is a characteristic of quadratic functions, and only of quadratic functions; in fact it can therefore be regarded as the defining characteristic of a quadratic function.
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What function represents the rate r of depth change at clock time t? What is the clock time halfway between t = 6.4 and t = 12.8, and what is the rate of depth change at this instant?
For the function y = .029 t2 + -1.4 t + 88, y` = r(t) = 2at + b where a = .029 and b = -1.4.
What is the clock time halfway between t = 6.4 and t = 12.8, and what is the rate of depth change at this instant?
(12.8 + 6.4) / 2 = 9.6 = the clock time halfway between t = 6.4 and t = 12.8 so the rate of depth change at this instant r(9.6) = 2*.029*9.6 - 1.4 = -8342.
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-.8432?
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If the function r(t) = .211 t + -1.8 represents the rate at which depth is changing at clock time t, then how much depth change will there be between clock times t = 6.4 and t = 12.8?
If the function r(t) = .211 t + -1.8 represents the rate at which depth is changing at clock time t, then depth change between clock times t = 6.4 and t = 12.8 will be (1/2*.211*12.8^2 - 1.8*12.8) - (1/2*.211*6.4^2 - 1.8*6.4) = 1.44384.
Using r(t) we have ((.211*6.4 + -1.8) + (.211*12.8 + -1.8))/2*6.4 = 1.44384
• What function represents the depth?
The function that represents the depth from integrating r(t) = .211 t + -1.8 is y = 1/2*2.11t^2 - 1.8t + c = .1055t^2 - 1.8t + c where c can be any starting position.
• What would this function be if it was known that at clock time t = 0 the depth is 80?
If it was known that at clock time t = 0 the depth is 80, this function would be y = .1055t^2 - 1.8t + 80.
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Excellent. Do check the notes I've inserted.
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