wee 2 quizz

#$&*

course Mth 271

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?Y1 = 0.015 * 13.9^ - 1.7 * 13.9 + 93 = 72.268

Y2 = 0.015 * 27.8^2 – 1.7 * 27.8 + 93 = 57.333

(57.333 – 72.268)/(27.8 – 13.9) = -14.935/13.9 = -1.074

R(20.85) = -1.074

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?

r(t) = 2 * 0.015 t + -1.7

r(t) = 0.03t + -1.7

13.9 + (27.8 – 13.9)/2 = 20.85

r (20.85) = -1.074

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?

R1 = 0.583

R2 = 3.265

r2 – r1 = 2.682

@& The difference of the two rates is not related to the average rate.

The approximate average rate is the average of the two rates.

This quantity then needs to be used, along with the time interval, to find the change in depth.*@

* What function represents the depth?

y = .015 t2 + -1.7 t + 93

@& The condition now is that the rate function is .193 t - 2.1.

What depth function corresponds to this rate function?*@

* What would this function be if it was known that at clock time t = 0 the depth is 130 ?

Y = .015t^2 + -1.7 + 130

130 is the y intercept when t = 0

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