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course Mth 173
6/11/13~7:40PM
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?The average rate of depth change is as follows:
Y=.015(13.9)^2 -1.7(13.9) +93 → y=72.6815 →(point (13.9,72.6815)
Y=.015(27.8)^2 -1.7(27.8) +93 → y= 57.3326 →(point (27.8,57.3326)
Slope of two points → y2-y1/x2-x1
(57.3326-72.6815)/ (27.8 - 13.9)→ 14.93555/13.9 = -1.07497 depth change per second.
The average rate of depth change at the clock time halfway between t=13.0 and t=27.8:
Y=.015t^2 +-1.7t +93 → Y’=2at + b
Y’=.03t -1.7 → plug in the midpoint((13.9 +27.8)/2 = 20.85 → Y’=.03t -1.7
Y’=.03(20.85) -1.7 = (-1.0745)
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?
The clock time halfway between t=13.9 and t=27.8 is ((13.9 +27.8)/2 = 20.85
Rate = .030t -1.7 = -1.0745 the average rate over an interval is always the instantaneous rate at the midpoint.
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?
y(t) = 1/2 * .193 t^2 - 2.1 t + c
y(t) = 1/2 * .193 t^2 - 2.1 t + c
y(t) = .0965 t^2 - 2.1 t + c
Average rate = change in depth/ change in time
(.5827 + 3.2654)/2 = 1.92405 = change in depth/13.9→ change in depth = 26.74
• What function represents the depth? d(t) = .097t^2 ·2.1 + c
What would this function be if it was known that at clock time t = 0 the depth is 130?
d(t) = .097t^2 ·2.1 + 130
&#Good responses. Let me know if you have questions. &#